Solutions Manual for Neural Engineering 3rd Edition by Bi He

SOLUTIONS MANUAL

1. 4 2. 2 3. 3 4. 4 5. 3 6. 2 7. 2 8. 1 9. 5 10. 4 11. 4 12. 3 13. 3 14. 5 15. 4 16. 5 17. 5 18. 4

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7

Biopotential Measurements and Electrodes

Solutions

1. Block diagram for a cochlear implant Tx Coil

Transmitter

Power Recovery

Data Demodulato r

Battery and External Power Management

N-channel Neurostimulator

Cochlea Flexible Electrode Array

Skin

Rx Coil

(a) Microphone: transduces audio signals to electrical impulses. (b) Audio AFE: a set of amplifiers that take the small amplitude microphone signal and amplify it to the full range of ADC filtering signals outside audio range. (c) ADC: digitizes signals (d) DSP: Further filtering in digital domain plus channelization to determine which cochlear electrode to stimulate. (e) Transmitter: Power amplifier for the appropriate frequency and supply required to drive the external coil. (f) Tx and Rx coils: provide inductive coupling for power and data transmission. (g) Power Recovery: Rectifier and Regulators, doing AC/DC power conversion. (h) Data demodulator: recover the data signals in the transmitted carrier wave. Amplifiers and Comparator or ADC. (i) N-channel Neurostimulator: N-programmable pulse current sources to stimulate hair cells on the cochlea. (j) Flexible Electrode array: insertable into the cochlea reaching areas of hair cells corresponding to different frequencies.

Biopotential Measurements and Electrodes

2. EMG Dipole Traveling Wave

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Biopotential Measurements and Electrodes

3. Bioamplifier design to specifications

4. Electroporation

Biopotential Measurements and Electrodes

5. Intracellular recording

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Biopotential Measurements and Electrodes

6. Electrode Model

Biopotential Measurements and Electrodes

7. Electrodes, redox reactions, and half-potentials:

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Biopotential Measurements and Electrodes

Biopotential Measurements and Electrodes

8. Skin-electrode model

(a) Impedance without sweat ZTotalDry = Cel||Rel + Rgel + Csk||Rsk + RB ZTotalDry =

Rel Rsk + Rgel + + RB 1+jωCelRel 1+jωCskRsk

(b) Sweating condition ZT otalSweat = Cel||Rel + Rgel + [(Csk||Rsk)||(Csw||Rsw)] + RB Rel

Z TotalSweat = 1+jωCelRel + R gel +

Rsk Rsw Rsk +Rsw R

R

1+jω(Csk +C sw) R sk+Rsw sk

sw

+ RB

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Biopotential Measurements and Electrodes

(c) Bode plot:

9. Sweat glands model of electrode skin impedance The solution is presented under different notation: Ehc, Cd, Rd = Eel, Cel, Rel; Rs, Ru = Rgel, RB; Esc, Ce, Re, Ep, Cp, Rp = Esk, Csk, Rsk, Esw, Csw, Rsw

Biopotential Measurements and Electrodes

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Biopotential Measurements and Electrodes

Biopotential Measurements and Electrodes

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Biopotential Measurements and Electrodes

10. Multi-channel biopotential amplification with common-mode compensation and DRL (a) the outputs of the 3 channels: OUT1 = IN1(1 + RR12 ) OUT2 = IN2(1 + RR12 ) R2 OUT3 = IN3(1 + R ) 1 (b) The DRL output is: Rf Rf Rf f RL = −R R1 IN1 − R1 IN2 − R1 IN3 = − R1 [IN1 + IN2 + IN3] 2+IN3 Since CM = IN1+IN thus: 3 Rf RL = −3 R1 CM (c) effective resistance from body to ground: RRL = RRL Rf RRLEF F ECT IV E = 1+A DRL

1+3

R1

(d) Ri and Ro are protection resistors, if an electrical fault were to occur current would be limited to the supply voltage divided by the protection resistance.

11. Amplifier common-mode rejection (a) Differential Gain: ' A = V R2 ' (1 + R2 ) − VB R2 ' R R DM

A R +R 1

'

1

2

1

2 ' R1+R2 VA R'R+R ( R1 ) − VB RR12

ADM = 1 2 to find differential gain we set: VCM = 0 = VA + VB and VDM = 1 = VA − VB = 1 therefore: VA = 1/2 and VB = −1/2 ' ' R +R 2 1 R2 1 R2 (R1+R2)+R2(R 1'+R '2) 2 '( 1 ADM = 21 R'R+R R1 ) + 2 R1 = 2 R1(R' +R' ) 1 2 1 2 if there is no resistor mismatch R′ = R1 & R′ = R2 1

2

ADM = RR21 for common mode: VA' = V' B = VCM = 1 R' (R1+R 2)−R2(R +R ) 1 2 ACM = 2 and without mismatch: ' ' R1(R1+R2)

ACM = 0 CMRR = AADM = CM without mismatch: CMRR → − ∞

'

'

'

2

1

2

1 R2 (R1+R2)+R2(R1 +R2 ) 2 R' (R1+R2)−R2(R' +R' )

(b) Input referred noise: RMS Vni,AMP RMS = Vno,AMP ADM Psig Vsig RMSADM 2 SNROUT = Pnoise = ( Vno,AMP RMS ) (c) Differential gain will be affected: ADM ≈ CMRReff ≈

R1 |ZA−ZB|

≈

R1 200kΩ

R2 R1+1MΩ

Biopotential Measurements and Electrodes

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Therefore in order to remedy this we must either increase the input impedance by increasing R1, although it might be difficult to proportionally increase R2 to maintain the same differential gain. A better solution is to use a high input impedance topology such as a 3-opamp instrumentation amplifier, a current balancing instrumentation amplifier, or a differential difference amplifier.

12. Current source cascoding

a)

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Biopotential Measurements and Electrodes

b)

Chapter 3 - Homework Questions & Solutions Questions: 1) What are the chief uses of the EEG? Divide these into clinical/diagnostic as well as purely functional or cognitive and behavioral.

Clinical uses: Epilepsy monitoring, sleep studies, test drugs for seizure effects, Locate areas of brain damage following head injury, stroke, tumor, etc. Test afferent pathways (by evoked potentials); Purely functional or cognitive & behavioral: Brain-machine interface, observation of vigilance states, EEG Biofeedback 2) What are the different bands of the EEG and what are their primary uses? Why is the power spectrum such an important tool for discovering the state of the EEG at any one time?

Delta (0-4 Hz) Delta waves are usually seen during sleep, in infancy, or in serious organic brain diseases.

Theta (4-8 Hz) In healthy and alert adults, such slow activity is generally inconspicuous or absent, but it does appear during emotional stress (during periods of disappointment or frustration) or during certain stages of sleep. Diffuse theta and other slower activity are commonly encountered shortly after a generalized seizure, as well as in patients with metabolic disorders, white matter encephalopathy, or extensive lesions of the upper brain stem.

Alpha (8—13 Hz) Brain activity in this frequency range is often recorded from the occipital region (and sometimes from the parietal and frontal regions as well) during consciousness, and is attenuated by visual and other sensory stimulus. Alpha waves are typically seen in an awake but relaxed person and when eyes are closed. The waves tend to disappear in sleeping or attentive patients. When eyes are closed and visual stimulation ceases, the alpha wave assumes a virtual sinusoidal shape. When the eyes reopen, the sinusoidal character disappears. This phenomenon is called alpha desynchronization as it is believed that neural ensembles are firing in synchrony during this prior rhythm.

Beta (13-30 Hz) . It can be further classified into two subclasses: beta I (13-20 Hz) and beta II (20-50 Hz). Beta I, almost twice the alpha wave in frequency, appear together with alpha wave. It is affected by mental activity in much the same way as the alpha wave.

Beta II appears only during intense mental activity and tension. Thus, one type of beta activity is elicited by mental activity, whereas the other is inhibited by it Gamma (30+ Hz)-- in select research and clinical applications, gamma frequencies are used as indicators of cognitive function and arousal. Signal power or modulation at these frequencies is also used in brain machine interfaces as well as cognitive studies.

The power spectrum is easily the most generalizable measure with detailed features used to characterize the EEG. The EEG’s reactivity is almost entirely characterized by the power spectrum and that power in individual bands is often a critical tool in understanding the state of the EEG. 3) Why is the EEG so effective for detecting epileptic seizures in living beings? What is the primary characteristic of the seizure signal that we measure in different anatomically connected areas of the brain?

The EEG in seizure states is quite deterministic and not random. Spikes as well as several seizure waveform have definite shapes that are easy to recognize and therefore computational methods are easy to develop to capture EEG based tests for seizure detection.

4) What are the differences between parametric (model-based e.g. autoregressive) and nonparametric (e.g. FFT-based) spectral methods?

The chief non-parametric method uses the fast Fourier transform (FFT). It is rapidly generalizable to signals arising from a variety of sources. The autoregressive method uses an independent noise source as input to an all-pole or low pass filter as a signal generating model.

5) What are the chief characteristics of the MUSIC method of spectral analysis?

This is a method of direct sinusoidal modeling in the time domain using the autocorrelation matrix. Once the autocorrelation matrix is estimated, the eigendecomposition of this matrix is used to calculate the spectral line components of the represented time series.

Depending on the size of eigenvalues, respective eigenvectors form part of the signal or noise subspace. Frequency estimates are generated with the reciprocal of the inner product of noise subspace eigenvectors with a frequency vector.

6) Define the normalized separation of the EEG? What spectral method is used to calculate it? How is it calculated? What is the optimum normailized separation?

Normalized separation is an index of EEG recovery after cardiac arrest. It is based on an autoregressive spectral analysis of recovering rat EEG under 30 Hz . These power levels are then normalized relative to their baseline levels to produce three power bands:

Pnlf = Pl Pbl Pnmf = Pm Pbm Pnhf = Ph Pbh

Now the information can be reduced to one quantitative number, the Normalized Separation shown in the equation below. This study showed that the hyp injury causes a dispersion or redistribution of power in the dominant frequencies. NS monitors the rate of recovery for each band with respect to baseline. A high NS represents a disproportionate recovery of power, and vice versa. A high NS implies a poor recovery of the electrical function. Fractional NS provides good indicator of

NS =

Plf − Pmf + Pmf − Phf + Plf − Phf Plf + Pmf + Phf

7) Define IQ or the information quantity? How does IQ reflect the total entropy in the EEG? How does IQ magnitude prognosticate outcome after cardiac arrest in rats?

IQ measures and sums the entropy across several wavelet scales. Since it M

IQ( n ) = − pn

wc

( m )log 2 p nwc ( m )

m=1

where M is the total number of scales, m is the particular scale wherein the n is the particular data segment.

Since it includes all the wavelet scales in the summation, it measures total entropy across the entire spectrum.

8) What are uses of the measurement index that we call wavelets? Is there any advantage that we glean from using wavelets from independent sinusoidal signals?

Wavelets provide excellent representation of short duration signals. These signal components are of finite duration and are dilated and contracted in order to afford the EEG signal processor the ability to detect and estimate transients such as electrographic spikes and burst of short time length with great accuracy 9) Why is it so important to monitor the brain after cardiac arrest occurs? Does the EEG offer any benefits for monitoring the brain after cardiac arrest?

The lack of oxygen caused by disruption in circulation affects the health of the brain which is quite sensitive to the effects of cardiac arrest. EEG represents the mass action or summative neural activity of coordinated neurons. The lack of oxygen deprives these cells of vital metabolic processes which can cause brain death in a contracted period of time.

10) What are primary effects of the stimulant and neuro-peptide, orexin on the EEG? What measures have been utilized to categorize the EEG after orexin treatment?

Orexin increases the gamma output. Overall there is an increase in entropy as measured by the IQ entropy algorithm. The contribution of gamma to the total entropy

as measured by IQ shows remarkable increases with administration of orexin to the animal.

Solution Hints to Homework – Chapter 4 1. Answer the following questions about the general aspects of BCI. 1.1) Define Brain-computer interface (BCI) in your own words.

1.2) Describe at least 3 examples of BCI according to different signal resources and explain their pros and cons.

EEG, Motor imagery based BCI EEG, SSVEP based BCI fNIRS, Motor imagery based BCI 1.3) Describe what the unique challenges of BCI research are.

1.4) If you want to decode people’s motor imagery movement, which brain areas you want to choose in order to build an EEG based BCI? Sensorimotor cortex 2. Answer the following questions about the BCI signal acquisition.

2.1) What is the spatial resolution of noninvasive techniques such as EEG, MEG, and fMRI? Centimeter to sub-centimeters for MEG/EEG, millimeter for fMRI 2.2) What is the spatial resolution of invasive technique such as ECoG, multi-unit recording?

2.3) What is the temporal resolution of noninvasive technique such as EEG, MEG, and fMRI? Milliseconds for EEG and MEG, seconds for fMRI. 2.4) For EEG based BCI, does increasing the electrode number help to improve the decoding accuracy of motor imagination, why?

It does to a certain degree, but increasing number of electrodes alone would not keep improving the decoding accuracy.

2.5) Does the combination of different noninvasive modalities help to improve the decoding accuracy such as the simultaneous acquisition of EEG and fMRI? Please explain why?

It could. Multimodal recordings would provide complementary information, although limitation in practical applications for non-portable device such as fMRI should be considered.

3. Answer the following questions about the BCI feature extraction.

3.1) What kind of features could be extracted to decode the event-related potentials (ERP)?

Features in time domain, e.g. the peak value around a certain period after the event is triggered or features in the frequency domain, e.g. the amplitude of a certain frequency point after the event is triggered.

1

3.2) Is it possible to decode the ERP in single-trials? Please explain.

Yes, it is.

3.3) What kind of features could be used to decode the motor imagery induced sensorimotor rhythms?

For example, event-related synchronization or desynchronization.

4. Answer the following questions about the SSVEP BCI. 4.1) What is the limitation to use a computer monitor as the display of the flicker in a steady state visual evoked potential (SSVEP) based BCI? The fresh rate of the computer monitor is usually a limited number, e.g. 60Hz, 120Hz. 4.2) Download one of the examples (shared data, e.g., S1.mat, http://thubci.org/en/?a=nr&id=100) from the shared data in the ‘Wang et al (2016). A benchmark dataset for SSVEP-based brain–computer interfaces. IEEE Transactions on Neural Systems and Rehabilitation Engineering, 25(10), 1746-1752.’ Plot the power spectrum of electrode Oz from any one of the 40 targets in a single block and the average from all of the six blocks.

5. Answer the following questions about the motor imagery based BCI. 5.1) Download one of the examples (shared data, e.g. S1_LR_20150130.mat) from the shared data in (Meng et al., 2016)and Readme file to learn the structure of the shared data.

5.2) Extract the multichannel signals of each trial; calculate the average feedback duration for the example session.

5.3) Calculate the average band power (8-13Hz) of channel C3 and C4 over all of the left trials, respectively.

5.4) Calculate the average band power (8-13Hz) of channel C3 and C4 over all of the right trials, respectively.

5.5) Compare the above average band power for left trials and right trials, describe the difference.

6. What kinds of classification algorithms are commonly used in the EEG based BCI? Linear classifier such as Linear discriminant analysis (LDA), support vector machine (SVM) etc. Nonlinear classifiers such as neural network, etc.

7. Answer the following questions about robotic arm control using BCI. 7.1) Please explain what are the pros and cons to control a prosthetic or robotic arm by using different types of noninvasive BCI, such as SSVEP based and sensorimotor rhythm based.

Using SSVEP based BCI to control a prosthetic might provide high information transfer rate, but it intensively occupies the visual system, making it challenging to provide continuous and high dimensional control simultaneously. Using sensorimotor rhythm based BCI could provide continuous and high dimensional control at the same time, 2

but the information transfer rate might be lower compared to SSVEP based strategy.

Also, the performance might vary greatly among different participants.

7.2) What is the challenge for control of a high degree of freedom (DoF) robotic arm by noninvasive BCI? Please describe your solution of controlling a high DoF robotic arm.

Control of a robotic arm with high degree of freedom requires a simultaneous coordination of multiple coordinates, thus the BCI output have to provide coordinated high DoF information timely. It’s shown that the neural firing of single neurons in the motor area is closely related to the movement parameters. However, in the macroscopic scale, the information might be contaminated and deteriorated greatly in such as the scalp EEG recording. Thus, it is challenging to directly decode the movement intention noninvasively.

8. Answer the following questions about BCI applications.

1) What could BCI be used as a tool? Please describe at least three examples.

BCI keyboard BCI control of a robotic arm BCI control of a wheelchair BCI for education 2) Please describe how BCI could be used to induce tactile sensation neurofeedback.

When subjects successfully produce a control signal, the BCI system might provide a vibration feedback on the wrist of subjects. The subjects feel the tactile sensation, thus the BCI system closes the loop.

9. Answer the following questions about hybrid BCI. 9.1) Please describe an example of hybrid BCI. Combining EEG based BCI with EMG signals for the purpose of neural prosthetic control.

9.2) Please describe your solution of driving a wheelchair mounting with an assistive robotic arm to help drinking and eating via a hybrid BCI. 10. Answer the following questions about information transfer rate of BCI. 10.1) What is the state-of-the-art information transfer rate (ITR) of different types of noninvasive based BCI? For SSVEP based BCI, the ITR is over 300 bits/min, for VEP based BCI, the ITR is over 100 bits/min, for motor imagery based BCI, the ITR is around 20-30 bits/min so far.

10.2) Please describe a possible solution of increasing the ITR of a noninvasive sensorimotor rhythm based BCI and explain why it might work.

3

Using the hybrid acquisition modality such as combining the EEG and MEG might improve the ITR since EEG and MEG could provide supplementary information for each modality.

11. Answer the following questions about BCI development.

11.1) Please list three most important questions to be solved in order to significantly improve the field of BCI. The signal quality, the control algorithm, the easy use of the BCI system, etc.

11.2) Please discuss the potential of BCI application in the clinical field.

BCI application in stroke rehabilitation.

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Intracortical BrainMachine Interfaces: Problem set with solutions 1. Consider designing a BMI to classify movement to the right or left, and we want to test how well it works with one neuron. If the BMI user intends to move right, the neuron’s firing rate is drawn from a Gaussian distribution with mean 𝜇𝑟𝑖𝑔ℎ𝑡 = 8𝑠𝑝𝑖𝑘𝑒𝑠/𝑠𝑒𝑐𝑜𝑛𝑑 and standard deviation 𝜎𝑟𝑖𝑔ℎ𝑡 = 5𝑠𝑝𝑖𝑘𝑒𝑠/𝑠𝑒𝑐𝑜𝑛𝑑. If the BMI user intends to move left, the neuron’s firing rate is drawn from a Gaussian distribution with mean 𝜇𝑙𝑒𝑓𝑡 = 12𝑠𝑝𝑖𝑘𝑒𝑠/𝑠𝑒𝑐𝑜𝑛𝑑 and standard deviation 𝜎𝑙𝑒𝑓𝑡 = 6𝑠𝑝𝑖𝑘𝑒𝑠/𝑠𝑒𝑐𝑜𝑛𝑑. a. Suppose we make one measurement of the firing rate, y, and we assume the prior probability of ‘left’ and ‘right’ are equal. For each of the cases below, would we classify ‘left’, or ‘right’? y (spikes/second)

2

5

8

11

14

17

classification

right

right

right

right

left

left

From Bayes Rule 𝑃(𝐶 = 𝑐|𝑌 = 𝑦) = 𝑃(F=𝑦|𝐶=𝑐)𝑃(𝐶=𝑐) we know that 𝑃(F=𝑦)

𝑃(𝐶 = 𝑐|𝑌 = 𝑦) ∝ 𝑃(𝐶 = 𝑐)𝑃(𝑌 = 𝑦|𝐶 = 𝑐). In this example, 𝑃(𝑟𝑖𝑔ℎ𝑡) = 𝑃(𝑙𝑒𝑓𝑡) = 0.5. So, we can simply compare 𝑃(𝑌 = 𝑦|𝐶 = 𝑟𝑖𝑔ℎ𝑡) to 𝑃(𝑌 = 𝑦|𝐶 = 𝑙𝑒𝑓𝑡).

2 The general form of a Gaussian distribution is 𝑓(𝑥) =

1

σ√2𝜋

1 𝑥−𝜇

– (

𝑒 2

σ

)

. In Matlab, we

can use the function ‘normpdf’ to return the Gaussian distribution with mean µ and standard deviation σ evaluated at the value x, 𝑛𝑜𝑟𝑚𝑝𝑑𝑓(𝑥, µ, 𝜎).

𝑃(𝐶 = 𝑟𝑖𝑔ℎ𝑡|𝑌 = 2) ∝ 𝑛𝑜𝑟𝑚𝑝𝑑𝑓(2,8,5) = 0.0388 𝑃(𝐶 = 𝑙𝑒𝑓𝑡|𝑌 = 2) ∝ 𝑛𝑜𝑟𝑚𝑝𝑑𝑓(2,12,6) = 0.0166 The likelihood of observing 2 spikes/second given that the intention was right is greater than the probability given that the intention was left, so we predict that the intention was right. Repeat for all other values of y in the table. b. Suppose we now assume that the BMI user moves ‘left’ twice as often as ‘right’ (i.e., 𝑃(𝑙𝑒𝑓𝑡) = 2/3 and 𝑃(𝑟𝑖𝑔ℎ𝑡) = 1/3. For each of the cases below, would we classify ‘left’, or ‘right’? y (spikes/second)

2

5

8

11

14

17

classification

right

left

left

left

left

left

Now, we must incorporate the different values for 𝑃(𝑟𝑖𝑔ℎ𝑡) and 𝑃(𝑙𝑒𝑓𝑡).

1 ( | ) ( ) 𝑃 𝐶 = 𝑟𝑖𝑔ℎ𝑡 𝑌 = 2 ∝ F G ∗ 𝑛𝑜𝑟𝑚𝑝𝑑𝑓 2,8,5 = 0.0129 3 2 ( | ) ( ) 𝑃 𝐶 = 𝑙𝑒𝑓𝑡 𝑌 = 2 ∝ F G ∗ 𝑛𝑜𝑟𝑚𝑝𝑑𝑓 2,12,6 = 0.0111 3

Because 𝑃(𝐶 = 𝑟𝑖𝑔ℎ𝑡|𝑌 = 2) > 𝑃(𝐶 = 𝑙𝑒𝑓𝑡|𝑌 = 2) , we would choose ‘right’ in this case. Repeat for all other values of y.

2. In section 3 we showed how to implement a classifier with Gaussian firing statistics, where the neural activity for class k is modeled as 𝒀~𝑁(𝝁𝑘, Z𝑘), where 𝝁𝑘 ∈ 𝑅𝐷 and Z𝑘 ∈ 𝑅𝐷𝑥𝐷 are the mean and covariance of the activity of a population of D neurons. Here we will assume that the covariance matrix is the same for each class 𝑘 = 1, … , 𝐾, and let Z = Z1 = Z2 = ⋯ = Z𝐾. a. First, suppose we have a new recording of neural activity, y. Also, suppose that

𝑃(𝐶 = 𝑘) = 𝜋𝑘. Using Bayes’rule, find 𝑙𝑜𝑔 𝑃(𝐶 = 𝑘|𝒀 = 𝒚), up to the normalizing constant.

1 𝑃(𝒀 = 𝒚|𝐶 = 𝑘) = (2𝜋)−𝐷/2|Z|−1/2 𝑒𝑥𝑝 I− (𝒚 − 𝝁 )𝖳Z−1(𝒚 − 𝝁𝑘)K 𝑘 2 𝑙𝑜𝑔 𝑃 ( 𝐶 = 𝑘 ∣ 𝒀 = 𝒚 ) ∝ 𝑙𝑜𝑔 𝑃 (𝒀 = 𝒚|𝐶 = 𝑘) + 𝑙𝑜𝑔𝑃(𝐶 = 𝑘) =−

𝐷

1 1 𝑙𝑜𝑔(2𝜋) − 𝑙𝑜𝑔|Z| − (𝒚 − 𝝁 )𝖳Z−1(𝒚 − 𝝁𝑘) + 𝑙𝑜𝑔 𝝅𝑘 𝑘 2 2 2

b. Find the decision boundary used for determining whether the point y came from class j or class k and simplify the expression.

We have more evidence that the point y belongs to class k rather than to class j whenever 𝑃(𝐶 = 𝑘|𝒀 = 𝒚) > 𝑃(𝐶 = 𝑗|𝒀 = 𝒚).

Or equivalently, whenever 𝑙𝑜𝑔 𝑃(𝐶 = 𝑘|𝒀 = 𝒚) > 𝑙𝑜𝑔 𝑃 (𝐶 = 𝑗|𝒀 = 𝒚).

After plugging in the expression we found in part a into the above rule, and cancelling the shared terms, we are left with:

1 − (𝒚 − 𝝁 )𝖳Z–1(𝒚 − 𝝁 ) + 𝑙𝑜𝑔 𝝅 𝑘 𝑘 𝑘 12 𝖳 –1

> − X𝒚 − 𝝁jY Z 2

X𝒚 − 𝝁jY + 𝑙𝑜𝑔 𝝅j 𝖳 −1

To simplify this further, we can now rewrite the term (𝒚 − 𝝁𝑘) Z

(𝒚 − 𝝁𝑘) =

𝒚𝖳Z−1𝒚 + 𝝁𝖳𝑘Z−1𝝁𝑘 − 2𝝁𝑘𝖳Z−1𝒚, and apply the similar procedure to the righthand side. This lets us cancel even more shared terms, leaving us with:

1 𝖳 –1 1 𝖳 –1 𝖳 –1 𝖳 –1 − 𝝁𝑘Z 𝝁𝑘 + 𝝁𝑘 Z 𝒚 + 𝑙𝑜𝑔 𝝅𝑘 > − 𝝁j Z 𝝁j + 𝝁j Z 𝒚 + 𝑙𝑜𝑔 𝝅j 2 2 Now let 𝒘𝑘 = Z−1𝝁 and 𝒘𝒌0 = − 1 𝝁𝖳Z−1𝝁 + 𝑙𝑜𝑔 𝜋𝑘, and similarly for 𝒘𝑗 and 𝑘

2

𝑘

𝑘

𝒘𝑗0. Then we can write our above decision rule as follows: 𝒘𝖳𝒚 + 𝒘𝑘0 > 𝒘𝖳𝒚 + 𝒘𝑗0 𝑘

𝑗

c. Is the decision boundary linear?

