P813 Suppose That the Following Continuous Time Signal XT 6 Cos42pi T is Sampled With
Chapter 1
Q. 1.8.26
Let the continuous-time signals:
x_{1}(t)=\cos (6\pi t),x_{2}(t)=\cos (14\pi t), x_{3}(t)=\cos (26\pi t) and y(t) = x_{1}(t)+x_{2}(t)+x_{3}(t).
a. For sampling frequency of 1KHz, sketch the samples of the given signals.
b. For sampling frequencies of 10Hz and 100Hz, sketch the samples of the given signals. Also, sketch the signal resulting from the reconstruction (based on the sampling theorem).
Step-by-Step
Verified Answer
a. The maximum frequency of the signal y(t) is 13Hz. We will sketch the samples of the given signals for sampling frequency 1000Hz, which is clearly greater than 2⋅13Hz. Because we chose a sampling frequency that exceeds the minimum limit set by Nyquist, we are able not only to reconstruct the original signal, but also to produce much better graphs.
The MATLAB code follows:
MATLAB Verified Solution
Script Files
fs = 1000; % set sampling frequency 1000Hz
ts = 1/fs; % compute sampling period
n = 0:ts:1; %set time interval
x1 = cos(6*pi*n);
x2 = cos(14*pi*n);
x3 = cos(26*pi*n);
y = x1+x2+x3;
subplot(411),plot(n,x1);
subplot(412),plot(n,x2);
subplot(413),plot(n,x3);
subplot(414),plot(n,y);title('Final signal y(t) – fs=1000Hz')
b. If we choose as a sampling frequency 10Hz < 2⋅13Hz, using the appropriate code in MATLAB, we will get a result that is going to differ from the previous one due to the aliasing effect.
fs1 = 10; % Set sampling frequency 10Hz
ts1 = 1/fs1; % compute sampling period
n1 = 0:ts1:1; %set time interval
x11 = cos(6*pi*n1);
x12 = cos(14*pi*n1);
x13 = cos(26*pi*n1);
y1 = x11+x12+x13;
subplot(411),plot(n1,x11);
subplot(412),plot(n1,x12);
subplot(413),plot(n1,x13);
subplot(414),plot(n1,y1);title('Final signal y1(t) – fs1=10Hz')
With the following commands we will sketch the original signal y(t) and the one produced from its reconstruction after setting 10Hz as the sampling frequency.
The original signal will be reconstructed based on the sampling theorem.
Having in mind the relation y_{a}(t)=\sum\limits_{n=-\infty }^{\infty }{y_{a}(nT_{s})\sin c[f_{s}(t-nT_{s})]} where y_{a}(t) is the analog reconstructed signal, and y_{a}[nT_{s}] is the sampled signal, we get:
for i = 1:length(y)
yrec(i) = sum(y1.*sinc(fs1*(n(i)-n1)));
end
subplot(211);plot(n,y);
subplot(212);plot(n,yrec);
We notice that the reconstructed signal has no resemblance to the original due to overlapping. The signal resulting from the reconstruction is a cosine of 3Hz, with an amplitude of 3.
The spectrum of the sampled signal is formed by shifting the original spectrum by integer multiples of the sampling frequency and by adding up the results, that is Y_{s}(f)=\frac{1}{T_{s}} \sum\limits_{k=-\infty }^{\infty }{Y(f+kf_{s})}. The function Y_{s}(f) is periodic with a period, fs. Consider an interval [–f_{s}/2,f_{s}/2] of Y_{s}( f ), namely [–5,5]. We will find out what spectral components are present in this interval, and because of the periodicity we will have absolutely defined Y_{s}(f).
Because of the component x_{1}(t)=\cos(6πt)=\cos(2⋅pi⋅3⋅t) in the original signal, there will be a frequency component with a frequency of 3Hz in the sampled signal.
This component corresponds to Y_{s}(f)=\frac{1}{T_{s}} \sum\limits_{k=-\infty }^{\infty }{Y(f+kf_{s})} for k = 0 (zero shift), and it is in [–5,5] interval.
The component x_{2}(t)=\cos(14πt)=\cos(2⋅pi⋅7⋅t)is out, but its shift \cos(2⋅pi⋅(10 – 7)⋅t) =\cos(2⋅pi⋅3⋅t) is in the interval [–5,5] and is of frequency 3Hz. The component x_{3}(t) = \cos(26πt)=\cos(2⋅pi⋅13⋅t) is out, but its shift \cos(2⋅pi⋅(13 – 7⋅t) =\cos(2⋅pi⋅3⋅t) is in the interval [–5,5] and is of frequency 3Hz.
So the wrong sampling with f_{s} = 10Hz < 2⋅13Hz turned the frequency signals of 7Hz and of 13Hz to signals of 3Hz. That is the reason that the signal at the output is a cosine of amplitude 3 (as a result of the superposition of these three signals).
Let us choose as sampling frequency 100Hz > 2⋅13Hz.
fs1 = 100; % set sampling frequency 10Hz
ts1 = 1/fs1; % compute sampling period
n1 = 0:ts1:1; %set time interval
x11 = cos(6*pi*n1);
x12 = cos(14*pi*n1);
x13 = cos(26*pi*n1);
y1 = x11+x12+x13;
subplot(411),plot(n1,x11);
subplot(412),plot(n1,x12);
subplot(413),plot(n1,x13);
subplot(414),plot(n1,y1);title('Final signal y1(t) – fs1=100Hz')
The reconstructed signal will be the same as the original:
for i = 1:length(y)
yrec(i) = sum(y1.*sinc(fs1*(n(i)-n1)));
end
subplot(211);plot(n,y);
subplot(212);plot(n,yrec);
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