The sliding-tape method is conceptually quite valuable in understanding the conv

Question

The sliding-tape method is conceptually quite valuable in understanding the convolu- tion mechanism. Numerical convolution can also be pertormed from the arrays using the sets f[0], f[1], f12], ..., and 9[O], g[1], g[2], ..., as depicted in Fig. P9.4-16. The nth element (element in the ith row and jth column) is given by g[i]f we add the elements of the array along its diagonals to produce c[k] = f[k] * gk]. For example, if we sum the elements corresponding to the first diagonal of the array, we obtain c[0] Similarly, if we sum along the second diagonal, we obtain cf1], and so on. Draw the array for the signals fik] and gle] in Example 9.9, and find fk] * g[k] 9.4-16 c [01 g [k] f [o] Fig. P9.4-16

Explanation / Answer

By convolution property: (matrix method)

c(k)=fk)*g(k)= { c(0), c(1), c(3),......... c(k-1) }

where,

c(0)=f(0)g(0)

c(1)=f(1)g(0)+f(0)g(1)

c(2)=f(2)g(0)+f(1)g(1)+f(0)g(2).

c(3)=f(3)g(0)+f(2)g(1)+f(1)g(2)+f(0)g(3)

..............

.............

if we consider i and j as row and column respectively and its value as corresponding row and column

then the value is

c(0)=f(0)g(0)=0*0=0

c(1)=f(1)g(0)+f(0)g(1)=1*0+0*1=0

c(2)=f(2)g(0)+f(1)g(1)+f(0)g(2)=2*0+1*1+0*2=1

c(3)=f(3)g(0)+f(2)g(1)+f(1)g(2)+f(0)g(3)=3*0+2*1+1*2+0*3=4

..............

.............

therefore, C(k)=f(k)*g(k)={ 0 , 0 , 1, 4 .............}

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