From the previous problem, we know that the decision boundary between class k and class j is the set of y’s where 𝖳

𝒘𝖳𝒚 + 𝒘𝑘0 = 𝒘𝖳𝒚 + 𝒘𝑗0 ⇔ (𝒘𝑘 − 𝒘𝒋) 𝒚 + (𝒘𝑘0 − 𝒘𝑗0) = 0 𝑘

𝑗

Because we have written the decision boundary in the form Ay + b = 0, this is clearly a linear decision boundary.

3. In this problem we will derive the equations to implement a classifier based on Poisson spike counts.

The spike count of neuron i given class k is Poisson-distributed with parameter 𝜆𝑖𝑘. We will assume that the D neurons, 𝑌1, … , 𝑌𝐷, are conditionally independent given the class j. In other words, given neural activity 𝒚 ∈ 𝑅𝐷, the probability that 𝒚 came from class k is as follows:

𝑃(𝒀 = 𝒚|𝐶 = 𝑘) = ∏𝐷𝑖=1 𝑃(𝑌𝑖 = 𝑦𝑖|𝐶 = 𝑘), where 𝑦𝑖/𝑦𝑖! 𝑃(𝑌𝑖 = 𝑦𝑖|𝐶 = 𝑘) = 𝑒𝑥𝑝(−𝜆𝑖𝑘) 𝜆𝑖𝑘

a. Let 𝑃(𝐶 = 𝑘) = 𝜋𝑘. Find 𝑃(𝐶 = 𝑘|𝒀 = 𝒚) using Bayes rule.

𝑃(𝐶 = 𝑘|𝒀 = 𝒚) ∝ 𝑃(𝐶 = 𝑘)𝑃(𝒀 = 𝒚|𝐶 = 𝑘) 𝐷

= 𝜋𝑘 𝖦 𝑒𝑥𝑝(−𝜆𝑖𝑘) 𝜆𝑦𝑖 𝑖𝑘 /𝑦𝑖! 𝑖=1

Now simplify the expression above by taking the log: 𝑙𝑜𝑔𝑃(𝐶 = 𝑘|𝒀 = 𝒚). 𝐷

𝑙𝑜𝑔𝑃(𝐶 = 𝑘|𝒀 = 𝒚) ∝ 𝑙𝑜𝑔𝜋𝑘 + V 𝑦𝑖 𝑙𝑜𝑔 𝜆𝑖𝑘 − 𝜆𝑖𝑘 − 𝑙𝑜𝑔(𝑦𝑖!) 𝑖=1

b. Given a new point y, we want to determine to which class this point belongs. Derive the decision boundary that determines whether we classify a new point y as belonging to either class j or class k. Use the expression that you derived in part a.

Given two different classes j and k, we classify y as class k if:

𝑃(𝒀 = 𝒚|𝐶 = 𝑘)𝑃(𝐶 = 𝑘) > 𝑃(𝒀 = 𝒚|𝐶 = 𝑗)𝑃(𝐶 = 𝑗) ⇒ 𝑙𝑜𝑔 𝑃(𝒀 = 𝒚|𝐶 = 𝑘)𝑃(𝐶 = 𝑘) > 𝑙𝑜𝑔 𝑃(𝒀 = 𝒚|𝐶 = 𝑗)𝑃(𝐶 = 𝑗) 𝐷

⇒ 𝑙𝑜𝑔 𝜋𝑘 + '[𝑦𝑖 𝑙𝑜𝑔 𝜆𝑖𝑘 − 𝜆𝑖𝑘 − 𝑙𝑜𝑔(𝑦𝑖!)] 𝑖=1 𝐷

> 𝑙𝑜𝑔 𝜋j + '1𝑦𝑖 𝑙𝑜𝑔 𝜆𝑖j − 𝜆𝑖j − 𝑙𝑜𝑔(𝑦𝑖!)2 𝑖=1

The decision boundary is the set of points where 𝑃(𝒀 = 𝒚|𝐶 = 𝑘)𝑃(𝐶 = 𝑘) =

𝑃(𝒀 = 𝒚|𝐶 = 𝑗)𝑃(𝐶 = 𝑗). Plugging in the above, we can simplify this to: 𝜆𝑖𝑘 𝜋𝑘 𝐷 + X𝜆 − 𝜆 Ya = 0 𝑙𝑜𝑔 + _ `𝑦 𝑙𝑜𝑔 𝑖 𝑖j 𝑖𝑘 𝜋j 𝜆𝑖j 𝑖=1

c. Is the decision boundary linear?

Because all terms involving y are linear, this is a linear decision boundary.

4. In Equation 9 of Section 3 we provided 1the following expressions for the training phase of a 𝑁 classifier: 𝝁 = 1 ∑𝑁 𝒚 and Z𝑘 = ∑ (𝒚 − 𝝁 ) (𝒚 − 𝝁 )𝖳, where 𝒚 ∈ 𝑅𝐷 for all 𝑖 = 𝑘

𝑁 𝑖=1

𝑖

𝑁 𝑖=1

𝑖

𝑘

𝑖

𝑘

𝑖

1, … , 𝑁 is the neural activity recorded with class k, 𝝁𝑘 ∈ 𝑅𝐷, and Z𝑘 ∈ 𝑅𝐷×𝐷. Show that these values of 𝝁𝑘 and Z𝑘 maximize the following equation for the likelihood: 𝐿Z𝝁𝑘, Z𝑘|𝒚1, … , 𝒚𝑁,𝑐𝑘[ = 𝑃(𝒚1, … , 𝒚𝑁|𝑐𝑘) 𝑁

= 𝖦(2𝜋)−𝐷/2|Z𝑘|−1/2𝑒𝑥𝑝 I− 𝑖=1

1 2

(𝒚𝑖 − 𝝁𝑘)𝖳Z𝑘−1(𝒚𝑖 − 𝝁𝑘)K

Instead of maximizing the likelihood, we can equivalently maximize the log of the likelihood, which is the following:

𝑁𝐷

1 𝑁 1 ) | | (𝒚 )𝖳𝛴 –1(𝒚 − 𝝁 ) − 𝝁 𝑙𝑜𝑔 2𝜋 – 𝑙𝑜𝑔 𝛴𝑘 − ' 𝑖 𝑘 𝑘 𝑖 𝑘 2 2 𝑖=1 2

𝒄 <=−

𝑙𝑜𝑔𝐿4𝝁 𝑘 , 𝛴𝑘 |𝒚 1, … , 𝒚𝑁, 𝑘

(

To find where this function attains its maxima, we will first take the derivative with respect to 𝝁𝑘 and Z𝑘, set each of these expressions to zero, and solve. Note that to take the derivative with respect to Z𝑘, we will use the fact that 𝒙𝖳Z−1𝒙 = 𝑡𝑟𝑎𝑐𝑒(Z−1𝒙𝒙𝖳).

𝜕 𝑙𝑜𝑔 𝐿 (𝝁 , Z ) 𝑘

𝜕𝝁𝑘

𝑘

𝑁

= _ Z–1(𝒚𝑖 − 𝝁𝑘) = 0 𝑖=1

𝑁

𝑁

𝑖=1

𝑖=1

1 ⟹ V(𝒚𝑖 − 𝝁𝑘) = 0 ⟹ 𝝁𝑘 = V 𝒚𝑖 𝑁 𝑁

𝜕 𝑙𝑜𝑔 𝐿 (𝝁 , Z ) 𝑘

𝑘

=_

𝜕Z𝑘

𝑖=1

1

1 2

Z𝑘–1(𝒚𝑖 − 𝝁𝑘)(𝒚𝑖 − 𝝁𝑘)𝖳Z𝑘–1 −

2

Z𝑘–1 = 0

𝑁

⟹ V Z𝑘−1(𝒚𝑖 − 𝝁𝑘)(𝒚𝑖 − 𝝁𝑘)𝖳Z𝑘−1 = 𝑁Z𝑘−1 𝑖=1 𝑁

1 ⟹ V(𝒚𝑖 − 𝝁𝑘)(𝒚𝑖 − 𝝁𝑘)𝖳 = Z𝑘 𝑁 𝑖=1

5. In section 4, we considered a two neuron example of the PVA decoder where the neurons had orthogonal preferred directions (e.g., one neuron preferred 90 while the other neuron preferred

180). Show that if the two neurons do not have orthogonal tuning directions, the directions decoded by PVA will be biased.

Suppose we have two neurons whose preferred directions are 45 degrees apart, 135 and 180 (Figure A). Suppose the BCI user attempted to make a movement towards a target located at 180.

In this case, the blue neuron will fire at its maximal rate and push the cursor towards its preferred direction, 180. However, the magenta neuron will also be near its maximal rate, and will push the cursor towards its preferred direction, 135. The result is that the cursor movement doesn’t point directly towards 180, but rather points up and to the left. Figure B shows the results of noiseless PVA decoding with the two neurons shown in A. The tail of each arrow is the intended movement direction, while the tip points to the decoded direction (with length proportional to the speed of the push). Since these arrows don’t point toward their intended targets, this decoder is biased.

6. Show that a neuron that exhibits cosine tuning also shows linear tuning to velocity. That is, suppose that given a reach in the θ direction with speed s, a neuron’s firing rate can be written as 𝑦 =

𝑏7 + 𝑚𝑠𝑐𝑜𝑠 `𝜃 − 𝜃𝑝→b, where 𝑏7is the neuron’s baseline firing rate, 𝑚 is its modulation depth, and 𝜃→ is its preferred direction. Show that this means we can also write 𝑦 = 𝑏0 + 𝒃𝑇𝒗, where v 𝑝

is the velocity of the reach and b and v are both 2D vectors.

The term 𝒃𝑇𝒗 is a dot product between two vectors. The dot product between two vectors b and v is |𝒃||𝒗| 𝑐𝑜𝑠 𝜑, where |𝒃| is the magnitude of b, |𝒗| is the magnitude of v, and 𝜑 is the angle between the two vectors. Let m be the magnitude of b, and 𝜃→ its direction. Let s be the magnitude 𝑝

of v (i.e., speed), and 𝜃 its direction. Then we have 𝑚 = |𝒃|, 𝑠 = |𝒗|, and 𝜑 = 𝜃 − 𝜃→𝑝. This lets us write:

𝑦 = 𝑏0 + 𝒃𝑇𝒗 = 𝑏0 + |𝒃||𝒗| 𝑐𝑜𝑠 𝜑 = 𝑏0 + 𝑚𝑠 𝑐𝑜𝑠(𝜃 − 𝜃→𝑝).

This shows that the two equations are equivalent.

7. Derive the expressions for the training phase of the Kalman filter in section 4:

𝐵 = (∑𝑇

𝒚𝑡𝒙𝑇)(∑𝑇

𝑡=1

𝑡

𝑡=1

𝒙𝑡𝒙𝑇𝑡)−1

1 Z = ∑ 𝑇𝑡=1 (𝒚𝑡 − 𝐵𝒙𝑡)(𝒚𝑡 − 𝐵𝒙𝑡)𝑇 𝑇

𝐴 = (∑𝑇

𝒙 𝒙𝑇 )(∑𝑇

𝑡=2

𝑄=

1

𝑇–1

𝑡 𝑡–1

𝒙

𝑡=2 𝑡–1

𝒙𝑇𝑡–1 )–1

∑𝑇𝑡=2 (𝒙𝑡 − 𝐴𝒙𝑡–1 )(𝒙𝑡 − 𝐴𝒙𝑡–1 )𝑇

𝑙𝑜𝑔𝐿(𝜃) = 𝑙𝑜𝑔 𝑃({𝒚}, {𝒙}|𝜃) 𝑇

𝑇

= 𝑙𝑜𝑔 𝑃(𝒙1) + V 𝑙𝑜𝑔 𝑃(𝒙𝑡|𝒙𝑡−1) + V 𝑙𝑜𝑔 𝑃(𝒚𝑡|𝒙𝑡) =−

𝑀

𝑡=2

1

1

𝑡=1

( ) | | (𝒙 − 𝛱)𝑇𝑉−1(𝒙 − 𝛱) 𝑙𝑜𝑔 2𝜋 − 𝑙𝑜𝑔 𝑉 − 1 2 2 2 1 𝑇 𝑀 1 + V n− 𝑙𝑜𝑔(2𝜋) − 𝑙𝑜𝑔|𝑄| 2 2 𝑡=2 1 − (𝒙 − 𝐴𝒙𝑡−1 )𝑇𝑄−1(𝒙𝑡 − 𝐴𝒙𝑡−1)q 2 𝑡 𝑇 𝐷 1 1 + V n− 𝑙𝑜𝑔(2𝜋) − 𝑙𝑜𝑔|Z| − (𝒚𝑡 − 𝐵𝒙𝑡)𝑇Z−1(𝒚𝑡 − 𝐵𝒙𝑡)q 2 2 2 𝑡=1

𝜕 𝑇 𝜕𝐿(𝜃) o_(−𝒙𝑇 𝐴𝑇𝑄–1𝒙 − 𝒙𝑇𝑄–1𝐴𝒙 )p + 𝒙𝑇 𝐴𝑇𝑄–1𝐴𝒙 = 𝑡–1 𝑡 𝑡 𝑡–1 𝑡–1 𝑡–1 𝜕𝐴 𝜕𝐴 𝑡=2

𝑇

𝑇

𝜕 {−𝑇𝑟 (𝐴𝑇𝑄−1 ∑ 𝒙 𝒙𝑇 ) − 𝑇𝑟 (𝐴 (∑ 𝒙 𝒙𝑇) 𝑄−1) = 𝑡 𝑡−1 𝑡−1 𝑡 𝜕𝐴 𝑡=2

𝑡=2

𝑇

+ 𝑇𝑟 (𝑄−1𝐴 (∑ 𝒙𝑡−1 𝒙𝑇𝑡−1 ) 𝐴𝑇)} 𝑡=2

𝑇

𝑇

𝑇

𝑇 F = −𝑄–1 D' 𝒙𝑡𝒙𝑇𝑡–1 F − 𝑄–1 D' 𝒙𝑡𝒙𝑇𝑡–1 F + 𝑄–1𝐴 D' 𝒙𝑡–1𝒙𝑡–1

𝑡=2

𝑡=2

𝑡=2

𝑇 𝑇 F + 𝑄–1𝐴 D' 𝒙𝑡–1𝒙𝑡–1

𝑡=2

= [0] –1

𝑇

𝑇

𝐴 = q_ 𝒙 𝒙𝑇

r q_ 𝒙

𝑡 𝑡–1 𝑡=2

(𝑇 − 1)

𝜕 =

𝜕𝑄

o−

2

(𝑇 − 1) =−

r

𝑡–1 𝑡–1 𝑡=2

𝜕𝐿(𝜃) 𝜕𝑄

𝒙𝑇

𝑛=2

𝑇

1 𝑄−1 −

2

𝑇

1 𝑙𝑜𝑔|𝑄| − 𝑇𝑟 q𝑄–1 _(𝒙𝑡 − 𝐴𝒙𝑡–1)(𝒙𝑡 − 𝐴𝒙𝑡–1)𝑇rp 2

2

s−𝑄−1 V(𝒙𝑡 − 𝐴𝒙𝑡−1)(𝒙𝑡 − 𝐴𝒙𝑡)𝑇𝑄−1t 𝑛=2

= [0] 𝑄=

1 𝑇−1

𝑇

_(𝒙𝑡 − 𝐴𝒙 𝑡–1 )(𝒙𝑡 − 𝐴𝒙𝑡–1 )𝑇 𝑡=2

𝜕 𝑇 𝜕𝐿(𝜃) o_(−𝒙𝑇 𝐵𝑇 Z–1𝒚 = 𝑡 𝜕𝐵 𝜕𝐵 𝑡=1

− 𝒚𝑇Z–1𝐵𝒙 + 𝒙𝑇𝐵𝑇Z–1𝐵𝒙 )p 𝑡

𝑡

𝑡

𝑡

𝑡

𝑇

=

𝜕 𝜕𝐵

𝑇

{−𝑇𝑟 (𝐵𝑇Z–1 ∑ 𝒚 𝒙𝑇) − 𝑇𝑟 (𝐵 (∑ 𝒙 𝒚𝑇) Z–1) 𝑡 𝑡

𝑡 𝑡

𝑡=1

𝑡=1 𝑇

+ 𝑇𝑟 (Z–1𝐵 (∑ 𝒙 𝒙𝑇) 𝐵𝑇)} 𝑡 𝑡 𝑡=1

𝑇

𝑇

𝑇

𝑇

= −Z−1 sV 𝒚𝑡𝒙𝑇t − Z−1 sV 𝒚𝑡𝒙𝑇t + Z−1𝐵 sV 𝒙𝑡𝒙𝑇t + Z−1𝐵 sV 𝒙𝑡𝒙𝑇t 𝑡

𝑡

𝑡=1

𝑡

𝑡=1

𝑡=1

𝑡 𝑡=1

= [0] 𝑇

−1

𝑇

𝐵 = sV 𝒚𝑡𝒙𝑇t sV 𝒙𝑡𝒙𝑇t 𝑡

𝑡=1

𝑡=1

𝜕𝐿(𝜃)

𝑡

𝑇

𝜕

𝑇 1 = v− 𝑙𝑜𝑔|Z| − 𝑇𝑟 sZ−1 V(𝒚 𝑡 − 𝐵𝒙𝑡)(𝒚𝑡 − 𝐵𝒙𝑡)𝑇tw 𝜕Z 𝜕Z 2 2 𝑚=1

𝑇

𝑇

1 = − Z−1 − s−Z−1 V(𝒚 𝑡 − 𝐵𝒙𝑡)(𝒚𝑡 − 𝐵𝒙𝑡)𝑇Z−1t 2 2 𝑚=1

= [0] 1 Z=

𝑇

𝑇

_(𝒚𝑡 − 𝐵𝒙𝑡)(𝒚𝑡 − 𝐵𝒙𝑡)𝑇 𝑡=1

Intuitively, the expressions for A and Q are entirely analogous to those in linear regression, where A is the slope and Q is the minimum mean squared error. Similarly, for B and Σ.

8. Consider using a BMI to play Pong with one neuron. That is, we will use a Kalman filter to decode position along a one dimensional axis from the firing rate of a single neuron. Let the state model be 𝑥𝑡 = 𝑥𝑡−1 + 𝜔𝑡, 𝜔𝑡~𝑁(0, 𝑞), and the observation model be 𝑦𝑡 = 𝑏𝑥𝑡 + 𝜀𝑡, 𝜀𝑡~𝑁(0, 𝜎).

a. Show that the estimate of the position on time step t, μt, can be written in the form:

𝑦 𝜇𝑡 = (1 − 𝛼)𝜇𝑡−1 + 𝛼 ( 𝑡 ) 𝑏 µ− = µ 𝑡

𝑡−1

µ𝑡 = µ𝑡−1 + 𝑘𝑡(𝑦𝑡 − 𝑏µ𝑡−1) 𝑤ℎ𝑒𝑟𝑒 𝑘𝑡 =

𝑏𝜑–2 𝑏2𝜑2– + 𝜎2

We can rewrite µ𝑡 = (1 − 𝑘𝑡𝑏)µ𝑡−1 + 𝑘𝑡𝑦𝑡 And including

𝑏 𝑏

𝑦

in the last term yields: µ𝑡 = (1 − 𝑘𝑡𝑏)µ 𝑡−1 + 𝑘𝑡𝑏 ( 𝑏𝑡)

Now let 𝛼 = 𝑘𝑡𝑏. Then we have:

𝑦 µ𝑡 = (1 − 𝛼)µ 𝑡−1 + 𝛼 ( 𝑡 ) 𝑏 b. Prove that 0≤α≤1.

𝛼 = 𝑘𝑡𝑏 =

𝑏2𝜑2– 𝑏2𝜑2– + 𝜎2

As 𝜎2 gets large 𝛼 approaches 0. As 𝜎2 goes to zero 𝛼 approaches 1. c. When does α approach 0? Under this case, why does it make sense for 𝜇𝑡 = 𝜇𝑡−1 ?

When 𝜎2 is large relative to 𝑏2𝜑2−. Then the state estimate at this time step will be really close to the mean at the last time step. The information from the neuron is unreliable, so the Kalman filter ignores it in favor of the information from the state evolution. d. When does α approach 1? Under this case, why does it make sense for 𝜇𝑡 = 𝑦𝑡/𝑏?

2 . The information from the neuron is quite reliable, When 𝜎2 is small relative to 𝑏2𝜑− so the Kalman filter weights that information more than the information from the state evolution.

9. You decide to speed up the implementation of your Kalman filter by skipping the one-step prediction.

Whereas normally you would solve the measurement update (Eq. 5.17) and one-step predictions iteratively on each timestep (5.16). You instead decide to just iterate the Measurement Update step, by directly plugging in the velocity estimate from the previous time step, P(𝒗𝑡−1|{𝒚}𝑡−1 1 ), without making a one-step prediction:

𝑡–1 𝑃(𝒗𝑡|{𝒚}𝑡 ) = 𝑃(𝒚𝑡 |𝒗𝑡 )𝑃(𝒗𝑡–1 |{𝒚} 1 ) 1 𝑃(𝒚𝑡|{𝒚}1𝑡–1)

Describe qualitatively what will happen to the velocity estimate over time if you do this.

(Hint: when in doubt, try (1) simulating it, or (2), solving the 1D case…)

If you skip the one-step prediction step, the covariance of your estimate from the measurement update will gradually converge to zero. In other words, your model will gradually get more certain about a state estimate that is incorrect, causing your decoded value to freeze at a particular value.

10. The goal of the measurement update for Kalman filter is to find 𝑃(𝒙𝑡|𝒚1, … , 𝒚𝑡). To do so, we adopted the strategy in Section 4 whereby we would first find the joint distribution

𝑃(𝒙𝑡, 𝒚𝑡|𝒚1, … , 𝒚𝑡−1), and then use the theorem of conditioning for jointly Gaussian random variables to find 𝑃(𝒙𝑡|𝒚1, … , 𝒚𝑡). Here we will derive the means and covariances of the joint distribution

𝒚𝑡|𝒚1, … , 𝒚𝑡−1 𝐵µ− 𝐵𝛷−𝐵𝑇 + Z 𝐵𝛷− 𝑡 𝑡 ÅK | } ~𝑁 I| −𝑡 } , É µ𝑡 𝛷𝑡− 𝛷𝑡−𝐵𝑇 𝒙𝑡|𝒚1, … , 𝒚𝑡−1 a. Find the mean of 𝒚𝑡|𝒚1, … , 𝒚𝑡−1

𝐸[𝒚𝑡|𝒚1, … , 𝒚𝑡−1] = 𝐸[𝐵𝒙𝑡|𝒚1, … , 𝒚𝑡−1] + 𝐸[𝜺𝑡|𝒚1, … , 𝒚𝑡−1] Taking this by parts, the second term 𝐸[𝜺𝑡|𝒚1, … , 𝒚𝑡−1] = 0. The first term

𝐸[𝐵𝒙𝑡|𝒚1, … , 𝒚𝑡−1] = 𝐵𝐸[𝒙𝑡|𝒚1, … , 𝒚𝑡−1].

So, 𝐸[𝒚𝑡|𝒚1, … , 𝒚𝑡−1] = 𝐵µ−𝑡 b. Find the variance of 𝒚𝑡|𝒚1, … , 𝒚𝑡−1 𝑉𝑎𝑟[𝒚𝑡|𝒚1, … , 𝒚𝑡–1] = 𝑉𝑎𝑟[𝐵𝒙𝑡|𝒚1, … , 𝒚𝑡–1] + 𝑉𝑎𝑟[𝜺𝑡|𝒚1, … , 𝒚𝑡–1] + 2𝐶𝑜𝑣[𝐵𝒙𝑡, 𝜺𝑡|𝒚1, … , 𝒚𝑡–1]

Again, taking this by parts, the third term 2𝐶𝑜𝑣[𝐵𝒙𝑡, 𝜺𝑡|𝒚1, … , 𝒚𝑡–1] = 0. The second term 𝑉𝑎𝑟[𝜺𝑡|𝒚1, … , 𝒚𝑡−1] = Z. The first term involves the covariance of our prior multiplied by a constant matrix, B. So, 𝑉𝑎𝑟[𝒚𝑡|𝒚1, … , 𝒚𝑡–1] = 𝐵𝛷𝑡–𝐵𝑇 + Z. c. Find the covariance of 𝒙𝑡, 𝒚𝑡 when both are conditioned on 𝒚𝑡, … , 𝒚𝑡−1 Recall the relationship between Covariance and Expectation:

𝐶𝑜𝑣[𝒙𝑡, 𝒚𝑡|𝒚1, … , 𝒚𝑡−1] = 𝐸[𝒚 𝒙𝑇|𝒚 , … , 𝒚 𝑡 𝑡

1

𝑡−1

] − 𝐸[𝒚 |𝒚 , … , 𝒚 𝑡

1

𝑡−1

]𝐸[𝒙 |𝒚 , … , 𝒚 𝑡

1

𝑡−1

]𝑇

𝑇

We know 𝐸[𝒙𝑡|𝒚1, … , 𝒚𝑡–1]𝑇 is (µ−) . We also know 𝐸[𝒚𝑡|𝒚1, … , 𝒚𝑡−1] is 𝐵µ−. 𝑡

𝑡

To get 𝐸[𝒚𝑡𝒙𝑇𝑡|𝒚1, … , 𝒚𝑡−1] we use the observation equation: 𝒚𝑡 = 𝐵𝒙𝑡 +

𝜺𝑡, 𝜺𝑡~𝑁(0, Z) 𝐸[𝒚 𝒙𝑇|𝒚 , … , 𝒚 𝑡 𝑡

1

] = 𝐸[(𝐵𝒙 + 𝜺 )𝒙𝑇|𝒚 , … , 𝒚 𝑡−1

𝑡

1

𝑡

1

] = 𝐵𝐸[𝒙 𝒙𝑇|𝒚 , … , 𝒚 𝑡−1

𝑡 𝑡

1

] + 𝐸[𝜺 𝒙𝑇|𝒚 , … , 𝒚 𝑡−1

𝑡 𝑡

1

] 𝑡−1

] = 0.

