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Therefore, the DFT coefficients are ;N
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Suppose that we are given a program to tind the DFT of a complex-valued sequence x ( n ) . How can this program be used to find the inverse DFT of X ( k )
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A program to find the DFT of a sequence x ( n ) evaluates the sum
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and produces the sequence of DFT coefficients X(k). What we would like to do is to use this program to find the inverse DFT of X ( k ) , which is
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Note that the only difference between the forward and the inverse DFT is the factor of 1 / N in the inverse DFT and the sign of the complex exponentials. Therefore, if we conjugate both sides of Eq. (6.20)and multiply by N , we have
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Comparing this to Eq. (6.19), we see that the sum on the right is the DFT of the sequence X * ( k ) . Thus, if X * ( k ) is used as the input in the DFT program, the output will be N x * ( n ) . Conjugating this output and dividing by N produces the sequence x ( n ) . Therefore, the procedure is as follows:
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Conjugate the DFT coefficients X ( k ) to produce the sequence X*(k). Use the program to find the DFT of the sequence X*(k). Conjugate the result obtained in step 2, and divide by N .
DFT Properties 6.10
Consider the finite-length sequence
x ( n ) = S(n)
+ 2S(n
(a) Find the 10-point discrete Fourier transform of x ( n ) . (6) Find the sequence that has a discrete Fourier transform
Y(k)=e j 2 k % ~ ( k )
where X ( k ) is the 10-point DFT of x ( n ) .
Find the 10-point sequence y ( n ) that has a discrete Fourier transform
Y ( k )=X(k)W(k)
where X ( k ) is the 10-point DFT of x ( n ) , and W ( k ) is the 10-point DFT of the sequence
w(n)= 1
Osn16 otherwise
CHAP. 61
( a ) The DFT of x(n) is easily seen to be
THE DFT
( b ) Multiplying X ( k ) by a complex exponential of the form w:' corresponds to a circular shift of x ( n ) by no. In this case, because no = -2, x ( n ) is circularly shifted to the left by 2, and we have
(c) Multiplying X ( k ) by W ( k )corresponds to the circular convolution of x ( n ) with w(n). To perform the circular
convolution, we may find the linear convolution and alias the result. The linear convolution of x ( n ) with w ( n ) is
and the circular convolution is
Because z(n) and z(n 10) are the only two sequences in the sum that have nonzero values for 0 5 n c 10, using a table to list the values of z(n) and z(n lo), and summing for n = 0, 1,2, . . . , 9 , we have
Thus, the 10-point circular convolution is
y(n)=I3.3,l,l.1,3.3,2,2,21
Consider the sequence x ( n ) = 4S(n) Let X ( k ) be the six-point DFT of x ( n ) .
+ 3S(n - 1) + 2S(n - 2 ) + 6 ( n
(a) Find the finite-length sequence y ( n ) that has a six-point IlFT
Y( k )= w i k x ( k ) (6) Find the finite-length sequence w ( n ) that has a six-point DFT that is equal to the real part of X ( k ) ,
W ( k )= R e ( X ( k ) )
(c) Find the finite-length sequence q ( n ) that has a three-point DFT
Q ( k )= X(2k)
k = 0, 1,2
( a ) The sequence y(n) is formed by multiplying the DFT of x(n) by the complex exponential W:k. Because this corresponds to a circular shift of x ( n ) by 4,
THE DFT The real part of X ( k ) is
[CHAP. 6
R e { X ( k ) J= f [ X ( k )
+ X*(k)l
To find the inverse DFT of R e { X ( k ) ) ,we need to evaluate the inverse DFT of X * ( k ) . Because
X * ( k ) is the DFT of .w*((N - n ) ) N .Therefore, the inverse DFT of R e { X ( k ) Jis
With N = 6, this becomes
w(n)=[4,
I. I. I ,
The sequence q ( n ) is of length three with a DFT Q ( k ) = X ( 2 k ) fork = 0, 1, 2 where X ( k ) is the six-point DFT of x ( n ) . Because the coefficients X ( k ) are samples of X ( z ) at six equally spaced points around the unit circle, X ( 2 k ) fork = 0. 1.2 corresponds to three equally spaced samples of X ( z ) around the unit circle. Therefore,
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