2.161 Signal Processing: Continuous and Discrete

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1 MIT OpenCourseWare Signal Processing: Continuous and Discrete Fall 8 For information about citing these materials or our Terms of Use, visit:

2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING.6 Signal Processing - Continuous and Discrete Fall Term 8 Solution of Problem Set 7 Assigned: October 3, 8 Due: November 6, 8 Problem : There are a number of ways to prove this here are two: (a) Start with an acausal filter of the same length H a (z) = (N)/ n= (N)/ n h n z with N odd, and with real, odd symmetric coefficients h n = h n so that the impulse response is a real, odd function. Since DTFT {h n } H(e jω ), from the properties of the DTFT, namely that a real, odd function has a imaginary, odd DTFT, then H a (e jω ) is imaginary and odd. H a (z) may be made causal by adding a delay of (N )/, that is H(z) = z (N)/ H a (z) so that Since H a (e jω ) is imaginary H(e jω ) = e jω(n)/ H a (e jω ) H(e jω ) = je jω(n)/ = (N ) Ω ± π/ which is linear with frequency Ω = ωδt (apart from possible jump discontinuities where H a (e jω ) changes sign. Note that in practical filters, any jump in the phase response will occur outside the pass-band, and is generally ignored.) The equivalent delay is found from the slope, and represents a delay of (N )/ samples. (b) Assume the causal filter and let H(z) = N n h n z n= = h z (N) + h z h N z with N odd, and with odd symmetric coefficients about the mid-point, h n = h N n. Group the symmetric components together ( ) ( ) ( ) (N) (N ) (N ) (N)/ H(z) = h z z + h z z + h z z h (N)/ z

3 (N)/ ( ) n (N n) (N)/ = h n z z + h (N)/ z n= (N)/ ( ) (N)/ = z h n z (N)/ n ((N/) n) z n= since, as noted in the problem statement, for an odd-symmetric function about the mid-point h (N)/ =. The frequency response is H(e jω ) = H(z) z=e jω (N)/ ( ) H(e jω jω(n)/ ) = e h n e jω((n)/ n) e jω((n/) n) n= (N)/ = je jω(n)/ h n sin (Ω((N )/ + n)) n= = je jω(n)/ H a (e jω ), The function H a (e jω ) is purely real, therefore H(e jω ) = je jω(n)/ = (N ) Ω ± π/ which is linear with frequency Ω = ωδt (apart from possible jump discontinuities of π at sign changes in H a (e jω ), which as noted above occur outside the pass-band) and represents a delay of (N )/ samples. Problem : (a) The system has 7 zeros and 7 poles at the origin. It is therefore a non-recursive FIR system. (b) The zeros are at: ( ( ) ( )) ( ( ) ( )) π π 3 π π,,, cos ± j sin, cos ± j sin With the help of MATLAB: z = ; z = -4/3; z3 = -3/4; z4 = (4/3)*(cos(pi/3) + i* sin(pi/3)); z5 = (4/3)*(cos(pi/3) - i* sin(pi/3)); z6 = (3/4)*(cos(pi/3) + i* sin(pi/3)); z7 = (3/4)*(cos(pi/3) - i* sin(pi/3)); z = [z z z3 z4 z5 z6 z7]; p = [ ]; K=; H = tf(zpk(z, p, K, -)) The transfer function is z 7 z z 4.79z 3 + z z H(z) = z7 = z z +.79z z + z z

4 so that this is an odd symmetric filter with N = 8 and {h n } = {,,,.79,.79,,, } Then H(ze jω ) has a linear phase shift with frequency and the delay Δ is the slope, that is (c) The MATLAB command ( ) H(e jω ) = je jω(n)/) 7 = Ω + π/ d Δ = H(e jω ) = 7/ steps. dω freqz([ ], []) generates the following plot Magnitude (db) Phase (degrees) which shows a crude band-pass characteristic. The phase plot shows a change in phase of 7π/ over a range of π in frequency, corresponding to a delay of 7/ steps. Problem 3: The following MATLAB script was used for this problem: Problem Set 6 -- Problem 3 Enter the signal and contaminate it: t = :.:; signal = sin(4*pi*t); noise = *(rand(size(signal))-.5); 3

