Kulliyyah of Engineering, IIUM Department of Electrical and Electronics Engineering
Digital Signal Processing Lab Semester II, II, 2016/2017, Due Date: March 21, 2017 2017 Lab 2. Continuous Time to Discrete Time Conversion (Sampling and Aliasing) Introduction:
In order to perform digital processing of analog signals, you need to do analog to discrete conversion. The first step in A/D conversion is CT to DT conversion, which is the objective of this lab. Nyquist-Shannon sampling theorem: States that perfect reconstruction of a signal is 1: possible when the sampling frequency is greater than the twice the maximum frequency of the signal being sampled. If lower sampling rates are used, the original signal’s information may not be completely recovered from the sampled signal. For example if a signal has an upper band limit of 100 hz, a sampling frequency greater than 200 hz will avoid aliasing and allow theoretically perfect reconstruction. Fs >= 2 Fmax Aliasing Effect: Aliasing is the phenomenon that results in loss of information when a 2: CT signal is reconstructed from its sampled values. In principle the original analog signal can be reconstructed back from the samples and aliasing can be avoided, provided that Nyquist sampling theorem was not violated while sampling i.e. the sampling rate was at least twice the frequency of the highest frequency components present in the signal. Sampling of Sinusoidal signal: %original analog signal t=0:0.0005:1; F=13; xa=cos(2*pi*F*t); subplot(2,1,1) plot(t,xa); grid on on;; xlabel('Time, xlabel( 'Time, sec') sec') ylabel('Amplitude' ylabel( 'Amplitude')) title('continuous title('continuous time original signal' ) axis([0 1 -1.2 1.2]) % sampled signal T=0.1; n=0:T:1; xs=cos(2*pi*F*n); k=0:length(n)-1; subplot(2,1,2) stem(k,xs); grid on on;; xlabel('Time xlabel( 'Time index n') n')
ylabel('Amplitude' ylabel( 'Amplitude')) title('Sampled title('Sampled signal') signal' ) axis([0 length(n)-1 length(n)-1 -1.2 1.2]); After execution the following subplot appears:
Verification of Nyquist Theorem by taking user input: %original analog signal f=input('Give f=input('Give original signal frequency in hertz, f='); f=' ); t=0:0.005:1; xa=cos(2*pi*f*t); subplot(2,1,1) plot(t,xa); grid on on;; xlabel('Time, xlabel( 'Time, sec') sec') ylabel('Amplitude' ylabel( 'Amplitude')) title('continuous title('continuous time original signal' ) axis([0 1 -1.2 1.2]) % sampled signal when sampling rate is less than Nyquist criteria fs1=input('Enter fs1=input('Enter sampling frequency, less than twice the maximum signal frequency in hertz, fs1=')) fs1=' T1=1/fs1; n=0:T1:1; w=2*pi*f; xs1=cos(w*n); k=0:length(n)-1;
subplot(2,1,2) stem(k,xs1); grid on on;; xlabel('Time xlabel( 'Time index n') n') ylabel('Amplitude' ylabel( 'Amplitude')) title('Sampled title('Sampled signal with Fs<=2Fma Fs<=2Fmax' x')) axis([0 length(n)-1 -1.2 1.2]); After execution enter following frequencies: Give original signal frequency in hertz, f=15 Enter sampling frequency, less than twice tw ice the maximum signal frequency in hertz, fs1=25
Aliasing Effect: %original analog signal t=0:0.0005:1; F=13; xa=cos(2*pi*F*t); subplot(2,1,1) plot(t,xa); grid on on;; xlabel('Time, xlabel( 'Time, sec') sec') ylabel('Amplitude' ylabel( 'Amplitude')) title('continuous title('continuous time original signal' ) axis([0 1 -1.2 1.2]) % Reconstructed signal with aliasing effect T=1/(13*1.5); n=(0:T:1)'; xs=cos(2*pi*F*n); t=(-0.5:0.004:1.5)'; %Reconstruction low pass filter, implemented in time domain
Ya=sinc((1/T)*t(:,ones(size(n)))-(1/T)*n(:,ones(size(t)))')*xs; subplot(2,1,2) plot(n,xs,'o' plot(n,xs,'o',t,Ya); ,t,Ya); grid on on;; xlabel('Time xlabel( 'Time seconds') seconds' ) ylabel('Amplitude' ylabel( 'Amplitude')) title('Reconstructed title('Reconstructed signal') signal' ) axis([0 1 -1.2 1.2]) After simulation, following plot will show up.