𝐸[𝜺 𝒙𝑇|𝒚 , … , 𝒚 𝑡 𝑡

𝑡

𝑡–1

So, 𝐶𝑜𝑣[𝒙𝑡, 𝒚𝑡|𝒚1, … , 𝒚𝑡–1] = 𝐵𝐸[𝒙 𝒙𝑇|𝒚1, … , 𝒚𝑡–1] − 𝐵µ–µ–𝑇 𝑡 𝑡

𝑡

𝑡

Using the definition of variance

𝑉𝑎𝑟[𝒙𝑡|𝒚1, … , 𝒚𝑡–1] = 𝐸[𝒙𝑡 𝒙𝑡𝑇|𝒚1, … , 𝒚𝑡–1] − 𝐸[𝒙𝑡|𝒚1, … , 𝒚𝑡–1]𝐸[𝒙𝑇𝑡|𝒚1, … , 𝒚𝑡–1] We can rewrite 𝐵𝐸[𝒙 𝒙𝑇|𝒚 , … , 𝒚 𝑡 𝑡

1

𝑡−1

] = 𝐵𝑉𝑎𝑟[𝒙 |𝒚 , … , 𝒚 𝑡

1

𝑡−1

]+

𝐵𝐸[𝒙𝑡|𝒚1, … , 𝒚𝑡−1]𝐸[𝒙𝑇𝑡|𝒚1, … , 𝒚𝑡−1]

𝑉𝑎𝑟[𝒙𝑡|𝒚1, … , 𝒚𝑡−1] = 𝛷−𝑡 𝐸[𝒙𝑡|𝒚1, … , 𝒚𝑡−1] = µ− and 𝐸[𝒙𝑇|𝒚1, … , 𝒚𝑡–1] = µ–𝑇 𝑡 1

𝑡

] = 𝐵𝛷– + 𝐵µ–µ–𝑇

So, 𝐵𝐸[𝒙 𝒙𝑇|𝒚 , … , 𝒚 𝑡 𝑡

𝑡

𝑡–1

𝑡

𝑡

𝑡

Putting this all together:

𝐶𝑜𝑣[𝒙𝑡, 𝒚𝑡|𝒚1, … , 𝒚𝑡–1] = 𝐵𝛷𝑡 – + 𝐵µ𝑡–µ𝑡–𝑇 − 𝐵µ𝑡–µ𝑡–𝑇 = 𝐵𝛷–𝑡

11-12. We have provided a dataset (https://github.com/emilyoby/bmi-data-set) consisting of kinematic data recorded during center-out reaches and neural activity from a Utah electrode array implanted in M1. The following describes the data format. The .mat file has two data structures: trainTrials contains 180 trials to be used as training data, and testTrials contains 8 trials to be used as test data.

Each data structure contains spikes, handPos, and handVel variables, representing the spiking activity, hand position, and hand velocity, respectively, on each trial in which a monkey reached to one of eight different targets. The spikes variable contains, for each trial, the number of threshold crossings recorded simultaneously from the 91 electrodes, and has dimensions (n time steps) x (91 electrodes), where n is the number of 50ms time steps within a particular trial. For example, trainTrials.spikes{i}(n,k) contains the number of threshold crossings recorded on the kth electrode in the nth time step of the ith trial. The handPos and handVel variables are structured similarly, and contain the 2D hand position (in mm) and velocity (in mm/sec), respectively, for the same time steps as in the spikes variable.

For the problems below, use the provided neural and kinematic data to implement the continuous decoders discussed in section 4.

11. Use PVA decoder to estimate the movement velocity during a center out task. a. Fit the parameters of the decoder using the 180 trials of training data. % All solutions below assume data has been loaded into workspace: % load('decodingData.mat', 'trainTrials', 'testTrials'); % Calculate preferred directions spk = trainTrials.spikes; pos = trainTrials.handPos; vel = trainTrials.handVel; X = vertcat(vel{:}); X = [X./(sqrt(sum(X.^2,2))*ones(1,2)) ones(size(X,1),1)]; spkTrain = vertcat(spk{:}); b = (X'*X)^(-1)*X'*spkTrain; md = sqrt(sum(b(1:2,:).^2,1)); %modulation depth pd = atan2d(b(2,:),b(1,:)); %preferred direction angle p = [cosd(pd);sind(pd)]; %preferred direction vector

b. Test the decoder on the 8 test trials. Plot this decoded trajectory and the actual movement trajectory on the same plot. spk = testTrials.spikes; pos = testTrials.handPos; vel = testTrials.handVel; dt = 50; figure; hold on for i = 1:8 m = (spk{i}-ones(size(spk{i},1),1)*b(3,:))*p'; %baseline correct m = m./(sqrt(sum(m.^2,2))*ones(1,2)); %find movement direction m decVel = m*400; %constant velocity scale matched by eye decPos = bsxfun(@plus,pos{i}(1,1:2),([[0 0];cumsum(decVel,1)]*dt/1000)); plot(pos{i}(:,1),pos{i}(:,2),'k') plot(decPos(:,1),decPos(:,2),'r') end xlabel('X Position (mm)'), ylabel('Y Position (mm)') title('PVA')

c. Try improving the decoding by smoothing the firing rates by using a running average of 250ms. figure; hold on for i = 1:8 for t=1:size(spk{i},1)-5 avgSpk{i}(t,:)=mean(spk{i}(t:(t+5),:)); end m = (avgSpk{i}-ones(size(avgSpk{i},1),1)*b(3,:))*p'; %baseline correct m = m./(sqrt(sum(m.^2,2))*ones(1,2)); %find movement direction m decVel = m*400; %constant velocity scale matched by eye decPos = bsxfun(@plus,pos{i}(1,1:2),([[0 0];cumsum(decVel,1)]*dt/1000)); plot(pos{i}(:,1),pos{i}(:,2),'k') plot(decPos(:,1),decPos(:,2),'r') end xlabel('X Position (mm)'), ylabel('Y Position (mm)') title('smoothedPVA')

12. Use a Kalman filter decoder to estimate the movement trajectory for each trial. a. Fit the parameters of the decoder using the 180 trials of training data. spk = trainTrials.spikes; pos = trainTrials.handPos; vel = trainTrials.handVel; dt = 50; x = vel; [Y,X] = deal(spk(:),x(:)); [D,M] = deal(size(Y{1},2),size(X{1},2)); [X1x1,Xx1,Xx,Q,V] = deal(zeros(M)); [Yx,R,pX] = deal(zeros(D,M),zeros(D),zeros(M,1)); for i=1:180 % calculations for A xt1=X{i}(1:(end-1),:); %x_{t-1}, t=2:T Xx1=Xx1+(X{i}(2:end,:)'*xt1); %x_t*x_{t-1}', t=2:T X1x1=X1x1+(xt1'*xt1); %x_{t-1}*x_{t-1}', t=2:T % calculations for B Yx=Yx+(Y{i}'*X{i}); %y_t*x_t', t=1:T Xx=Xx+(X{i}'*X{i}); %x_t*x_t', t=1:T % calculations for pi pX=pX+(1/180)*X{i}(1,:)'; %sample mean of x_1 end A=Xx1*(X1x1^-1); B=Yx*(Xx^-1); for i=1:180 q=X{i}(2:end,:)'-A*X{i}(1:(end-1),:)'; %x_t-A*x_{t-1}, t=2:T Q=Q+(q*q'); %variance around model x_t estimate R=R+((Y{i}'-B*X{i}')*(Y{i}'-B*X{i}')'); %variance around estimate V=V+(1/180)*((X{i}(1,:)'-pX)*(X{i}(1,:)'-pX)'); %sample covar of x_1 Tn(i) = length(X{i}); end

Q=1/(sum(Tn)-180)*Q; %divide by total number of points in calculation R=1/sum(Tn)*R; %divide by total number of points in calculation

b. Test the decoder on the 8 test trials. Plot this decoded trajectory and the actual movement trajectory on the same plot. pos = testTrials.handPos; Ytst = testTrials.spikes; Xtst = testTrials.handVel; [xHat,sigHat,err]=deal(cell(size(Ytst))); for i=1:length(Ytst) T=size(Ytst{i},1); [xHat{i},sigHat{i}]=deal(zeros(M,T),zeros(M,M,T)); [u,s]=deal(pX,V); for t=1:T Kgain=s*B'*(B*s*B'+R)^(-1); %Kalman gain xHat{i}(:,t)=u+Kgain*(Ytst{i}(t,:)'-B*u); %u^t_t sigHat{i}(:,:,t)=(eye(M)-Kgain*B)*s; %sig^t_t u=A*xHat{i}(:,t); %u^{t-1}_t s=A*sigHat{i}(:,:,t)*A'+Q; %sig^{t-1}_t end err{i}=sqrt(sum((xHat{i}-Xtst{i}').^2)); end meanErr = mean([err{:}]); fN = length(findobj('type','figure')); c = jet(8); figure; hold on for i = 1:8 decPos = bsxfun(@plus,pos{i}(1,1:2),([[0 0];cumsum(xHat{i},2)']*dt/1000)); plot(pos{i}(:,1),pos{i}(:,2),'k') plot(decPos(:,1),decPos(:,2),'c') end AX = axis; xlabel('X Position (mm)'), ylabel('Y Position (mm)') title('vKF')

Note: shown in problems 11 and 12 are the offline decoding results of arm movements using basic implementations of the PVA and Kalman filter. In a closed-loop BMI, decoded trajectories are typically substantially better than shown here for two reasons: 1. When the BMI user gets moment-by-moment feedback about the BMI cursor or robotic limb movements, they are able to correct for errors (cf. Chase et al., Neural Networks, 2009).

2. There are variants on these decoders that are used in practice but not included here for the sake of simplicity. For example, the observation model in the Kalman filter typically includes an additive offset that accounts for the mean firing rate of each neuron being greater than zero, which is not included here and can substantially improve the accuracy of the decoded movements.

Answers 1. Assume you want to do a controlled experiment between low-frequency DBS at 50Hz and high-frequency DBS at 100Hz by delivering the same amount of charge into the tissue. If you keep your pulse width the same, what would be the ratio of amplitudes in these two stimulation schemes with a current-controlled pulse generator?

As charge is given by current by time, the ratio of the amplitudes for a 50 Hz stimulation to a 100 Hz stimulation would have to be 2:1.

2. In the scenario above, how would you adjust your pulse width if that was the only parameter you could change?

Similarly, the ratio of pulse widths, would have to be 2:1 for an equivalent amount of charge.

3. If in the above scenario, safety limits only allow you to increase your amplitude by 25%, what kind of adjustment would you have to make to your pulse width?

If the ratio of amplitudes is 1.25:1, the ratio of pulse widths would have to 1.6:1 to yield an overall ratio of 2:1 to offset the frequency ratio of 1:2.

4. A recent study [82] determined the total electrical energy delivered (TEED) by nonrechargeable implantable voltage-controlled pulse generators based on the stimulation parameters is given by: (𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 (𝑉)2 × 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝐻𝑧)< × 𝑝𝑢𝑙𝑠𝑒 𝑤𝑖𝑑𝑡ℎ 𝑇𝐸𝐸𝐷 =

𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒

× 1𝑠𝑒𝑐

Redo Problem 1 to attain equal TEED for the two different stimulation frequencies.

As the amplitude is squared an approximate ratio of 1.41:1 yields equal TEED values.

5. Redo Problem 2 to attain equal TEED for the two different stimulation frequencies.

A pulse width ratio of 2:1 compensates for a frequency ratio of 1:2.

6. Redo Problem 3 to attain equal TEED for the two different stimulation frequencies.

If the ratio of amplitudes are 1.25:1, the pulse width ratios would have to 1.28:1 to yield an overall ratio of 2:1 to offset the frequency ratio of 1:2.

7. What would be the amount of current drain off of a voltage-controlled pulse generator if the tissue impedance reduced by an average of 10% across time?

To keep the voltage constant, the amount of current would increase 10%.

8. What would be the amount of current drain off of a voltage-controlled pulse generator if the tissue impedance increased by an average of 10% across time?

Similarly, the amount of current would decrease 10% across time.

9. What would be the amount of current drain off of a current-controlled pulse generator if the tissue impedance reduced by an average of 10% across time?

The current drain from a current-controlled pulse generator is independent of tissue impedance.

10. What would be the amount of current drain off of a current-controlled pulse generator if the tissue impedance increased by an average of 10% across time?

Again, the current drain from a current-controlled pulse generator is independent of tissue impedance.

Solutions of problems in Chapter Transcranial Magnetic Stimulation: Principles and Applications in “Neural Engineering” Authors: Dr. Lari M. Koponen, Dr. Angel V. Peterchev 1.1 PROBLEM 1 From Eq. (7.16), we can derive the magnetic field in the origin of an 𝑁-turn circular loop: 𝑩 (𝒓, 𝑡 ) =

𝜇0 4𝜋

𝐼(𝑡) ф 𝑝

𝑑𝒓′ × (𝒓 − 𝒓′) |𝒓 − 𝒓′|3

.

We can sum up the contribution from 𝑁 turns by substituting a constant current 𝐼(𝑡) = 𝑁 ⋅ 𝐼. For a loop with radius 𝑅, we can transform the closed path integral to be a 1-dimensional integral in cylindrical coordinates (𝑟, 𝜃, 𝑧) with 𝑑𝒓′ = 𝜽: 𝑅𝑑𝜃 , where the angle 𝜃 going from 0 to 2𝜋. Thus 𝜇0 𝜽: × (−𝑅 𝒓= ) 𝜇0 𝒛= 𝜇0𝐼 𝑩(𝒓, 𝑡) = 𝑁 ⋅ 𝐼 ф 𝑅𝑑𝜃 = 𝑁 ⋅ 𝐼 ф 𝑑𝜃 = 𝑁 ⋅ 𝒛= .

4𝜋 2𝜋 4𝜋 2𝜋 𝑅 2𝑅 𝑅3 Substituting the given numerical values, we obtain H ⋅ 5000 A m = 0.91 T 2 ⋅ 0.045 m

4𝜋 ⋅ 10−7 𝐵 = 13 ⋅ which is indeed approximately one tesla.

The more difficult additional part of the problem can be solved with at least two different methods.

Outside the coil windings, a semi-analytical method is to solve the generalized off-axis version of the Eq. (1), which involves elliptic functions, and then (numerically) integrate this expression over the cross section of each individual turn. A fully numerical method is to model the system in a (2-d axisymmetric) finite element method (FEM) solver. Note that due to the quasi-static nature of the problem, also the FEM can also be solved for the static case.

By modelling the geometry, we obtain about 0.93 T field in the center of the coil. However, the maximum field in the coil surface slightly exceeds 1.4 T (bottom surface of the coil), or 1.6 T (inner wall of the hole inside the circular windings). Interpretation: The approximations made in part (a) provide the

(1)

correct order of magnitude for the magnetic field and correctly predict the value in the center for the coil; but, for computing the maximum field value at coil surface, they are at best questionable.

Figure 1: An example solution for Problem 1b using COMSOL to solve a two-dimensional axisymmetric FEM problem. The problem can also be solved without FEM (as the wording of the problem does not require being able to solve the magnetic field within the coil windings).

1.2 PROBLEM 2 1.2.1 Part A The energy stored in a capacitor is 1 𝑈 = 𝐶𝑉2 = 0.5 ⋅ 0.185 mF ⋅ (2.8 kV)2 = 0.5 ⋅ 185 ⋅ 2.82 J = 725.2 J .

2

A smartphone battery, circa year 2019, typically has a rated capacity of around 10 Wh (Apple iPhone XS, 10.13 Wh), which is about 36,000 J. Thus, a single battery contains about s 10.13 Wh ⋅ 3600 h = 50.3 725.2 J times more energy than a single TMS pulse at maximum stimulator output. At typical outputs, the device is operated at around half this voltage, and the energy is reduced by a factor of 4, making the battery contain more than 200 such pulses worth of energy.

The javelin has kinetic energy of around

2

⎛ km ⎞ 100 1 h 2 2 𝑈 = 𝑚𝑣 = 0.5 ⋅ 0.8 kg ⋅ ⎜ ⎟ = 0.5 ⋅ 0.8 ⋅ 27.7 J = 309 J m 2 [ \ ⎜ s ⎟ 3.6 km [ h\ ⎠ 𝗁 less than half of that of the TMS pulse! Still, it is already enough to cause serious injury should it hit a living being. The important point is the rate at which this energy is released, i.e., the instantaneous power.

The most important observation is that just knowing the absolute amount of energy stored is not that important. A second important observation is that there is enough energy to cause serious damage should the device fail catastrophically.

1.2.2 Part B Dry air has dielectric breakdown voltage of about 3 kV/mm. Thus, the current might leap up to about 1 mm from an exposed conductor. Typical insulators have several times larger breakdown voltages (glass >10 kV/mm, typical engineering plastics >20 kV/mm). In case of polyethylene, the dielectric breakdown voltage varies from 20 to over 100 kV/mm. So, under ideal conditions, about 0.15 mm of such insulator material suffices to prevent a breakdown.

Even the thinner of the two standard electrical tapes (7 mil vs 10 mil, i.e. 0.18 mm and 0.25 mm tapes) is sufficiently thick to provide such rated breakdown voltage. However, to ensure safety, a safety margin should always be applied. Thus, such electrical tapes are usually rated at 600 V at both thicknesses.

Consequently, the answer is ultimately no, the tape is not adequate as a sole means of insulation. (Note that there are also sufficiently thick tapes for the high voltages, but the common variety is not amongst them.)

1.3 PROBLEM 3 Losses are obtained from Eq. (7.7): 𝑇

𝑊𝑙𝑜𝑠𝑠 = a 𝑟𝐼2𝐿(𝑡)𝑑𝑡 , 0

Where the pulse waveform is from Eq. (7.1): 𝑉𝐶(0) 2𝜋 𝐼 (𝑡) = sin(𝜔𝑡) exp(−𝛼𝑡) , 𝑡 ≥0&𝑡 ≤𝑇 = , 𝐿 𝜔𝐿 𝜔 2 where 𝛼 = 𝑟/2𝐿 and 𝜔 = o1/𝐿𝐶 − 𝛼 . The integral is easiest to compute with suitable numerical quadrature, like the trapezoidal rule. About 67% or 2/3rds of the energy are lost to heat per pulse.

We obtain that each pulse losses 39 J. So, to sustain 10 Hz rTMS, we need to charge the energy storage at 390 W. Considering the 80% efficiency of the power supply, we need to draw 389 W 0.8 from the wall socket.

= 487 W

An example code, in MATLAB: r = 50e-3; % Coil resistance (ohm) L = 16e-6; % Coil inductance (H) C = 185e-6; % Energy storage capacitance (F) V_C_0 = 800; % Stimulator output (V) alpha = r / (2 * L); omega = sqrt(1 / (L * C) - alpha^2); T = 2 * pi / omega; % Pulse duration t = linspace(0, T, ceil(1e6 * T)); % A time vector, with about 1 µs resolution I_L_t = V_C_0 / (omega * L) * (sin(omega * t) .* exp(-alpha * t)); % The coil current at time t w_loss_t = r * I_L_t.^2; % The instantaneous power loss at time t.

W_loss = trapz(t, w_loss_t); % A trapetsoidal rule integral of the loss. fprintf('The pulse duration is %.1f µs', 1e6 * T); fprintf(', and the energy loss during the pulse is %.1f J.\n', W_loss); U_C_0 = 0.5 * C * V_C_0^2; fprintf('There was %.1f J, and %.1f%% of those were lost.\n', U_C_0, 100 * W_loss / U_C_0);

The example code should output: The pulse duration is 343.1 µs, and the energy loss during the pulse is 38.9 J. There was 59.2 J, and 65.8% of those were lost.

The losses over time can be further visualized with, for example, following figure: figure(1); clf; subplot(2, 1, 1); plot(1e6 * t, I_L_t); xlabel('Time (µs)'); ylabel('Coil current (A)'); subplot(2, 1, 2); plot(1e6 * t, 1e-3 * w_loss_t); xlabel('Time (µs)'); ylabel('Instantaneous loss (kW)');

1.4 PROBLEM 4 We assume a worst case, where the magnetic field from the TMS coil points directly at the loop.

Omitting the negative sign, the magnitude of the electromotive force through the loop is 𝜕𝐵 5 cm 2 2π x = 61.7 V |ℰ| = q−𝐴 ⋅ q = t3 ⋅ π ⋅ v x y ⋅ v0.5 T ⋅ 𝜕𝑡 2 300 µs where we inserted the largest rate of change for the magnetic field (which is about 10.4 kT/s).

As 𝐼 = 𝑉 = ℰ, this EMF drives about 62 mA through the 1 kΩ impedance, which is more than 60 times 𝑅

𝑅

the normal DBS electrode current.

In part (b), the driving voltage is reduced by 5 V, to 57 V, and the current is about 57 mA. Still more than 50 times the current under normal DBS operation.

For part (c), the surgeon can try to minimize the size of the wire loops between the skull and the scalp.

Alternatively, the surgeon can alternate the winding direction of each turn, so the voltages induced in the individual turns largely cancel out. For example, should the third loop be in the opposite direction to the first two, the induced voltage would drop from about 60 V to 20 V for part (a). With the 5 V block in part (b), the current would be reduced to about 15 mA.

1.5 PROBLEM 5 1.5.1 Part A We can model the disk as a set of (independent) circular loops of varying diameter. At a radius 𝑟 such a loop has a cross-section of

𝑑𝐴(𝑟) = ℎ ⋅ 𝑑𝑟 , length of 𝑙(𝑟) = 2𝜋 ⋅ 𝑟 , resistance of 𝑙(𝑟)

𝑅(𝑟) =

𝑑𝐴(𝑟) ⋅ 𝜎

=

2𝜋 ⋅ 𝑟 , ℎ ⋅ 𝑑𝑟 ⋅ 𝜎

and electromotive force of ℰ(𝑟, 𝑡) = −𝐴(𝑟) ⋅

𝜕𝐵(𝑡)

= −𝜋 ⋅ 𝑟2 ⋅

𝜕𝑡

2𝜋

2𝜋𝑡 ⋅ 𝐵 ⋅ cos v x, 𝑇 𝑇

where 𝑇 is the pulse duration, and where we assume that the thin disk has negligible effect on the magnetic field. The current is 𝐼(𝑟, 𝑡) =

ℰ 𝑅

=

−𝜋 ⋅ 𝑟2 ⋅

2𝜋 2𝜋𝑡 ⋅ 𝐵 ⋅ cos [ \ 𝑇 𝑇 = − ℎ ⋅ 𝑑𝑟 ⋅ 𝜎 ⋅ 𝜋 ⋅ 𝑟 ⋅ 𝐵 ⋅ cos v2𝜋𝑡 x, 2𝜋 ⋅ 𝑟 𝑇 𝑇 ℎ ⋅ 𝑑𝑟 ⋅ 𝜎

and the maximum current density (at the outer edge of the electrode) 𝐼 =−𝜎⋅𝜋⋅𝑟 2𝜋𝑡x = − 𝐽= ⋅ 𝐵 ⋅ cos v 𝑑𝐴 𝑇 𝑇

MS 63 m ⋅ 𝜋 ⋅ 5 mm 300 µs

kA 2𝜋𝑡 ⋅ 0.5 T ⋅ cos v 2𝜋𝑡 x = −1.6 ⋅ cos v x. 𝑇 mm2 𝑇

The maximum absolute value is obtained at the beginning of the pulse, and it is about 1.6 kA/mm2. This result can also be obtained by noticing that electric field within each such loop is 𝐸(𝑟) = ℰ(𝑟), and 𝑙(𝑟)

applying differential form of Ohm’s law 𝐽(𝑟) = 𝜎𝐸(𝑟). (Should one include the effect the silver disk has to surrounding magnetic field and use an FEM solver to solve the current density, the peak current density will be slightly less, on the order of 1.4 kA/mm3.)

1.5.2 Part B Continuing with the results from first part of the problem, we know that the current density at an arbitrary point in the electrode is

𝐽(𝑟, 𝑡) = −

𝜎⋅𝜋⋅𝑟 𝑇

2𝜋𝑡 ⋅ 𝐵 ⋅ cos v x. 𝑇

The power density of the Joule heating𝐽(𝑟, is 𝑡)2 𝑝(𝑟, 𝑧, 𝑡) = =𝜎⋅[

𝜎

2

𝜋⋅𝑟 𝑇

2

⋅ 𝐵\ ⋅ cos v

2𝜋𝑡 𝑇

x,

Integrating this over time (for full sinusoid, the squared integral is half times the duration), we obtain the resistive loss density as 2 𝑇 𝜋⋅𝑟 𝜎 𝑤(𝑟, 𝑧) = 𝜎 ⋅ [ ⋅ (𝜋 ⋅ 𝑟 ⋅ 𝐵)2. ⋅ 𝐵\ ⋅ = 𝑇 2 2𝑇 Which, when integrated over rotation around z-axis, radii, and thickness gives us 𝑧

𝑟𝑜𝑢𝑡

𝑊=a a 0

2𝜋𝑟 ⋅ 𝑤(𝑟, 𝑧) 𝑑𝑟 𝑑𝑧 =

𝑟𝑖𝑛

𝜎

𝑟4 − 𝑟4 𝑖𝑛 , ⋅ (𝜋 ⋅ 𝐵)2 ⋅ 𝑧 ⋅ 𝑜𝑢𝑡 2𝑇 4

where 𝑧 is the thickness of the electrode. Substituting the numerical values:

𝑊 = 2𝜋 ⋅

MS 63 m 2 ⋅ 300 µs

⋅ (𝜋 ⋅ 0.5 T)2 ⋅ 0.5 mm ⋅

(5 mm)4 − (1 mm)4 4

= 0.13 J.

