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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 />

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

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