HW13 Solutions. Pr (a) The DTFT of c[n] is. C(e jω ) = 0.5 m e jωm e jω + 1

Transcription

1 HW3 Solutions Pr..8 (a) The DTFT of c[n] is C(e jω ) = = =.5 n e jωn + n=.5e jω + n= m=.5 n e jωn.5 m e jωm.5e jω +.5e jω.75 =.5 cos(ω) }{{}.5e jω /(.5e jω ) C(e jω ) is the power spectral density. (b) The z-transform C(z) = = =.5z +.5z =.5z +.5z.5z.5z +.5z.5 (.5z )(.5z).75z (z.5)(.5z) so there is a zero at and poles at.5 and. (c) C(e j ) =.75.5 = 3 To obtain the phase, consider that C(e jω ) is real and that its denominator.5 cos(ω) > since cos(ω) for ω [ π, π), thus the phase is zero.

2 Pr.. (a) The DTFT of x[n] = sin(.πn) = (e j.πn e j.πn )/j is X(e jω ) = π [δ(ω.π) δ(ω +.π)] j The period of x[n] is since the frequency is π/. The signal y[n] can be written y[n] = sin(.πn) (u[n] u[n ]) = w[n] sin(.πn)[n] = w[n] ej.πn e j.πn }{{} j w[n] so that by the frequency shifting property Y (e jω ) = j ( ) W(e j(ω.π) ) W(e j(ω+.π) ) i.e., the spectrum of the rectangular pulse shifted to ±.π. If instead of we use a very large value, Y (e jω ) would look similar to the DFT of x[n]. The following script computes and plots the DTFTs of y[n]. If you increase the value of q to a large value, e.g.,, the Y (e jω ) looks like X(e jω ) as many more values are used, that is the frequency resolution is larger. %% Pr. q=; N=*q; w=ones(,n); n=:n-;x=sin(.*pi*n); y=w.*x; M=5; Y=fft(y,M); omega=[:m-]**pi/m-pi; figure() subplot() plot(omega,fftshift(abs(y)));axis([min(omega) max(omega) 5]) ylabel( Y ) subplot() plot(omega,fftshift(unwrap(angle(y))));axis([min(omega) max(omega) -3 ]) ylabel( <Y );xlabel( \omega ) (b) The following script is used to compute and plot the window and the spectrum of the windowed signal clear all N=;n=-5:5; w=+cos(*pi*n/n); x=sin(.*pi*n); y=w.*x; M=5; Y=fft(y,M); omega=[:m-]**pi/m-pi; figure()

3 subplot(3) stem(n,w); axis([-5 5.*max(w)]);ylabel( w_[n] );xlabel( n ) subplot(3) plot(omega,fftshift(abs(y)));axis([min(omega) max(omega) 5]) ylabel( Y_ ) subplot(33) plot(omega,fftshift(unwrap(angle(y))));axis([min(omega) max(omega) -3 ]) ylabel( <Y_ );xlabel( \omega ) 5 w [n] Y n 3 3 Y 3 3 <Y ω <Y ω Figure.5: Magnitude and phase spectra of y[n] using a rectangular window (left). Raised cosine window and magnitude and phase spectra of y [n].

4 Pr..7 (a) When the signal is aperiodic, padding it with zeros increases the number of frequencies, i.e., increases the frequency resolution while the original amplitude values remain the same. %% Pr.7 clear all; clf x=ones(,); x=[x zeros(,)]; M=length(x); X=fft(x); x=[x zeros(,)]; M=length(x);X=fft(x,M); w=[:m-]**pi/m;n=:length(x)-; w=[:m-]**pi/m;n=:length(x)-; figure() subplot() stem(n,x); axis([ max(n).]);ylabel( x_[n] ) subplot() stem(n,x); axis([ max(n).]);ylabel( x_[n] ) subplot(3) stem(w,abs(x));axis([ *pi ]);ylabel( X_ ) subplot() stem(w,abs(x)); axis([ *pi ]) (b) When the signal is periodic, the fundamental frequencies do not change by considering several periods, zeros appear in between them. The amplitudes at the fundamental frequencies increases as the number of periods being considered increases. When padding zeros to a period, the signal is corrupted and its spectrum changes. % part (b) clear all N=; M=N; n=:m-; x=cos(pi*n/5); X=fft(x); w=[:m-]**pi/m; x=[x x x x x x x x x x]; M=*N;w=[:M-]**pi/M; X=fft(x) x3=[x zeros(,)]; X3=fft(x3); M=length(x3);w3=[:M-]**pi/M; figure() subplot(3) stem(w,abs(x));axis([ *pi 3.]);ylabel( X_ ) subplot(3) stem(w,abs(x));;axis([ *pi 3]);ylabel( X_ ) subplot(33) stem(w3,abs(x3));axis([ *pi 3.]);ylabel( X_3 ) xlabel( \omega (rad) )

5 x [n] X x [n] X X X X ω (rad) Figure.: Improving the frequency resolution of an aperiodic signal (left) and of a periodic signal.