Each pulse delivers about 0.13 joules into the electrode. Thus, the average power delivered into the electrode is 0.13 W (one pulse per second). (An FEM model in which the eddy currents in the silver disk are allowed to reduce the surrounding magnetic field will give slightly lower value, similarly to it having lower peak current density. For such model, expect a value around 0.090 J.) 1.5.3 Part C The mass of the electrode is

𝑚 = 𝜌 ⋅ 𝑧 ⋅ 𝜋 ⋅ (𝑟2 𝑜𝑢𝑡

− 𝑟2 á = 10500 𝑖𝑛

kg

⋅ 0.5 mm ⋅ 𝜋 ⋅ ((5 mm)2 − (1 mm)2) = 0.396 g .

m3

J . Thus, each pulse increases the At room temperature, silver has heat capacity of 𝑐 = 0.24 g°C

temperature of the electrode by Δ𝑇 =

𝑊 𝑐⋅𝑚

=

0.13 J = 1.4 °C J 0.24 ⋅ 0.40 g g°C

Omitting any cooling, e.g., via the contact to the scalp, a train of 1000 pulses will heat the electrode by hundreds of degrees. In reality, the thermal resistance to the skin will limit the temperature, and it will remain close to the skin temperature of about 37 °C. We must estimate this effect. There are several valid approaches. As our electrodes are relatively small compared to the head, the simplest approach is to model the head as an isotropic half space of water (thermal conductivity 0.6 W/m°C). For a small circular heat source touching an isotropic half space, the

1 , where 𝑟 is the radius of the source. We can safely omit the effect of the thermal resistance is 𝑅𝑠 = 4𝑘𝑟

hole in the electrode, and obtain a thermal resistance of 𝑅𝑠 =

1

4⋅0.6 W/m°C⋅0.005 m

= 83 °C/W.

The steady-state temperature rise for the electrode will be Δ𝑇 = 𝑅𝑠 ⋅ 𝑃 = 𝑅𝑠 ⋅ 𝑊 ⋅ 𝑓, where 𝑓 is the pulse repetition rate. This is 83 °C/W ⋅ 0.13 J ⋅ 1 Hz = 10.6 °C ≈ 11 °C above the skin temperature, i.e., about 48 °C. In addition to this, each pulse will momentarily bump the temperature by the 1.4 °C computed above. Thus, the peak temperature during the pulse train, and at the end of the train, will be about 50 °C. 1.5.4 Part D The heating is likely not safe. The electrode should remain at 41 °C or below.

1.5.5 Part E The electrode cannot be allowed to heat up by more than 4 degrees. This can be achieved with notched electrodes that preclude a highly conductive pathway around the electrode, or with electrodes made of a material with a much lower electrical conductivity.

1.6 PROBLEM 6 We can (approximately) compute this with the simpler of the two coil models of Problem 1.

The vector torque is 𝝉 = (𝑁𝐼𝐴)𝒏 × 𝑩 , where 𝑁 is the number of turns, 𝐼 is the coil current, 𝐴 is the (mean area) of the turns, 𝒏 is the unit normal vector of the TMS coil, and 𝑩 is the external magnetic field (due to the MRI device).

The worst case torque occurs when TMS coil is at 90 degree angle to the static magnetic field of the MRI device. Substituting 0.090 m 2 𝜏 = 𝑁𝐼𝐴𝐵 = 13 ⋅ 5000 A ⋅ tπ v x y ⋅ 3 T = 1241 Nm .

2 To put this into perspective, the car tire bolts are usually tightened to around 110–120 Nm. In TMS, however, the torque is applied for a very brief amount of time, and then it is reversed during the later parts of a biphasic TMS pulse.

1.7 PROBLEM 7 Coil inductance is proportional to the square of the number of turns: 𝐿 ∝ 𝑁2 .

Thus, to reduce the inductance from 16 to 10 µH, we need to drop the number of turns to

𝑁10 = 𝑁16 ⋅ ï

𝐿10 𝐿16

= 18 ⋅ ï

10 16

= 14.2 .

That is, we reduced 4 turns, or about 22% of the turns. The coil inductance is now 𝑁10 2 14 2 𝐿10 = 𝐿16 ⋅ v x = 16 µH ⋅ v x = 9.7 µH . 𝑁16 18 As the coil size and shape are preserved, our coupling to the brain was also reduced by (18 − 14)/18 = 4/18 = 22%, and we can expect to need about 1/(1 − 4/18) − 1 = 18/14 − 1 = 29% larger rate-ofchange of the coil current to get same induced electric field. (a, b) The pulse waveform is 𝐼 (𝑡) = 𝐿

𝑉𝐶(0)

sin(𝜔𝑡) exp(−𝛼𝑡)

𝜔𝐿

where 𝑉𝐶(0) is the initial capacitor voltage, 𝛼 = 𝑟/2𝐿 characterizes the damping from the circuit series resistance 𝑟, 𝜔 = o1/𝐿𝐶 − 𝛼2 is the damped resonance frequency of the pulse, and 𝐿 and 𝐶 are respectively the coil inductance and energy-storage capacitance. Therefore, the induced electric field is proportional to underdamped sinusoidal current, 𝑉𝐶(0) 𝛼 ( ) ( ) ( ) 𝐸(𝑡) ∝ 𝑁 𝑡 ≥ 0, ócos 𝜔𝑡 − sin 𝜔𝑡 ò exp −𝛼𝑡 , 𝐿 𝜔 where 𝑁 is the number of turns in the (same shape and size) coil.

For parts a & b, we will omit the exponentially decaying term from the end and its contribution to the angular frequency.

The brain stimulation is roughly proportional to the membrane depolarization of a passive membrane, obtained as (Eq. 7.32): 𝑡 − 𝑡′ exp v− x 𝜏 𝑡 Δ𝑉(𝑡) ∝ a 𝐸(𝑡′) 𝑑𝑡′ . 𝜏 0 We are interested in the peak (absolute) value of this depolarization 𝛥𝑉max = max |Δ𝑉(𝑡)|, and must 𝑡

write a short script to solve this integral at an arbitrary time 𝑡 for the given four cases (two reference cases and two modified cases). The required peak voltage is then obtained as 𝑉𝐶,new =

𝛥𝑉max,old 𝛥𝑉max,new

⋅ 𝑉𝐶,old .

In (a), we obtain very slightly higher maximum depolarization for the new coil with less turns. Instead of 1600 V, we can do away with 1580 V, about 3% less energy. We will, however, now have much higher maximum current in the coil.

In (b), we obtain considerably lower maximum depolarization for the smaller capacitor, and we must increase the maximum voltage from 1600 to 2000 V. Because of the smaller capacitor, however, we still need 16% less energy.

In part (b), if we omit the resistance similarly to part (a), the maximum voltage must be increased to about 2140 V, which requires only 3% less energy than the initial design, which is similar to result of part (a). That is, most of the gain in efficiency in part (b) is not due to a shorter pulse having better efficiency in coupling with the neurons, but the shorter pulse being more sinusoidal due to lower resistive losses during the pulse. If the coil resistance could be reduced even by a few tens of percent, the device efficiency would be greatly increased even without needing shorter pulses (in addition to reducing the heating and power consumption due to lower resistive losses).

1.8 PROBLEM 8 This problem is partially open ended, and there are several ways to arrive at the correct conclusion. We present one of them. This problem can, in principle, also be solved by noticing that 0.5 cm is considerably less than 2.5 cm (and by doing no additional math). We prefer the method below, in which we include first order corrections to the comparison of the two distances, to ensure that our factor of 5 difference is not diminished by including more details to the way the two numbers interact.

Simplify the numbers, assume that resting motor threshold (RMT) is exactly 100 V/m. This does not change the result but will allow having the field gradient in units of V/m2.

Simplify the E-field distribution, assume the E-field magnitude to be approximately a parabola around the stimulation target.

The E-field magnitude attenuates slowest in the direction of the E-field. Thus, we should use the 2.5 cm for the distance to drop to 70%. Thus, the E-field magnitude behaves approximately as V m) ù100 − m √2 V

𝐸(𝑥) ≈ 100

V

−

m

100

(2.5 cm)2

𝑥2,

and the field gradient is V m) ù100 − m √2 V

𝑑𝐸 𝑑𝑡

(𝑥) = −2

100

(2.5 cm)2

𝑥.

This function obviously increases without bounds, and we need to see a reasonable cut of point where the approximation of a parabola is still somewhat reasonable. This is at about 0.025 m. At this point, the gradient is about 2,300 V/m2.

Assume a pyramidal axon near, but not quite at, the top of a gyrus. The initial segment of this axon is close to parallel with the E-field, and the axon enters white matter at the centerline of the gyrus. At this

point, the axon is perpendicular to the E-field. Eq. 29 only considers the component of E-field parallel with the axon. Thus, the effective E-field must reduce from 100 V/m to 0 V/m in less than half the width of the gyrus. This gradient at its maximum must be at least 100 V/m / 0.005 m = 20,000 V/m 2. In the cortex, the axon geometry causes far larger local gradients than the E-field gradient.

1.9 PROBLEM 9 The human arm span is approximately equal to their height, and we can further assume that the size difference of the hands is small compared to the length difference of the arms. Similarly, the width difference in the torso is small compared to length difference in the arms. Thus, the 190 cm tall person has 35 cm longer arms (in total) than the 155 cm tall person.

Consequently, to reach from the primary motor cortex to the contralateral hand, the signal must travel 35 cm / 2 = 17.5 cm longer distance. At 7 cm/ms, this takes a 2.5 ms longer time. We can reasonably expect around 2–3 ms longer latency for the MEPs for the tall subject.

1.10 PROBLEM 10 A Class 2 study means that we expect at most indirect benefits and can tolerate at most moderate risks (“the study is anticipated to yield valuable data for the development of treatments”). Note that if we should expect direct benefits from increased stimulation dose, a Class I study would allow exceeding the limits.

The guideline gives two tables for rTMS safety for Class 2 and 3 studies. Table 5 for multiple trains at short inter-train intervals (ITI, from 0.25 to 5.0 seconds) and Table 4 for individual rTMS trains. Because we are delivering multiple trains, we should use the former. a. A 10 Hz rTMS pulse train at 120% RMT can last up to 3.2 seconds (32 pulses, note that the original treatment is expected to produce direct benefits and does not comply with the safety limit for our study; it is closer to the limit of single rTMS train (4.2 seconds or 42 pulses in Table 4).

The new 130% RMT train can last 2.2 seconds (22 pulses).

Should we upgrade this to a Class I study with similar dose to that of the original treatment protocol, we might argue for closer to 2.9 seconds (29 pulse) at 130% RMT, based on Table 4. b. The 20 Hz rTMS pulse train at 120% RMT can last 0.8 seconds (16 pulses), less than half the original 4 s and 40 pulses.

Even in Table 4, doubling the rate reduces the duration to 1.0 seconds (20 pulses), half that of the original 4 s and 40 pulses. c. The reduction in inter-train interval forces us to use the top part of Table 5. According to it, there is insufficient data on safety of rTMS at 120% RMT with 5 s ITI. Reduction to 110% RMT is considered safe, for at most 10 trains (each of which containing at most 50 pulses).

Answer Key 1) Dose is defined by electrode montage (place and shape of each electrodes) and waveform applied (intensity, waveform shape, frequency etc…)

2) Multiple answers possible including transcranial Direct Current Stimulation (tDCS), transcranial Alternating Current Stimulation (tACS), transcranial Pulses Stimulation (tPCS), transcranial Random Noise Stimulation (tRNS), Cranial Electrotherapy Stimulation (CES). An invasive technique would not be tES. 3) Third electrode is cathode passing 1.5 mA out of the body. Sum of total current in and total out must equal zero.

4) For ECT about 800 mA (about 1 mA) is provided. For tDCS and tACS about 1-2 mA (a few mA) is provided.

5) 13 kOhm. After adjustment 4 kOhm 6) When only using two electrodes, placing electrodes close together will cause shunting of current between the electrodes (mostly through skin) and therefore significant current won’t arrive at the brain. In contrast to conventional tES where large electrodes are used, High-Definition tES uses small electrodes, a higher number of electrodes, arranged in arrays. With HD-tES (or other HD techniques like HD-tDCS, HD-tACS) placing HD electrodes close together can be rational.

7) Electrode area is the surface are between the electrolyte and the skin. In this case, with 1 cm radius, the area is 3.14 cm2.

8) Models predict the intensity of current flow in the brain (inside the cranium), which brain regions are targeted by current and which are spared. Models predict the current density or electric field in reach brain region.

9) Electrolyte 1) prevent electrochemical products produced at the metal/rubber from reaching (diffusing to) the skin and 2) support even (uniform) current flow across the skin (from the electrolyte into the body).

10) A majority of current in tES will be “shunted” through the scalp. Not important for this question, a small fraction may go through the skull or through cerebral spinal fluid.

Answers: Q1: 1 Q2: 4 Q3: 2 Q4: 4 Q5: 3 Q6: 3 Q7: 2 Q8: 1 Q9: 1 Q10: 4

Answers to Homework Questions 1. Intraneural electrodes: Broadly speaking, these electrodes offer intimate connections to the nerve and the ability to recover single action potentials. In the case of the Utah slanted array, recordings from up to 100 individual fibers are possible, with the TIME and LIFE electrodes recording from up to several fibers. The drawback to these approaches is the difficulty with longevity. Electrodes which penetrate the perineurium tend to elicit ongoing foreign body responses, and ultimately cause electrode failure. Extraneural electrods on the other hand offer a high degree of biocompatibility, lasting for years in the body. The main disadvantage is the low signal to noise ratio and the low number of extracted signals.

In traditional nerve cuff electrodes, only a single signal is obtained per implant. The FINE approach can increase this number to several independent signals, but still of lower signal to noise ratio and lower temporal resolution as compared to intraneural electrodes.

2. Create a set of 16 instrumentation amplifiers in parallel, each with a gain of 1000. Connect all the + inputs together and all the – inputs together. Then add all 16 outputs using a summing circuit (i.e. inverter circuit with 16 inputs). Connect a voltage source (0.5mV 1Kz) and a source resistance Rs (1K) at the + input and connect another source(-0.5mV, 1KHz, with a similar Rs at the negative input. Verify that the output is 1Vpp. Carryout out a noise analysis to measure the input thermal referred noise. Then cut down the number of amplifiers to 8, then 4 then 2 and 1/ Plot RMS versus the number of amplifiers. The noise increases as the square root of the number of amplifiers 3. Reduce voltage noise • • • •

Reduce the gain of the amplifier Increase the size of the recording electrode to decrease the source resistance.

Choose an amplifier with small In and small Vn Add an transformer between the source and amplifier such that Rs=Vn/In (power matching)

4. Standard Cuff Electrode – <1kOhm Similar design to that described in this document. Current noise will not affect the recording as much as voltage noise.

Utah Electrode – >100kOhm Current noise begins to dominate due to high impedance, and thus needs to be controlled more than voltage noise.

LIFE – 10kOhm Both current and voltage noise will affect the recorded signal.

5. There are several methods to calculate this current. Using NEURON, download the MRG model and run: https://senselab.med.yale.edu/ModelDB/showmodel.cshtml?model=3810#tabs-1

Once loaded, sum up the relevant currents for a given node (i.e. node 48) and compare to figure 9.

This can also be implemented in Matlab using the software package Neurocal: https://www.mathworks.com/matlabcentral/fileexchange/1588-neurocal Run one of the myelinated axon models (i.e. model 7) and sum up the relevant nodal currents.

6. Use the equation for calculating the potential at a point for a known current source: 𝐼 𝑉= 4𝜋𝜎𝑟 where r is the distance between a given node and the recording electrode. Use a distance of 1mm from the axon to the recording electrode, and simulate nodes starting directly below the recording electrode (such that r = 1mm) and up to 20mm on either side in 2mm steps to simulate the potential for a large 20um axon.

0.0040

0.0044 0.0050 0.0057 0.0066 0.0079 0.0099 0.0131 0.0193 0.0356 0.0796 These numbers show half of the matrix, from 20mm away to directly under the electrode.

The other side identical due to symmetry. The current and conductivity (sigma) are assumed to be one for these numbers.

7. This can be achieved simply by using the principle of superposition. First scale the membrane current at each node by the potential found in question 6, i.e create 21 scaled versions of the membrane current from question 5. These represent the recorded potentials from each individual node. Then sum these potentials together, adding in a delay between each potential equal to the propagating action potential speed. Propagation velocity is assumed to be 6 times the axon size, or 120m/s for this example. Given that each node is 2mm apart, the time delay between successive current at each node is 16.7µs.

8. This simulated signal is created by first creating trains action potentials (with a variable delay between each action potential) to create a single fiber recording. This process can then be repeated within a loop to simulate hundreds of action potentials occurring simultaneously. Noise (typically white Gaussian noise) must also be added to either each action potential or the resulting summed potential to generate realistic signals.

9. For help in calculating the power spectra, visit the site below: https://www.mathworks.com/help/signal/ref/pwelch.html The overall spectra should resemble those shown in Figure 12, but not figure 13. The main reason is that these recordings simulate a monopolar recording, which will not show the spatial filtering properties of the cuff electrode, which are exhibited in Figure 13.

2

Homework Solutions: 1. A 2. A 3. B C D 4. A B C 5. A B C D 6. A B 7. A C 8. B 9. A 10. A 11. B 12. A 13. A B 14. A B

Homework solutions 1. Show that the units of the pressure and the energy density are the same.

Solution: J / m3 = N m / m3 = N / m2 = Pa

1

2. Estimate the temperature and the initial pressure rises upon short-pulsed laser excitation of whole arterial blood at the body temperature, with an optical fluence of 1 mJ cm-2 at 532 nm.

Solution: The molar extinction coefficient for oxy-hemoglobin at 532 nm is 43876 cm-1/M, and the concentration of hemoglobin in whole blood is 2.3 mM, the optical absorption of whole arterial blood is 𝜇𝑎 = 43876 cm−1 M−1 × 2.3 mM × ln10 = 232.4 cm−1. 𝐴𝑒 = 232.4 cm−1 × 1 mJ cm−2 = 232.4 mJ cm−3, 𝑇=

𝐴𝑒 𝜌𝐶𝑣

=

232.4 m J cm−3

= 58 mK,

1 g cm−3 ×4 J g−1K−1

𝑝0 = Γ𝐴𝑒 = 0.2 × 232.4 mJ cm−3 = 4.6 × 104 Pa.

2

3. Estimate the local initial pressure rise per one-degree local temperature rise at the body temperature.

Solution: According to Homework Problem 2, 58 mK temperature rise would yield 4.6 × 104 Pa (460 mbar), thus for one-degree local temperature rise, an 8-bar pressure rise is produced.

3

4. In water, estimate the fractional PA amplitude change upon one-degree local temperature rise with the baseline temperature of (a) 20 °C and (b) 37 °C. Solution: The Grueneisen coefficient of water is 𝛤4(𝑇0) = 0.0043 + 0.0053𝛵0 , where T0 is the water temperature. Thus, we can obtain: ∆Γ5 0.0053Δ𝑇0 = . Γ4 0.0043 + 0.0053𝑇0 (a) When T0 = 20 °C, Γ4 varies 4.8% per degree. (b) When T0 = 37 °C, Γ4 varies 2.6% per degree

4

5. Given 𝑑𝑐 = 1 mm or 0.01 mm, compute 𝜏𝑡ℎand 𝜏𝑠 in soft tissue.

Solution: For 𝑑𝑐 = 1 mm: (0.1 cm)2

𝜏𝑡ℎ = 1.3 ×10−3 cm2/s = 7.7 s, 𝜏𝑠 =

0.1 cm = 0.67 µs.

0.15 cm/μs

For 𝑑𝑐 = 0.01 mm: (0.01 cm)2

𝜏𝑡ℎ = 1.3 ×10−3 cm2/s = 7.7 × 10−4 s, 𝜏𝑠 =

0.01 cm = 0.067 µs.

0.15 cm/μs

5

6. Derive the photoacoustic equation shown in Eq. (10).

Solution: There are basic equations responsible for photoacoustic generation: the thermal expansion equation (Eq. 6.1) and the linear inviscid force equation (Eq. 6.2). They are expressed below:

(6.1)

∇ ∙ 𝜉⃗ (𝑟 , 𝑡) = −𝜅𝑝(𝑟 , 𝑡) + 𝛽𝑇(𝑟 , 𝑡), 2

(6.2)

ρ 𝛛 𝜉⃗ (𝑟 , 𝑡) = −∇𝑝(𝑟 , 𝑡), 𝛛𝑡 2

where vector 𝜉⃗ denotes the medium displacement.

Taking the divergence of Eq. (6.2), we get 2

(6.3)

ρ 𝛛 [∇𝜉⃗ (𝑟 , 𝑡)] = −∇2𝑝(𝑟 , 𝑡), 𝛛𝑡 2

Substituting Eq. (6.1) into Eq. (6.3) yields to Eq. (10), where 𝜈 𝑠 =

6

1

.

√𝜌𝜿

7. Show that the time reversal of the temporal function is equivalent to the complex conjugation of the temporal spectrum.

Solution: Using the following Fourier transformation equation t convert pressure between the time and frequency domains: +∞ 𝓕{𝐹(𝑡)} = 𝐹Z (𝑘) = ∫−∞ 𝐹(𝑡) exp( 𝑖𝑘𝑡) 𝑑𝑡,

we have the complex conjugate as follows +∞ 𝓕{𝐹(𝑡)}∗ = ∫−∞ 𝐹(𝑡) exp( − 𝑖𝑘𝑡)𝑑𝑡 = 𝐹Z (−𝑘).

In addition, using the scaling property of Fourier transformation, 𝓕{𝐹(𝑎𝑡)} = 1 𝐹Z b𝑘c, |𝑎|

𝑎

we obtain { }∗, { } ~( ) 𝓕 𝐹(−𝑡) = 𝐹 −𝑘 = 𝓕 𝐹(𝑡) where 𝐹(−𝑡) is the time reversal of 𝐹(𝑡).

7

8. Use Eq. (17) to derive and plot the PA pressure wave as a function of time observed outside a sphere excited by (a) a delta pulse and (b) a Gaussian pulse.

Solution: (a) Since the book has provided the derivation, we simply rewrite the pressure as a function of time observed outside a sphere excited by a delta pulse as follows: (1) When 𝑣𝑠𝑡 < 𝑟 − 𝑅𝑠, 𝑝(𝑟, 𝑡) = 0.

(2) When 𝑟 − 𝑅𝑠 ≤ 𝑣𝑠𝑡 < 𝑟 + 𝑅𝑠, 𝑝(𝑟, 𝑡) = 𝑝0 𝛛 [𝑡(1 − 𝑐𝑜𝑠 𝜃)] = 𝑝0 (𝑟 − 𝑣𝑠𝑡).

2 𝛛𝑡

2𝑟

(3) When 𝑣𝑠𝑡 > 𝑟 + 𝑅𝑠, 𝑝(𝑟, 𝑡) = 0. where 𝑅𝑠 is the radius of the sphere and 𝑝0 is the initial pressure generated inside the sphere.

MATLAB code (PATpr08.m):

close all clear all vs = 1500; p0 = 1; Rs = 0.5E-3; rd = 2*Rs; % Location of detector t = linspace(0, (rd + 2*Rs)/vs, 1000); figure(1) pin1 = p0/2*(1+vs*t./rd).*heaviside(rd+vs*t).*heaviside(Rs-rd-vs*t); pinr = p0/2*(1-vs*t./rd).*heaviside(-rd+vs*t).*heaviside(Rs+rd-vs*t); pout = p0/2*(1-vs*t./rd).*heaviside(rd-vs*t).*heaviside(Rs-rd+vs*t); p = pin1 + pinr + pout; plot(vs*t/Rs, p, 'k', 'LineWidth',2) tick = [.015 .025]; set(0,'DefaultAxesTickLength',tick) xlabel('Normalized time: \itv_s t\rm/\itR_s') ylabel('Normalized pressure: \itp\rm/\itp\rm_0') title('Pressure at \itr \rm= 2\itR_s')

8

Pressure at r = 2R s

0.3

Normalized pressure: p/p

0

0.2

0.1

0

-0.1

-0.2

0

0.5

1

1.5 2 2.5 Normalized time: v t/R s

3

3.5

4

s

(b) The response pressure to a Gaussian pulse can be computed numerically by convolution: +∞

𝑝 𝑆

𝑅(𝑡) = m 𝑑𝑡′𝑝(𝑟, 𝑡′)𝑆(𝑡 − 𝑡′) = −∞

0 0

2

2

where 𝑆(𝑡) = 𝑆0 exp b− 𝑡 2c.

2σ

9

𝑟+𝑅𝑠 2

m p1 − 𝑟−𝑅𝑠 𝑣𝑠

𝑣 𝑡′ 𝑠

𝑟

(𝑡 − 𝑡′)2 q exp(−

2𝜎2

)𝑑𝑡′,

9. Use Eq. (17) to derive and plot the pressure wave as a function of time observed outside a line object excited by (a) a delta pulse and (b) a Gaussian pulse.

