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Problem 1:

Signal Processing Continuous And Discrete

A Mini-project: An Audio Scrambler .Text-messaging is out-of-date! You and I have been communicating by sending voice-messages to each other as “.wav” files. It’s not that I am paranoid, but everybody is out to get me! In particular, “they” have been intercepting all of our voice messages. But we are going to outwit “them” by using MATLAB to scramble our voice messages before sending them, making them unintelligible using a secret encoding scheme, and then we will descrambling the messages when we get them. your task is to design and implement the speech encoder and decoder. the requirement is that the scrambled message should occupy the same audio bandwidth (approximately 300 – 3000 Hz) as the original message, and the descrambled message should have good fidelity.

We will use a frequency domain “mirroring” scheme, where the frequency components are flipped, so that high frequencies are translated to low frequencies and vice-versa. A spectral com­ ponent with frequency Ω1 will be translated to a new frequency Ωa − Ω1 , where Ωa is known to the sender and recipient. The resulting waveform is unintelligible until the frequency relationships have been restored. The scheme is shown below: F j   

F j(

a



 

s c ra m b le o rig in a l s p e e c

s c ra m b le d s p e e c h

Thus, a spectrum as shown on the left will be translated to that shown on the right. The decoder will restore the correct frequency relationships.

Consider a component A sin(Ωt), we want to translate it to A sin ((Ωa − Ω) t). From elementary trigonometry A sin ((Ωa − Ω) t) = A (sin(Ωat) cos(Ωt) − cos(Ωat) sin(Ωt)) = A sin(Ωa t) sin(Ωt + π/2) − A sin(Ωa t + π/2) sin(Ωt) which leads to the following time-domain processing to shift a single frequency component.

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A π/2 phase-shifter is easy to build for a single frequency, such as sin(Ωat) but when the input signal, such as a speech signal, contains more than just a single component each spectral component must be shifted by 90◦. This requires an all-pass filter with a constant phase shift, which is a difficult design task for a continuous filter.

Approximations as active or passive filters are available.

The Hilbert transformer is an all-pass filter with a transfer function

which is exactly what we want. The following shows the scrambler implemented with a Hilbert transformer

Then G(jΩ) = F(j(Ωa − Ω).

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(a) Practical implementations of Hilbert transformers are approximations. For example, suppose implementation has ripple in its passband so that at a particular frequency Ωo the transfer function is

an

where δ is a perturbation from the ideal response. Find the output g(t) when the input is f (t) = A sin(Ωot).

Have any “spurious” spectral components been introduced? (You can do this with simple trigonometric identities.) (b) Your task is to design, implement, and test an audio scrambler and descrambler using this encoding technique: • You should write a pair of MATLAB functions as follows: y_scramble = encode(f_audio, Fs, Fa) y_descramble = decode(y_scramble, Fs , Fa) where f audio is a MATLAB array containing the audio file to be processed, Fs is the sampling frequency (Hz) of the data in f audio, and Fa is the frequency (Hz) around which the spectrum is reflected. The first function should take encode the file, the second restore the scrambled version to the original.

• We will supply you with two audio files (.wav) that you can download and use. One is an ordinary audio file (PS8Raw.wav) that is to be to be scrambled and descrambled. The π second is one that we have already scrambled (PS8Scrambled.wav). These may be imported into your MATLAB workspace using the wavread() function [f_audio,Fs,Nbits]

= wavread(’myfi le’)

see the MATLAB help.

• You should use the Parks-McClellan design function fi rpm() to design a Hilbert trans­ former that covers the frequency span 300 – 3000 Hz. (The help on fi rpm() has an example) Be aware that the quality of you scrambling will be affected by the pass-band ripple that you allow.

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• Don’t forget that the FIR Hilbert transformer is a causal linear-phase system, and that it MUST have an odd length impulse response. You will need to add an equivalent delay filter in the other arm. Design an appropriate FIR impulse response.

• The audio arrays will be large. Use fft fi l t ( ) to do the actual filtering operations.

• Let’s standardize on a reflection frequency F a = 3500 Hz. • To test your scrambler, load the audio file, then compute and plot its magnitude spec­ trum. Then scramble the file and plot its magnitude spectrum. Make sure it is what you expect. If there are any spurious components, try to find why they are there, and fix them.

You should submit the following: (a)Listings of your two functions. (b) Details of your Hilbert transformer design, including plots of the pass-band ripple, and details of the phase response. (c) A summary of your approach, including the compromises you made in any design choices. (d)Plots of the magnitude spectra of 1. The audio file in PS8Raw.wav.

2. Your scrambled version of PS8Raw.wav.

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3. The result of descrambling your scrambled version of PS8Raw.wav. Use fft s h i ft ( ) on the spectra to make them more readable. (e)You should upload three .wav files to the 2.161 MIT Server site in the homework section: 1. Your scrambled version of PS8Scrambled.wav.

2. Your descrambled version of PS8RawScrambled.wav.

3. The result of scrambling then descrambling PS8Raw.wav.

Be sure to name the files so that they are clearly identified as yours...

