Assignment 2 and Submission Guidelines School School of Information Technology and Engineering Course Name Master of Engineering (Telecommunications) Unit Code ME606 Unit Title Digital Signal...

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Answered Same DayMay 21, 2021ME606

Answer To: Assignment 2 and Submission Guidelines School School of Information Technology and Engineering...

Kshitij answered on May 30 2021
143 Votes
Ans.1
MATLAB code
wp=pi/8; % lowpass filter bandwidth
M=121;
n=-(M-1)/2:(M-1)/2; % selection time window
h0=(wp/pi)*sinc((wp/pi)*n); % truncated impulse response
h=h0.*rectwin(M)'; % windowing

% h=h0.*hamming(M)';% windowing
% h=h0.*hanning(M)';% windowing
% h=h0.*bartlett(M)';% windowing
% h=h0.*blackman(M)';% windowing
figure;plot(h)
ylabel('Impulse response')
xlabel('Samples')
% spectrum
FFTsize=512;
pxx=20*log10(abs(fft(h,FFTsize)));
fxx=(0:(FFTsize/2)-1)*(pi/(FFTsize/2));
figure;plot(fxx,pxx(1:FFTsize/2));
ylabel('Amplitude[dB]');
xlabel('frequency [Radian]');grid on
1) Plot the impulse response and the amplitude spectrum of the filter:-
2.Frequecy measured at wp= pi/8 = 0.3927, is approximate -5.91 db which is very close to expected
value of -6 DB.
Maximum ripple is -85.45 db at frequency 1.559
3.
The ratio of the transient band to the pass band is approximate 3
4.
When we increases impulse response length m= 255 ,its ripple amplitude increase to -152.5
5.
When we are using Hanning window with the filter impulse response length, M=121
Maximum ripple value we obtaines is -173.7 db
When we are using hamming window with the filter impulse response length ,M=121
Maximum ripple measured value= 122.4
6.
When we have taken wp =pi/4
7.
MATLAB Program
%%
wp=pi/16; % lowpass filter bandwidth
M=121;
n=-(M-1)/2:(M-1)/2; % selection time window
h0=(wp/pi)*sinc((wp/pi)*n); % truncated impulse response
h1=h0.*rectwin(M)'; % windowing
w0=pi/4; % BPF center frequency
h=h1.*exp(j*w0*n); % Lowpass to bandpass conversion
figure;
subplot(211);plot(real(h));ylabel('Real part of impulse response')
xlabel('samples')
subplot(212);plot(imag(h));ylabel('imag part of impulse response')
xlabel('samples')
% spectrum
FFTsize=512;
pxx=20*log10(abs(fft(h,FFTsize)));
fxx=(0:(FFTsize/2)-1)*(pi/(FFTsize/2));
figure;plot(fxx,pxx(1:FFTsize/2));ylabel('Amplitude[dB]');xlabel('frequency
[Radian]');grid on
Ans7.
Ans.8
the passband edges wp1=0.5522 , wp2= 1.019...
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