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where is the correlation of hd(n).
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The inverse of a linear shift-invariant system with unit sample response g(n) and system function G ( z ) is the system that has a unit sample response, h(n),such that
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solution. One of the reasons is that, In most applications, the system function H ( z ) = I/G(z) is not a v~able unless G ( z )is minimum phase, I / G ( z )cannot be both causal and stable. Another consideration comes from the fact that, in some applications, it may be necessary to constrain H ( 2 ) to be an FIR filter. Because l / G ( z )will be infinite in length unless G ( z )is an all-pole filter, constraining h ( n )to be FIR would only be an approximation to the inverse filter. In the FIR least-squares inverse filter design problem. the goal is to find the FIR filter h ( n )of length N such that h ( n )* ~ ( n ) S(n) The filter that minimizes the squared error
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where may be found by solving the linear equations
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where In many cases, constraining the least-squares inverse filter to minimize the difference between h ( n )* ~ ( n ) and S(n) is overly restrictive. For example. if a delay may be tolerated, we may consider finding the filter h ( n ) so that h ( n )* ~ ( n ) 6(n - no) for some delay no. In most cases, a nonzero delay will produce a better approximate inverse filter and, in many cases, the improvement will be substantial. The least-squares inverse filter with delay is found by solving the linear equations
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FIR Filter Design 9.1
Use the window design method to design a linear phase FIR filter of order N = 24 to approximate the following ideal frequency response magnitude:
The ideal filter that we would like to approximate is a low-pass filter with a cutoff frequency w, = 0 . 2 ~ .With N = 24, the frequency response of the filter that is to be designed has the form
Therefore, the delay of h ( n ) is a = N / 2 = 12. and the ideal unit sample response that is to be windowed is
All that is left to do in the design is to select a window. With the length of the window fixed, there is a trade-off between the width of the transition band and the amplitude of the passband and stopband ripple. With a rectangular window, which provides the smallest transition band,
and the filter is
otherwise
However. the stopband attenuation is only 21 dB. which is equivalent to a ripple of 6, = 0.089. With a Hamming window, on the other hand,
and the stopband attenuation is 53 dB, or 6 , = 0.0022. However, the width of the transition band increases to
which, for most designs, would be too wide.
Use the window design method to design a minimum-order high-pass filter with a stopband cutoff frequency w, = 0.22n, a passband cutoff frequency w,, = 0.28n, and a stopband ripple 6 , = 0.003.
A stopband ripple of 6, = 0.003 corresponds to a stopband attenuation of a,= -20 log 6, = 50.46. For the minimumorder filter. we use a Kaiser window with
Because the transition width is Aw = 0.06n, or A f = 0.03. the required window length is
CHAP. 91
FILTER DESIGN
38 1
Rounding this up to N = 99 results in a type I1 linear phase filter, which will have a zero in its system function at z = -1. Because this produces a null in the frequency response at w = n , this is not acceptable. Therefore, we increase the order by I to obtain a type 1 linear phase filter with N =: 100. to In order to have a transition band that extends from o,= 0 . 2 2 ~ w, = 0.28n, we set the cutoff frequency of the ideal high-pass filter equal to the midpoint:
The unit sample response of an ideal zero-phase high-pass filter with a cutoff frequency w,. = 0 . 2 5 ~ is
where the second term is a low-pass filter with a cutoff frequency w,. = 0.25n. Delaying hh,(n) by N / 2 = 50, we have
and the resulting FIR high-pass filter is h(n) = hd(n). w(n) where w(n) is a Kaiser window with N = I00 and B = 4.6.
Given a desired frequency response Hd(eJ"), show that the rectangular window design minimizes the least-squares error
For this problem, we use Parseval's theorem to express the least-squares error ELs in the time domain:
If we assume that h(n) is of order N , with h(n) = 0 for n < 0 and n z N ,
Because the last two terms are constants that are not affected by the filler h(n), the least-squares errorELs is minimized by minimizing the first term, which is done by setting h(n) = hd(n) for n = 0, I . . . . N (i.e., using a rectangular window in the window design method).
If hd(n) is the unit sample response of an ideal filter, and h ( n ) is an N th-order FIR filter, the least-squares error
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