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Filter design by impulse invariance
In the impulse variance design procedure the impulse response of the impulse response of the discrete time system is proportional to equally spaced samples of the continues time filter, i.e.,
where T_{d} represents a sampling interval, since the specifications of the filter are given in discrete time domain, it turns out that T_{d} has no role to play in design of the filter. From the sampling theorem we know that the frequency response of the discrete time filter is given by
Since any practical continuous time filter is not strictly bandlimited there issome aliasing. However, if the continuous time filter approaches zero at high frequencies, the aliasing may be negligible. Then the frequency response of the discrete time filter is
We first convert digital filter specifications to continuous time filter specifications. Neglecting aliasing, we get specification by applying the relation
(9.2)
where is transferred to the designed filter H(z), we again use equation (9.2) and the parameter T_{d}cancels out.
Let us assume that the poles of the continuous time filter are simple, then
The corresponding impulse response is
Then
The system function for this is
We see that a pole at s= s_{k} in the splane is transformed to a pole at z = e ^{sk }T_{d} in the zplane. If the continuous time filter is stable, that is Re {s_{k}} < 0 then the magnitude of e^{s}k^{T}d will be less than 1, so the pole will be inside unit circle. Thus the causal discrete time filter is stable. The mapping of zeros is not so straight forward.
Example:
Design a lowpass IIR digital filter H(z) with maximally flat magnitude characteristics. The passband edge frequency with a passband ripple not exceeding 0.5dB. The minimum stopband attenuation at the stopband edge frequency is 15 dB.
We assume that no aliasing occurs. Taking T_{d} = 1 , the analog filter has , the passband ripple is 0.5dB, and minimum stopped attenuation is 15dB. For maximally flat frequency response we choose Butterworth filter characteristics. From passband ripple of 0.5 dB we get
at passband edge.
From this we get
From minimum stopband attenuation of 15 dB we get
at stopped edge A^{2} = 31.62
The inverse discrimination ratio is given by
and inverse transition ratio 1/k is given by
Since N must be integer we get N=4. By we get
The normalized Butterworth transfer function of order 4 is given by
This is for normalized frequency of 1 rad/s. Replace s by from this we get
Bilinear Transformation
This technique avoids the problem of aliasing by mapping axis in the splane to one revaluation of the unit circle in the zplane.
If H_{a}(s) is the continues time transfer function the discrete time transfer function is detained by replacing s with
(9.3)
Rearranging terms in equation (9.3) we obtain.
Substituting , we get
If , it is then magnitude of the real part in denominator is more than that of the numerator and so. Similarly if , than for all. Thus poles in the left half of the splane will get mapped to the poles inside the unit circle in zplane. If then
So, we get
rearranging we get
or
(9.5)
or
(9.6)
The compression of frequency axis represented by (9.5) is nonlinear. This is illustrated in figure 9.4.
Because of the nonlinear compression of the frequency axis, there is considerable phase distortion in the bilinear transformation.
Example
We use the specifications given in the previous example. Using equation (9.5) with T_{d} = 2 we get
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