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Therefore. the bandwidth of the input signal is limited to 18.8 kHz (i.e., X,( )must be zero for If I z 18.8 kHz). f
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( b ) Sampling speech at 8 kHz produces 8000 samples per second. Therefore, we have T, = 1/8000 = 0.125 ms to compute each output. This allows for M = (0.125 x 10-"/10-' = 1250 instruction cycles. Thus, we may implement an FIR filter of order N = 1250 - I 1 = 1239.
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Find the unit sample response, h(n), of the network drawn below and find the 64-point DFT of h(n).
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IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
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[CHAP 8
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This is a frequency sampling structure for an FIR filter with a unit sample response of length N = 64. Because the gain of the first-order section with a pole at z = I is equal to 1, H(0) = I. With second-order networks of the form
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A(k) = H (k) B(k) =
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- k)
+ H(N - k)eJ2nkJN
we see that H(1) and H(2) are nonzero, along with H(62) and H(63). We may therefore solve these equations for H(1), H(2), H(62), and H(63) as follows. Because A(I) = 2,
and because B(1) = 2cos(rc/32),
Thus, we have two equations in two unknowns, which may be written in matrix form as follows:
Solving these equations we find that H(1) = H(63) = 1. Similarly, with A(2) = 2 and B(2) = 2cos(n/32), we find that H(2) = H(62) = 1. Therefore, the 64-point DFT of h(n) is
H(k) =
k = 0. I . 2, 62.63
else
and the unit sample response is
h(') =
H(k) eJ2nt1"lh -
, + & cos nrr + & cos nrr ,
64 ,=o
Consider the FIR filter with unit sample response
The 64-point DFT of the unit sample response is
otherwise
Draw the frequency sampling structure for this filter and compare the computational complexity of this structure to a direct form realization.
H(k)=[;i
k=O k=l,63 else
Therefore, for the frequency sampling structure, we write the system function in the following form,
CHAP. 83
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
This leads to the frequency sampling structure shown below.
This implementation has 67 delays (4 more than the minimum), and it requires 3 multiplies and 6 adds to evaluate each output y(n). A direct-form realization, on the other hand. has 63 delays and, because h ( n ) has linear phase, requires 32 niultiplies and 63 adds to compute each output value.
The frequency sampling structure for an FIR filter is based on expressing the system function in the form
where H ( k ) are samples of the frequency response at wk = 2 7 r k / N . If h ( n ) is real, the symmetry of the DFT may be used to simplify this structure so that all of the coefficients are real. For example, Eq. (8.2) specifies a structure when N is even. Derive the corresponding structure when N is odd. If N is odd, we may write H(z) as follows:
where Note that, due to the conjugate symmetry of the DFT, H(k) = H*(N - k), the coefficients A(k) and B(k) are real.
As discussed in Chap. 3, sample rate reduction may be realized by cascading a low-pass filter with a down-sampler as shown in the following figure:
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
Because the down-sampler only retains one out of every M outputs from the low-pass filter H ( z ) , if M is large, most of the filter outputs will be discarded. Therefore, if H ( z ) is an FIR filter, it is not necessary to evaluate the discarded values, and efficient implementations of the decimator are possible.
(a) Assume that H ( z ) is an FIR filter with h ( n ) = 0 for n < 0 and n 2 N. If H ( z ) is implemented in direct form, draw a flowgraph for the decimator, and determine how many multiplications and additions are necessary to compute each output value y(n).
(b) Exploit the fact that only one out of every M values of w ( n )is saved by the down-sampler to derive a more efficient implementation of this system, and determine how many multiplications and additions are necessary to compute each value of y(n).
(c) If H ( z ) is an IIR filter, are efficient implementations of the decimator still possible If so, for which structures, and by what factor are they more efficient
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