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for a total of 45 DFTs and 44 inverse DFTs.
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Suppose that we are given 10 s of speech that has been sampled at a rate of 8 kHz and that we would like to filter it with an FIR filter h ( n ) of length L = 64. Using the overlap-save method with 1024-point DFTs, how many DFTs and inverse DFTs are necessary to perform the convolution
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Sampling LO s of speech with a sampling frequency of 8 kHz generates N = 10 .8OOO = 8 . lo4 samples x(n). If we segment the speech into records of length 1024 and perform the circular convolution of these segments with h(n), the first 63 values will be aliased, and the last 1024 - 63 = 961 values will be equivalent to the linear convolution. Therefore, each circular convolution generates 961 valid data points. Because the filtered signal y(n) = x(n) * h(n) is of length 8 . lo4 63 = 80,063, x(n) will be segmented into
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or 84 overlapping sections. Therefore, to perform the convolution, we need 85 DFTs and 84 inverse DFTs.
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If an 8-point sequence x ( n ) is convolved (linearly) with a 3-point sequence h ( n ) , the result is a 10-point sequence y(n) = x ( n ) r h(n). Suppose that we would like to construct the entire output y ( n ) from two 6-point circular convolutions:
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n = 0 , 1.2.3 n =4,5
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If the values of y l ( n ) and y2(n) are as tabulated below,
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find the sequence y ( n ) .
CHAP. 61
THE DFT
In this problem, note that x ( n ) has been partitioned into two 4-point sequences, x l ( n ) and x2(n). Because h ( n ) is of length three, the linear convolution of h ( n ) with x l ( n ) and x 2 ( n ) are both of length six. Therefore, the six-point circular convolutions are equal to the linear convolution, and y ( n ) = x ( n ) * h ( n ) is given by
which is tabulated below.
Applications
Consider a linear shift-invariant system characterized by the linear constant coefficient difference equation
Describe a method that may be used to plot N samples of the frequency response H(eJw) using N-point DFTs.
The unit sample response of the linear shift-invariant system that is described by this difference equation is infinite in duration. Finding N samples of the frequency response can be accomplished by computing the N-point DFT of the time-aliased signal
However, an easier method is as follows. The system function is
k= l
Therefore, samples of H ( z ) at N equally spaced points around the unit circle may be found as follows:
B(k) H (k) = A@)
where A ( k ) and B ( k ) are the N-point DFTs of the denominator and numerator sequences, respectively, that is,
In many applications, it is necessary to multiply a sequence by a window w ( n ) . Let x ( n ) be an N-point sequence, and let w ( n ) be a Hamming window:
How would you find the DFT of the windowed sequence, x ( n ) w ( n ) , from the DFT of the unwindowed sequence
THE DIT
Let us express the Hamming window in terms of complex exponentials:
[CHAP. 6
Therefore,
I x(n)w(n) = i x ( n ) - ,e j $ , t x(n) - :eplgnx(n)
Because the DFT of e ~ ~ ~ " l ~ xX((k)is ( n
and the DFT of e-j""lNx(n) is X((k
+l ) ) ~ ,
A signal x,(t) that is bandlimited to 10 kHz is sampled with a sampling frequency of 20 kHz. The DFT of N = 1000 samples of x ( n ) is then computed, that is.
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