5 noisy_signal = signal + noise; Plot the waveforms figure() plot(t,signal, b ) hold on plot(t,noisy_signal, r ) title( Signal and Additive Noise ) hold off Design the filter w_c =.4; w_stop =.48; passband_ripple =.; stopband_ripple =. [N,Wn,BETA,FILTYPE] = kaiserord([w_c w_stop],[ ], [passband_ripple stopband_ripple],) B = fir(n,wn,filtype,kaiser(n+,beta)); A = ; Filter the noisy data and plot the output filtered_signal = filter(b,a,noisy_signal); figure() plot(t,filtered_signal, g ) title( Filtered Signal ) Plot the frequency response and pole-zero plots figure(3) freqz(b,a); figure(4) zplane(b,a) df = /.; f = -5:df:-5+*df; figure(5) plot(f,fftshift(abs(fft(noisy_signal)))) xlabel( Frequency (Hz) ) ylabel( Magnitude F ) title( Magnitude Spectrum of Noisy Signal ) figure(6) plot(f,fftshift(abs(fft(filtered_signal)))) xlabel( Frequency (Hz) ) ylabel( Magnitude F ) title( Magnitude Spectrum of Low-Pass Filtered Signal ) This script gave the filter parameters from kaiserord() as: N = 75, β = 3.395, ω n =.45. The following plots were produced: 4

6 Signal and Additive Noise Filtered Signal The following two spectra have been prepared using fftshift() to center them on Hz, 5 Magnitude Spectrum of Noisy Signal Magnitude F Frequency (Hz) 5

7 5 Magnitude Spectrum of Low Pass Filtered Signal Magnitude F Frequency (Hz) 5 Magnitude (db) Phase (degrees) Imaginary Part Real Part Note that the pole-zero plot is composed with zeros on the unit-circle at angles Ω corresponding to frequencies above the cut-off frequency thus forcing the magnitude response to zero at those 6

8 frequencies - this is therefore a low-pass filter. It is interesting to note that below the cut-off frequency, the zeros appear to be in pairs where their magnitudes are reciprocals - one inside and one outside the unit circle (in addition to be in complex conjugate pairs as well). Think about what this means (it also is evident in the pole-zero plot given in Problem, and we will see the same thing in the results of Problems 4 and 5). Problem 4: The MATLAB script from Problem 3 was modified slightly for this problem to design a band-pass filter: Problem Set 6 -- Problem 4 Enter the signal and contaminate it: t = :.:; signal = sin(4*pi*t); noise = *(rand(size(signal))-.5); noisy_signal = signal + noise; Plot the waveforms figure() plot(t,signal, b ) hold on plot(t,noisy_signal, r ) title( Signal and Additive Noise ) hold off Design the filter Frequency specs - normalized to the Nyquist frequency w_sl=.35; w_cl=.39; w_cu =.4; w_su =.45; Pass and stop band ripple p_rip =.; s_rip =.; [N,Wn,BETA,TYPE] = kaiserord([w_sl w_cl w_cu w_su],[ ], [s_rip p_rip s_rip],) B = fir(n,wn,type, kaiser(n+,beta)); A = ; Filter the noisy data and plot the output filtered_signal = filter(b,a,noisy_signal); figure() plot(t,filtered_signal, g ) title( Filtered Signal ) Plot the frequency response and pole-zero plots figure(3) freqz(b,a); figure(4) zplane(b,a) df = /.; 7

9 f = -5:df:-5+*df; figure(5) plot(f,fftshift(abs(fft(noisy_signal)))) xlabel( Frequency (Hz) ) ylabel( Magnitude F ) title( Magnitude Spectrum of Noisy Signal ) figure(6) plot(f,fftshift(abs(fft(filtered_signal)))) xlabel( Frequency (Hz) ) ylabel( Magnitude F ) title( Magnitude Spectrum of Band-Pass Filtered Signal ) This script gave the filter parameters fromkaiserord() as: N =, β = 3.395, ω n = [.37.43]. The following plots were produced: Signal and Additive Noise Filtered Signal

10 The following two spectra have been prepared using fftshift() to center them on Hz, 5 Magnitude Spectrum of Noisy Signal Magnitude F Frequency (Hz) 45 Magnitude Spectrum of Band Pass Filtered Signal Magnitude F Frequency (Hz) Magnitude (db) Phase (degrees)

