EXAM 1 Solution A 14-bit analog-to-digital converter (ADC) has an input range of 1V to 6 V. If the o

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EXAM 1 Solution Problem 1 (20 points): #Q1. (3 points) A 14-bit analog-to-digital converter (ADC) has an input range of 1V to 6 V. If the output of the ADC is 74, what is the voltage at its input? Ignore quantization error.

6−1

74× 214 −1 +1=1.02258 V #Q1. (3 points) A 12-bit analog-to-digital converter (ADC) has an input range of 2V to 5 V. If the output of the ADC is 47, what is the voltage at its input? Ignore quantization error.

47×

6−15−2 212 −1

+2= 2.0344 V

#Q2. (3 points) Briefly state the purpose of using voltage followers.

Used as a buffer between two stages since it has infinite input impedance and zero output impedance.

#Q3. (3 points) An analog signal contains three sinusoidal components with frequencies 5, 7 and 12 Hz. What is the minimum sampling rate required in order to avoid aliasing?

24 sps #Q3. (3 points) An analog signal contains three sinusoidal components with frequencies 5, 7 and 24 Hz. What is the minimum sampling rate required in order to avoid aliasing?

48 SPS

1

#Q4. (3 points) A6-bit ADC has an input range of 1 to 4.15 V. If the analog input changes by 82 mV over two samples (which are between 2 to 3V), the estimated voltage after ADC will change by (circle one) a) b) c) d) e) f)

25 mV or 50 mV 100 mV or 150 mV 50 mV or 100 mV 20 mV or 40 mV 40 mV or 60 mV 50 mV or 75 mV

4.15 − 1 = 50 𝑚𝑉 26 − 1

#Q4. (3 points) A6-bit ADC has an input range of 1 to 4.15 V. If the analog input changes by 108 mV over two samples (which are between 2 to 3V), the estimated voltage after ADC will change by (circle one) a) 25 mV or 50 mV b) 100 mV or 150 mV c) 50 mV or 100 mV d) 100 mV or 120 mV e) 40 mV or 60 mV f) 50 mV or 75 mV

#Q5. (2 points) How many comparators are required to build a 3-bit direct conversion analog-to-digital converter (ADC)? (circle one) 3

4

5

7

8

9

15

16

17

31

#Q5. (2 points) How many comparators are required to build a 4-bit direct conversion analog-to-digital converter (ADC)? (circle one) 3

4

5

7

8

9

15

16

17

31

#Q6. (2 points) For a given input signal, if the sampling rate of an ADC is increased, the signal-toquantization-noise ratio (SQNR)(circle one) a) b) c) d)

must increase must decrease may or may not change will not change

2

#Q7. (4 points) A 10 Hz signal accompanied by a 60 Hz noise is passed through a filter with the transfer function, 45 𝑠 + 10 The signal-to-noise ratio (SNR) at the output of the filter is 54 dB. Find the SNR in dB at the input of the filter. 𝐻 𝑠 =

Input SNR+ SNR gain at filter=Output SNR Input SNR+20 log10

| H j2π.10 | H j2π.60

= 54

Input SNR = (54-15.45) = 38.55 dB

Problem 2 (35 points): Answer any five; 7 points each # P2.1:The two filters shown below have transfer functions H1(s) and H2(s). Their input impedances are 15 kΩ and 10 kΩ, respectively, while their output impedances are 2 kΩ and 3 kΩ, respectively. The output of the second filter is connected to a voltage divider as shown. Find Hac(s), the transfer function of the entire system.

2

Hac(s) = H1.H2.

.

10

.

3

3 10+2 3+3

3

=

5 18

H1. H2

# P2.2:A sinusoidal voltage with frequency,𝑓 Hz, is sampled using an ADC at a constant sampling rate.

1

Fs=3

v[n]

0.5

Apparent frequency=1 Possible values of F =M.3 ±1

0

-0.5

-1

0

0.5

1

1.5 time (seconds)

2

2.5

3

The first 10 samples of the digitized signal are shown above. Ignore any quantization error. Which of the following are some possible values of 𝑓(circle all that apply) ?

