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 />
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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