We will be available to help if you run into trouble.

Problem 2: An analog integrator has a transfer function H(s) = 1/s. Use the bilinear transform to find (a) the discrete time transfer function H(z), and (b) the difference equation for this form of digital integrator.

Plot the frequency response functions of the resulting digital integrator.

Problem 3:

A continuous band-pass filter is described by the transfer function

Derive a recursive computing formula (difference equation) for (1) a step-invariant, (2) a root matching, and (c) a bilinear transform digital filter based on this system. Assume Δ T = 0.1s.

Problem 4: A digital filter is to be designed to the following specifications:

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(a)What is the order of the continuous filter if a Chebyshev design is used. (b) Prewarp the critical frequencies to find the equivalent frequencies for use in the bilinear trans­ form with a sampling frequency of 50,000 samples/sec. (c) Use MATLAB to design a continuous Chebyshev Type 1 filter with the prewarped crical frequencies. Then use the b i l i ne a r( ) function to compute the discrete time transfer function from the continuous prototype. (d)Make frequency response and pole-zero plots for your resulting filter. (e) Use MATLAB to convert the prototype design to a band-pass digital filter with a passband of 5-15 kHz. .

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Solutions Problem 1: The standard Hilbert transformer is defined as:

MAT LAB uses the same definition and this can be verified by finding the phase of ω → 0 + . The implemented filter’s phase has a linear component added to the −π/2 value (for ω > 0).

Our filter has −π/2 phase shift and this corresponds to: A sin ((Ω a − Ω) t) = A (sin(Ω a t) cos(Ωt) − cos(Ωa t) sin(Ωt)) = −A sin(Ω a t) sin(Ωt − π/2) − A cos(Ω a t) sin(Ωt) We have to be careful in implementing the scrambler. The scrambler diagram corresponds to the below figure.

However, note that the audio play is not affected by sign of output (i.e. a signal is played the same as its negative version).

(a) The delay in implementation does not affect this discussion. So we ignore it for this analysis. We consider a sinusoidal component of input as f(t) = sin(Ωot). The output of filter would be |H(jΩo)|sin(Ωot+6 H(jΩo)) = (1+δ)sin(Ωot− π 2 ). We follow this through above diagram to compute g(t):

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As a result due to practical implementation, some “spurious” components are introduced at Ωa + Ωo .

The original signal has a (desired) audio frequency content of Ωo = 300 − 3000 Hz. In the beginning, we can preprocess the audio file with a band-pass filter passing 300 − 3000 Hz contents and then follow it with the scrambling process.

We use a Hilbert transformer with a pass-band equal to 300 to (F s /2 − 300) Hz, where F s is the sampling frequency . The extended pass-band of the filter, is because a high-order (low-ripple) “firpm-built” Hilbert filter requires a symmetric pass-band (see MAT LAB doc­ umentation). After passing the audio file through our transformer with Ωa = 3500 Hz, the output will have a large (desired) spectrum content at 500 − 3200 Hz and a spurious content at 3800 − 6500 Hz (which could be folded to some lower frequency). If δ is not small enough, we might wish to post-process the filter output with a pass-band filter passing 500 − 3200 Hz contents and then save it as scrambled signal. This post processing step of scrambling, can be considered as a pre-processing step of desrcambling procedure.

Theoretically, descrambling is the same as scramblibg. However, the practical realization of the Hilbert filter is bandwidth limited (while the ideal Hilbert filter is an all-pass filter). The Hilbert Filter for desrcambling, requires a pass-band equal to 500 to (F s /2 − 500) Hz. However, if we pre-process the srcambled signal with a pass-band filter passing 500 − 3200 Hz contents, then we can use the same Hilbert filter for scrambling and descrambling. The output of desrcambling can be further post-processed with a pass-band filter passing audio contents (300 − 300 Hz).

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(b) This is not the complete solution to all parts of the problem. This simply serves to show the approach. To study the full solution start with attached Main.m file and follow other scripts. Note that the encoding and decoding functions can be identical (if we use proper pre/post filters). Here is the encoding script: functi on fout = encode(fi n,Fs,Fc) %% Design the Hilber transformer % Note: fi rpm() seems to need the band edges to be symmetric about 0.5 h1=firpm(150,[300*2/Fs 1-300*2/Fs],[1 1 ] ,’ H i l b e r t ’ ) ; % h1=firpm(150,[500*2/Fs 1-500*2/Fs],[1 1 ] ,’ H i l b e r t ’ ) ; % change i n the decode fi l e y 1 = fft fi l t ( h 1 , fi n ) ; %% Delay: Approach 1 N=length(h1); %N i s odd y2=zeros(size(fi n));y2((N-1)/2+1:end)=fi n(1:end-(N1)/2); %% Delay: Design the delay fi l t e r, Approach 2 % h2 = zeros(1,length(h1)); % h2((N-1)/2+1) = 1; % F i l t e r the data % y 2 = fft fi l t ( h 2 , fi n ) ; %% F i l t e r Output t=(0:length(fi n)-1)*(1/Fs); y3=-y1.*sin(2*pi*Fc*t ’); y4=-y2.*cos(2*pi*Fc*t ’); fout= y3+y4; The following graph shows one the encoder’s Hilbert transformer (order 150).

