Digital Signal Processing Reference
In-Depth Information
Taking the z-transform on both sides of Equation (8.38) yields the digital filter as
HðzÞ¼Z½T$hðnÞ
(8.39)
The effect of the scale factor T in Equation (8.38) can be explained as follows. We approximate the
area under the curve specified by the analog impulse function hðtÞ using a digital sum given by
Z N
area ¼
hðtÞdt z T$hð 0 ÞþT$hð 1 ÞþT$hð 2 Þþ /
(8.40)
0
Note that the area under the curve indicates the DC gain of the analog filter while the digital sum in
Equation (8.40) is the DC gain of the digital filter.
The rectangular approximation is used, since each sample amplitude is multiplied by the sampling
interval T . Due to the interval size for approximation in practice, we cannot guarantee that the digital
sum has exactly the same value as the one from the integration unless the sampling interval T in
Equation (8.40) approaches zero. This means that the higher the sampling rate d that is, the smaller the
sampling interval d the more accurately the digital filter gain matches the analog filter gain. Hence, in
practice, we need to further apply gain scaling for adjustment if it is a requirement.
EXAMPLE 8.15
Consider the following Laplace transfer function:
HðsÞ¼ 2
s þ 2
a. Determine HðzÞ using the impulse-invariant method if the sampling rate f s ¼ 10 Hz.
b. Use MATLAB to plot the following:
1. The magnitude response jHðf Þj and phase response 4ðf Þ with respect to HðsÞ for the
frequency range from 0 to f s =2 Hz;
2. The magnitude response Hðe jU Þ ¼jHðe j2pfT
Þj and phase response 4ðf Þ with respect to HðzÞ for the
frequency range from 0 to f s =2 Hz.
Solution:
a. Taking the inverse Laplace transform of the analog transfer function, the impulse response is found to be
hðtÞ¼L 1 2
s þ 2
¼ 2e 2t uðtÞ
Sampling the impulse response hðtÞ with T ¼ 1=f s ¼ 0:1 second, we have
Th n ¼ T2e 2nT u n ¼ 0:2e 0:2n u n
Using the z-transform table in Chapter 5, we yield
Z h e an u n i
z
z e a
¼
And noting that e a
¼ e 0:2 ¼ 0:8187, the digital filter transfer function HðzÞ is finally given by
H z ¼
0:2z
z 0:8187 ¼
0:2
1 0:8187z 1
 
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