6 Pr..8(a) The impulse response of the IIR filter appears to taper down around n = 3 so the approximation of the impulse response seems good. %% Pr.8 clear all;clf x=ones(,5); b=[.5 ]; a=[ ]; delta=[ zeros(,3)]; h=filter(b,a,delta); N=8; X=fft(x,N);H=fft(h,N); Y=X.*H; y=real(ifft(y)); figure() subplot(3) stem([x zeros(,)]) subplot(3) stem([h zeros(,)]) subplot(33) stem(y) (b)(c) In this case we do not truncate the impulse response, and the results y [n] are better than before y[n], (the control is the solution obtained using filter, y c [n]). % part X=fft(x,N);A=fft(a,N);B=fft(b,N); Y=X.*B./A; y=ifft(y); yc=filter(b,a,x); figure() subplot() plot(y) hold on plot(yc, r );grid hold off subplot() plot(y) hold on plot(yc, r ); grid hold off

7 x[n] h[n] y [n], y c [n] y[n] 3 8 n y [n],y c [n] 8 n Figure.:.

8 Pr..9(a)(b) Plotting x[m] and y[n m] (y[m] reversed and shifted to the right n samples), multiplying them and adding the overlapping values gives z[n] = (x y)[n], the linear convolution. See results of the script below. The values are also obtained by multiplying the z-transforms X(z) and Y (z) giving Z(z) = X(z)Y (z) = z + 3z + z 3 + 5z + 3z 5 which is the DTFT of z[n] when z = e jω (the coefficients of Z(z) coincide with the values from the linear convolution. (c) For N the circular convolution is graphically done as follows. The inner circle corresponds to x[m] and the outer circle to y[n m] (reversed and shifted clockwise), and at each shift the corresponding values are multiplied and added. The results are z[] = 5, z[] =, z[] =, z[3] = 3 and z[] =, which do not N =,,,,,, 3,,, z[] = 5 z[] = z[] = 3 z[3] =,,, Figure.: Circular convolution for N =. coincide with the values from the linear convolution, this is because N = < length[x[n]] + length[y[n]] =. For the other values of N, bigger than, a similar graphical procedure is used, but in these cases the results of the circular and the linear convolutions coincide. (d) If we let z = e jπk/n we obtain from the result in (a) that X(k)Y (k) = e j(πk/n) + 3e j(πk/n) + e j3(πk/n) + 5e j(πk/n) + 3e j5(πk/n) Now if N =, e j(πk/) = e j(πk/n) and e j5(πk/n) = e j(πk/n) so that the above can be written as X(k)Y (k) = 5 + ( + 3)e j(πk/n) + 3e j(πk/n) + e j3(πk/n) The coefficients of the DFT correspond to the values given by the circular convolution of length N =. If N = 7 > we have X(k)Y (k) = e j(πk/7) + 3e j(πk/7) + e j3(πk/7) + 5e j(πk/7) + 3e j5(πk/7) and its coefficients correspond to the circular convolution results with N = 7 and also coincide with the ones from the linear convolution. Likewise, for N = > we get X(k)Y (k) = e j(πk/) + 3e j(πk/) + e j3(πk/) + 5e j(πk/) + 3e j5(πk/)

9 In the cases when N > there is no simplification of the exponentials, as in the case for N <, and as such the circular and the linear convolutions coincide. The implementation using the fft function is shown below. clear all; clf x=[ 3 ]; y=[ ]; z=conv(x,y); % linear convolution n=:9; figure() subplot(3) stem(n,[x zeros(,)]);title( x[n] );grid subplot(3) stem(n,[y zeros(,)]);title( y[n] );grid subplot(33) stem(n,[z zeros(,3)]);title( z[n]=(x*y)[n] );grid % circular convolutions figure() for i=:3, if i==, N=;subplot(3) elseif i==, N=7;subplot(3) else N=;subplot(33) end X=fft(x,N); Y=fft(y,N); Z=X.*Y; z=ifft(z);m=length(z); stem(n,[z zeros(,-m)]) if i==, title( circular conv N= );grid elseif i==, title( circular conv N=7 );grid else title( circular conv N= );grid end end

10 3 x[n] circular conv N= y[n] z[n]=(x*y)[n] circular conv N= circular conv N= Figure.3: Linear convolution of x[n] and y[n] (left). Results of circular convolutions of x[n] and y[n] for lengths N = (right top), N = 7 and N = (right bottom).