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Encoder Hilbert Filter for F s=11.025 KHz

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We passed the input original audio through a band-pass filter (300-3000 Hz) to minimize the folding of high frequencies. We used the same filter for the decoder output. We also used a band pass filter (500-3200 Hz) for the encoder output or for the decoder input. These pass-band filters are implemented as a series of low and high pass filters (all order 200). Otherwise, we could not build a nice pass band filter in a single step. All these filters are shown on the next page Note that following the original signal through all filters (audio filter, Hilbert, pre-decode, Hilbert and audio filter), results in a delay equal to 750 samples ( 3 ∗ (200/2 + 200/2) + 2 ∗ (150/2) ). Although 750 might be a large number, but even for our lowest F s = 11.02 KHz, this means less then 0.07 second delay. This small delay is hardly noticeable by our hearing system and hence our filter’s orders are not too large.

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Audio Filter (Post−Decoding or Pre−Encoding)

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The following plots resulted from working on PS8Raw.wav:

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8000

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Comparing the input and decoded spectra shows them to be the same. In your solutions, we will be looking for stray spectral components that may result from ripple in your Hilbert transformer response etc.

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Problem 2: (a) If H(s) = 1/s, then the bilinear transform (using Tustin’s approximation) gives:

(b) From above

which is the trapezoidal numerical integration rule.

The following MATLAB commands generate the frequency response: b = [0.5 0.5]; a = [1 -1]; freqz(b,a) title(’Bilinear Integrator’)

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Note that while the phase is constant at −π/2, the magnitude plot does not show a constant slope of -20 dB/decade as does the analog integrator.

Problem 3: (a) The plant

has (1) a zero at s = 0, and (2) a pair of coincident poles at s = −1. The root-matching (matched z-transform) discrete-time system is

and with T = 0.1 sec.,

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To create a minimum delay filter, make the order of the numerator and denominator equal by adding a zero at the origin:

The gain factor K must be determined empirically. This is normally done using the finalvalue theorem, but in this case lims→0 H(s) = 0 and the final values cannot be compared. Some other amplitude criterion must be used; in this case the peak value of the two step responses were compared and found to be 0.367 for the continuous system H(s) and 4.018 for the discrete system H(z). Then K = 0.367/4.018 = 0.0913.

Alternatively you might compare the response of the analog system to a ramp input r(t) = t with the analogous discrete response to a ramp rn = 0.1n and use their final values to match the gains resulting in K = 0.097. The step responses with K = 0.0913 are compared in the plot below

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Note the slight delay in the digital response. (b) The system has repeated poles at s = −1. Then H(s) may be expressed in partial fractions as

and

As we discussed in class, this form of impulse invariant simulation suffers from a gain error, and a correction

is more frequently used. The step responses of H(s) and H ′ (z) are compared in the following plot.

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and we note: • The final value is incorrect (due to aliasing in the transfer function), and • Further empirical gain adjustment is necessary to match the peak responses. (c) With the bilinear transform

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The step responses of H(s) and H(z) are compared in the following plot.

Bilinear Simulation − Step Response

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bilinear continuous

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The frequency response of these three discrete approximations of our continuous filter, are plotted and compared to the frequency response of the continuous filter in the next page. In general, root matching and impulse invariant filters are approximately the same and show almost exactly the same curves.

The magnitude plot shows that below 1 Hz, all of the three filters match very well with the continuous filter.

However as the frequency increases toward the Nyquist frequency the bilinear filters deviates strongly from the continuos filter and the two others almost follow the continuos curve.

On the other hand, in the phase plot, the bilinear filter almost exactly matches the continuos filter, while the two others deviate strongly from the continuous filter and only match it at very low frequencies (i.e. below 0.3 Hz).

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Problem 4: (a) From the specifications

For a continuous filter (no pre-warping)

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Therefore take N = 4. (b) In the pre-warped analog filter the critical frequencies will be

and the order is given by

Therefore take N = 3. (c) MATLAB gave me a lot of problems with ill-conditioning while trying to do this! I had to scale the sampling rate back – for example by a factor of 100, to 500 samples/sec.:

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(d) The pole-zero plot and frequency response plots are shown on the next page.

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(e) We have to pre-warp the band limits to use them in converting the prototype continuous low pass filter (with cut-off frequency of 1 rad/sec) to our prototype continuous band pass filter.

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T = 1/500; wc1 = 2*pi*50; wc2 = 2*pi*150; wc1_p = (2/T)*tan(wc1*T/2); wc2_p = (2/T)*tan(wc2*T/2); [ b , a] = cheby1(3, 3, 1 ,’s ’ ) ; % Design LP with wc=1 rad/sec [b,a]=lp2bp(b,a,sqrt(wc1_p*wc2_p),(wc2_p-wc1_p)); %Change to BP [bz ,az ] = bi l i nea r(b, a , 500);

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