# barcode generator for ssrs which is the sequence defined above. Therefore, the M-point DFT of y ( n ) is in Software Drawing Code 128 Code Set B in Software which is the sequence defined above. Therefore, the M-point DFT of y ( n ) is

which is the sequence defined above. Therefore, the M-point DFT of y ( n ) is
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The unit sample response of a single pole filter is h(n) = (\$)"u(n) The frequency response of this filter is sampled at wk = 2 n k / 1 6 for k = 0 , 1, . . . , 15. The resulting samples are G ( k ) = ~ ( e j ~ ) l ~ , z , k ~ l s k = 0 , 1, 2,. Find g(n), the 16-point inverse D F T of G ( k ) .
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The straightforward but tedious way to solve this problem would be to find the DTFT of h(n),
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sample it at the given frequencies,
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and then find the inverse DFT. Another approach is to use thefrequency sampling theorem given in Eq. (6.13),which states that if the DTFT of a sequence h(n) is sampled at N equally spaced frequencies between zero and 2rr, the sequence g ( n ) that has these samples as its DFT coefficients is the time-aliased sequence
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Therefore, with h(n) = (+)"u(n),it follows that
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Because u(n - k N ) is equal to zero for 0 5 n 5 N - 1 when k 2 0. this sum may be simplified to
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CHAP. 61 Evaluating the sum, we find
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Therefore,
The DFT of a sequence x(n) corresponds to N equally spaced samples of its z-transform, X(z), around the unit circle starting at z = 1. ( a ) If we want to sample the z-transform on a circle of radius r , how should x ( n ) be modified so that the DFT will correspond to samples of X(z) at the desired radius
(6) Suppose that we would like to shift the samples around the unit circle. In particular, consider the N samples that are equally spaced around the unit circle with the first sample at z = exp{jn/N). How should the sequence x(n) be modified so that the DFT will correspond to samples of X(z) at these points
( a ) The z-transform of x ( n ) is
If we sample X ( z ) at N equally spaced points around a circle of radius r , we have
which is the z-transform of r - " x ( n ) sampled at N equally spaced points around the unit circle. Therefore. N equally spaced samples of X ( z ) around a circle of radius r. may be found by computing the N-point DFT of r-"x(n).
( b ) Here, we want to rotate the DFT samples by an amount equal to T / N . In other words. we would like to find
Therefore, to find N samples of X ( z ) that are equally spaced around the unit circle, with the first sample at z = exp{jrr/N], we multiply x ( n ) by e-Jn"lN and find the N-point DFT of the resulting sequence.
Linear Convolution Using the DFT
Two finite-length sequences, x l ( n ) and xz(n), that are zero outside the interval [O, 991 are circularly convolved to form a new sequence y(n),
where N = 100. If xl (n)is nonzero only for 10 5 n 5 39, determine the values of n for which y(n) is guaranteed lo be equal to the linear convolution of x l (n) and .xz(n).
Because the values of n for which y ( n ) is equal to the linear convolution of x l ( n ) with .rz(n) are those values of n in the o interval 10.991 for which the circular shift x l ( ( n - k ) ) l ~ lis equal to the linear shift x l ( n - k ) . With x l ( n ) nonzero only over the interval [ l o , 391 we see that x , ( ( n - k))loO = x l ( n - k ) for n in the interval [39.99]. Therefore, the circular convolution and the linear convolution are equal for 39 5 n 6 99.
THE DFT
[CHAP. 6
We would like to linearly convolve a 3000-point sequence with a linear shift-invariant filter whose unit sample response is 60 points long. To utilize the computational efficiency of the fast Fourier transform algorithm, the filter is to be implemented using 128-point discrete Fourier transforms and inverse discrete Fourier transforms. If the overlap-add method is used, how many DFTs are needed to complete the filtering operation
With overlap-add, x(n) is partitioned into nonoverlapping sequences of length M . If h(n) is of length L, x,(n) * h(n) is of length L M - 1. Therefore, we must use a DFT of length N > L M - 1. Here, we have set N = 128, and h(n) is of length L = 60. Therefore, x(n) must be partitioned into sequences of length
Because x(n) is 3000 points long, we will have 44 sequences (with the last sequence containing only 33 nonzero values). Thus, to perform the convolution we need: 1. One DFT to compute H (k) 44 DFTs for X,(k) 44 inverse DFTs for Y,(k) = H(k)Xi(k)