barcode generator for ssrs THE DFT Evaluating X(k) at k = 0,we have the desired result in Software

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THE DFT Evaluating X(k) at k = 0,we have the desired result
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[CHAP 6
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Evaluate the sum
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The N -point DFT of x l (n) is zero, except fork = kl and k = N - kl. when it is equal to N/2. Similarly, the N-point DFT of x2(n) is zero, except fork = k2 and k = N - k2, when it is equal to N/2. Using the results of Prob. 6.20,
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we see that if kl = k2 (or k I = N
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and the sum is equal to zero otherwise.
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L e t x ( n ) be an N-point sequence with an N-point DFT X ( k ) . Derive Parseval's theorem,
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This property is an immediate consequence of the property derived in Prob. 6.20. Specifically, Parseval's theorem follows from Eq. (6.22) by setting xl(n) = x2(n) = x(n).
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Let x ( n ) be a sequence that is zero outside the interval [0, N - I ] with a z-transform X ( z ) . Listed below are four sequences of length 2N that are derived from x ( n ) . Find an expression for the DFT of each sequence in terms of samples of X ( z ) .
(c) y3(n) =
IX(')
n even n odd
(a) The 2N-point sequence y,(n) is formed from x(n) by padding with zeros. Therefore, Yl(k) corresponds to 2N equally spaced samples of X(z) around the unit circle:
(b) The sequence y2(n) is formed by adding x(n - N) to x(n) (i.e., a delayed version of x(n)). If X(z) is the ~ z-transform of x(n), the z-transform of x(n - N) is z - X(Z). Therefore,
CHAP. 61
and Y2(k) is
THE DFT
(c) The third sequence is formed by up-sampling by a factor of 2 (i.e., by stretching x(n) in time by a factor of 2 and inserting a zero between each sample). The z-transform of y3(n) is
Therefore, the 2N-point DFT is
Thus, the DFT coefficients Y3(k)correspond to two periods of the coefficients X(k).
Sampling the DTFT
Let h ( n ) be a finite-length sequence of length N with h ( n ) = 0 for n < 0 and n Fourier transform of h ( n ) is sampled at 3N equally spaced points:
> N. The discrete-time
Find the sequence g ( n ) that is the inverse DFT of the 3N samples H ( k ) = H ( e J w k ) .
Because h(n) is a finite-length sequence of length N, it may be recovered from its N-point DFT, which corresponds to N equally spaced samples of H (ei"). A sequence of length N may also be considered to be a sequence of length 3N, with the last 2N samples having a value of zero. The inverse DFT of the 3N equally spaced samples of H(eJU) corresponds to this 3N-point sequence. Thus, g(n) =
n=0, I....,N - 1 else
Consider the finite-length sequence
-4n) = [ I , 1, 1, 1 , 1, 11
and let X ( z ) be its z-transform. If we sample X ( z ) at zk = e x p ( j % k ) fork = 0, 1 , 2 , 3 , we obtain a set of DFT coefficients X ( k ) . Find the sequence, y ( n ) , that has a four-point DFT equal to these samples.
Sampling X(z) at four equally spaced points around the unit circle produces an aliased version of x(n):
Using the tabular method to evaluate this sum, noting that x(n) and x(n +4) are the only sequences that have nonzero values for 0 5 n 5 3, we have
Therefore.
THE D m
[CHAP. 6
Consider a finite-length sequence x ( n ) that is zero outside the interval 10. N - 11. Suppose that we form a new sequence i ( n ) as follows:
where M < N. Find the M-point DFT of the sequence i(n) O l n c M otherwise expressing the answer in terms of the D T F T of x(n). This problem is most easily solved if we take advantage of what we know about the DFT. Recall that the M-point , DFT corresponds to M samples of the DTFT, X ( e J U )at wi = 2 n k l M for k = 0, 1, . . ., M - 1. In addition, if these samples are used for the DFT coefficients of a sequence y ( n ) of length M, y ( n ) is related t o x ( n ) as follows:
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