Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
<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 />
In general cases, a system can be described by the difference equation<br />
N<br />
<br />
k=<br />
0<br />
a y[<br />
n − k]<br />
=<br />
k<br />
M<br />
<br />
k=<br />
0<br />
b x[<br />
n − k]<br />
k<br />
Take Fourier transform for both sides,<br />
N<br />
<br />
k=<br />
0<br />
a<br />
k<br />
exp( − jkΩ)<br />
Y ( Ω)<br />
=<br />
M<br />
<br />
k=<br />
0<br />
b exp( − jkΩ)<br />
X ( Ω)<br />
k<br />
or<br />
Y ( Ω)<br />
N<br />
<br />
k = 0<br />
a<br />
k<br />
exp( − jkΩ)<br />
= X ( Ω)<br />
M<br />
<br />
k = 0<br />
b<br />
k<br />
exp( − jkΩ)<br />
Therefore, the frequency response can be obtained by<br />
Y ( W)<br />
H ( W)<br />
= =<br />
X ( W)<br />
M<br />
∑<br />
k=<br />
0<br />
N<br />
∑<br />
k=<br />
0<br />
b exp( − jk W)<br />
Example 4.7: A digital high-pass filter is described by<br />
a<br />
k<br />
k<br />
exp( − jk W)<br />
(4.8)<br />
y[n]= - 0.8 y[n-1] + x[n]- x[n-1]<br />
Find the frequency response <strong>and</strong> sketch its magnitude <strong>and</strong> phase over the range 0