Unformatted text preview: UNIVERSITY OF ILLINOIS AT URBANACHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 8 Solutions Due Wednesday, October 22, 2008 Prof. Bresler / Prof. Jones 1. (25 points) Consider the system shown in the diagram below: where 2 1 and . a) Find the impulse response, of the overall system. We can find the overall system response using the ztransform approach as follows: 1 1 0.5 Taking the inverse ztransform, we get: 1 2 1 1 , 1 0.5   1 0 1 4 1 0.5 1 0.5 b) Determine the difference equation representation of the overall system that relates the output to the input. Convolving the input with the impulse response, we obtain: 1 2 1 c) Is the overall system causal? Is it BIBO stable? Explain your reasoning for full credit. Yes, and Yes. The system is causal because 0 for response consists of only two samples (finite duration), the system is BIBO stable. d) Determine the frequency response, Computing the DTFT of the system: 1 2 1 1 1 2 0. Moreover, since the impulse is also absolutely summable. Thus, , of the overall system. Notice that the above answer can be easily obtained just by evaluating e) Plot the magnitude and phase response of the overall system for 0 at . using MATLAB. 2. (15 points) Determine the frequency response function their impulse response. Plot the magnitude response  interval ,. a) 0.9
 for each of the LSI systems described by  and the phase response over the Using the DTFT definition, we compute 0.9 1 0.9 1 :
 0.9 1 0.9 1 0.9 1 1 0.19 1.81 1.8 cos Plotting the frequency response: b) 0.4 0.5 : Using the DTFT definition, we compute 0.4 1 0.4 1 0.5 0.5 2 0.9 0.9 0.4 0.5 1 1 1 0.2 Plotting the frequency response: c) 0.5  cos 0.1 0.5  cos 0.1 1 2 0.5
 . 0.5  . 0.5 0.75 1.25 cos 0.1 3/8 1.25 cos 0.1 0.5 0.75 1.25 cos 0.1 3/8 1.25 cos 0.1 Notice that we used the result from part (a) to avoid similar computations. Plotting the frequency response: 3. (30 points) A LSI system can be described by the following difference equation: a) Create a function in MATLAB that takes inputs vectors , and , where is the vector of the coefficients , is the vector of the coefficients and is a vector of the frequencies on the interval , , that outputs the frequency response, , in a vector. The frequency response, , should only be evaluated at the frequencies that the vector provides. First of all, notice that taking the DTFT of both sides of the difference equation above and solving for , we have: ∑ 1 ∑ Thus, implementing this in MATLAB, we have the following function:
function [H] = diffeq2freqresp(b,a,w) % Function outputs the frequency response evaluated at the frequencies % given, using the difference equation. % % Inputs: % a = filter coefficient array a (denominator) % b = filter coefficient array b (numerator) % w = frequency location vector % % Outputs: % H = frequency response vector % if a(1)~=1, error('Must have a(1)=1.'); end L = length(a); M = length(b); % b a w reshape vector into row vectors = reshape(b,1,length(b)); = reshape(a,1,length(a)); = reshape(w,1,length(w)); m = [0:M1]; l = [0:L1]; % index arrays num = b*exp(j*m'*w); denom = a*exp(j*l'*w); H = num./denom; end b) Use the function developed in part (a) to plot the magnitude and phase for each of the following systems: i) ∑ 1, ,,,,. In this case, ii) By inspection, , , 2 0.95 1 0.9025 , , 2 1,0, 1 . 1, 0.95,0.9025 and 4. (15 points) A LSI system is described by the following difference equation: 2 0.81 2 Determine the steadystate response of this system to the following input signals: First of all, it is useful to use the given difference equation to find a mathematical form for the LSI system’s frequency response: ∑ ∑ 0.81 a) x n 5 10 1 5 10cos nπ 0 and The form of x[n] tells us that we only need to compute the frequency response at : 0 ∑ ∑ 0.81 ∑ ∑ 0.81 1 0.81 1.6885, 4 0.81 0.81 1.6885 1 Thus, the steadystate response is given by: 5 0 10 8.4425 b) 2 sin and : ∑ 4 3 4 ∑ 0.81 ∑ ∑ 0.81 0 3 cos  cos 16.885 5 1.6885 1 10 1.6885 1 Following a similar procedure here, we only care about how the system responds to frequencies 0 Thus, the steadystate response is trivially given by: 0 ∑ 1 cos , 0, … ,5: 3 3 4 0 0 0 c) Similarly, in this part, we care the frequency response evaluated at ∑ ∑ 0.81 1.6885 2 4 0 1.6885 3 4 5 4 4 0 Thus, the steadystate response is given by: 1 0 5  cos 1.6885 8.4425 1 5. (15 points) Solution 1: A first–order feedback system is shown below with a constant feedback gain, K. x[n] +  p[n] y[n] K The plant’s impulse response is given by 2 . For what values of K is the system stable? Starting from writing an equation for the system in the time domain: Then, we take the ztransform of both sides and simplify: 1 Then, we take the ztransform of p[n]: 1 1 2 ,   2 Notice that p[n], itself corresponds to an unstable system (treating K=0 means there’s no feedback) because the pole z = 2 pole is not inside the unit circle. Then, the H(z) is simply P(z) and the system is unstable. The idea is that using a feedback path will allow us to make the overall system stable, and to do that, we need to properly choose K. Plugging in P(z) into H(z), we have: 1 1 Thus, there is a pole at circle. Thus, we require: 2 1 1 1. 1 2 1 1 2 2 1 2 2 1 2 . To make the system stable, we simply need this pole to be inside the unit This can happen only if the gain K is greater than 1. Thus, we need If we consider negative feedback as well, we could also satisfy the above criterion with any Solution 2: A first–order feedback system is shown below with a feedback gain of Kz1. 3. The plant’s impulse response is given by 2 . For what values of K is the system stable? Starting from writing an equation for the system in the time domain: 1 Then, we take the ztransform of both sides and simplify: 1 Then, we take the ztransform of p[n]: 1 1 2 ,   2 Notice that p[n], itself corresponds to an unstable system (treating K=0 means there’s no feedback) because the pole z = 2 pole is not inside the unit circle. Then, the H(z) is simply P(z) and the system is unstable. The idea is that using a feedback path will allow us to make the overall system stable, and to do that, we need to properly choose K. Plugging in P(z) into H(z), we have: 1 1 Thus, there is a pole at 2 unit circle. Thus, we require: 1 2 1 1 2 2 1 2 2 2 . To make the system stable, we simply need this pole to be inside the 2  1 3. This can happen only if the gain K is between 1 and 3. Thus, we need 1 ...
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 Fall '09
 Digital Signal Processing, Signal Processing, University of Illinois, Prof. Jones, Department of Electrical and Computer Engineering ECE, 2 pole, Prof. Bresler

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