University of Waterloo Department of Electrical and Computer Engineering ECE 413 Digital Signal Processing. Spring Home Assignment 2 Solutions

Transcription

1 University of Waterloo Department of Electrical and Computer Engineering ECE 13 Digital Signal Processing Spring 017 Home Assignment Solutions Due on June 8, 017 Exercise 1 An LTI system is described by the difference equation y[n] = b x[n] 0.81 y[n ]. Solution 1. Determine the frequency response H(e jω ) of the system. Frequency response: H(e jω ) = Magnitude response: H(e jω ) = = Phase response: H(e jω ) = tan 1 b e = b jω cos(ω) 0.81jsin(ω) b ( cos(ω)) sin (ω) b ( cos(ω) 0.81sin(ω) cos(ω). Determine the value of b such that max π ω<π H(e jω ) = 1. Plot the resulting magnitude and phase responses (both in its wrapped and unwrapped form). The magnitude response is maximized when its denominator is minimized, i.e., at ω = π/. In order to constrain the maximum magnitude response to be equal to 1, we have b = = The unwrapped phase response will be the same as the wrapped phase response. 1

2 w = linspace(-pi,pi); hz = freqz( [0.19], [ ],w); plot(w/pi,angle(hz)); plot(w/pi,unwrap(angle(hz))); 3. Determine analytically the response y[n] to the input x[n] = cos(0.5πn + π/3). We will use the eigenvalue-eigenfunction concept for LTI systems. x[n] = e 0.5πn+π/3 + e 0.5πn+π/3 y[n] = e 0.5πn+π/3 H(e jπ/ ) e H(ejπ/) + e 0.5πn+π/3 H(e jπ/ ) e H(e jπ/ ) y[n] = e 0.5πn+π/ e0 + e 0.5πn+π/ e0 y[n] = e 0.5πn+π/3 + e 0.5πn+π/3 y[n] = cos(0.5πn + π/3) = x[n] n = 0:100; xn = *cos((pi/).*n+pi/3); yn = filter( [0.19], [ ], xn); plot(xn, r ) hold on plot(yn, g )

3 Exercise Determine the system function H(z), magnitude response H(e jω ), and phase response H(e jω ) of the following systems. Use the pole-zero pattern to explain the shape of their magnitude responses. Solution 1. y[n] = 1 (x[n] x[n 1]). System function: H(z) = Y (z) X(z) = 1 z 1 Frequency response: H(e jω ) = 1 e jω = 1 (1 cos(ω) + jsin(ω)) (1 Magnitude response: H(e jω cos(ω)) + sin ) = (ω) = sin( ω ) Phase response: H(e jω ) = tan 1 sin(ω) 1 cos(ω) = π ω w = linspace(0,pi); hz = freqz([1/ -1/],1,w); zplane(roots([1/ -1/])); 3

4 . y[n] = 1 (x[n] x[n ]). System function: H(z) = Y (z) X(z) = 1 z Frequency response: H(e jω ) = 1 e jω = 1 (1 cos(ω) + jsin(ω)) (1 Magnitude response: H(e jω cos(ω)) + sin ) = (ω) = sin(ω) Phase response: H(e jω ) = tan 1 sin(ω) 1 cos(ω) = π ω w = linspace(0,pi); hz = freqz([1/ 0-1/],1,w); zplane(roots([1/ 0-1/])); 3. y[n] = 1 (x[n] x[n 1] x[n ] x[n 3]). System function: H(z) = Y (z) X(z) = 1 z 1 z z 3 Frequency response: H(e jω ) = 1 e jω e jω e 3jω = 1 (1 cos(ω) cos(ω) cos(3ω) + j(sin(ω) + sin(ω) + sin(3ω))) Magnitude response: H(e jω ) = 1 (1 cos(ω) cos(ω) cos(3ω)) + (sin(ω) + sin(ω) + sin(3ω)) Phase response: H(e jω ) = tan 1 sin(ω) + sin(ω) + sin(3ω) 1 cos(ω) cos(ω) cos(3ω)

5 w = linspace(0,pi); hz = freqz([1/ -1/ -1/ -1/],1,w); zplane(roots([1/ -1/ -1/ -1/]));. y[n] = 1 (x[n] x[n 1] x[n 3] x[n ]). System function: H(z) = Y (z) X(z) = 1 z 1 z 3 z Frequency response: H(e jω ) = 1 e jω e 3jω e jω = 1 (1 cos(ω) cos(3ω) cos(ω) + j(sin(ω) + sin(3ω) + sin(ω))) Magnitude response: H(e jω ) = 1 (1 cos(ω) cos(3ω) cos(ω)) + (sin(ω) + sin(3ω) + sin(ω)) Phase response: H(e jω ) = tan 1 w = linspace(0,pi); hz = freqz([1/ -1/ -1/ 0-1/],1,w); zplane(roots([1/ -1/ -1/ 0-1/])); sin(ω) + sin(3ω) + sin(ω) 1 cos(ω) cos(3ω) cos(ω) 5