Solution: We first construct a coordinate system by using the line as the 𝑧- axis. The position of a point in this system is denoted by (𝑥, 𝑦, 𝑧). Then the heating function of a line object by delta pulse illumination is described by 𝐻(𝑟, 𝑡) = 𝑝0𝛿(𝑥)𝛿(𝑦)𝛿(𝑡). a) For a delta pulse source and position (𝜌, 𝜃, 𝑧), we have 𝑝0 𝜕 1 ′ ′ ′ 𝑝(𝑟, 𝑡) = | m 𝑑𝑥 𝑑𝑦 ′𝑑𝑧′ 𝛿}𝑥 ~𝛿}𝑦 ~𝛿(𝑣𝑠𝑡 − |𝑟′ − 𝑟|)Ä 4𝜋𝑣𝑠2 𝜕𝑡 𝑡 𝑝0 𝜕 1 = | m 𝑑𝑧 ′ 𝛿(𝑣𝑠𝑡 − |𝑟′ − 𝑟|)Ä 4𝜋𝑣𝑠2 𝜕𝑡 𝑡 The delta function can be rewritten explicitly as 𝛿(𝑣𝑠𝑡 − |𝑟 ′ − 𝑟 |) = 𝛿 b𝑣𝑠𝑡 − Å(𝑥 − 𝑥′)2 + (𝑦 − 𝑦′)2 + (𝑧 − 𝑧′)2c = 𝛿 b𝑣𝑠𝑡 − Å𝑥2 + 𝑦2 + (𝑧 − 𝑧′)2c = 𝛿 b𝑣𝑠𝑡 − Å𝜌2 + (𝑧 − 𝑧′)2c 2𝑣𝑠𝑡𝛿(𝜌2 + (𝑧 − 𝑧′)2 − (𝑣𝑠𝑡)2), if 𝑣𝑠𝑡 > 𝜌 =Ç 0, 𝑖𝑓 𝑣𝑠𝑡 < 𝜌 𝑣𝑠𝑡

(𝛿 b𝑧 − 𝑧′ + Å(𝑣 𝑡)2 − 𝜌2c + 𝛿 b𝑧 − 𝑧′ − Å(𝑣 𝑡)2 − 𝜌2c), if 𝑣 𝑡 > 𝜌 𝑠

= ÜÅ(𝑣𝑠𝑡)2

𝑠

0, if 𝑣𝑠𝑡 < 𝜌 where 𝜌 = Å𝑥2 + 𝑦2. Then we have 𝑝0 𝜕 1 2𝑣𝑠𝑡 𝑝(𝑟, 𝑡) = 𝑈(𝑣𝑠𝑡 − 𝜌)â á 4𝜋𝑣2 𝜕𝑡 𝑡 Å(𝑣 𝑡)2 − 𝜌2 𝑠

𝑠

𝑝 𝛿(𝑣𝑠 𝑡 − 𝜌) − = 0( 2𝜋 Å(𝑣𝑠𝑡)2 − 𝜌2

𝑣𝑠 𝑡 3 ((𝑣𝑠𝑡)2 − 𝜌2)2

MATLAB code (PATpr09.m): % Photoacoustic signal from a homogeneously heated line excited by a % delta pulse % Use SI units clear all vs = 1500; p0 = 1; rho = 1E-3; % Location of detector

10

𝑈(𝑣 𝑡 − 𝜌)) 𝑠

𝑠

dvt = rho/200; vt1 = 0:dvt:rho; vt2 = rho+dvt:dvt:3*rho; t= [vt1 vt2]; clf p1 = vt1.*0; [m1,n1]=size(p1); p1(n1)=5e7/2/pi; p2=-vt2./(vt2.^2-rho^2).^(1.5)/2/pi; p=[p1 p2]; plot(t/rho, p, 'k') tick = [.015 .025]; set(0,'DefaultAxesTickLength',tick) ylim([-1e7 1e7]); xlabel('Normalized time: \itv_st\rm/\rho') ylabel('Normalized pressure: \itp\rm/\itp\rm_0') title('Pressure at \rho')

x 10

7

Pressure at 

1 0.8

Normalized pressure: p/p

0

0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0

0.5

1

1.5 2 Normalized time: v t/ 

2.5

3

s

(b) The response pressure to a Gaussian pulse can be computed numerically by convolution: +∞

𝑅(𝑡) = ∫−∞ 𝑑𝑡′𝑝(𝑟, 𝑡′)𝑆(𝑡 − 𝑡′), 2

where 𝑆(𝑡) = 𝑆0 exp b− 𝑡 2c.

2σ

11

10. The line in Question 9 has a finite length; please simulate the PA pressure wave detected by (a) a linear transducer array and (b) a ring array (see the geometry below). Please reconstruct the PA image using the forward data from the linear array and the ring array. Hint: please use the Matlab k-wave toolbox for both the forward and reconstruction simulations. Please download the k-wave toolbox from http://www.k-wave.org/.

Solution: (a)

MATLAB code (PATpr10_linear_array.m)

clear all; close all; % create the computational grid PML_size = 20; % size of the PML in grid points Nx = 600; % number of grid points in the x (row) direction Ny = 600; % number of grid points in the y (column) direction dx = 0.05e-3; % grid point spacing in the x direction [m] dy = 0.05e-3; % grid point spacing in the y direction [m] kgrid = makeGrid(Nx, dx, Ny, dy); % define the object disc_magnitude = 5; % [au] disc_1 = zeros(Nx,Ny); disc_1(150:450,299:301) = disc_magnitude;% % disc_1(10:end,99:101) = 5;

% smooth the initial pressure distribution and restore the magnitude p0 = smoothk(kgrid, disc_1, true);

12

% define the properties of the propagation medium medium.sound_speed = 1500; % [m/s] sos_cons = 1500; medium.density = 1000; % [kg/m^3] medium.sound_speed = sos_cons*ones(Nx, Ny); medium.density = 1000*ones(Nx, Ny); % [kg/m^3] % assign to the source structure source.p0 = p0; % define a binary line sensor center_freq = 5e6; % [Hz] bandwidth = 100; % [%] sensor.frequency_response = [center_freq, bandwidth]; sensor.mask = zeros(Nx, Ny); sensor.mask(1, Nx/2-255:Nx/2+256) = 1; %256 elements % create the time array [kgrid.t_array dt] = makeTime(kgrid, medium.sound_speed); % set the input arguements: force the PML to be outside the computational % grid; switch off p0 smoothing within kspaceFirstOrder2D input_args = {'PMLInside', false,'DataCast', ... 'gpuArray-single', 'PMLSize', PML_size,'Smooth', false, 'PlotPML',... false,'RecordMovie', false, 'PlotSim', true}; % run the simulation sensor_data = kspaceFirstOrder2D(kgrid, medium, source, sensor, input_args{:}); sensor_data0 = gather(sensor_data); figure;imshow(sensor_data0,[]); ax = gca; ax.FontSize = 16; xlabel('Time'); ylabel('Transducer index'); % reset the initial pressure source.p0 = 0;

% assign the time reversal data sensor.time_reversal_boundary_data = sensor_data;

13

% run the time reversal reconstruction p0_recon = kspaceFirstOrder2D(kgrid, medium, source, sensor, input_args{:}); % add first order compensation for only recording over a half plane p0_recon = p0_recon; % plot the reconstructed initial pressure figure;imshow(double(p0_recon),[])

(b) MATLAB code (PATpr10_ring_array.m) clear all; close all; % assign the grid size and create the computational grid Nx = 1300; % number of grid points in the x (row) direction Ny = 1300; % number of grid points in the y (column) direction x = 52e-3; % grid size [m] y = 52e-3; dx = x/Nx; % grid point spacing in the x direction [m] dy = y/Ny; % grid point spacing in the y direction [m] kgrid = makeGrid(Nx, dx, Ny, dy); %% sos_cons = 1500; % m/s medium.density = 1000; % [kg/m^3] medium.sound_speed = sos_cons*ones(Nx, Ny);

14

% define the object p0_magnitude = 5; % [au] p0 = zeros(Nx,Ny); p0(Ny/2-149:Ny/2+140,Nx/2-1:Nx/2+1) = p0_magnitude; %% % smooth the initial pressure distribution and restore the magnitude p0 = smoothk(kgrid, p0, true); % assign to the source structure source.p0 = p0; % define a centered Cartesian circular sensor sensor_radius =25e-3; % [m] sensor_angle = 4*pi/2; % [rad] sensor_pos = [0, 0]; % [m] num_sensor_points = 512; % assign to sensor structure ini_angle = 0; angle_step1 = ini_angle/180*pi; angle_per_step = 2*pi/num_sensor_points; detector_angle = (0:num_sensor_points-1)*angle_per_step+angle_step1; x_receive = cos(detector_angle)*sensor_radius; y_receive = sin(detector_angle)*sensor_radius; cart_sensor_mask = makeCartCircle(sensor_radius, num_sensor_points, sensor_pos, sensor_angle); sensor.mask = cart_sensor_mask; center_freq = 5e6; % [Hz] bandwidth = 100; % [%] sensor.frequency_response = [center_freq, bandwidth]; dt = 2.5e-8; %s kgrid.t_array = ((1:1300)-1)*2.5e-8; % set the input options input_args = {'Smooth', false, 'PMLInside', false,'DataCast', ... 'gpuArray-single','RecordMovie', true, 'PlotSim', false}; % run the simulation sensor_data = kspaceFirstOrder2D(kgrid, medium, source, sensor, input_args{:}); %% sensor_data0 = gather(sensor_data); %% sensor_data = sensor_data0(:,1:1300); figure;imshow(sensor_data,[]); ax = gca; ax.FontSize = 16; xlabel('Time');

15

ylabel('Transducer index');

source.p0 = 0; % assign to sensor structure cart_sensor_mask = makeCartCircle(sensor_radius, num_sensor_points, sensor_pos, sensor_angle); % sensor_radius_grid_points = round(sensor_radius/kgrid_recon.dx); sensor_radius_grid_points = round(sensor_radius/kgrid.dx); binary_sensor_mask = makeCircle(kgrid.Nx, kgrid.Ny, kgrid.Nx/2, kgrid.Ny/2, sensor_radius_grid_points, sensor_angle); sensor.mask = binary_sensor_mask; sensor.time_reversal_boundary_data = interpCartData(kgrid, sensor_data, cart_sensor_mask, binary_sensor_mask); %% input_args = {'Smooth', false, 'PMLInside', false,'DataCast', ... 'gpuArray-single','RecordMovie', true, 'PlotSim', false}; p0_recon = kspaceFirstOrder2D(kgrid, medium, source, sensor, input_args{:}); p0_recon0 = gather(p0_recon); figure;imshow(double(p0_recon0),[]);

16

17

11. For the circular geometry, if the designed imaging FOV is 25 mm in diameter, to satisfy the spatial Nyquist sampling requirement, what is the minimum number of sampling channels for detection at a cutoff frequency of (a) 2.25 MHz and (b) 15 MHz. Solution: For the circular geometry, the spatial Nyquist sampling requirement should follow the equation below: 𝜆 𝑁 = 𝜋𝐷 2 where N is the number of sampling channels, λ is the wavelength corresponding to the cutoff frequency, D is the diameter of the FOV, within which the spatial Nyquist sampling requirement is satisfied. (a) If the cutoff frequency is 2.25 MHz, assume speed of sound in water is 1500 m/s, then the sampling channel N is 262. (b) If the cutoff frequency is 15 MHz, assume speed of sound in water is 1500 m/s, then the sampling channel N is 1570.

18

12. Under the same conditions in Q.10, please calculate the minimum number of sampling channels for the full spherical geometry.

Solution: For the full spherical geometry, the spatial Nyquist sampling requirement can be estimated by the equation below: 𝜆 2 𝑁 ç é = 𝜋𝐷2 2 where N is the number of sampling channels, λ is the wavelength corresponding to the cutoff frequency, D is the diameter of the FOV, within which the spatial Nyquist sampling requirement is satisfied. (c) If the cutoff frequency is 2.25 MHz, assume speed of sound in water is 1500 m/s, then the sampling channel N is 21806. (d) If the cutoff frequency is 15 MHz, assume speed of sound in water is 1500 m/s, then the sampling channel N is 785000.

19

13. Derive Eq. (31), assuming the impulse response of the ultrasonic transducer is approximated as a Gaussian modulated sinusoid.

Solution: When the impulse response 𝑝(𝑡) of the ultrasonic transducer is approximated as a Gaussian modulated sinusoid: 2

𝑝(𝑡) = 𝐴 exp b− (𝑡−𝑡0) c cos(𝜔 𝑡 + 𝜑),

(13.1)

0

2ẟ2

where A is the amplitude of the impulse response, t is time, 𝜔0 = 2𝜋𝑓0 and f0 is the central frequency of the ultrasonic transducer, t0, σ and φ are constants. The axial resolution Raxial is determined by the distance of the FWHM of the temporal Gaussian envelope:

(13.1)

𝑅axial = 2√2 ln 2𝛿 ∙ 𝑐, where c is the speed of sound.

The Fourier transformation of 𝑝(𝑡) is: 𝑃î (𝜔) = 𝐴ï𝜋 𝛿 exp ñ−𝑖𝜔 b𝑡0 − 𝜙 có ∙ 2

𝜔0

[exp b− (𝜔−𝜔0)

2 ẟ2

2ẟ2

c + exp (− (𝜔+𝜔0)

2

(13.3)

)],

2

where ω is the angular frequency. The acoustic -6 dB bandwidth B can be approximated by the FWHM of the Gaussian peak of ò𝑃î (𝜔)ò at positive frequency:

(13.4)

𝐵 = √2 ln 2 (Hz), 𝜋ẟ

Combining Eqs. (13.2) and (13.4) leads to: 𝑅

= axial

4 ln 2 𝑐 𝑐 ∙ ≈ 0.88 𝜋 𝐵 𝐵

20

(13.4)

14. Assuming you are engineering a ‘perfect’ PA contrast agent for molecular imaging, please list all the desired key characteristics and explain the reasons.

Solution: This is an open question; the desired characteristics may include but not limited to the followings: 1. Its absorption should peak at the near-infrared wavelength region to maximize the tissue penetration.

2. Its absorption coefficient should be much higher than that of hemoglobin to provide high imaging contrast.

3. A low fluorescence quantum yield of the contrast agent is preferred for providing high PA signals.

4. It should be non-toxic and biocompatible with the tissue.

21

Chapter 13 Homework Solutions– Electrophysiological Mapping and Source Imaging

1. Nyquist theorem states conditions for a perfect reconstruction of a continuous signal from its discrete samples. Simply put, it states that a bandlimited signal can be reconstructed from its discrete samples if it is sampled with a frequency which is higher than a critical frequency, i.e. the Nyquist frequency. In other words, the inter-sample interval cannot be too large. The Nyquist frequency is 2 times the largest frequency of a band-limited signal.

2. If the highest frequency in EEG/MEG recordings is 50 Hz then a minimum sampling rate of 100 Hz is required for perfect reconstruction. In practice, an anti-aliasing filter has to be applied first to ensure that the signal under study is truly bandlimited. In practice, a sampling frequency of 3-5 times the Nyquist frequency is, typically, used.

3. Similarly, EEG (MEG) sensors placed on the scalp can be thought of as spatial samples of the scalp potential (magnetic field). If the maximum spatial frequency in these signals can be determined, the spatial Nyquist frequency and subsequently the minimum spatial distance between sensors can be determined, in theory (For instance, refer to Nunez and Srinivasan, 1989).

4. For instance, we could simply say that if both the elements of 𝒱𝜙 are positive, this is condition A, and if both elements are negative are, this is condition B. However, this partitioning, does not consider other cases, for instance if one of the elements of 𝒱𝜙 is positive and one negative, this partitioning cannot determine if condition A has occurred or condition B. i.

A better classification system will partition the 𝒱𝜙 space into two mutually exclusive regions so that depending on the location of a given sample of 𝒱𝜙, the underlying condition can be determined. For instance, a line might separate the space such that if a point lies above the line, it is considered condition A, and it is assigned as condition B if it lies below it.

1

ii.

Based on the classification described in (i & ii) we need to minimize the error of such a classification system. There might be sample points that lie above the line we have drawn and yet pertain to condition B and vice versa. In practice, an analysis named linear discriminant analysis (LDA) can be performed with given data points, i.e. training samples, to determine the most optimal parameters for the line (in higher dimensions a hyperplane can be used which partitions the space in two, corresponding to each condition). For the purpose of this exercise we will adopt a more analytical approach given that the probability density function of conditions A and B are given. We propose to minimize the following: 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑒 ℰ𝐼 + ℰ𝐼𝐼 ℰ𝐼 = Probability of identifying a point as pertaining to condition B while it really belongs to condition A ℰ𝐼𝐼 = Probability of identifying a point as pertaining to condition A while it really belongs to condition B

In other words, ℰ𝐼 = 𝑝𝑟𝑜𝑏/𝒱𝜙 𝑖𝑠 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒5𝒱𝜙 ∈ 𝐴8 = ∬ 𝑝(𝑥 + 1, 𝑦 + 1) 𝑑𝑥𝑑𝑦 for −∞ < 𝑥 < +∞ , 𝑦 < 𝛼𝑥 + 𝛿 ℰ𝐼𝐼 = 𝑝𝑟𝑜𝑏/𝒱𝜙 𝑖𝑠 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒5𝒱𝜙 ∈ 𝐵8 = ∬ 𝑝(𝑥 − 1, 𝑦 − 1)𝑑𝑥𝑑𝑦 for −∞ < 𝑥 < +∞ , 𝑦 > 𝛼𝑥 + 𝛿 Ultimately our goal is to solve the following optimization problem: 𝛼𝑥+ẟ

argmin ( P 𝛼,ẟ

iii.

+∞

𝑝(𝑥 + 1, 𝑦 + 1)𝑑𝑥𝑑𝑦 + P

𝑝(𝑥 − 1, 𝑦 − 1)𝑑𝑥𝑑𝑦 ) 𝛼𝑥+ẟ

–∞

By selecting the optimal values of (𝛼, 𝛿) we seek to minimize the misclassification error, i.e. ℰ𝐼 + ℰ𝐼𝐼.

In order to solve the optimization problem proposed in (ii) we will differentiate with respect to 𝛼 and 𝛿 separately and set the derivatives to zero and solve for 𝛼 and 𝛿.

If we re-parametrize the integrals in the goal function, we get: 𝛼𝑥–𝛼+ẟ+1

𝐺(𝛼, 𝛿) = P

+∞

𝑝(𝑥, 𝑦)𝑑𝑥𝑑𝑦 + P –∞

𝑝(𝑥, 𝑦)𝑑𝑥𝑑𝑦 𝛼𝑥+𝛼+ẟ–1

Differentiating with respect to 𝛿 and setting to zero we get: 𝛛𝐺 (𝛼, 𝛿) = 0, which leads to

𝛛ẟ

+∞

+∞

S 𝑝(𝑥, 𝛼𝑥 − 𝛼 + 𝛿 + 1)𝑑𝑥 − S 𝑝(𝑥, 𝛼𝑥 + 𝛼 + 𝛿 − 1)𝑑𝑥 = 0 –∞

–∞

If we use the fact that 𝑝(𝑥, 𝑦) = 𝑝(−𝑥, −𝑦)we have

2

+∞

+∞

S 𝑝(𝑥, 𝛼𝑥 − 𝛼 + 𝛿 + 1)𝑑𝑥 − S 𝑝(−𝑥, −𝛼𝑥 − 𝛼 − 𝛿 + 1)𝑑𝑥 = 0 –∞

–∞

Using the change of variable (taking -x to be the new parameter in the second integral) we get: +∞

+∞

S 𝑝(𝑥, 𝛼𝑥 − 𝛼 + 𝛿 + 1)𝑑𝑥 − S 𝑝(𝑥, 𝛼𝑥 − 𝛼 − 𝛿 + 1)𝑑𝑥 = 0 –∞

–∞

The only solution to ensure that the equality holds in its most general form is for the arguments of the two integrands to be the same, that is 𝛼𝑥 − 𝛼 + 𝛿 + 1 = 𝛼𝑥 − 𝛼 − 𝛿 + 1 which inadvertently results in 𝛿 = 0. Now, the goal function to be optimized is as follows: 𝛼𝑥

𝐺(𝛼, 𝛿) = P

+∞

𝑝(𝑥 + 1, 𝑦 + 1)𝑑𝑥𝑑𝑦 + P –∞

𝑝(𝑥 − 1, 𝑦 − 1)𝑑𝑥𝑑𝑦 𝛼𝑥

Differentiating with respect to 𝛼 and setting to zero we get: +∞

+∞

S 𝑥𝑝(𝑥 + 1, 𝛼𝑥 + 1)𝑑𝑥 − S 𝑥𝑝(𝑥 − 1, 𝛼𝑥 − 1)𝑑𝑥 = 0 –∞

–∞

If we use the fact that 𝑝(𝑥, 𝑦) = 𝑝(−𝑥, −𝑦), and reparametrize the integral with a change of variable (𝑥′ = −𝑥) we get: +∞

+∞

S 𝑥𝑝(𝑥 + 1, 𝛼𝑥 + 1)𝑑𝑥 + S 𝑥𝑝(𝑥 + 1, 𝛼𝑥 + 1)𝑑𝑥 = 0 –∞

–∞

In other words, we have to ensure that the following integral is equal to zero: +∞

2 S 𝑥𝑝(𝑥 + 1, 𝛼𝑥 + 1)𝑑𝑥 = 0 –∞

This means that the expectation of 𝑝(𝑥 + 1, 𝛼𝑥 + 1), i.e. the integral above, must be equal to zero. One way for this equality to hold, is for the distribution 𝑝(𝑥 + 1, 𝛼𝑥 + 1) to be symmetric or for 𝑝(𝑥 + 1, 𝛼𝑥 + 1) to be an even distribution (in x): 𝑝(𝑥 + 1, 𝛼𝑥 + 1) = 𝑝(1 − 𝑥, 1 − 𝛼𝑥) Using the fact that 𝑝(𝑥, 𝑦) = 𝑝(𝑦, 𝑥), we get: 𝑝(1 − 𝛼𝑥, 1 − 𝑥) = 𝑝(1 + 𝑥, 1 + 𝛼𝑥)

3

This equality holds under general conditions if 𝛼 = −1. Thus, the line 𝑦 + 𝑥 = 0 seems to be the solution. However, can you estimate the Hessian of the goal function and substitute 𝛼 = −1 and 𝛿 = 0 and confirm if it is a positive definite matrix and ensure that the obtained values indeed are a minimum? What happens? Is this solution an optimal one under all circumstances or does it depend on the probability distribution’s properties? This is left as an exercise for the reader. For more general functions and practices refer to support vector machine (SVM) and linear discriminant analysis (LDA) algorithms.

5. Following simple differentiation rules, we can solve this problem. We must keep in mind with respect to which set of variables we are differentiating, i.e. primed versus unprimed variables. i.

Following simple differentiating rules, we get: 1 ∇V W = 𝑟 ii.

3

((𝑥 − 𝑥′)2 + (𝑦 − 𝑦′)2 + (𝑧 − 𝑧′)2)2

Similarly, 1 𝑟

∇′ V W =

iii.

−(𝑥 − 𝑥′, 𝑦 − 𝑦′, 𝑧 − 𝑧′)

−(𝑥′ − 𝑥, 𝑦′ − 𝑦, 𝑧′ − 𝑧) 3

((𝑥 − 𝑥′)2 + (𝑦 − 𝑦′)2 + (𝑧 − 𝑧′)2)2

It is easy to see that, 1 ∇′ V W = 𝑟

−(𝑥′ − 𝑥, 𝑦′ − 𝑦, 𝑧′ − 𝑧) 3

((𝑥 − 𝑥′)2 + (𝑦 − 𝑦′)2 + (𝑧 − 𝑧′)2)2 1 (𝑥 − 𝑥′, 𝑦 − 𝑦′, 𝑧 − 𝑧′) = −∇ . = V 𝑟W 3 ((𝑥 − 𝑥′)2 + (𝑦 − 𝑦′)2 + (𝑧 − 𝑧′)2)2

6. We will solve this problem following the formula given in Eq. 3 and the hint given in (i): i.

ii.

Following the identities derived in problem (5) and substituting x = y = z = 0, we can easily see that, 1 (𝑥′, 𝑦′, 𝑧′) ∇V W = 3 𝑟 ((𝑥′)2 + (𝑦′)2 + (𝑧′)2)2 Given that Z𝒥ZZл= (0,0,1)T and using the hint we can easily calculate the field strength at any given point as follows: 1 1 1 𝑧′ Z𝒥ZZл = 𝛟(𝑥′, 𝑦′, 𝑧′) = . ∇ V W .

4𝜋𝜎 4𝜋𝜎 𝑟 3 ((𝑥′)2 + (𝑦′)2 + (𝑧′)2)2 Substitute (𝑥′, 𝑦′, 𝑧′) = (0,0,1) in formula derived in (i) we get, 1 1 1 1 ZZZл 1 𝛟(0,0,1) = .𝒥 = .

3 = 4𝜋𝜎 ∇ V W 4𝜋𝜎 𝑟 4𝜋𝜎 ((0)2 + (0)2 + (1)2)2

4

iii.

iv.

Substitute (𝑥′, 𝑦′, 𝑧′) = (1,0,0) in formula derived in (i) we get, 1 1 0 1 ZZZл 𝛟(1,0,0) = .𝒥 = . ∇ V W 4𝜋𝜎 𝑟 4𝜋𝜎 3 = 0 2 2 2 ((1) + (0) + (0) )2 Following the procedure in (iii) we have, 1 1 0 1 ZZZл 𝛟(0, −1,0) = . 𝒥 = .

4𝜋𝜎 ∇ V𝑟W 4𝜋𝜎 3 = 0 ((0)2 + (−1)2 + (0)2)2

7. Following the same procedure as problem (6) we can simply calculate the magnetic field at any given point as follows: i.

Given that ∇ (1` for a dipole located at origin is calculated in problem 6 and using the 𝑟

Green theorem, we have, ZZZл B(𝑥′, 𝑦′, 𝑧′) = 𝒥

1 ZZZл × ∇ ) 1 𝜇 (𝒥 × ∇V W = VW 𝑟 4𝜋 𝑟 4𝜋 𝜇

ZZZл = (0,0,1), Using the definition of cross product and the fact that 𝒥 ′ ′ (−𝑦 , 𝑥 , 0) 1 ZZZл × ∇ V W = 𝒥 3 𝑟 ((𝑥′)2 + (𝑦′)2 + (𝑧′)2)2 𝜇

B(𝑥′, 𝑦′, 𝑧′) =

4𝜋

(−𝑦′, 𝑥′, 0)

. ′ 2

′ 2

′ 2 3

((𝑥 ) + (𝑦 ) + (𝑧 ) )2 ii.