2

3

4

5

6

7

8

# P2.2: A sinusoidal voltage with frequency,𝑓 Hz, is sampled using an ADC at a constant sampling rate.

1

v[n]

0.5

0

-0.5

-1

0

0.5

1

1.5 time (seconds)

2

2.5

3

The first 13samples of the digitized signal are shown above. Ignore any quantization error. Which of the following are some possible values of 𝑓(circle all that apply) ?

0

2

6

8

10

4

12

16

P2.3:What is the root mean square (rms) value of the periodic voltage waveform shown below?

2 4 1 ( 12 𝑑𝑡 + 3 (−2)2 𝑑𝑡) = 1.0954 5 0

# P2.3:What is the root mean square (rms) value of the periodic voltage waveform shown below?

1 4 1 ( 32 𝑑𝑡 + 1 (−1)2 𝑑𝑡) = 1.55 v 5 0

5

# P2.4:In the circuit shown below, 𝑣𝑖𝑛 and 𝑣𝑜𝑢𝑡 are the input and output voltages, respectively. Find the input and output impedances of this circuit.

Zin= 8 Ω Zout= 32Ω

6

#P2.5: In the instrumentation amplifier (IA) shown below, 5𝑅1 = 𝑅2 . The common-mode and difference mode gains of the second stage are 0.01 and 10, respectively. a) [5 points] What is the common mode rejection ratio (CMRR) of the entire IA? b) [2 points] The difference mode component of the input is the desired signal while the common mode component is considered noise. If the input signal-to-noise ratio (SNR) = −46 dB, what is the SNR at the output of the IA?

a) CMRR=20log10 (

11×10 1×0.01

) = 80.83 dB

b) -46+80.83=34.83 dB P2.5: In the instrumentation amplifier (IA) shown below, 5𝑅1 = 𝑅2 . The common-mode and difference mode gains of the second stage are 0.1 and 10, respectively. c) [5 points] What is the common mode rejection ratio (CMRR) of the entire IA? d) [2 points] The difference mode component of the input is the desired signal while the common mode component is considered noise. If the input signal-to-noise ratio (SNR) = −46 dB, what is the SNR at the output of the IA?

a) CMRR=20log10 (

11×10 1×0.1

) = 60.83 dB

b) -46+60.83=14.83 dB 7

# P2.6: A bioelectric signal fluctuates between 2 V and 3 V (it can be modeled as a uniformly distributed random variable within this range). It is sampled using two different ADCs without any amplification:

(a) 5-bit ADC with an input range of 2V to 3V (b) 7-bit ADC with an input range of 2V to 4V Find the signal-to-quantization-noise ratio (SQNR) for each of these ADCs.

a) 6.02N dB=6.02×5 dB=30.1 dB b)

3−2 12 𝑄 12

1

1

𝑄

4−2 (2)7 −1

= =

= 63.5 ~ 36.05 dB

# P2.6: A bioelectric signal fluctuates between 2 V and 3 V (it can be modeled as a uniformly distributed random variable within this range). It is sampled using two different ADCs without any amplification:

(c) 5-bit ADC with an input range of 0V to 3V (d) 7-bit ADC with an input range of 2V to 3V Find the signal-to-quantization-noise ratio (SQNR) for each of these ADCs.

c)

3−2 12 𝑄 12

1

1

𝑄

3−0 (2)5 −1

= =

= 10.33 ~ 20.28 dB

d) 6.02N dB=6.02×5 dB=30.1 dB

8

Problem 3 (45 points): Answer any three; 15 points each # P3.1:You want to amplify (with or without inversion) an EKG signal that has arange from −20 mV to+60mV. Design a dc-coupled one-op-amp amplifier circuit that will have the maximal gain possible without exceeding the typical guaranteed linear output range (bounded by the op amp’s supply voltages +10V and -10V). You can choose resistors of any values you want. If you need an additional voltage source, use 10V or -10V. Clearly draw your amplifier circuit and calculate the required values.

