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Hz = 37.6 kHz in Software
Hz = 37.6 kHz Decoding ANSI/AIM Code 128 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128 Encoder In None Using Barcode encoder for Software Control to generate, create Code 128 Code Set A image in Software applications. 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 Code 128 Recognizer In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Generate Code 128A In C# Using Barcode encoder for .NET framework Control to generate, create USS Code 128 image in .NET framework applications. ( 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. Generate Code 128C In VS .NET Using Barcode generator for ASP.NET Control to generate, create Code 128 Code Set A image in ASP.NET applications. Code 128 Code Set B Generation In Visual Studio .NET Using Barcode generator for .NET framework Control to generate, create Code 128 Code Set A image in .NET applications. Find the unit sample response, h(n), of the network drawn below and find the 64point DFT of h(n). Drawing Code 128 Code Set A In VB.NET Using Barcode printer for .NET framework Control to generate, create Code 128B image in Visual Studio .NET applications. Generate Barcode In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. IMPLEMENTATION OF DISCRETETIME SYSTEMS
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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 64point 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 64point 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 DISCRETETIME 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 directform 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 lowpass filter with a downsampler as shown in the following figure: IMPLEMENTATION OF DISCRETETIME SYSTEMS
[CHAP. 8
Because the downsampler only retains one out of every M outputs from the lowpass 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 downsampler 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