Substitute (𝑥′, 𝑦′, 𝑧′) = (0,0,1) in formula derived in (i) we get, 𝜇

B(0,0,1) =

4𝜋 iii.

(−0,0,0) 3 ((0)2 + (0)2 + (1)2)2

= d0

Substitute (𝑥′, 𝑦′, 𝑧′) = (1,0,0) in formula derived in (i) we get,

B(1,0,0) =

𝜇 4𝜋

iv.

.

.

(−0,1,0)

𝜇 d𝑒d

3= 4𝜋 ((1)2 + (0)2 + (0)2)2

𝑦

Substitute (𝑥′, 𝑦′, 𝑧′) = (0, −1,0) in formula derived in (i) we get,

B(0, −1,0) =

𝜇 4𝜋

.

(−(−1),0,0)

𝜇 d𝑒d

3= 4𝜋 𝜄 2 2 2 ((0) + (−1) + (0) )2

8. As it can be seen in problems (6) and (7), EEG sensors are most sensitive to normal sources, for instance when sensor location was in line with dipole orientation. MEG sensors are least sensitive to such sources. Mathematically speaking this can be understood by referring to the hints

5

provided in problems (6) and (7); that is the field strength is related either to the inner product (cross product) of dipole orientation and sensor location for EEG (MEG). Inner product (cross product) of two vectors achieves its maximum when two vectors are in line (perpendicular) to each other, hence the sensitivity (insensitivity) of EEG (MEG) to normal sources (normal to the surface of a sphere that models the cortex and in line with EEG sensors above it; also placed on a sphere representing the scalp; under ideal spherical assumptions).

9. Using simple linear algebra rules, we have: i. 𝐵 = 𝐴𝑆 1 1 1 3 ii. 𝐵 = g3 1 6h i1j = g2h 1 6 1

iii.

𝐵 = g3 1 6 1

iv.

𝐵 = g3 1 6

v.

vi.

vii.

1

1 3 1

3

3

2

1 2 2 1 6h i−1j = k0l 1 0 1 2 3 1 3 3 1 6h i0j = g2h 1 3 1 3

𝐵 = 𝐴((1,1,1)𝑇

+

1

2

𝓉(2, −1,2)𝑇) = 𝐴 i1j + 𝓉𝐴 i−1j =

3 g2h + 𝓉 k0l =

3

0

3

g2

3

+ 𝓉0

3

h = g2h

3 + 𝓉0 1 2 2 2 2 1 1 𝑥 2𝑥 + 6𝑦 + 𝑧 = 0 𝑥−𝑧=0 g3 1 6h o𝑦p = 0 => q => (𝑥, 𝑦, 𝑧) = => q 1 1 3𝑥 + 6𝑦 = 0 𝑥 + 6𝑦 + 2𝑧 = 0 1 6 3 𝑧 2𝑡 (2𝑡, −𝑡, 2𝑡) is the null space of A such that 𝐴 i−𝑡j = 0 for all t ∈ ℛ. 𝑡 The number of sensors is always less than the number of possible sources in practice, hence, the null space is not empty for all lead-field matrices. This means that for any given source configuration 𝑆∗ that satisfies 𝐴𝑆∗ = 𝐵∗ we can find infinitely many vectors such as 𝑆𝑛𝑒w = 𝑆∗ + 𝑡𝑋, where 𝐴𝑋 = 0 and t ∈ ℛ, that also fit the given measurements, that is 𝐴𝑆𝑛𝑒w = 𝐴(𝑆∗ + 𝑋) = 𝐴𝑆∗ + 𝑡𝐴𝑋 = 𝐵∗ + 𝑡0 = 𝐵∗. No unique solution exists for a given vector of measurements, 𝐵∗.

10. We substitute Eq. (8) in Eq. (5) to get i. 𝑋 = 𝐵𝜙 = 𝐵(𝐴𝑆) = (𝐵𝐴)𝑆 ii.

Mathematically speaking iff (if and only if) 𝐵𝐴 = 𝐼, or basically an identity matrix, then the estimated source will be equal to the true underlying source. iii.

The answer is no; as described in (ii) for linear methods to achieve this goal, 𝐵𝐴 = 𝐼. This condition can never be satisfied in an underdetermined system, i.e. the number of measurements are less than the number of unknowns. We know that matrix A is not a square matrix and has many more columns than rows. As such the null space of matrix A is not empty, that is, there exists non-zero vectors such as X that 𝐴𝑋 = 0. Multiplying B to the left-hand side we get 𝐵𝐴𝑋 = 0; so the vector X cannot be retrieved, which means that

6

𝐵𝐴 = 𝐼 cannot hold for vectors lying in null space of A such as X; thus, linear methods, without appropriate constraints, cannot retrieve underlying sources fully.

11. We have the following objective function: 2

argmin (5|𝜙 − 𝐴𝑥|5 + 𝜆||𝐺𝑥||2) 2

2

𝑥

In order to optimize this goal function, we need to differentiate the goal function with respect to 𝑥 and set to zero and solve for 𝑥: 2 𝜕5|𝜙 − 𝐴𝑥|5 + 𝜆||𝐺𝑥||2 2

2 = 0

𝜕𝑥 2

𝜕5|𝜙 − 𝐴𝑥|5

2 +𝜆

𝜕||𝐺𝑥||2

2

𝜕𝑥 Following matrix differentiation rules, we get:

=0

𝜕𝑥

2

𝜕5|𝜙 − 𝐴𝑥|5

= 2𝐴 𝐴𝑥 − 2 𝐴 𝜙 𝜕𝑥 𝜕||𝐺𝑥||2 2 = 2𝐺 𝑇 𝐺𝑥 𝜕𝑥 So, combining the two terms we have 2𝐴𝑇𝐴𝑥 − 2 𝐴𝑇𝜙 + 𝜆 2𝐺𝑇𝐺𝑥 = 0 2(𝐴𝑇𝐴 + 𝜆𝐺𝑇𝐺)𝑥 − 2 𝐴𝑇𝜙 = 0 Which results in, 𝑥 = (𝐴𝑇𝐴 + 𝜆𝐺𝑇𝐺)–1𝐴𝑇𝜙 Thus, the inverse filter is: 𝐵 = (𝐴𝑇𝐴 + 𝜆𝐺𝑇𝐺)–1𝐴𝑇 Assuming matrix G to be a diagonal matrix, it is easy to show (left as an exercise) that: 𝐵 = (𝐴𝑇𝐴 + 𝜆𝐺𝑇𝐺)–1𝐴𝑇 = 𝐴𝑇(𝐴𝐴𝑇 + 𝜆𝐺𝐺𝑇)–1 2

𝑇

𝑇

12. We will use basic probability theorems and identities to solve this problem.

–

1

1

||(𝑛–0)||2

i.

𝑝(𝑛) = σ √2𝜋 𝑒 2σ𝑛 2

ii.

The distribution 𝑝(𝜙|𝑥) is the likelihood of measurements if the underlying sources were given. This represents our belief of how the random variable 𝜙, i.e. measurements, would behave, i.e. its distribution, if we knew the underlying sources. As in our model, the variability in the measurements are due to white additive noise and underlying sources’ variability, i.e. 𝜙 = 𝐴𝑥 + 𝑛, and given that the underlying sources are assumed to be given, i.e. 𝑥 is known, the only source of variability is due to noise, i.e. 𝑛, so we have: 𝑝(𝜙|𝑥) = 𝑝(𝑛) and given that 𝜙 = 𝐴𝑥 + 𝑛, so we know that 𝑛 = 𝜙 − 𝐴𝑥, so it follows

𝑛

1

easily that 𝑝(𝜙|𝑥) = 𝑝(𝑛) = σ √2𝜋 𝑒 iii.

𝑝(𝑥) =

1 σ𝑠√2𝜋

𝑒

1 – ||(𝑥–0)||2 2σ𝑠 2

𝑛

–

1 ||𝜙–𝐴𝑥||2 2σ𝑛 2

.

7

iv.

𝑝(𝑥|𝜙) = 𝑝(

𝜙|𝑥)𝑝(𝑥)

, which is difficult to evaluate analytically, and as

𝑝(

∫ 𝑝(𝜙|𝑥)𝑝(𝑥)𝑑𝑥

𝑝(𝜙)

v. vi.

𝜙|𝑥)𝑝(𝑥)

=

we will see in the following, unnecessary to directly calculate. ℒ(𝑥|𝜙) = ln 𝑝(𝑥|𝜙) = ln/𝑝(𝜙|𝑥)8 + ln/𝑝(𝑥)8 − ln (𝑝(𝜙)). 𝑥| = argmax ℒ(𝑥|𝜙) = argmax (ln/𝑝(𝜙|𝑥)8 + ln/𝑝(𝑥)8 − ln (𝑝(𝜙))) 𝑥

𝑥

As ln (𝑝(𝜙)) does not depend on x, it can be dropped in our optimization problem, we also have: ln/𝑝(𝜙|𝑥)8 = ln(

1

) + ln(𝑒

–

1 1 ||𝜙–𝐴𝑥||2 ) = − 2 2σ𝑛2 2𝜎 2 5|𝜙 − 𝐴𝑥|5 + ln(

𝜎𝑛√2𝜋 𝑛 1 1 1 2 2 1 – ||𝑥|| ) = − 2𝜎 2 5|𝑥|5 + ln( ) ln(𝑝(𝑥)) = ln( ) + ln(𝑒 2σ𝑠2 𝜎𝑠√2𝜋 𝜎𝑠√2𝜋 𝑠

1

)

𝜎𝑛√2𝜋

Dropping the constant terms and noticing that argmax ℒ(𝑥|𝜙) = argmin −ℒ(𝑥|𝜙), we 𝑥

𝑥

have: 𝑥| = argmin −ℒ(𝑥|𝜙) 𝑥

1 5|𝜙 − 𝐴𝑥|52 + 1 2 5|𝑥|5 2𝜎𝑠2 𝑥 2𝜎𝑛2 2 2 = argmin5|𝜙 − 𝐴𝑥|5 + 𝜆25|𝑥|5 = argmin

2

𝑥

Where 𝜆2 = σ𝑛 2. This is the minimum norm estimation which is equivalent to an MAP σ𝑠

solution.

13. Following simple calculus rules, we have: i.

We will have a unit-circle and a unit-square as follows

ii.

Following simple algebraic rules we get the following: When K0 increases the line moves towards the upper-right corner and when K0 decreases the line moves towards the lower-left corner of the plane.

8

14. This problem provides some intuition into solving the L1-norm minimization problem: i.

This optimization problem is minimizing K0, which is to move the line towards the lowerleft corner of the plane, but it has two constraints, first, it has to touch the unit-square, and secondly, it has to be in the first quadrant. It has a unique solution (depicted in (ii)). ii.

Look at the following graph,

iii.

Absolutely. For instance, we can start with a relatively large K0 and continue decreasing K0 until we touch the unit-square. As the tangent of the goal function is -2, the line will touch the unit-square at its corner at (0,1). For a line with a tangent smaller than -1, it would touch the unit-square at (1,0) as we decrease K0 (the problem has no unique solutions if the line tangent is equal to -1; why?). Based on these explanations it is evident that 𝑥∗ = 1, 𝑦∗ = 0 and we also have 𝐾0∗ = 𝑦∗ + 2𝑥∗ = 2.

15. This problem provides some intuition into solving the minimum norm (L2-norm minimization) problem: i.

This optimization problem is minimizing K0, which is to move the line towards the lowerleft corner of the plane, but it has two constraints, first, it has to touch the unit-circle, and secondly, it has to be in the first quadrant. It has a unique solution (depicted in (ii)). ii.

Look at the following graph,

9

iii.

For instance, we can start with a relatively large K0 and continue decreasing K0 until we touch the unit-circle. In this example, we can use calculus to exactly solve for the problem by finding the point on the unit circle (with positive x and y), where its derivative is equal to -2 (the lines tangent): 𝑦 = Ä1 − 𝑥2, 𝑥 > 0 Differentiate with respect to 𝑥, and setting to -2 we have 𝑑𝑦 −𝑥 2 1 = = −2 => 𝑥∗ = , 𝑦∗ = 𝑑𝑥 √1 − 𝑥2 √5 5 √5 𝐾 ∗ = 𝑦∗ + 2𝑥∗ = = √5.

0

√5

16. As it can be seen in the examples provided in problems (15) and (16), the optimal solution of an L1-norm optimization problem is more likely to occur at the corners of the unit-square because of the geometry of the L1-norm (as long as the goal function is a convex function, like a line or a parabola). In problem 14 this meant that either 𝑥∗ or 𝑦∗ would be 0 almost surely (disregarding the degenerate cases). This is not the case for an L2-norm problem where most likely, the solution is dense, i.e. both 𝑥∗ and 𝑦∗ are most likely non-zero. This briefly demonstrates why L1-norm estimations induce sparse solutions as opposed to L2-norm estimates.

10

Homework Solution Keys 1. Based on the information provided, 𝑋(𝑡) is a stationary process. 𝜇(3) = 𝐸(𝑋(3)* = 0 and 𝜇(9) = 𝐸(𝑋(9)* = 0. 𝜎2(3) = 𝜎2(9) = 1. 𝑅𝑥𝑥(3,9) = 2–6.

2. First, 𝜇(𝑡) = 𝐸(𝑋(𝑡)) = 𝐸(𝐴) cos(𝜔𝑡) + 𝐸(𝐵) sin(𝜔𝑡) = 0. Namely, 𝜇(𝑡) is a constant and therefore independent of time, a requirement of stationarity. Next, 𝑅𝑥𝑥(𝑡1, 𝑡2) = 𝐸 =(𝐴𝑐𝑜𝑠(𝜔𝑡1) + 𝐵𝑠𝑖𝑛(𝜔𝑡1)*(𝐴𝑐𝑜𝑠(𝜔𝑡2) + 𝐵𝑠𝑖𝑛(𝜔𝑡2)*C = 𝐸(𝐴2)𝑐𝑜𝑠(𝜔𝑡1)𝑐𝑜𝑠(𝜔𝑡2) + 𝐸(𝐵2)𝑠𝑖𝑛(𝜔𝑡1)𝑠𝑖𝑛(𝜔𝑡2) = 𝜎2 cos(𝜔(𝑡2 − 𝑡1)* = 𝑅𝑥𝑥(𝑡2 − 𝑡1).

Namely, the autocovariance function is only dependent on the difference between 𝑡2 and 𝑡1, another requirement of stationarity. Thus, 𝑋(𝑡) is stationary.

3. This is a simulation exercise. The student will write Matlab code to generate data by simulating the model, estimate the autocovariance function from the simulated data and compare it with the theoretical autocovariance function.

4. 𝑅𝑥𝑦(𝜏) = 𝐸(𝑋(𝑡)𝑌(𝑡 + 𝜏)) = 𝐸((𝑍(𝑡) + 0.4𝑍(𝑡 − 1))(𝑍(𝑡 + 𝜏) − 0.4𝑍(𝑡 + 𝜏 − 1)) = 𝐸(𝑍(𝑡)𝑍(𝑡 + 𝜏)+0.4𝑍(𝑡 − 1)𝑍(𝑡 + 𝜏) − 0.4𝑍(𝑡)𝑍(𝑡 + 𝜏 − 1) − 0.16𝑍(𝑡 − 1)𝑍(𝑡 + 𝜏 − 1)) 0.4𝜎𝑧2 =J 0.84𝜎𝑧2 −0.4𝜎𝑧2

𝜏 = −1 𝜏=0 𝜏=1

𝑅𝑥𝑦(𝜏) = 0 for other values of τ.

5. This is a simulation exercise. The student will carry out the simulation according to the instructions.

6. Based on the information provided, 𝑋(𝑡) is a stationary process with the autocovariance function being: 𝑅𝑥𝑥(𝜏) = 4𝑒–0.2|𝜏|. The power spectral density is 4(1 − 𝑒–0.4)

∞

𝛤𝑥𝑥(𝑓) = 𝑅𝑥𝑥(0) + 2 O 𝑅𝑥𝑥(𝜏) cos(2𝜋𝑓𝜏) = 𝜏=1 σ

1 + 𝑒–0.4 − 2𝑒–0.2 cos(2𝜋𝑓)

2

7. The autocovariance function is: 𝑅𝑥𝑥(𝜏) = 1–𝛼𝑧 2 𝛼2. (a) The power spectral density is ∞

𝛤𝑥𝑥(𝑓) = 𝑅 𝑥𝑥(0) + 2 O 𝑅 𝑥𝑥(𝜏) cos(2𝜋𝑓𝜏) = 𝜏=1

𝜎𝑧2

1 + 𝛼2 − 2𝛼 cos(2𝜋𝑓)

(b)-(e) The student will carry out the numerical simulation portion of the problem according to the instructions.

8. The student will carry out the simulations according to the instructions.

9. Refer to Problem 4 for more detailed procedures. Briefly, 𝑅𝑥𝑦(𝜏) = 𝐸(𝑋(𝑡)𝑌(𝑡 + 𝜏)) 0.4𝜎𝑈2 𝜏=0 2 =J 0.84𝜎𝑈 𝜏=1 2 −0.4𝜎𝑈 𝜏=2 𝑅𝑥𝑦(𝜏) = 0 for other values of τ.

The cross spectrum is: ∞

𝛤𝑥𝑦(𝑓) = O 𝑅𝑥𝑦(𝜏) 𝑒–𝑖2𝜋𝑓𝜏 = 0.4𝜎2 + 0.84𝜎2 𝑒–𝑖2𝜋𝑓 − 0.4𝜎2𝑒–𝑖4𝜋𝑓 𝑈

𝑈

𝑈

𝜏=–∞

= 0.4𝜎𝑈2 + 0.84 cos(2𝜋𝑓) − 0.4 cos(4𝜋𝑓) + 𝑖(0.84 sin(2𝜋𝑓) − 0.4 sin(4𝜋𝑓)) To compute the coherence function, we need to obtain the power spectra of X and Y processes, respectively. They are 𝛤𝑥𝑥(𝑓) = (1.16 + 0.8 cos(2𝜋𝑓))𝜎2𝑈 and 𝛤𝑦𝑦(𝑓) = (1.16 − 0.8 cos(2𝜋𝑓))𝜎2𝑈 The spectral coherence function, after some algebra, is found to be 𝐶𝑥𝑦(𝑓) =

|𝛤𝑥𝑦(𝑓)| T𝛤𝑥𝑥(𝑓)T𝛤𝑦𝑦 (𝑓)

=1

10. When the two processes are examined individually, we have 𝑋(𝑡) = 𝜀1(𝑡) 𝑌(𝑡) = 𝛼𝑌(𝑡 − 1) + 𝜂1(t) where 𝜀1(𝑡) = 𝑈(𝑡) and 𝜂1(𝑡) = 𝑈(𝑡 − 1) + 𝑉(𝑡). Thus, Z1 = 𝜎2 and 𝛤1 = 𝜎2 + 𝜎2. 𝑈

When the two processes are examined jointly, we have 𝑋(𝑡) = 𝜀2(𝑡) 𝑌(𝑡) = 𝛼𝑌(𝑡 − 1) + 𝑋(𝑡 − 1) + 𝜂2(𝑡) where 𝜀2(𝑡) = 𝑈(𝑡) and 𝜂2(𝑡) = 𝑉(𝑡). Thus, Z2 = 𝜎2 and 𝛤2 = 𝜎2. 𝑈

Based on the definition of time-domain Granger causality

𝑉

𝑈

𝑉

𝛤1

σ2 +σ2

𝛤𝑥→𝑦 = ln 𝛤 = ln 𝑈σ2 7 > 0 2

7

Z1

σ2

𝛤𝑦→𝑥 = ln Z = ln σ2𝑈 = 0 2

𝑈

These Granger causality results are in agreement with the causal influence patterns in the model in which causal influence flows unidirectionally from X to Y.

Deep Learning Models with Applications to Brain Image Analysis Dinggang Shen1, Luping Zhou2, and Mingxia Liu1 1Department of Radiology and BRIC, University of North Carolina at Chapel Hill,

NC 27599, USA 2School of Electrical and Information Engineering, The University of Sydney,

Australia

1

Exercises

1. In practice, we usually have only limited number of brain images to train deep models for analysis. Please explain the problem and list at least two strategies to deal with this situation.

Possible Answer: Deep learning models usually involve many parameters to learn, which requires sufficiently large number of training samples for a reliable estimation. Otherwise, the problem of “overfitting” may occur. That is, the test error increases although the training error keeps decreasing. Multiple strategies could be used to deal with this situation in brain image analysis, for example, using data augmentation or patch-based approaches to increase the number of training samples.

2. Consider the following MLP model with two hidden layers and the loss function shown in Table A. Please calculate ∂h1,i and ∂J . ∂w1,2

∂w1,2

Table A. An MLP model with two hidden layers Input x1,i, x2,i, where i = 1, · · · N and N is the number of samples Layer 1 h1,i = max(w1,1x1,i + w1,2x2,i, 0), h2,i = max(w2,1x1,i + w2,2x2,i, 0) Layer 2 pi = w5h1,iΣ+ w6h2,i J(p, w) = (pi − yi)2 Loss i

Possible Answer: ∂h1,i ∂w1,2 = u(w1,1x1,i + w1,2x2,i)x2,i, where u(x) = 1 if x > 0, or 0 otherwise. Σ Σ Σ ∂J ∂pi ∂hj,i ∂J ∂pi = ∆pi = ∂w1,2 ∂pi ∂hj,i ∂w1,2 ∂w1,2 i i j Σ = 2(pi − yi)w5u(w1,1x1,i + w1,2x2,i)x2,1. i

3. What are the benefits to use CNN to analyze brain images, compared with MLP? Possible Answer: When analyzing brain images, without flattening an image into MLP, CNN models can better preserve local spatial information.

Also, by using local region connection and weights sharing, CNN models can significantly reduce the number of parameters to learn, compared with MLP. 4. What’s key difference and advantages of the fully convolutional networks (FCNs) over convolutional neural networks (CNNs) in brain image segmentation (pixel-level prediction)?

Possible Answer: The key difference between FCN and CNN is that FCN converts the fully-connected layers at the end of CNN into convolutional layers. In this way, FCN has the advantages of i) accepting input images of arbitrary size, ii) better keeping spatial information, and iii) no need to go through pixels one by one for prediction.

5. Please describe the network structure of U-net. What are the benefits to use skip connections in U-net?

Possible Answer: U-net consists of the contracting and the expanding paths. The contracting path consists of convolutional layers while the expanding path consists of deconvolutional layers. Between these two paths, multiple skip connections are used to link the corresponding convolutional and deconvolutional layers. The skip connections could help mitigate the common gradient vanishing problem during the training of deep models as the gradient of the deeper layers could be directly backpropagated to the shallower layers via the skip connections. Moreover, this structure can preserve the information from the shallow layers, forming a hierarchy of visual clues.

6. What’s difference between RNNs and CNNs? What is the advantage of LSTM over RNN? Possible Answer: CNNs assume that all inputs (and outputs) are independent of each other, and thus cannot handle problems with sequential data.

RNNs are designed to address this issue. RNNs perform the same task for every element of a sequence (with the output being depended on the previous computations), and have a “memory” which captures information about what has been calculated so far.

7. The Auto-Encoder (AE) model can extract imaging features in an unsupervised manner. What is the principle of AE? Possible Answer: A typical Auto-Encoder (AE) model contains an encoder to first transform the input into its low-dimensional latent representation space and a decoder to reconstruct the initial data from the representation by closing the distance between the input and the output. Once the models are well trained, the latent representations can be leveraged as the extracted

features in the following tasks.

8. What’s the purpose of using the generator and discriminator in GANs? Possible Answer: The goal of the generator is to learn an embedding function G(·) that transforms the random inputs z to the output synthetic images G(z) whose distribution matches the data distribution Pdata. Meanwhile, the discriminator D is trained to learn an embedding function D(·) to maximize the probability of the correct label assignment to discriminate the real and the fake samples.

9. What’s the advantage of CycleGAN over GAN? Possible Answer: A GAN model transforms training images between two domains with one-to-one correspondence. Using CycleGAN with a cycleconsistent loss, the samples x and y in two domains do not need to have one-to-one correspondence. Thus, CycleGAN is particularly useful in applications with unpaired data in two domains.

10. Please list possible deep learning models that could be used for the following brain image analysis tasks: (a) brain disease diagnosis, (b) brain lesion segmentation, (c) brain network analysis based on fMRI images, (d) brain image transferring across modality, (e) brain disease diagnosis without labeled training samples, (f) brain image generation without paired training samples.

Possible Answer: Possible choices are: (a) CNN, (b) FCN, (c) RNN/LSTM, (d) GAN, (e) Auto-encoder, and (f) Cycle-GAN.

Exercise Solutions Exercise 1 The Nernst equation (Eqn. 10) can be used to calculate the reversal potentials of the individual ions 145 8.314 × 310 𝐸𝑁𝑎 = 𝑙𝑛 / 1 = 66.6 mV 4 1 × 9.65 × 10 12 8.314 × 310 4 𝐸 = 𝑙𝑛 / 1 = −96.8 mV 4 𝐾 1 × 9.65 × 10 150 𝐸 = 𝐶𝑙

8.314 × 310 −1 × 9.65 × 104

𝑙𝑛 /

100 1 = −80 mV 5

Exercise 2 As indicated in the equivalent circuit shown in Fig. 16, the current flow due to the activity of a particular ion species depends on the reversal potential of the ion and its conductance, and is given by 𝐼𝑖 = 𝐺𝑖(𝑉𝑚 − 𝐸𝑖). The reversal potential is modeled as an electric cell (battery) and the conductance as a resistor. The membrane also has a capacitance element that is modeled as an electrical capacitor. The capacitive current is given by 𝐼𝑐 = 𝐶𝑚 d𝑉𝑚⁄d𝑡.