11 .5 Imaginary Part Real Part Note that the pole-zero plot is composed with zeros on the unit-circle at angles Ω corresponding to frequencies above and below the cut-off frequency thus forcing the magnitude response to zero at those frequencies - this is therefore a band-pass filter. As in Problem 3, it is interesting to note that within the pass-band the four zeros appear to be in pairs where their magnitudes are reciprocals one inside and one outside the unit circle (in addition to be in complex conjugate pairs as well). Problem 5: The following MATLAB script was used to design and test a band-stop filter: Problem Set 6 -- Problem 5 Enter the signal and contaminate it: Dt = /3; t = :Dt:999*Dt; f_sample = /Dt; f_nyquist = f_sample/; signal =.5*sin(*pi*3*t) + *cos(*pi*9*t); noise = sin(*pi*6*t); noisy_signal = signal + noise; Plot the magnitude spectrum of the contaminated signal f = -5:/(*Dt):-5+999/(*Dt); figure() plot(f,fftshift(abs(fft(noisy_signal)))) title( Magnitude Spectrum of Signal with 6 Hz Contamination ) xlabel( Frequency (Hz) ) ylabel( Magnitude F_m ) f_norm = 6/f_nyquist; Design the filter w_sl = f_norm*.9; w_cl = f_norm*.7; w_cu = f_norm*.3; w_su = f_norm*.;

12 p_ripple =.; s_ripple =.; [N,Wn,BETA,TYPE] = kaiserord([w_cl w_sl w_su w_cu],[ ], [p_rip s_rip p_rip],) B = fir(n,wn,type, kaiser(n+,beta)); A = ; Filter the noisy data and plot the output magnitude spectrum figure() filtered_signal = filter(b,a,noisy_signal); plot(f, fftshift(abs(fft(filtered_signal)))) title( Magnitude Spectrum of the Filtered Signal ) xlabel( Frequency (Hz) ) ylabel( Magnitude F_m ) Plot the frequency response and pole-zero plots figure(3) freqz(b,a); figure(4) zplane(b,a) Extra plots - not asked for! Plot segments of the input and filtered signals First interpolate the signals ant time by a factor of eight to make the plots more readable. s_8 = resample(signal,8,); ns_8 = resample(noisy_signal,8,); fs_8 = resample(filtered_signal,8,); t_8 = resample(t,8,); Plot the original signal and the contaminated signal figure(5) plot(t_8(3:5), s_8(3:5), b--,t_8(3:5), ns_8(3:5), r ) xlabel( Time (sec) ) ylabel( Signal ) title( Segment of Signal and Contaminated Signal ) figure(6) plot(t_8(3:5), s_8(3:5), b--,t_8(3:5), fs_8(3:5), r ) xlabel( Time (sec) ) ylabel( Signal ) title( Segment of Uncontaminated Signal and Filter Output ) figure(7) plot(t_8(:5), s_8(:5), b--,t_8(:5), fs_8(:5), r ) xlabel( Time (sec) ) ylabel( Signal ) title( Initial Segment of Uncontaminated Signal and Filter Output ) This script gave the filter parameters from kaiserord() as: N = 56, β = 3.395, ω cl = 36 rad/s, ω cu = 48 rad/s. The following plots were produced

13 5 Magnitude Spectrum of Signal with 6 Hz Contamination Phase (degrees) Magnitude F Magnitude F m m Magnitude (db) Frequency (Hz) Magnitude Spectrum of the Filtered Signal Frequency (Hz)

14 .5 Imaginary Part Real Part The following plots were not asked for in the problem. To look at the waveforms it was necessary to interpolate the data records to show details of the waveforms, using resample(). See the script for details. 4 Segment of Signal and Contaminated Signal 3 Signal Time (sec) Desired signal Signal contaminated with 6 Hz component Segment of Uncontaminated Signal and Filter Output 3 Signal 3 Uncontaminated input signal Filter output Time (sec) 3

15 4 Initial Segment of Uncontaminated Signal and Filter Output 3 Signal Time (sec) Uncontaminated input signal Filter output Note the delay in the filter output. 4