Introduction to Digital Signal and System Analysis - Tutorsindia

Introduction to Digital Signal and System Analysis
Z Domain Analysis
2
y [ n]
= 2r
cosq
y[
n −1]
− r y[
n − 2] + x[
n]
By evaluating the impulse response, it can be known that when the numbers of poles and zeros are equal, there is no
longer a time delay.
5.7 Evaluation of the Fourier transform in the z-plane
Let z = exp(jW), i.e. as the phase angle W varies, but its modulus z = 1 keeps on the unit circle and takes values on the
circle. From this viewpoint, the Fourier transform is a special z-transform.
Example 5.8 A z-transform function
Its frequency response is
The magnitude
Therefore,
H ( W)
=
exp
H ( W)
=
exp
is given by
( jW)
− 0.8
( jW) + 0. 8
2
2
{( cosW − 0.8) + ( sin W)
}
2
2
( cosW + 0.8) + ( sin W)
{ } 1/ 2
z − 0.8
H ( z)
= , z + 0.8
1/ 2
1 − 0.8
H ( 0) = H (2p
) = = 0.1.1111
1 + 0.8
and
−1
− 0.8
H ( p ) = = 9.0
−1
+ 0.8
Graphically, referring to the following Figure 5.6 (b), let
z
1
p
1
= exp( jW)
− 0.8
= exp( jW)
+ 0.8
it can be seen that the quotient ( z
H =
p
. This has explained the frequency selective property of a high pass.
W ) 1
1
reaches the minimum at = 1
0
1
W , and reaches the maximum at = p
W 1
75
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