Circuit to be designed

Vi

Vout

Vout=250 Vin -5 Using a +10v source: Vout =250 Vin – 0.5 × 10

differential amp

Using a -10v source: Vout = 250 Vin + 0.5 × (-10 ) non-inverting summing amp

# P3.1: You want to amplify (with or without inversion) an EKG signal that has a range from −30 mV to+70mV. Design a dc-coupled one-op-amp amplifier circuit that will have the maximal gain possible without exceeding the typical guaranteed linear output range (bounded by the op amp’s supply voltages +10V and -10V). You can choose resistors of any values you want. If you need an additional voltage source, use 10V or -10V. Clearly draw your amplifier circuit and calculate the required values.

Vi

Circuit to be designed

Vout

Vout = 200Vin - 4

9

# P3.2:In the difference amplifier circuit shown below, the op amp is ideal. (a) (3 points) If the resistors are all 1 kΩ each:  Input impedance of the circuit as seen by the source, 𝑉𝑎 = 1K

 

Input impedance of the circuit as seen by the source, 𝑉𝑏 = 2K Common mode rejection ratio (CMRR) of the circuit = ∞

(b) (12 points) When the resistors are not equal, let 𝑣𝑜𝑢𝑡 = 5𝑣𝑏 − 4𝑣𝑎 . Find the CMRR of the difference amplifier.

Vout= Acm. 𝐴𝑐𝑚 2 𝐴𝑐𝑚 2

𝑉𝑎 +𝑉𝑏 2

+ 𝐴𝑑𝑚( 𝑉𝑏 − 𝑉𝑎)

+ Adm = 5

– Adm = -4

SO :Acm=1 , Adm=4.5 CMRR = 4.5 ~ 13.06 dB

10

# P3.2: In the difference amplifier circuit shown below, the op amp is ideal. (c) (3 points) If the resistors are all 2 kΩ each:  Input impedance of the circuit as seen by the source, 𝑉𝑎 = 2KΩ

 

Input impedance of the circuit as seen by the source, 𝑉𝑏 = 4KΩ Common mode rejection ratio (CMRR) of the circuit =∞

(d) (12 points) When the resistors are not equal, let 𝑣𝑜𝑢𝑡 = 4𝑣𝑏 − 5𝑣𝑎 . Find the CMRR of the difference amplifier.

Vout= Acm. 𝐴𝑐𝑚 2 𝐴𝑐𝑚 2

𝑉𝑎 +𝑉𝑏 2

+ 𝐴𝑑𝑚( 𝑉𝑏 − 𝑉𝑎)

+ Adm = 4

– Adm = -5

SO : Acm=-1 , Adm=4.5 CMRR = 4.5 ~ 13.06 dB

11

# P3.3: A biopotential voltage, 𝑣𝑖𝑛 , is applied to the input of the circuit shown below. The output of the circuit, 𝑣𝑂 ,is connected to the input of an 8-bit analog-to-digital converter (ADC). The ADC has an input range of 1 V to 5 V. The raw output of the ADC, denoted as 𝑥, is an integer between 0 and 𝐾.

Ignore any quantization error.

8 a) (3 points) Find the valueof𝐾.(2) -1= 255 b) (12 points) Find the value of 𝑣𝑖𝑛 when 𝑥 = 100.

𝑉𝑖𝑛 −2 2

=

2−𝑉𝑜 18

20−𝑉𝑜

Vin =

9

5−1

Vo = 1+ xQ = 1+ 100 ×

255

𝑉𝑖𝑛 =

= 2.57

20 − 2.57 = 1.94 𝑉 9

12

# P3.3: A biopotential voltage, 𝑣𝑖𝑛 , is applied to the input of the circuit shown below. The output of the circuit, 𝑣𝑂 , is connected to the input of an 𝑁-bit analog-to-digital converter (ADC). The ADC has an input range of 1 V to 3 V. The raw output of the ADC, denoted as 𝑥, is an integer between 0 and 511.

