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It is not necessarily true that the convolution of an infinite-length sequence with a finite-length sequence will be infinite in length. It may be either. Clearly, if x ( n ) = 6 ( n ) and h ( n ) = (OS)"u(n), the convolution will be an infinite-length sequence. However, it is possible for the finite-length sequence to remove the infinite-length tail of an infinite-length sequence. For example, note that

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Therefore, the convolution of x ( n ) = 6 ( n ) - f S ( n - I) with h ( n ) = (OS)"u(n) will be finite in length:

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Find the convolution of the two finite-length sequences:

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Shown in the figure below are the sequences x ( k ) and h ( k ) .

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Because h ( n ) is equal to zero outside the interval [-3, 31, and x ( n ) is zero outside the interval [ l , 51, the convolution y ( n ) = x ( n ) * h ( n ) is zero outside the interval 1-2, 81. One way to perform the convolution is to use the slide rule approach. Listing x ( k ) and h ( - k ) across two pieces of paper, aligning them at k = 0 , we have the picture as shown below (the sequence h ( - k ) is in front).

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Forming the sum of the products x(k)h(-k), we obtain the value of y ( n ) at time n = 0 , which is y ( 0 ) = 2 . Shifting h(-k) to the left by one, multiplying and adding, we obtain the value of y ( n ) at n = -1, which is y ( - I ) = 2. Shifting one more time to the left, forming the sum of products, we find y ( - 2 ) = 1, which is the last nonzero value of y ( n ) for n < 0 . Repeating the process by shifting h ( - k ) to the right, we obtain the values of y ( n ) for n > 0 , which are

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Another way to perform the convolution is to use the fact that

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x ( n ) S(n - no) = x ( n - n o )

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Writing h ( n ) as

CHAP. I ] we may evaluate y ( n ) as follows

SIGNALS AND SYSTEMS

y(n) = 2x(n

+ 3 ) - 2 x ( n + 1 ) + 2 x ( n - I) - 2x(n - 3 )

Making a table of these shifted sequences.

and adding down the columns, we obtain the sequence y ( n ).

Derive a closed-form expression for the convolution of x ( n ) and h ( n ) where

x(n) = (6)

I N-6

u(n)

h ( n ) = ( f ) " u ( n - 3)

Because both sequences are infinite In length. it is easier to evaluate the convolution sum directly:

Note that because x ( n ) = 0 for n < 0 and h ( n ) = 0 for n < 3 , y ( n ) will be equal to zero for n < 3. Substituting x ( n ) and h ( n ) into the convolution sum, we have

Due to the step u ( k ) , the lower limit on the sum may be changed to k = 0, and because u ( n - k - 3) is zero for k > n - 3 , the upper limit may be changed to k = n - 3 . Thus. for n 2 3 the convolution sum becomes

Using the geometric series to evaluate the sum, we have

A linear shift-invariant system has a unit sample response

Find the output if the input is

x ( n ) = -n3"u(-n)

Shown below are the sequences x ( n ) and h ( n ) .

SIGNALS AND SYSTEMS

[CHAP. I

Because x ( n ) is zero for n > -1, and h ( n ) is equal to zero for n > - 1 , the convolution will be equal to zero for n z -2. Evaluating the convolution sum directly, we have