Theorem 

If the discrete-time system linear shift-invariant, T[.], has the unit sample response T[δ(n)] = h(n) then the output y(n) in correspondence to any input x(n) which is given by

The second summation is achived by setting m = n-k; then for k = -∞ we have m = +∞ , and for k = ∞ we have m = -∞. Thus

Example Linear Convolution 

Given the input {x(n)} ={1, 2, 3, 1} and unit sample response {h(n)} = {4, 3, 2, 1} find out the response y(n) = x(n) * h(n).

Answer Since x(k) = 0 for k < 0 and h(n - k) = 0 for k > n, the sum of convolution becomes

Now y(n) is evaluated for various values of n; for instance, setting n = 0 gives y(0). See the table below. The product terms shown in bold italics are not required be calculated; they are zero as the signal values involved are zero.

  Linear Convolution of {x(n)} ={1, 2, 3, 1} and {h(n)} = {4, 3, 2, 1}

Sketches of the 2 sequences x(n) and h(n) are shown below.

To do the convolution we require the sequences x(k) and h(n-k), k being the independent variable. Of these x(k) is simply x(n) with k replacing n, is shown below.

The sequence h(-k) is reflected version of h(k). If h(-k) is delayed by n samples we get h(-(k-n)) i.e., h(n-k), shown above.

For each value of n sequences x(k) and h(n-k) are multiplied point by point and the products of them are added, yielding the value of y (.) for corresponding n.

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