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In addition to the direct method, convolutions may also be performed graphically. The steps involved in using the graphical approach are as follows: Plot both sequences, x(k) and h(k), as functions of k. 2. Choose one of the sequences, say h(k), and time-reverse it to form the sequence h(-k). 3. Shift the time-reversed sequence by n. [Note: If n > 0, this corresponds to a shift to the right (delay), whereas if n < 0, this corresponds to a shift to the left (advance).] 4. Multiply the two sequences x(k) and h(n - k) and sum the product for all values of k. The resulting value will be equal to y(n). This process is repeated for all possible shifts, n.
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EXAMPLE 1.4.2 To illustrate the graphical approach to convolution, let us evaluate y ( n ) = x ( n ) * h ( n ) where x ( n ) and h ( n ) are the sequences shown in Fig. 1-6 ( a ) and (b), respectively.To perform this convolution, we follow the steps listed above:
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Because x ( k ) and h ( k ) are both plotted as a function of k in Fig. 1-6 ( a ) and (b), we next choose one of the sequences to reverse in time. In this example, we time-reverse h ( k ) , which is shown in Fig. 1-6(c). Forming the product, x ( k ) h ( - k ) , and summing over k , we find that y ( 0 ) = 1. Shifting h ( k ) to the right by one results in the sequence h ( l - k ) shown in Fig. 1-6(d). Forming the product, x ( k ) h ( l - k ) , and summing over k , we find that y ( 1 ) = 3. Shifting h ( l - k ) to the right again gives the sequence h(2 - k ) shown in Fig. 1-6(e). Forming the product, x ( k ) h ( 2 - k ) , and summing over k , we find that y ( 2 ) = 6 . Continuing in this manner, we find that y ( 3 ) = 5. y ( 4 ) = 3, and y ( n ) = 0 for n > 4 . We next take h ( - k ) and shift it to the left by one as shown in Fig. 1-6 (f ). Because the product, x ( k ) h ( - 1 - k ) , is equal to zero for all k , we find that y(- I ) = 0. In fact. y ( n ) = 0 for all n < 0 .
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Figure 1-6 (g) shows the convolution for all n .
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Fig. 1-6. The graphical approach to convolution.
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A useful fact to remember in performing the convolution of two finite-length sequences is that if x ( n ) is of length L 1 and h ( n ) is of length L 2 , y ( n ) = x ( n ) * h ( n ) will be of length
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Furthermore, if the nonzero values of x ( n ) are contained in the interval [ M,, N,] and the nonzero values of h ( n ) are contained in the interval [Mh, Nh], the nonzero values of y ( n ) will be confined to the interval [ M , Mh, N, Nh].
EXAMPLE 1.4.3 Consider the convolution of the sequence
x(n) =
0 n 0
L o n 5 2 0 otherwise -55n55 otherwise
with
h(n) =
Because x ( n ) is zero outside the interval [ l o , 201, and h ( n ) is zero outside the interval [ - 5 , 51, the nonzero values of the convolution, y(n) = x ( n ) * h ( n ) ,will be contained in the interval [ 5 , 251.
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