# HW8sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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Unformatted text preview: UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 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 z-transform approach as follows: 1 1 0.5 Taking the inverse z-transform, 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:M-1]; l = [0:L-1]; % 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 steady-state 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 steady-state 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 steady-state 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 steady-state 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 z-transform of both sides and simplify: 1 Then, we take the z-transform 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 Kz-1. 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 z-transform of both sides and simplify: 1 Then, we take the z-transform 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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