6 Exercise 3 Show that the group delay of an LTI system with frequency response H(e jω ) = H R (ω) + jh I (ω) can be expressed as τ gd (ω) = H R(ω)G R (ω) + H I (ω)g I (ω) H(e jω ), where G(e jω ) = G R (ω) + jg I (ω) is the DTFT of nh[n]. Solution Now, τ gd (ω) = dψ(ω) Ψ(ω) = tan 1 H I(ω) H R (ω) dψ(ω) = d ( tan 1 H I(ω) H R (ω) + kπ + kπ, Ψ(ω) is continuous ) DT F T {nh[n]} = j d(h R(ω) + jh I (ω)) = dh R(ω) = G I (ω) and dh I(ω) = j dh R(ω) = G R (ω) and H(e jw ) = H R(ω) + H I (ω) = τ gd (ω) = H R(ω)G R (ω) + H I (ω)g I (ω) H(e jω ) dh I(ω) 6

7 Exercise An ECG signal s[n] has been sampled at sampling frequency F s = 00 Hz. The signal is known to be contaminated by a baseline drift d[n] and a power line interference p[n]. Both the drift and the interference signals may be assumed to be sinusoidal signals with fundamental frequencies 0.03 Hz and 50 Hz, respectively. Our goal is to recover s[n] from its noisy measurement x[n], where x[n] = s[n] + d[n] + p[n]. To this end, we are going to use a causal IIR filter with the following system function where D is a positive integer and 0 a < 1. H(z) = b(1 z D ) 1 az D, Solution 1. Determine the value of D such as to make the filter suppress the drift and the interference noises simultaneously. We need D = to supress both frequency components simultaneously.. Determine the value of b such that max π ω<π H(e jω ) = 1. The magnitude response is maximized when its numerator is maximized and the denominator is minimized. In order to constrain the maximum magnitude response to be equal to 1, we get 1 = b a = b =. 1+a 3. Plot the zero-plot diagram and the frequency response of the filter for a = 0.95 and a = Comment on your results. (a) a=0.95 (b) a=0.98 7

8 (c) a=0.95 (d) a=0.98 (Phase and Magnitude distortion is lesser than a=0.95) r=0.98; b=[ ]; a=[ r]; c=(1+r)/; w = linspace(-pi,pi); hz=freqz(b,a,w) plot(w/pi,angle(hz)); figure, zplane(roots(b), roots(a)); 8

9 . For both values of a above, apply the filter to the noisy data (sequence x in data.mat) and plot the filtered output against the original ECG signal (sequence s in data.mat). Does they change as a function of a? Are there any amplitude/phase distortions? (a) a=0.95 (b) a= Let b and a be the feed-forward and feed-back coefficients of filter H, respectively. A better filtering result can be produced by the following code: y = flip( filter(b, a, flip( filter(b, a, x) ) ) ). Plot the system function, magnitude response, and the impulse response of the above filter. Why does this filter perform better? Does it have any phase distortions, and, if not, why? This filter presents better characteristic than the previous; the phase distortion that was visible in the first filter has been improved. r=0.98; b=[ ]; a=[ r]; c=(1+r)/; y1=c*filter(b,a,x); y=c*flip(filter(b,a,flip(y1))); t=(0:n-1); plot(t,s,t,y1) plot(t,s,t,y) (a) filter in part () (b) filter in part (5) 9

10 Exercise 5 Consider a causal system given by the system function: Solution H(z) = z 1 0.9z z. 1. Compute and plot the magnitude and phase responses of the system. Frequency response: H(e jω ) = = Magnitude response: H(e jω ) = e jω 1 0.9e jω e jω cos(ω) jsin(ω) 1 0.9cos(ω) cos(ω) + j(0.9sin(ω) 0.81sin(ω)) ( cos(ω)) sin (ω) (1 0.9cos(ω) cos(ω)) + (0.9sin(ω) 0.81sin(ω)) Phase response: H(e jω ) = tan sin(ω) cos(ω) 0.9sin(ω) 0.81sin(ω) tan cos(ω) cos(ω) w = linspace(-pi,pi); hz=freqz([ ],[ ],w); plot(w/pi,angle(hz));. Determine the minimum-phase system H min (z) corresponding to H(z) and plot its magnitude and phase responses H(z) = z 1 0.9z z = (1 + 5 jz 1 )(1 5 jz 1 ) 1 0.9z z H min (z) = 5 8 (1 + 5 jz 1 )(1 5 jz 1 ) 1 0.9z z = z 1 0.9z z

11 w = linspace(-pi,pi); hz=freqz([ ],[ ],w); plot(w/pi,angle(hz)); 3. Determine the equalizer system H eq (z) corresponding to H(z) and choose a gain G so that the overall system H(z)H eq (z) = 1. Plot the magnitude and phase responses of H eq (z). H eq (z) = Choose G = 1 and n d = 0 H eq (z) = Gz n d H min (z) 1 H min (z) = 1 0.9z z z w = linspace(-pi,pi); hz=freqz([ ],[ ],w); plot(w/pi,angle(hz)); 11