Ignore any quantization error. a) (3 points) Find the value 𝑁. b) (12 points) Find the value of 𝑣𝑖𝑛 when 𝑥 = 150.

𝑉𝑖𝑛 −3 2

=

3−𝑉𝑜 18

21−𝑉𝑜

Vin =

Vo = 1+ xQ = 1+ 150 ×

3−1 511

6

= 1.59 𝑉𝑖𝑛 =

13

21−1.59 6

= 3.23 𝑉

# P3.4:The analog biopotential signal, 𝑣(𝑡), shown below is sampled using an𝑁-bit direct conversion analog-to-digital converter (ADC) starting at time 𝑡 = 0. The digitized values (𝑣𝑒𝑠𝑡 ) are shown for 𝑛 = 0 to 6, where 𝑛 is the sample index. The ADC has uniform quantization.

4 3.5 3 2.5 2 1.5 1

v (t) vest [n]

0.5 0

0

0.5

1

1.5

2

2.5

3

3.5

t (a) (3 points) What is the sampling rate? 2 sps

(b) (3 points) Find 𝑄, the voltage resolution of the ADC.0.5 V

(c) (3 points) What is the minimum possible value of𝑁? 3

(d) (3 points) What is the maximum magnitude of quantization error in this graph? For which sample?n=4 error=0.22 or, n=5 error=0.2

(e) (3 points) Instead of𝑣(𝑡), if you sample 2𝑣(𝑡) using this ADC, the value of𝑣𝑒𝑠𝑡 (0) will be 2. What will 𝑣𝑒𝑠𝑡 (5) be? 3.5

14

Bonus Problem (10 points): #B1:(3 points)Forthe instrumentation amplifier circuit in Problem P2.5, let 𝑣1 = 𝑣2 + 0.5 = 2 V and 5𝑅1 = 𝑅2 = 5 kΩ. Find the voltage at the output of the top left op amp.

+4.5 V #B1:(3 points)For the instrumentation amplifier circuit in Problem P2.5, let 𝑣1 = 𝑣2 − 0.5 = 2 V and 5𝑅1 = 𝑅2 = 5 kΩ. Find the voltage at the output of the top left op amp.

-0.5 V

#B2:(2 points)When a voltage source 𝑣𝑠 (with no source resistance) is connected to the input of a certain amplifier circuit, the current leaving the source is, 𝑖 =

2𝑣𝑠 −3 , where 𝑖 and 𝑣𝑠 are in Amperes and 6

Volts, respectively. What is the input impedance of the amplifier circuit? 𝑑𝑣𝑠 =3Ω 𝑑𝑖

#B2:(2 points)When a voltage source 𝑣𝑠 (with no source resistance) is connected to the input of a certain amplifier circuit, the current leaving the source is, 𝑖 =

3𝑣𝑠 −2 , where 𝑖 and 𝑣𝑠 are in Amperes and 6

Volts, respectively. What is the input impedance of the amplifier circuit? 𝑑𝑣𝑠 =2Ω 𝑑𝑖

15

#B3: (5 points)A voltmeter has an input resistance of 18 kΩ. If it is connected between points ‘a’ and ‘b’ in the circuit shown below, what voltage will it read?

Vab =5 mA (3k|| 6k) = 10V Zout =2 kΩ Voltmeter reads= 10.

18 18+2

=9v

#B3: (5 points)A voltmeter has an input resistance of 12 kΩ. If it is connected between points ‘a’ and ‘b’ in the circuit shown below, what voltage will it read?

Vab =2 mA (15k|| 10k) = 12 V Zout =6 kΩ Voltmeter reads= 12.

12 12+6

16

=8v