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q ( n )= 56(n)
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+ 36(n - 1 ) + 2S(n - 2 )
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x ( n ) = S(n)
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+ 2 6 ( n - 2 ) + S(n - 3 )
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(a) Find the four-point DFT of x ( n ) .
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(b) If y ( n ) is the four-point circular convolution of x ( n ) with itself, find y ( n ) and the four-point DFT Y(k).
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(c) With h ( n ) = S(n)
+ S(n - I) + 2 6 ( n - 3 ) , find the four-point circular convolution of x ( n ) with h ( n ) .
( a ) The four-point DFT of x ( n ) is
X ( I )= C x ( n ) w ; ' L I
,,=O
+~ w F ~ , 7 " +
(h) With y ( n ) = x ( n ) @ x ( n ) , it follows that Y ( k ) = x 2 ( k ) : Y ( k ) = (1 + 2 ~ , 2 k + ~ ; ' ) ( l + 2 ~ J 2 k + ~ J 3 1 )
+ 4w,Zk + 2wy + 4w44k+ 4wqS1 + wp
the expression for Y ( k ) may be simplified to
Therefore,
y ( n ) = 5 6 ( n ) 4 6 ( n - I)
+ 56(n - 2 ) + 26(n
CHAP. 61
( c ) With h(n) = 6(n)
T H E DFT
6(n - I ) 26(n - 3), the four-point circular convolution of x ( n ) with h ( n ) may be found using the tabular method. Because, the linear convolution of x ( n ) with h ( n ) is
then
Let x ( n ) be Lhe sequence
x ( n ) = 26(n)
+ S(n - 1 ) + tS(n - 3 )
The five-point DFT of x ( n ) is computed and the resulting sequence is squared:
Y(k) = ~ ' ( k )
A five-point inverse DFT is then computed to produce the sequence y ( n ) . Find the sequence y ( n ) .
The sequence y ( n ) has a five-point DFT that is equal to the product Y ( k ) = X ( k ) X ( k ) .Therefore, y ( n ) is the tivepoint circular convolution of x ( n ) with itself:
A simple way to evaluate this circular convolution is to perform the h e a r convolution y f ( n )= x ( n ) * x ( n ) and alias the result:
The linear convolution of x ( n ) with itself is easily seen to be
Using the tabular method for computing the circular convolution, we have
Therefore,
y(n) = 4S(n)
+ 56(n
+ S(n
- 2)
+ 4iS(n
+ 260) - 4 )
Consider the two sequences
x ( n ) = 6(n)
+ 36(n - 1) + 36(n 2 ) + 26(n h ( n ) = 6 ( n ) + S(n - 1) + S(n - 2 ) + S(n 3 )
If we form the product
Y(k)= X(k)H(k)
where X ( k ) and H ( k ) are the five-point DFTs of x ( n ) and h ( n ) , respectively, and take the inverse DFT to form the sequence y ( n ) , find the sequence y ( n ) .
THE DFT
[CHAP. 6
Because Y(k) is the product of two 5-point DFTs, H(k) and X(k), y(n) is the five-point circular convolution of h(n) with x(n). We may find ,y(n) by performing the circular convolution analytically (or graphically) or by finding the linear convolution and aliasing the result or by multiplying DFTs and finding the inverse DFT. In this problem, because h(n) is a simple sequence, we will use the analytic approach. The five-point circular convolution of .r(n) with h(n) is
Because h(n) = I for n = 0, 1. 2, 3, and h(4) = 0, the five-point convolution is
Therefore, the circular convolution is equal to the sum of the values of the circularly shifted sequence x((n from k = 0 to k = 3. Because a ( n ) is
k))~
(recall lhat .u(n) is considered to be a sequence of length five), .r((-n))s is formed by reading the sequence values
backward, beginning with n = 0:
Thus, y(0) is the sum of the first four values of ~ ( ( - n ) ) which gives v ( O ) = 6 . Circularly shifting this sequence to ~, the right by I, we have
and summing the first four values gives y ( l ) = 6. Continuing with this process, we find y(2) = 7, y(3) = 9, and y(4) = 8.
Let x(n) and h ( n ) be finite-length sequences that are six points long, and let X(k) and H ( k ) be the eight-point DFTs of x ( n ) and h(n), respectively. If we form the product
and take the inverse DFT to form the sequence y(n), find the values of n for which y ( n ) is equal to the linear convolution
If the linear convolution of two sequences is M points long, for an N-point circular convolution with N < M, the ) first M - N points will be aliased. With ~ ( nand h(n) both of length six. z(n) = .r(n) * h(n) will be 1 I points long. Therefore, with an eight-point circular convolution. the first three points will be aliased, and the last five will be equal to the linear convolution.
If Y(k) = H ( k ) X ( k ) where H ( k ) and X(k) are the N-point DFTs of the finite-length sequences h(n) and x(n), respectively, show that
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