To find the resting membrane potential we use the chord conductance equation (Eqn. 12), which is derived using Kirchhoff’s current law at equilibrium. At equilibrium, the capacitive current given by 𝐼𝑐 = 𝐶𝑚 d𝑉𝑚⁄d𝑡 is zero since the membrane potential is not changing. Since no current is being injected into the membrane from an external source, the sum of all ionic currents adds up to zero. This results in: 𝐺𝑁𝑎𝐸𝑁𝑎 + 𝐺𝐾𝐸𝐾 + 𝐺𝐶𝑙𝐸𝐶𝑙 𝑉𝑚 = 𝐺𝑁𝑎 + 𝐺𝐾 + 𝐺𝐶𝑙 66.6 × 10–5 − 96.8 × 10–4 − 80 × 2 × 10–5 = = −81.6 mV 10–5 + 10–4 + 2 × 10–5 The resting membrane potential is determined largely by the reversal potential of potassium and its high relative membrane conductance.

Doubling the external potassium concentration changes its reversal potential to 𝐸𝐾 = −78.3mV, which we then use to calculate the resting potential. The new resting potential is 𝑉𝑚 = −67.4mV. Thus, by increasing the external concentration of potassium, we observe a depolarization of the cell.

2 Fig. 16 Circuit representation of membrane having sodium, potassium, and chloride channels.

Exercise 3 An action potential (AP) is generated and propagated as a result of opening and closing of voltage-gated Na+ and K+ ion channels. In neurons, an AP is initiated at the axon hillock, a region of the neuron between the axon and soma that expresses a higher density of sodium channels. The Hodgkin Huxley (HH) model has 3 gating variables m, n, and h that model the sodium activation, potassium activation, and sodium inactivation gates, respectively.

We start with a neuron resting at a typical resting potential of -70 mV. At this resting potential, m is close to zero indicating that the Na+ activation gate is closed. In contrast, the inactivation gate of Na+ channels, is about half open as indicated by h being near 0.5. Together, this results in a negligible sodium conductance and current.

Excitatory inputs to the neuron result in a depolarization at the soma. At this more depolarized voltage, some of the sodium channels open resulting in a sodium current influx that further depolarizes the membrane. The sodium activation gate has a small time constant and hence Na+ channels open far faster than K+ channels. If the excitatory input is large enough and leads to a critical number of sodium channels opening, the positive feedback of the sodium current results in a rapid depolarization towards the reversal potential of sodium, near 60 mV. This process is the upstroke of the action potential, and is opposed by the inactivation of sodium through a decreased h (closing of the sodium inactivation gates) and the activation of the potassium conductance through an increasing n. This decrease in the sodium

3

conductance and increase in the potassium conductance results in the hyperpolarization of the membrane towards the reversal potential of potassium and constitutes the down stroke of the action potential. The slower kinetics of the h and n variables result in an undershoot of the resting potential before returning to the resting potential.

If the time constant of the sodium activation m is increased such that its kinetics become slower than the sodium inactivation h and potassium inactivation n, then the neuron would not be excitable and action potentials cannot be generated. However, if the kinetics of m remain faster than those of h and n, action potentials can still be generated but would require more input current as the avalanche effect, the positive feedback generated by the opening of sodium channel, of sodium conductance is slowed down.

Exercise 4 Without any external input, the neuron would be resting at Vreset of -70 mV. The neuron would start firing once it depolarizes to V thresh. The rheobase current would be calculated as ratio of this depolarization to the membrane resistance. 𝑉𝑡ℎ𝑟𝑒𝑠ℎ − 𝑉𝑟𝑒𝑠𝑒𝑡 20 mV 𝐼𝑟ℎ𝑒𝑜 = = = 0.4 nA 𝑅𝑚 50 MΩ A neuron firing continuously at a rate of 10 spikes / s has an interspike interval (Tisi) of 100 ms. We can find the current required by using algebraic manipulation of Eqn. 41 to express Iinj in terms of Vthresh, Vreset, Rm, and F. This gives us 𝐼

=

𝑖𝑛j

𝑉𝑡ℎ𝑟𝑒𝑠ℎ − 𝑉𝑟𝑒𝑠𝑒𝑡 𝑇𝑖𝑠𝑖

𝑅𝑚 /1 − 𝑒 – 𝜏 1

=

0.4

nA ≈0.633 nA

(1 − 𝑒–1)

Exercise 5 MATLAB is used simulate the LIF neuron. The output of the code below is shown in Fig. 17.

Rm = 50; Cm = 2000; Vt = -50; Vr = -70; Vp = 20; dt=0.5e-3; tau = 0.1; % Part 1 t1 = 0:dt:2; I1 = 0*t1;

% membrane resistance in Mohms % membrane capacitance in pF % threshold voltage in mV % reset voltage in mV % peak voltage in mV % time step in s % membrane time constant in s

4

I1(1000:3000) = 640; % input current in pA v = Vr; % inintialize the voltage Vm1(1)= Vr; for ind=2:length(t1) %simulate for 2000 ms v = v + dt*(-(v-Vr)/tau+ 1000*I1(ind)/Cm); if v>Vt Vm1(ind)=Vp; v=Vr; else Vm1(ind) = v; end end figure subplot(2,1,1); plot(t1,Vm1,'-k','linewidth',2) ylabel('Vm (mV)') subplot(2,1,2); plot(t1,I1,'-k','linewidth',2); xlabel ('time (s)') ylabel('I_{inj} (pA)') % Part 2 t2 = 0:dt:5; I2 = linspace(0,1000, length(t2)); % input current in pA v = Vr; % inintialize the voltage Vm2(1)= Vr; for ind=2:length(t2) %simulate for 5000 ms v = v + dt*(-(v-Vr)/tau+ 1000*I2(ind)/Cm); if v>Vt Vm2(ind)=Vp; v=Vr; else Vm2(ind) = v; end end figure subplot(2,1,1); plot(t2,Vm2,'-k','linewidth',2) ylabel('Vm (mV)') subplot(2,1,2); plot(t2,I2,'-k','linewidth',2); xlabel ('time (s)') ylabel('I_{inj} (pA)')

5

Fig. 17 Response of LIF neuron to a step current (top) and a ramp current (bottom).

Exercise 6 To find the peaks of the post-synaptic responses, we find the derivative of the response and find the time at which it is zero.

For the difference of two exponentials synapse, the derivative of Eqn. 53 gives the following equation that we have to solve 𝑡–𝑡𝑠𝑝𝑖𝑘𝑒 – 𝜏r𝑖𝑠𝑒

𝑒

𝜏𝑟𝑖𝑠𝑒

−

𝑡–𝑡 – 𝜏𝑠𝑝𝑖𝑘𝑒 𝑓𝑎𝑙𝑙

𝑒

𝜏𝑓𝑎𝑙𝑙

=0

and which results in 𝑡

𝑝𝑒𝑎𝑘

=𝑡

𝑠𝑝𝑖𝑘𝑒

+

𝜏𝑟𝑖𝑠𝑒𝜏𝑓𝑎𝑙𝑙 𝜏𝑓𝑎𝑙𝑙 ln / 1 𝜏𝑓𝑎𝑙𝑙 − 𝜏𝑟𝑖𝑠𝑒 𝜏𝑟𝑖𝑠𝑒

For an alpha synapse, taking the derivative of Eqn. 54 and solving

6

results in

𝑑 D𝑡 ⋅ 𝑒–𝛼(𝑡–𝑡𝑠𝑝𝑖𝑘𝑒) F = 𝑒–𝛼(𝑡–𝑡𝑠𝑝𝑖𝑘𝑒) − 𝑡 ⋅ 𝛼𝑒–𝛼(𝑡–𝑡𝑠𝑝𝑖𝑘𝑒) 𝑑𝑡

= 0

1 𝑡𝑝𝑒𝑎𝑘 =

𝛼

To derive Eqn. 41, we can solve the ordinary differential equation of the LIF model neuron, which results in the membrane potential expressed as a function of time and the injected current. Injected a neuron resting at V reset with a sufficiently large I inj would cause it to periodically depolarize past Vthresh and fire at intervals of TISI. The firing frequency is defined as the inverse of T ISI. Starting with a neuron resting at Vreset at t=0, the solution to the LIF differential equation is –

𝑡

𝑉𝑚(𝑡) = 𝐼𝑖𝑛j𝑅𝑚 + H𝑉𝑟𝑒𝑠𝑒𝑡 − 𝐼𝑖𝑛j𝑅𝑚I𝑒 𝜏 By rearranging the expression for Vm(t) = Vthresh when t=TISI, we find the expression for TISI 𝑉𝑡ℎ𝑟𝑒𝑠ℎ − 𝐼𝑖𝑛j𝑅𝑚 L 𝑇𝐼𝑆𝐼 = −𝜏 ln K 𝑉𝑟𝑒𝑠𝑒𝑡 − 𝐼𝑖𝑛j𝑅𝑚 Inverting it results in the firing frequency –1 𝑉𝑡ℎ𝑟𝑒𝑠ℎ − 𝐼𝑖𝑛j𝑅𝑚 𝐹H𝐼𝑖𝑛j I = K−𝜏 lnK LL 𝑉𝑟𝑒𝑠𝑒𝑡 − 𝐼𝑖𝑛j𝑅𝑚 Exercise 8 MATLAB is used simulate the Izhikevich neuron. The output of the code below is shown in Fig. 18. The model can be explored in further detail at www.izhikevich.org. The parameters generating the different firing patterns are in the code. dt=0.2; % time step in ms % parameters for the different modes pars=[0.02 0.02 0.02 0.02

0.2 0.25 0.2 0.25

-65 -65 -50 -55

mode =4; % mode selection a=pars(mode,1); b=pars(mode,2); c=pars(mode,3); d=pars(mode,4); t = 0:dt:500; I = 0*t; % input current

6 6 2 0.05

14 ;...

0.5 ;...

15 ;...

0.6];

% tonic spiking % phasic spiking % tonic bursting % phasic bursting

7

I(500:end) = pars(mode,5); v(1)= -70; u = -14;

% initialize the voltage % initialize u

for ind=2:length(t) %simulation v(ind)=v(ind-1)+dt*(0.04*v(ind-1)^2+5*v(ind-1)+140-u+I(ind-1)); u=u+dt*a*(b*v(ind)-u); if v(ind)>30 v(ind)=c; u=u+d; end end figure subplot(2,1,1); plot(t,v,'-k','linewidth',2) ylabel('Vm (mV)') subplot(2,1,2); plot(t,I,'-k','linewidth',2); xlabel ('time (ms)') ylabel('I_{inj}')

Fig. 18 Response of Izhikevich model neuron to a step current. By changing the parameters the neuron can exhibit various firing patterns.

8

Exercise 9 MATLAB is used simulate the effect of synaptic facilitation. The output of the code below is shown in Fig. 19.

Rm = 50; Cm = 2000; Vt = -50; Vr = -70; Vp = 20; dt=0.5e-3;

% membrane resistance in Mohms % membrane capacitance in pF % threshold voltage in mV % reset voltage in mV % peak voltage in mV % time step in s

tau = 0.1; % membrane time constant in s tau_syn = 0.02; % synaptic time constant in s Esyn = 0; % reversal potential of synapse Gmax = 25; % maximum conductance in nS P0 = 0.25; % steady state facilitation tau_p = 0.1; % facilitation time constant fF = 0.3; % facilitation factor % Without facilitation t = 0:dt:1; spikeInput = zeros(size(t)); spikeInput(200:50:400)=1; % 8 spikes at 40 Hz starting at t=100ms v = Vr; % inintialize the voltage Vm1(1)= Vr; Isyn(1)=0; % inintialize synaptic current Gsyn = 0; % inintialize synaptic conductance for ind=2:length(t) %simulate for 1000 ms if spikeInput(ind)==1 Gsyn = Gsyn+Gmax; end Isyn(ind) = -P0*Gsyn * (Vm1(ind-1) - Esyn); Gsyn = Gsyn * exp(-dt/tau_syn); v = v + dt*(-(v-Vr)/tau+ 1000*Isyn(ind)/Cm); if v>Vt Vm1(ind)=Vp; v=Vr; else Vm1(ind) = v; end end figure subplot(2,1,1); plot(t,Vm1,'-k','linewidth',1.5) ylabel('Vm (mV)'); subplot(2,1,2); plot(t,spikeInput,'-k','linewidth',1.5);

9

xlabel ('time (s)') ylabel('Presynaptic Firing') %% With facilitation v = Vr; % inintialize the voltage Vm2(1)= Vr; Isyn2(1)=0; % inintialize synaptic current Gsyn = 0; % inintialize synaptic conductance p(1) = P0; for ind=2:length(t) %simulate for 1000 ms if spikeInput(ind)==1 Gsyn = Gsyn+Gmax; p(ind) = p(ind-1) +fF*(1-p(ind-1)); else p(ind) = p(ind-1)+dt*(-(p(ind-1)-P0))/tau_p; Gsyn = Gsyn * exp(-dt/tau_syn); end Isyn2(ind) = -p(ind-1)*Gsyn * (Vm2(ind-1) - Esyn); v = v + dt*(-(v-Vr)/tau+ 1000*Isyn2(ind)/Cm); if v>Vt Vm2(ind)=Vp; v=Vr; else Vm2(ind) = v; end end figure subplot(2,1,1); plot(t,Vm2,'-k','linewidth',1.5) ylabel('Vm (mV)') subplot(2,1,2); plot(t,spikeInput,'-k','linewidth',1.5); xlabel ('time (s)') ylabel('Presynaptic Firing')

Exercise 10 The propagation of a signal along an axon is determined the electric potential of the axonal membrane, which is spatially and temporally dependent. At steady state conditions the temporal variations are null, and the response is solely attributed to spatial effect. Thus, Eqn. 32 can be simplified to 𝜕2𝑉𝑚(𝑥, 𝑡) = 2𝜋𝑎𝑟𝑖 ⋅ 𝑖𝑖𝑜𝑛 (𝑥, 𝑡 ) 𝜕𝑥2 The ionic current iion can be expressed in terms of the membrane potential as 𝑉𝑚(𝑥, 𝑡) 𝑖 (𝑥, 𝑡) = 𝐺 𝑉 (𝑥, 𝑡) + 𝑐𝑜𝑛𝑠𝑡 = + 𝑐𝑜𝑛𝑠𝑡 𝑖𝑜𝑛 𝑒𝑞 𝑚 𝑅𝑚

10

where Geq represents the total membrane conductance that is inverted to get the membrane resistance Rm. The constant is the summation of the ionic conductances multiplied by the reversal potentials. The differential equation becomes 𝜕2𝑉𝑚(𝑥, 𝑡) 2𝜋𝑎𝑟𝑖 = 𝑉𝑚( 𝑥, 𝑡 ) + 𝑐𝑜𝑛𝑠𝑡 𝜕𝑥2 𝑅𝑚 By expressing the intracellular resistance per unit length, ri, as the ratio of the resistivity to the area 𝜌𝑖 𝑟𝑖 = 2 𝜋𝑎 the differential equation can be expressed as 𝜕2𝑉𝑚(𝑥, 𝑡) 1 = 𝑉𝑚 (𝑥, 𝑡 ) + 𝑐𝑜𝑛𝑠𝑡 𝜕𝑥2 𝜆2 where 𝜆=Z

𝑎 𝑅𝑚 2𝜌

is the space constant.

Solving this second order differential equation results in solution with the following form 𝑥 –

𝑥

𝑉𝑚(𝑥, 𝑡) = 𝑘1𝑒 𝜆 + 𝑘2𝑒𝜆 + 𝑘3 where ki are constants depending on the conditions. Since the membrane potential is finite, k2 is zero, and the voltage is governed𝑥 by – 𝑉𝑚(𝑥, 𝑡) = 𝑘1𝑒 𝜆 + 𝑘3 Thus, a signal is attenuated exponentially as it propagates along the length of the fiber. The radius of the fiber determines the magnitude of the space constant. Thus, fibers with a larger radius have lower signal attenuation compared to narrower fibers of the same length.

11

Fig. 19 Synaptic facilitation result in larger postsynaptic potentials leading to the generation of action potentials.

26

Jason Z. Kim and Danielle S. Bassett

9 Solutions Problem 1: Linearize the following system about point x∗ = 1, x∗ = –1, x∗ = 0, 1

x˙1(t) x˙ (t) =

2

–x2(t) – 2x2(t) + x3(t) – 1 2x2(t) + 2x (t) 21 x1(t) – 2 3 x 1 (t)x 2 (t) – x3(t) + 1

2

x˙3(t)

3

.

and demonstrate that this point is a fixed point where x˙1 = x˙2 = x˙3 = 0.

Solution 1: We see that the the equation is in the form x˙ = ƒ (x), where ƒ is a nonlinear function. For clarity, we write each of the three differential equations separately x˙1(t) = ƒ1(x) = –x12 – 2x2 + x3 – 1 x˙2(t) = ƒ2(x) = 2x1 – 2x22 + 2x3 x˙3(t) = ƒ3(x) = x1x2 – x3 + 1.

From the chapter, we recall that linearization involves two components: the evaluation of the functions at the operating point ƒ (x∗), and the partial derivatives of the functions to approximate the vector field along small deviations from the operating point b ƒ /bxi. The first component is written ƒ (x∗) =

ƒ1(x∗) ƒ2(x∗)

=

–x∗2 – 2x∗ + x∗ – 1 1 2 3 2x∗ – 2x∗ 2 + 2x ∗ 1

ƒ3(x∗)

2

x∗x1∗ 2– x∗ 3+ 1

=

3

–(1)2 – 2(–1) + (0) – 1 2(1) – 2(–1) 2 + 2(0)

0 = 0 = 0,

(1)(–1) – (0) + 1

0

and because x∗1 = 1, x∗2 = –1, x∗3 = 0 is a fixed point, this first component becomes 0. Hence, we have demonstrated that x∗ is a fixed point. For the second component, we evaluate the matrix of partial derivatives evaluated at the fixed point

A=

61ƒ 6x1 6 ƒ2 6x1 6 ƒ3 6x1

6 ƒ1 6x2 6 ƒ2 6x2 6 ƒ3 6x2

6 ƒ1 6x3 6 ƒ2 6x3 6 ƒ3 6x3

= x=x∗

–2x1 –2 1 2 –4x2 2 x2 x 1 –1

= x=x∗

Hence, the linearized system is written x˙(t) ≈ ƒ (x∗) + A(x (t) – x∗) x˙1(t) –2 –2 1 x˙2(t) ≈ 2 4 2 x˙3(t) –1 1 –1

x1(t) – 1 x2(t) + 1 x3(t)

–2 –2 1 2 4 2 .

–1 1 –1

Linear Dynamics & Control of Brain Networks

27

Problem 2: Prove that the matrix exponential of A = eA =

a0 is 0b

ea 0 , 0 eb

using the Taylor series of the scalar and matrix exponentials.

Solution 2: Looking at the Taylor series of the matrix exponential, we have 1 1 e A = I + A + A2 + · · · + Ak + · · · , 2! k! and we notice that the main contributer is the matrix A raised to some power k. If we square the matrix through matrix multiplication, we obtain A2 = A ∗ A =

a0 0b

a0 a2 0 = .

0b 0 b2

If we cube the matrix through matrix multiplication, we obtain A3 = A2 ∗ A =

a2 0 0 b2

a0 a3 0 = .

0b 0 b3

It is obvious to see then that Ak becomes Ak =

ak 0 .

0 bk

Hence, if we evaluate the matrix exponential using element-wise matrix addition, we obtain A e =

1 + a + 1 2!a2 + · · · + 1k!ak + · · · 0 1 + b + 12!b2 + · · · + 1k!bk + · · · 0

and notice that the top left and bottom right entries are simply the Taylor series of ea and eb, respectively, to yield eA =

ea 0 .

0 eb

28

Jason Z. Kim and Danielle S. Bassett

Problem 3: Prove that the system response to control ∫ t e A(t–r) Bu(r)dr, x (t) = e At x0 + 0

satisfies the dynamical equation x˙(t) = Ax (t) + Bu(t) by substitution.

Solution 3: Let us begin with the left hand side of the differential equation, involving the time derivative of x (t). ∫ t d d A(t–r) Bu(r)dr At x0 + x (t) = substitute expression for x (t), e dt dt 0 e d ∫ t A(t–r) d Bu(r)dr distribute time derivative, = dt e At x0 + dt e 0

= Ae At x0 + = Ae At x0 +

d ∫ t dt ∫ t 0

= Ae At x0 +

∫ t

e A(t–r) Bu(r)dr

evaluate derivative for e At ,

b A(t–r) Bu(r)dr + Bu(t) e bt

Leibniz integral rule (below),

Ae A(t–r) Bu(r)dr + Bu(t)

evaluate partial derivative,

0

0

∫ t = A e At x0 + 0 e A (t – r ) Bu(r)dr +Bu(t) x ˛z __________ x ___________________________________

factor out A,

x (t)

= Ax (t) + Bu(t)

substitute x (t),

At the step where we apply Leibniz integral rule, we must take care to correctly evaluate the derivative of the integral. For some function ƒ (t, r), the derivative of the integral obeys Leibniz integral rule ∫ t ∫ t d b ƒ (t, r)dr = ƒ (t, r)dr + ƒ (t, t). dt 0 0 bt In our specific example, we notice that the function is ƒ (t, r) = e A(t–r) Bu(r), and ∫ t ∫ t d b A ( t –r ) e A(t–r) Bu(r)dr = e Bu(r)dr + e A(t–t) Bu(t). dt 0 0 bt Because the second term e A(t–t) = e A·0 = e0 = I, we have the final derivative ∫ t ∫ t d b A(t–r) e A(t–r) Bu(r)dr = e Bu(r)dr + Bu(t). dt 0 0 bt

Linear Dynamics & Control of Brain Networks

29

Problem 4: Prove that the convolution operator L(u(t)) =

∫ T

e A(T –r) Bu(r)dr

0

is linear according to Eq. 6; that is, if L(u 1 (t)) = c1, and L(u 2 (t)) = c2, then demonstrate that L(au 1 (t) + bu2(t)) = ac1 + bc2.

Solution 4: Here, we simply substitute au1(t) + bu2(t) into the operator equation, and use the distributive property on the right hand side to separate the two inputs u1(t) and u2(t) into separate terms.

L(au 1 (t) + bu2(t)) =

∫ T

e A(T –r) B(au 1 (r) + bu 2 (r))dr

0

∫ T =

e A(T –r) Bau1(r) + e A(T –r) Bbu 2 (r))dr

0

∫ T

ae A(T –r) Bu1(r) + be A(T –r) Bu 2 (r))dr ∫ T ∫ T = ae A(T –r) Bu 1 (r)dr + be A(T –r) Bu 2 (r))dr 0 0 ∫ T ∫ T =a e A(T –r) Bu 1 (r)dr + b e A(T –r) Bu 2 (r))dr =

0

0

= aL(u 1 (t)) + bL(u 2 (t)) = ac1 + bc2

0

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Jason Z. Kim and Danielle S. Bassett

Problem 5: Determine if the following system is controllable x˙1(t) 010 x˙2(t) = 0 0 1 x˙3(t) 100

1 x1(t) x2(t) + 0 u(t), x3(t) 0

by constructing the controllability matrix.

Solution 5: The controllability matrix is constructed as . .

C = B AB A2B By matrix multiplication, we have 010 1 0 AB = 0 0 1 0 = 0 , 100 0 1 and 010 A B = A( AB) = 0 0 1 100 2

0 0 0 = 1 , 1 0

yielding the controllability matrix C=

100 001 .

010

We notice that the columns of C are linearly independent of each other, because no column can be written as a weighted sum of the remaining columns. Hence, C has full rank, and the system is therefore controllable.

Linear Dynamics & Control of Brain Networks

31

Problem 6: Determine for what value of a the system is not controllable x˙1(t) 000 x˙2(t) = 1 1 0 x˙3(t) 10a

1 x1(t) x2(t) + 0 u(t), x3(t) 0

by constructing the controllability matrix.

Solution 6: We begin by constructing the controllability matrix . .

C = B AB A2B .

Through matrix multiplication, the matrices AB and A2B are written 0

0 0 A2 B = A( AB) = A 1 = 1 , 1 a

AB = 1 1 to yield the controllability matrix C=

100 011 .

01a

Here, we notice that for a = 1, the second and third columns of C would be identical, and the controllability matrix loses rank. Hence, the system is not controllable for a = 1.

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Jason Z. Kim and Danielle S. Bassett

Problem 7: Derive the minimum energy equation Eq. 10 Emin = (xT – e AT x0)T(LL∗)–1(xT – e AT x0), by substituting the minimum input u∗ (t) into the control energy Eq. 8 E =< u(t), u(t) > .

Solution 7: From the text, we defined the minimum input and energy to be ∫ T

u∗ (t) = L∗ (LL ∗) –1 (xT – e AT x0),

E =< u(t), u(t) >=

0

u(t) T u(t)dt.

If we substitute the minimum input into the energy expression, we get ∫ T T ∗

∗

∗

< u (t), u (t) > = 0

∗ –1

∗

AT

L (LL ) (xT – e

x0)

∗ –1

L (LL ) (xT – e

AT

x0)dt

∫ T (xT – e AT x0)T(LL∗)–1L∗TL∗ (LL∗)–1(xT – e AT x0)dt ∫ T AT T (LL∗)–1L∗TL∗ (LL∗)–1dt (xT – e AT x0), = (xT – e x0) =

0

0

as we can factor out the term xT – e AT x0 because it is simply a constant vector with no time dependence. Next, we consider the term from Eq. 11 ∫ T T ∗ e A(T –t) BBTe A (T –t) dt, LL = 0

and notice that the time variable t has already been integrated such that LL∗ has no time dependence, allowing us to factor (LL∗)–1 out of the integral ∗

∗

< u (t), u (t) >= (xT – e

AT

∗ –1

x0) (LL ) T

∫ T

L ∗TL ∗dt (LL ∗) –1 (xT – e AT x0).

