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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 />

Table 4.1: Fourier Transform Properties <strong>and</strong> Pairs<br />

<strong>Signal</strong> x [n]<br />

Fourier Transform X (Ω)<br />

Property<br />

ax [ n]<br />

+ by[<br />

n]<br />

aX ( Ω)<br />

+ bY ( Ω)<br />

Linearity<br />

x n − n ]<br />

X ( Ω )exp( − jΩn<br />

)<br />

Time-shift<br />

[<br />

0<br />

x[ n]<br />

∗ y[<br />

n]<br />

X ( Ω)<br />

Y ( Ω)<br />

Convolution<br />

x [ n]<br />

y[<br />

n]<br />

X ( Ω) * Y ( Ω)<br />

Modulation<br />

δ [n]<br />

1<br />

δ [ n − n0<br />

]<br />

exp( − jΩn 0<br />

)<br />

u [n]<br />

−1<br />

( ) ∞ 1−<br />

exp( − Ω)<br />

+<br />

= −∞<br />

j πδ ( Ω − 2πk<br />

)<br />

a n −<br />

u[ n]<br />

a < 1<br />

( 1−<br />

a exp( − jΩ)<br />

) 1<br />

x[ n]<br />

= 1 n ≤ m<br />

sin{ ( m + 1/ 2)<br />

Ω}<br />

x[<br />

n]<br />

= 0 n > m<br />

sin( Ω / 2)<br />

k<br />

0<br />

Example 4.6: Find the frequency response from the impulse responses:<br />

a) h[n]=0.2 {d[n-2]+ d[n-1]+ d[n]+ d[n+1]+ d[n+2]}<br />

Taking the Fourier transform, the frequency response is<br />

H ( W)<br />

=<br />

∑ ∞<br />

n=<br />

−∞<br />

h[<br />

n]exp<br />

( − jWn)<br />

= 0.2{exp( − j2W)<br />

+ exp( − jW)<br />

+ 1+<br />

exp( jW)<br />

+ exp( j2W)}<br />

= 0.2{1 + 2cosW + 2cos 2W}<br />

H ( W)<br />

= 0.21+<br />

2 cos W + 2 cos 2W<br />

<strong>and</strong> the phase Φ H<br />

( W) = 0 .<br />

51<br />

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