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barcode generator for ssrs With x ( n ) = 0 outside the interval 0 5 in Software
With x ( n ) = 0 outside the interval 0 5 Scan Code 128A In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128 Code Set A Creator In None Using Barcode creation for Software Control to generate, create Code 128 Code Set C image in Software applications. 9 3, it follows that
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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 fourpoint 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 fivepoint DFT of x ( n ) is computed and the resulting sequence is squared: Y(k) = ~ ' ( k ) A fivepoint inverse DFT is then computed to produce the sequence y ( n ) . Find the sequence y ( n ) . The sequence y ( n ) has a fivepoint 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 fivepoint 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 5point DFTs, H(k) and X(k), y(n) is the fivepoint 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 fivepoint circular convolution of .r(n) with h(n) is Because h(n) = I for n = 0, 1. 2, 3, and h(4) = 0, the fivepoint 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 finitelength sequences that are six points long, and let X(k) and H ( k ) be the eightpoint 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 Npoint 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 eightpoint 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 Npoint DFTs of the finitelength sequences h(n) and x(n), respectively, show that