Question 1: In the above sampling topic plot the sampled signal for the remaining two cases. 1: Take sampling frequency exact 2 times the original frequency. 2: Take sampling frequency 2.5 times the maximum frequency. Question 2: Reconstruct the original signal without aliasing. Take sampling frequency of your own choice.
Question 3: I.
Oversampling.
a. Using MATLAB, create a sinusoidal signal with the frequency 1000 Hz. b. Sample it at 8000 Hz and collect 512 (28) samples. c. Using a subplot with 2 rows and 2 columns, plot the waveform in the subplot on the first row and first column. d. Set the axis so that you get three waves. e. Convert from the time domain to frequency domain by taking the Fourier Transform. f. Plot the FFT vs. frequency.
f=1000; fs=8000; Ts=1/fs; Ns=512; t= [0:Ts:Ts*(Ns-1)]; [0:Ts:Ts*(Ns-1)] ; x=sin(2*pi*f*t); x=sin(2*pi* f*t); subplot(2,2,1),plot(t,x) grid on; axis([0,.003,-1.2,1.2]) axis([0,.003,-1.2,1.2]) X=(abs(fft(x,Ns))); X=(abs(fft(x ,Ns))); y=X(1:length(X)/2); f=(1:length(y)); subplot(2,1,2),plot(f*fs/Ns,y); grid on;
%frequency of the sine wave %sampling frequency %sampling time %512 points %50/8000~6.25ms %50/8000~6.25 ms %sine function for 50 samples %plot of the sine wave %set the x,y axis %take the FFT
II. Undersampling.
1. 2. 3. 4. 5.
Repeat part(I) with the frequency 7000hz. Are the graphs the same? Can you explain the reason? Which plot shows the effect of aliasing? a liasing? How can you tell?
Question 4:
1. Store in the vector x the samples of the chirp signal on the interval 0≤t≤1, and let T is the same value as above. 2. Use sound(x,fs) to play the chirp signal contained in x. Can you explain what you just heard? 3. Can you predict the time at which the played signal has zero (or very low) frequency? Use a longer x sequence to verify your prediction.
f=3000; %frequency of the sine wave fs=8192; %sampling frequency Ts=1/fs; %sampling time Ns=512; %512 points t= [0:Ts:Ts*20*(fs-1)]; [0:Ts:Ts*20*(fs-1 )]; %50/8000~6.25ms %50/8000~6.25 ms x=sin(2*pi*f*t+0.5*2000*t.^2);%sine function for 50 samples sound(x,fs) %play the wave
Question 5: A sinusoid type signal can be generated as follows: % Generates a 40Hz cosine wave % N = 64; % 64 data points fs = 1000; % 1kHz sampling frequency T = 1/fs; % sample period f = 40; % signal frequency n = 0:N-1; % discrete time vector x = cos(2*pi*f*n*T); %signal definition %plot the signal against time stem(n*T,x); title('40Hz cosine') xlabel('Time') ylabel('x(nT)')
5.1 Enter the code in a script file named sineGen.m and verify that what you get on the plot really is a sampled 40Hz cosine wave. You should be able to identify the outline of the sinusoid. 5.2 Now plot the signal using the plot function. This connects the points with a straight line and can be considered to be a form of “signal reconstruction”. It is not optimal but it is sufficient here. 5.3 Modify the code to generate sampled cosines at signal frequencies of 200Hz, 400Hz, 600Hz and 800Hz, with durations of 1 second but still a sampling frequency, fs, of 1kHz. In each case, generate plots of the first few oscillations of the signals using stem and plot on the same figure. 5.4 Add a plot of the signal sampled at 10000Hz (this is highly over sampled and can be viewed as a good approximation to the analogue signal). You can also ‘listen’ to the signal using the command: soundsc(interp(x,8),8*fs); This plays the signal ‘x’ assuming a sampling frequency of fs (the code actually interpolates the signal and uses a sampling frequency of 8*fs). Can you explain what you hear?