0

Finally, we integrate L∗TL∗ to yield the term from Eq. 11 ∫ T ∫ T L∗T L∗dt = e A(T –t) BBTe A(T –t) dt = LL∗, 0

0

and substitute this term into our minimum energy equation to yield Emin = < u∗ (t), u∗ (t) > = (xT – e AT x 0 ) T (LL ∗ ) – 1 (LL ∗ )(LL ∗ ) – 1 (x T – e AT x0) = (xT – e AT x0)T(LL∗)–1(xT – e AT x0).

Linear Dynamics & Control of Brain Networks

33

Problem 8: Show that the controllability Gramian can be written WC (T) =

∫ T

e A(T –t) BBTe A (T –t) dt = T

0

∫ T

T

e Ar BBTe A r dr, 0

using the substitution r = T – t.

Solution 8: With the substitution r = T – t, where r and t are variables and T is a constant, the differential becomes dr = –dt.

As for the limits of integration, we observe that t = 0 when r = T, and that t = T when r = 0. Hence, integrating the variable t from 0 to T is equivalent to integrating the variable r from T to 0.

Substituting the variables r = T – t, we obtain ∫ T T e Ar BBTe A r dt.

WC (T) = 0

Then, substituting the differential dr = –dt and replacing the limits of integration, we obtain ∫ 0 T e Ar BBTe A r dr.

WC (T) = – T

Finally, we can perform the integration in the reverse order to generate another negative sign WC (T) =

∫ T

T

e Ar BBTe A r dr, 0

to yield the desired form of the Gramian.

34

Jason Z. Kim and Danielle S. Bassett

Problem 9: Show that the controllability Gramian for system A=

a0 , 0b

B=

10 01

is WC (T) =

1 2a

e2aT – 1 _____ 0 1 e2bT – 1 0 2b

Solution 9: The Gramian is written WC (T) =

∫ T

T

e At BBTe A t dt.

0

From problem 2, we recall that the matrix exponential can be written e At =

eat 0 .

0 ebt

Substituting this term into the Gramian, and noticing that BBT = I, we obtain WC (T) =

∫ T 0

eat 0 0 ebt

eat 0 dt = 0 ebt

∫ T 0

e2at 0 dt 0 e2bt

An element-wise integration yields ∫ T

1 2aT 1 e2at T = e dt = 2a 2a ∫ T 0 2bt 1 T 2bT 1 2bt e = e e dt = 2b 2b 0 0 2at

0

e

1 1

to yield the final answer WC (T) =

∫ T 0

e2at 0 dt = 0 e2bt

1 2a

e2aT – 1 _____ 0 1 2bT – 1 0 2b e

.

Linear Dynamics & Control of Brain Networks

35

Problem 10: Compute the minimum energy required for the system 1 0 10 A= 2 , B= , 02 01 to transition from initial state x (0) =

0 1 to final state x (T) = in time T = 1.

0 2

Solution 10: From problem 7, the minimum energy is given by Emin = (xT – e AT x0)T(LL∗)–1(xT – e AT x0).

Because x (0) = x0 = 0, the expression simply becomes Emin = xTT (LL∗)–1xT .

From problem 9, we have that the controllability Gramian can be written WC (T) =

1 2a

e2aT – 1 0 1 2bT – 1 e 0 2b

(e – 1) 0 , 1 4 0 e –1 4

=

and because WC (T) is a diagonal matrix, its inverse is the element-wise reciprocal WC (T)–1 =

1 e–1

0

0 4

.

e4 –1

Hence, through matrix multiplication, the minimum energy comes out to Emin

. . = 12

1 e–1

0

0 4 e4 –1

16 .

1 1 = + 4 2 e–1 e –1

Answers: 1.

2.

3.

4.

5.

6.

7.

D A D D B B Describe the difference between correlation and synchrony Correlation simply describes a statistical interaction between two neurons (or any two signals), and is itself agnostic to time scale. Synchrony is a special case of correlation, in which pairs of neurons interact on a very fast time scale, typically faster than ten milliseconds.

8. Explain the concept of surrogate data and how it is used in correcting synchrony metrics Surrogate data is made by taking real data from pairs of neurons, and shuffling or transforming it in some way (e.g., shuffling the trial indices, or moving the times of individual spikes) such that it preserves some features of the data and destroys others. When a correlation metric is calculated on surrogate data, it can be compared to that same metric calculated on the raw, original data. Thus, the contribution of the feature of the data that is shuffled to the correlation among neurons can be assessed.

9. What is the difference between PCA and FA as applied to neural spiking data?

PCA finds the dimensions of greatest variability in the data. FA finds the dimensions with the greatest shared variability, given a noise model in which independent gaussian noise is assumed for each neuron. This, it separates correlated variability into that which can be explained by a latent variable shared among neurons vs. variability that can be accounted for by a variable process (e.g., Poisson-like spiking variability) independent to each neuron.

10. What advantage do GPFA and LFADS have over PCA and FA PCA and FA treat each bin of the data independently, and do not have a concept of time within a trial. GPFA and LFADS can be used when there are multiple data points within or between trials that have a temporal dependency that can be exploited.

Solutions: 1. Some very useful categories of features include: 1) Frequency band power, particularly in the HFO (>100Hz) or VHFO (>600Hz). These can be extracted by filters, Fourier transforms (stationary) or short-time Fourier transforms, wavelet coefficients (dynamic). Ratios to other bands, dynamic fluctuation rates, baseline normalizations, Hilbert transforms, and other derivatives of these features can be helpful.

2) Phase coherence, particularly as defined by wavelet phase coherence.

3) Phase-amplitude coupling, particularly in the form of wavelet-based comodulograms.

2. a) Several suggestions are possible; however, one approach is to allocate 10-20% of data for the test set, divided evenly between interictal, preictal, and ictal data. Another suggestion can be to utilize kfold cross-validation (or leave-one-out approach) to boost validation/testing strength with small data. b) The main benefit is that a good performance on an unseen patient (as opposed to unseen seizure data from patient used in training) suggests that the trained algorithm generalizes across patients and the detection/prediction task is not patient-specific. The main drawback is a drastic reduction in the amount of data available for algorithm training given the small dataset.

3. The problem in question 2 has a small dataset and is a multiclass classification, so the most likely algorithm is either a decision tree or a random forest. Unless transfer learning is used, it is likely (though not guaranteed) that there will not be enough data to properly train a deep learning algorithm.

SVMs would also be inappropriate as they deal with binary classification. A) If only two classes are present, SVMs are a likely contender, since they work well with small data. B) With a large dataset, a deep learning approach can be beneficial, since it will be able to learn the underlying data patterns that RF might miss. C) In general, random forests – due to smaller number of parameters and parallel training – perform faster than deep learning approaches, so RF would be the algorithm of choice.

4. Common noise sources are environmental noise such as 50Hz/60Hz AC powerline noise and electronic noise such as thermal and shot noise. These can be removed using band pass or notch filtering, trial averaging, and adaptive filtering using a reference signal.

Common artifacts in EEG are other unwanted physiological energy sources such as cardiac, respiratory, eye movement, and EMG signals. These artifacts may be eliminated using notch filtering, adaptive filtering if frequency content does not overlap with signal of interest. If spectral overlapping occurs, their impact can be reduced using independent component analysis (ICA) or isolation of features using a cross-frequency coupling signal (SCFC).

5. Depending on how imbalanced the data is, it can significantly reduce classification accuracy of the algorithm. In an extreme example, where 80% of data belongs to one class, the algorithm can default to that class in all cases and still have 80% accuracy. One way to overcome the issue is by using a performance metric that penalizes false positive classification. Another is to rebalance the data by either repeating the under-represented classes or by removing the over-represented class.

6. 1) Recurrent Neural Network. Since it is used in sequential data, it can be used with time-series EEG data for sequence-dependent state classification.

2) Autoencoder. Would likely use it to generate or reduce feature selection to represent EEG data.

7. a) The performance (as the number of hidden units increase) depends on the problem and the nature of the data set. More hidden units in the MLP will increase the complexity of the network and allow it to encode more features, however the number of weights (and accordingly the amount of data necessary for training) will increase, leading to higher likelihood of poor performance. b) As the number of hidden layers increases, so does the complexity of the decision boundary between different states. Having a decision boundary that is too complex may lead to overfitting of

1

the training set and decrease the neural network’s ability to generalize. However, not enough hidden layers may result in never reaching the mean squared error goal. In practice, an increase in the number of hidden layers shows only a small improvement in performance. c) In this case – probably yes. Using a CNN will likely reduce the number of parameters that need to be trained, while maintaining or improving algorithm performance.

8. a) Large step size is not always better in training because it may overshoot local minima. In contrast, a small step size may take a long time to train. There is no direct correlation between step size and testing performance (once the ANN training phase has completed and the network has reached a local minimum for the chosen goal). b) Momentum prevents getting stuck in narrow ridges on the error surface by changing the ANN weights using the changes of the previous iteration. A large momentum however can cause overshooting local minima and a non-decreasing mean squared error.

9. While the exact performance of the algorithm will depend on the implementation, it is extremely likely that it will perform significantly worse than both seizure detection task and benchmarks shown in Table 2. Likely reasons are small amount of data and lack of complexity in the MLP and features.

10. Multiple improvements can be suggested, some of them include: better feature selection (perhaps the features for seizure detection were too simple), transfer learning using other EEG data (to increase the effective amount of data used for training), using different algorithms better suited for small data, or utilizing ensemble models to incorporate different algorithms. Understanding the sequential state changes of seizure episodes can suggest incorporating recurrent neural networks into the prediction algorithm.

2

Answers 1. A retinal prostheses must 1) detect emitted or reflected light from the surrounding environment 2) convert the detected light into a pattern of electrical stimulation and 3) apply the pattern of electrical stimulation to the retina.

2. Photoreceptors and bipolar cells respond to light in a graded fashion and membrane potential of these cells can take on many values, roughly proportional to the strength of light excitation.

Keep in mind that there are four types of photoreceptors (in primates) and many types of bipolar cells, but in general, they respond in a graded fashion. They are analog because they can take on many values. In contrast, retinal ganglion cells are “digital” in a way, because they fire action potentials, which can be thought of as bits of information. More action potentials means a stronger excitation, but the shape of the action potential does not change significantly. Again, there are many types of ganglion cells, and they respond to different types of excitation, but their response is typically a train of action potentials.

3. Retinal cell density is greater at the center of the retina and less at the periphery, thus an electrode array that matches this anatomy may be more effective.

4. Epiretinal – can stimulate the final output cells of the retina, bypassing damaged neural circuit of the inner retina. Subretinal – device positioning puts it closer to the next cell after photoreceptors in the visual process (I.e., bipolar cell).

5. At the fovea, retinal ganglion cells are stacked 5-7 cells high. Thus, to achieve single cell resolution, an electrode array would need to penetrate into the fovea to reach these cells. The high cell-density and 3D nature of the fovea make such an electrode array challenging.

6. Age-related macular degeneration and retinitis pigmentosa can be treated since these are primarily diseases of the photoreceptor and the remaining bipolar and ganglion cells can be stimulated to create light perception. Optic nerve disease cannot be treated with a retinal prosthesis since the connection between the retina and brain is severed.

7. A silicon retina is an electronic circuit that emulates retinal processing via hard-wired circuitry. Such a device may be more useful for an epiretinal implant because epiretinal implants that activate ganglion cells bypass retinal processing. A silicon retina can re-create this processing to allow more natural ganglion cell patterns to be created.

8. Photolithography on curved surfaces is a fabrication challenge. Each person’s eye is curved differently, so matching an electrode array curvature to each person’s individual curvature.

9. The incision for insertion of an electrode array must be small to prevent loss of eye pressure during surgery. Thus, electrode array size is limited and current human implants are below 20 degrees of visual field, vs. 180 degrees in healthy vision.

10. Computer vision can perform object and face recognition and highlight objects that are closer or more important to the user.

Answers to Problems in Chapter 21: 1. This problem can be done with either of the conventional ways of relating conductances or permeabilities to resting membrane potential, that is, either the Goldman-Hodgkin-Katz (GHK) equation or the Chord Conductance (CC) equation. There are reasons to prefer the latter, as described here under Resting Potential by GHK and CC, but we compute it both ways. Em is the resting potential, and ENa and EK are the equilibrium potentials.

Ej = (RT/F) ln ([j]out/[j]in) = 2.3 (RT/F) log ([j]out/[j]in) We will assume a body temperature of 37 ⁰C, or T = 310 ⁰K; R = 0.002 Kcal/(mole-oK); F = 23 Kcal/(mole-V). 2.3 RT/F = 0.062V Let [Na+]out = 142 mM; [Na+]in = 10 mM; [K+]out = 5 mM; [K+]in = 140 mM (Reasonable values near these will yield similar final answers.)

So ENa is 0.062 V • log 142/10 = 0.0714 V and EK is 0.062 V • log 5/140 = - 0.0897 V

By CC: The general equation is Em = (ENagNa + EKgK + EClgCl) / (gNa + gK + gCl), where Es are equilibrium potentials, and g’s are conductances. If Cl- is at equilibrium, the net Cl- current is zero and the development of CC eliminates the Cl- terms in both numerator and denominator: Em = (ENagNa + EKgK) / (gNa + gK) We are only interested in the ratio of gNa to gK, so divide numerator and denominator by gK and do the algebra: Em = (gNa/gK ENa+ EK) / (gNa/gK +1) Em (gNa/gK +1) = ((gNa/gK) ENa+ EK) Em (gNa/gK) + Em = (gNa/gK) ENa+ EK Em - EK = (ENa – Em) (gNa/gK) gNa/gK = (Em - EK)/ (ENa – Em) = (-0.03 - -0.0897)/( 0.0714 - -0.03) = 0.59

By GHK (without Cl-): Em = (RT/F) ln (PNa [Na+]out + PK [K+]out)/ (PNa [Na+]in + PK [K+]in) As above we are only interested in PNa/PK, which is the same as the conductance ratio, so we divide numerator and denominator by PK. We’ll also convert to base10 log units:

1

Em = 0.062 log (PNa/PK [Na+]out + [K+]out)/ (PNa/PK [Na+]in + [K+]in) 10^(-0.03/0.062) = (PNa/PK [Na+]out + [K+]out)/ (PNa/PK [Na+]in + [K+]in) (0.328) • (PNa/PK [Na+]in + [K+]in) = (PNa/PK [Na+]out + [K+]out) (0.328) • (PNa/PK [Na+]in) - PNa/PK [Na+]out = [K+]out – 0.328 [K+]in PNa/PK = ([K+]out – 0.328 [K+]in)/(0.328 [Na+]in - [Na+]out ) = 0.29 With either solution, the sodium conductance or permeability is larger than in a typical neuron at rest, where gNa/gK would be closer to 0.1 or 0.15 2. a. First express the current of 30 pA in terms of ions per second. 1 amp is 1 coulomb/sec, and there are 6.24•1018 charges per coulomb.

30 pA = 30•10-12 coulomb/sec • 6.24•1018 charges/coulomb = 1.87•108 charges/sec.

This is all due to Na+, so the amount of ATP/min used just for Na+ pumping by one rod is at least: ATP/min = 1.87•108 Na+/sec • 60 sec/min • 0.33 ATP/Na+ = 3.74•109 ATP/rod-min b. There are 180,000 rods/mm2, and each inner segment (IS) is 25 µm long, so the volume occupied by those IS is 1 mm2 • 0.025 mm = 0.025 mm3 = 2.5•10-5 cm3 = 2.5•10-5 ml and there are 6 ATP/O2, so in terms of O2: molecules of O2-ml tissue-1-min-1 = 1.8•105 rods x 3.74•109 ATP/rod-min • (1 O2/6 ATP) /(2.5•10-5 ml) = 4.49•1018 molecules O2-ml-1-min-1 If we convert this to moles, it is (4.49•1018 molecules O2-ml-1-min-1)/ (6.02•1023 molecules/mole) = 0.74•10-5 moles O2-ml-1-min-1 = 7.4 µmoles O2-ml-1-min-1 Or

= 7.4 µmoles O2-ml-1-min-1 • 22.4 µL/µmole =166.88 µL O2-g-1-min-1 = 16688 µL O2-100g-1-min-1 = 17 ml O2-100g-1-min-1

This shows that the inner segments have an extraordinarily high oxygen consumption compared to almost any other tissue. However, because the rest of the photoreceptor uses no oxygen, the overall oxygen consumption per gram of photoreceptor is about a quarter of this value, and comparable to other nervous tissue.

3. In Figure 21.18 the center and surround signals are treated as being out of phase, so they are antagonistic. In Fig. 21.20, the relative phases of the center and surround are shown to change with the temporal frequency of the stimulus, here a grating. The behavior for certain temporal 2

frequencies is shown more clearly in Fig. 21.21. At low temporal frequency (e.g. 2 Hz), for all spatial frequencies, the phase of the response is shown as being close to zero, which means that the response follows the stimulus with no lag. At high spatial frequency, the response is completely due to the center. There is no spatial frequency at which the response is purely due to the surround, but the fact that the surround is subtractive is what allows the responsivity to decline at low spatial frequency while the phase does not change. The surround here must essentially have a phase of -180, that is, it is subtractive. At higher temporal frequency, 40 Hz, the phase is about -375 degrees at high spatial frequency, when only the center is involved in the response. The cell is no longer able to follow the stimulus, but is responding about a whole stimulus cycle late. At low spatial frequency, the response phase is about -250, which must mean that the center and surround are no longer 180 degrees apart. (The actual phase of the surround can’t be determined just from the data shown.) The fact that the surround is not purely subtractive means that there is no fall in responsivity at low temporal frequency. At even higher temporal frequency, the contrast sensitivity is higher at low spatial frequency than the center alone would allow, so essentially the surround signal is at a phase that partly adds to the center response. Again, it is not possible to determine from the graph exactly what the surround phase is and the original paper indicates that the surround phase is about -470 degrees and the center phase is -570 degrees at 60 Hz for this cell.

4. First you need some method to characterize the coverage of the visual field by the cell mosaic as it actually exists. Because the receptive field is not equally sensitive at all points, there will actually have to be a decision about what part of the receptive field counts as belonging to it.

One might choose the point where the receptive field sensitivity falls to some particular value, such as falling to half the peak sensitivity, or where the sensitivity is indistinguishable from noise. Gauthier et al chose a value that was between these, and showed that the conclusion about tiling did not depend critically on that choice. If one simply took the receptive field profiles in Fig. 21.24, that decision had already been made.

The second point is to determine how to characterize visual field coverage. If one took just the cell RF area divided by the total area, that would be one possible metric, but it would also be good to subtract areas where the RFs overlap, because these represent inefficiencies in coverage, just like holes in the coverage do.

The third point is to determine how much the coverage deteriorates with some permutation of the receptive fields. Using a method devised the previous paragraph, one could keep the center of the RFs in place, but alter them in some way. The things that Gauthier et al. tried were rotating each RF by different angles and replacing each RF by its mirror image. All their permutations suggested that the way the RFs actually are arranged is about the best. One could also keep the area of each RF the same, but make them circular or some other shape.

Another point that was addressed was knowing whether you have recorded from all the cells in a particular area. The figure on the right side of Fig 21.24 is a situation where they only did their manipulations of the RF for certain regions, because they believed that they had simply not recorded from some cells. There is an analysis of this in Gauthier et al.

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5. The elements in the stimulus array are smaller in the middle probably because the density of neurons is higher in the middle, and these regions might be expected to generate more current that is seen in remote (corneal) recordings as larger voltages. Even though the elements are smaller in the middle, the voltages from this region are still larger.

One hypothesis might be that equal numbers of neurons produce equal responses. If one knew the density of photoreceptors or bipolar cells (photoreceptor maps exist – see Curcio et al. Human photoreceptor topography, J Comp. Neurol. 292:497-523 (1990)), one could scale the voltage in each element by the relative number of neurons in that area, and test whether this equalized the responses. The density of bipolar cells is probably not known exactly, but one could count the cell bodies in the inner nuclear layer, and roughly determine, from known circuitry, how many bipolar cells there are. In this layer will also be cell bodies of horizontal, amacrine, and Muller cells, so this could not be an exact calculation. We know that this hypothesis is an oversimplification, because the fovea has no bipolar cells, and this is where the largest response is. It might be useful to know how large that center pixel is, and change its size.

Another possibility is that the responses in the center are faster, and the more peripheral regions give smaller responses because one flash interferes with another, causing more light adaptation in the periphery than the center. One might slow down the rate of flickering of the elements to test whether this increases more peripheral responses.

Another possibility is that currents sum differently in different parts of the retina, because the retinal or cellular resistances are different in different areas. This would be hard to test.

6. a. We will assume that the cell with the smallest center also has the smallest surround, and the cell with the largest center has the largest surround.

The smallest seems to have parameters of about rc =.025; Kc = 800; rs = 0.1; Ks = 30 The largest has parameters of rc = 0.2; Kc = 8; rs = 2; Ks = 0.01 From the text, the receptive field can be described in spatial coordinates as: W(r) = Wc(r) – Ws(r) = Kc exp [-(r/rc)2] – Ks exp [-(r/rs)2] The first term is for the center and the second is for the surround. The graphs, from -r to r are below. Graphs from 0 to r would be just the right half of each plot. In contrast to the schematic pictures of the center and surround, the actual peak surround sensitivity is rather low compared to the center. (Exact answers to the rest of the questions will depend on what values are chosen for the parameters above, but any choice should illustrate the same principles.) It is possible to graph both on the same set of axes, but the peripheral cell almost disappears then because its peak sensitivity is so much lower.

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b. Despite the very different peak sensitivities, the integrated center sensitivities are nearly equal, 0.5 and 0.32 impulses-sec-1-%contrast-1. Also, despite the very different shapes of the center and surround, and the fact that the surround is very weak in the middle of the receptive field, the integrated surround is 0.6 of center strength in the central cell and 0.25 in the peripheral cell. These similarities in the organization of the receptive field seem to be a strong property of the way receptive fields have relatively constant properties with eccentricity. c. To calculate the maximum spatial frequency, where 100% contrast is needed, we could plot the graphs with the center term of this equation. We want to find the frequency (), where S() = 0.01. (You can calculate smaller values of S, but they are not meaningful.)

S () = Sc() – Ss () = Kcrc2 exp [–(rc)2] – Ksrs2 exp [–(rs)2] (The surround makes no contribution to high spatial frequencies.)

S () = Sc() = Kcrc2 exp [–(rc)2] S ()/Kcrc2 = exp [–(rc)2] Log (0.01/ Kcrc2) = –(rc)2 ([-Log (0.01/ Kcrc2)]0.5)/ rc =  For the smaller receptive field, this is 10.6 cycles per degree. The graph is below. These are typically plotted on a log-log axes. Contrast sensitivities above 1 mean that less than 1% contrast is needed.

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7. Any eye disease which results in photoreceptor degeneration is a candidate for a retinal prosthesis; e.g., age-related macular degeneration or retinitis pigmentosa. For a retinal prosthesis to work the retinal ganglion cells must survive so that they can be stimulated to provide input to the brain. Diseases such as glaucoma, which lead to retinal ganglion cell death, are therefore untreatable with a retinal prosthesis.

On an eye chart, failure to have a visual acuity of 20/200 in the best eye after optical correction is considered legal blindness. This means that at 20 feet away, a subject cannot read the largest letter on a typical eye chart. An alternative way of defining 20/200 is having a logMAR equal to 1.0. There are grades of vision worse than 20/200 that might still be useful to a person with low vision. Being able to count fingers is the next step down; below that is being able to detect hand motion, and below that is the ability of light perception, but without any spatial vision, before one is completely blind Acuity depends on having photoreceptors or, for a retinal prosthesis, stimulation sites comparatively close together. A cost of making a stimulation site small, a necessity for closely packed arrays, is a high charge density on the electrode surface, which increases tissuedamaging heat production and often also the generation of chemical entities toxic to tissue.

This problem can be mitigated partially by reducing the distance between a stimulating electrode and the target of stimulation, but this is often difficult to accomplish.

8. There is no one correct answer but the student should be encouraged to propose some neural circuit that would provide a computation like that proposed in Figure 7 of Barlow and Levick (1965). Then they can be given access to a paper by Fried, Munch and Werblin paper (Fried, S.L., Munch, T.A. and Werblin, F.S. Mechanisms and circuitry underlying directional selectivity in the retina. Nature 420: 411- 414, 2002) It is hoped that through this exercise they will gain a better understanding for how modeling in retinal circuitry is conducted in real life.

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ANSWER KEY 1. Peripheral Nervous System (includes somatic and autonomic nervous systems) & Central Nervous System (includes brain and spinal cord) 2. These regulate blood supply to the CNS & maintain an immune-privileged zone.

3. They help maintain neural structure (especially BBB integrity) and play important roles in injury and inflammatory response. Reactive astrocytes can produce both beneficial and inhibitory molecules in the tissue remodeling process.

4. Immune-privilege (including its own immune sentinels, the microglia); existence of a blood-brain barrier; cerebral spinal fluid; regional diversity in structure and cellular milieu; presence of glial cells and/or neurons; segmental structure in the spinal cord; presence of meningeal layers; etc.

5. Injuries can lead to severe disabilities including paralysis, loss of motor function, cognitive disorders, depression, neurodegenerative disorders (e.g., Alzheimer’s, Parkinson’s, etc), or even death.

6. Schwann Cells 7. Deep Brain Stimulation; Vagal Stimulation; or Brain-Computer Interfaces 8. By bolstering fibrin cable formation, and for longer gaps as a substitute fibrin cable to provide topographical cues for guiding Schwann cell and axonal migration.

9. They are the natural progenitors of neural tissue, have the propensity to replace lost or injured tissue, possess immunomodulatory effects, and can induce neuroprotection and remyelination cascades.

10. Materials should be biocompatible, have facilitating structural, topographical and biochemical features, be non/minimally-immunogenic, and when appropriate be biodegradable or bioresorbable, provide in situ gelling, enable ECM formation and cell attachment, and promote outgrowth of axonal processes.

11. When cellular delivery is not possible either due to inaccessible target location, or lack of appropriate cell time; when the problem has a clear genetic cause or target in the host cells; when the therapy needs to be rendered permanently; or, when immunogenicity to implanted cells is a concern.