CHAP. 91

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The stopband cutoff frequency is w, = 0.4n, and the passband cutoff frequency is o,,= 0 . 5 ~ ~ . In addition, the stopband ripple is 8, = 0.0574, and the passband ripple is 6, = 0.1722. used Determine the weighting function, W ( e J W ) , to design this filter, and find the length of the unit sample response. Describe approximately where the zeros of the system function of this filter lie in the z-plane.

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To detennine the weighting function, we observe that

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Therefore, the weighting function used to design the filter has a value in the stopband that is 3 times larger than the value in the passband. This makes the errors in the stopband more costly and, therefore, smaller by a factor of 3. So, a weighting function that could have been used to design this filter is as follows:

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To determine the length of the unit sample response. recall that a type I equiripple high-pass (or low-pass) filter must either have L 2 or L 3 alternations where L = N / 2 . Therefore, the order of the filter may be detennined by counting the alternations. For this filter, we have nine alternations, which are labeled in the figure below.

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[CHAP. 9

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Thus, L = 7 or. in the case of an extraripple filter, L = 6. However, in order for h ( n ) to be an extraripple filter, w = 0 and w = IT must both be extremal frequencies (see Prob. 9.15). Because w = 0 is not an extremal frequency. this is no1 an extraripple filter. Therefore, L = 7 and N = 14.

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(b) Because the order of this filter is N = 14, the system function has 14 zeros. For a linear phase filter, we know that the zeros of the system function may lie on the unit circle, or they may occur in conjugate reciprocal pairs. From the plot of the frequency response magnitude, we see that I H ( e J ' " ) (= 0 at w l w 0.175n, w2 w 0.3n, . and w3 0 . 3 9 ~ Therefore, there are three zeros on the unit circle at these frequencies. Because there must also be zeros at the conjugate positions, z = e-'"'1 for i = 1, 2,3, these unit circle zeros account for six of the fourteen zeros. In addition to these. there must be a conjugate pair of zeros at z = re*'", where w4 0.71~. These zeros account for the dip in I H ( e l " ) I at w = 0.71~. Because the filter has linear phase, in addition to this pair of complex zeros, there must be a pair at the reciprocal locations, z - ' = re*jW4. For the same reason, there will be zeros on the real axis at z = a land z = I/cul,as well as zeros on the real axis at z = -a2 and z = -I/a2,where a , and a2 are positive real numbers that are less than I. These four zeros account for the minima in I H ( e J w at w = 0 and w = n . A plot showing the actual positions of the 14 zeros of H ( z ) is given ) below.

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With the frequency sampling method, the frequency samples match the ideal frequency response exactly. Derive an interpolation formula that shows how the frequency samples H (k) are interpolated. The frequency response of an FIR filter of length N is

If h ( n ) is designed using the frequency sampling method.

H ( k )=

,r=o

h(n)e-"n"klN - ~ , , ( e ~ ' " ~ / k = 0 , I . ~)