TTT4120 Digital Signal Processing Suggested Solutions for Problem Set 2

Transcription

1 Norwegian University of Science and Technology Department of Electronics and Telecommunications TTT42 Digital Signal Processing Suggested Solutions for Problem Set 2 Problem (a) The spectrum X(ω) can be found as follows. X(ω) x(n)e jωn n e jω e jω cos ω It is shown in Figure X(w) Figure : The spectrum X(ω)

2 (b) The spectrum Y (ω) can be found as follows. Y (ω) y(n)e jωn n M e jωn n M 2M e jω(l M) l e jωm 2M l e jωl l n + M jωm e jω(2m+) e e jω ejωm e jω(m+) e (e jω jω(m+ 2 ) e jω(m+ )) 2 jω e 2 e jω 2 sin ( ω(m + 2 )) sin( ω 2 ) The sketch is shown in Figure 2. ( ) e jω 2 e jω Y(w) Figure 2: The spectrum Y (ω) for M (c) Because they are even signals. (d) A sketch of z(n) for N5 is shown in Figure 3. The Fourier coefficients are given by: c k N N n z(n)e j2πkn/n, k,, N. 2

3 Figure 3: The signal z(n), periodic extension of x(n) Note that we sum from up to N. Thus, the first two samples are 2 and respectively, and the last sample is. All other samples are. The coefficients could be calculated over any other period. c k N N n z(n)e j2πkn/n N (2 + e j2πk/n + e j2πk(n )/N ) N (2 + e j2πk/n + e j2πk e j2πk/n ) N (2 + e j2πk/n + e j2πk/n ) (2 + 2 cos(2πk/n)) N The Fourier coefficients are displayed in Figure 4. (e) We have the following. X(f) cos(2πf) c k (2 + 2 cos(2πk/n)) N 3

4 Fourier coefficients Figure 4: The Fourier coefficients c k of z(n) for k 5,..., 5 Thus, we see that c k N X ( k N ). This means that the Fourier coefficients are (scaled) samples of the continuous spectrum X(f). This always holds true: a periodic extension in the time domain equals sampling in the frequency domain. Problem 2 (a) For the first case, we use the time-shift property of the DTFT, and get X (ω) e j3ω X(ω) (b) For the second case, we use the time-reversal property of the DTFT, and it follows that X 2 (ω) X( ω) (c) For the third case notice that: x 3 (n) x(3 n) x( (n 3)) x 2 (n 3) so that by the time-reversal and time-shift properties, it follows that X 3 (ω) e j3ω X 2 (ω) e j3ω X( ω) (d) For the last case, we have that X 4 (ω) DTFT{x(n) w(n)} X(ω)W (ω). 4

5 Problem 3 (a) By taking the DTFT of both sides of the first difference equation, we get Y (ω) X(ω) + 2e jω X(ω) + e 2jω X(ω) H (ω) Y (ω) X(ω) + 2e jω + e 2jω And for the second case, we get e jω (e jω e jω ) e jω (2 + 2 cos ω). Y (ω).9y (ω)e jω + X(ω) H 2 (ω) Y (ω) X(ω) +.9e jω. (b) We already have the frequency response H (ω) on polar form. Thus, the magnitude is simply H (ω) cos ω. Since cos ω for all ω, the phase is simply Θ (ω) H (ω) ω. The magnitude response of the second system can be found as follows. H 2 (ω) +.9e jω +.9e jω ( +.9 cos ω) 2 + (.9 sin ω) cos ω +.8 To find the phase, we can write H 2 (ω) as H 2 (ω) W (ω), where W (ω) +.9e jω. Then, the phase is given by Θ 2 (ω) H 2 (ω) W (ω). Since Re{W (ω)} > for all ω, we have ( ).9 sin ω H 2 (ω) tan +.9 cos ω ( ).9 sin ω tan. +.9 cos ω 5

6 We notice that all magnitude functions are even and that all phase functions are odd. This is a property of real signals. (c) The frequency response of the first filter can be found and plotted by the following code. [H_, w] freqz([ 2 ], []); subplot(2,, ); plot(w, abs(h_)); xlabel( ); ylabel( Magnitude ); subplot(2,, 2); plot(w, angle(h_)); xlabel( ); ylabel( Phase ); For the second filter, we change the freqz command as follows. [H_2, w] freqz([], [.9]); This gives the plots shown in Figures 5 and Magnitude Phase Figure 5: Magnitude and phase response of H (ω) (d) From the plots of the magnitude responses, we can see that the first filter attenuates high frequencies more than low frequencies. Thus, this is a lowpass filter. The second filter attenuates low frequencies more than high frequencies. Thus, this is a highpass filter. 6

7 8 Magnitude Phase Figure 6: Magnitude and phase response of H 2 (ω) (e) The response of a LTI-system H(ω) H(ω) e jθ(ω) to a sinusoidal input signal x(n) A cos(ω n + θ) equals y(n) A H(ω ) cos(ω n + θ + Θ(ω )). Thus, the output of the first system is y (n) 2 H ( π 2 ) cos(π 2 n + π 4 + Θ ( π 2 )) 2 2 cos(π 2 n + π 4 π 2 ) cos( π 2 n π 4 ). Likewise, the output of the second system is y 2 (n) 2 H 2( π 2 ) cos(π 2 n + π 4 + Θ 2( π 2 )) cos( π 2 ) cos(π 2 n + π sin( π tan 2 ( ) +.9 cos( π 2 )) cos( π n + π 4 + tan ( 9 )) cos( π n +.52)). 7

8 Problem 4 (a) The spectra of the sampled signals are shown in Figures 7 and 8. The latter has a wider range of frequencies than the required f [ 2, 2 ] to help making difference between alias components and signal components. The theory behind this is in ch.6. Figure 7: Spectrum of the signal x(n) when F s 4Hz Figure 8: Spectrum of the signal x(n) when F s 5Hz (b) Matlab-code for generating the signal corresponding to F s 4: t [:/4:-/4]; 8

9 cos4 cos(*2*pi*t); And for the signal corresponding to F s 5: t [:/5:-/5]; cos5 cos(5*2*pi*t); The sounds can be played with the commands: sound(cos4,4); pause(); sound(cos5,5); They sound different because the signal incurred aliasing in the sampling. To be able to reconstruct x a (t) from a sampled signal, the sampling theorem requires that F s > 2F max, where F max is the highest frequency component of the signal. In this case,the signal has only one frequency component, at Hz. Thus, we require: F s > 2Hz 9