Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
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<strong>Introduction</strong> <strong>to</strong> <strong>Digital</strong> <strong>Signal</strong> <strong>and</strong> <strong>System</strong> <strong>Analysis</strong><br />
Frequency Domain <strong>Analysis</strong><br />
|H(W)|<br />
12<br />
10<br />
8<br />
6<br />
4<br />
2<br />
0<br />
0 0.5 1 1.5 2 2.5 3<br />
2<br />
1.5<br />
Phase<br />
1<br />
0.5<br />
0<br />
0 0.5 1 1.5 2 2.5 3<br />
0 ≤ W ≤ p<br />
Figure 4.7 Modulus <strong>and</strong> phase of frequency response<br />
4.5 Frequency correspondence when sampling rate is given<br />
Let x[n] be a signal discretized with a sampling rate of<br />
harmonic component with frequency<br />
transform is given by<br />
f 0<br />
Hz<br />
⎛ 2pf<br />
exp<br />
⎜ j<br />
⎝ f<br />
s<br />
f Hz, we are about <strong>to</strong> find the position of peak for a complex<br />
s<br />
0<br />
⎟ ⎞<br />
n in the frequency Ω (rad/sample) domain. Its Fourier<br />
⎠<br />
X ( W)<br />
=<br />
=<br />
i.e.<br />
∞<br />
∑<br />
n=−∞<br />
∞<br />
∑<br />
n=−∞<br />
⎛ ⎛ 2pf<br />
exp⎜<br />
j<br />
⎜<br />
⎝ ⎝ f<br />
s<br />
⎛ 2pf<br />
exp<br />
⎜ j<br />
⎝ f<br />
s<br />
0<br />
0<br />
⎞ ⎞<br />
− W ⎟<br />
⎟n<br />
⎠ ⎠<br />
⎞<br />
n<br />
⎟exp<br />
⎠<br />
( − jWn)<br />
2πf<br />
X ( Ω)<br />
= 2πδ<br />
Ω −<br />
f s<br />
0<br />
<br />
<br />
<br />
Therefore, the peak appears at<br />
f W = 2p 0<br />
f s<br />
f<br />
s<br />
/ 2<br />
the signal is f<br />
0<br />
= f<br />
s<br />
/ 2 , corresponding <strong>to</strong> W = 2 p = p (rad/sample).<br />
f<br />
. According <strong>to</strong> the Nyquist sampling theorem, the maximum frequency in<br />
s<br />
56<br />
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