EDITOR: B. SOMANATHAN
NAIR
1. INTRODUCTION
Consider
the situation in which both x(n) and h(n) are periodic sequences. The resulting convolution is then called
the periodic or circular convolution.
We have already seen that conventional convolution is
the output of a given system resulting from its reaction (or impulse response)
to an external input. In this case, even though the external input x(n)
can be periodic, the impulse response h(n)
of the system can not and will not be
periodic. For example, the gain of an amplifier or the attenuation of a
filter will not be periodic for obvious reasons. Therefore, we conclude that in
conventional convolution, the impulse response of the system will not be
periodic. Then what is periodic or circular convolution?
The answer to the above question is obvious: circular
convolution is just a convolution-like operation between two periodic sequences
of equal lengths. To distinguish it from the conventional convolution, we use
the following rules:
·
Circular
convolution of two sequences x(n) and h(n) is given by the
expression
where x1(n) and x2(n), or x(n)
and h(n), respectively, are the input
sequences and y(n) is the convolution sum.
·
To distinguish
circular convolution from linear convolution, we use the notation
y(n) = x(n)
© h(n) (2)
· The convolution
sum of two periodic sequences is also found to be periodic. Hence the
convolution need be done over one period only.
It can be seen that the circular or
periodic convolution can also be obtained through our tabular method quite easily,
even though in many other textbooks, they use a circular shift method using
circles to represent the sequences. In this blog, only the tabular method of
circular convolution will be described, as this is much faster, clearer, and
easily understandable than other methods. Further, it can be implemented easily
using computer algorithms. We now illustrate the computation using a numerical
example.
ILLUSTRATIVE
EXAMPLE 1: Obtain the circular
convolution of the sequences
x(n) = (…,
1, 2, 1, 2, 1, 2, …) (1)
h(n) = (…, 2,
1, 2, 1, 2, 1, …) (2)
Solution: The tabulation for the example is shown in Tables 1,
2, and 3, respectively. Table 1 shows x(k), h(k), h(‒k).
Table 2 shows x(k), h(‒k), and their columnwise product y(0). Notice that we have shown only one
period (containing two terms only) of the product y(0). This will be periodically repeated in other columns.
Table 3 shows the column entries of Table 2 shifted
by one cell to the right to yield h(1‒k), and using this we find y(1). Since there are only two terms each
in x(k and h(k), we stop the computation at this
point. It can be seen that the convolution sum is periodic with the terms (8, 7)
in it, i.e., (..., 8,7, 8, 7, 8, 7, ...).
It can be seen that tabulation method is very
general and easy to use in all forms of convolution. It can be used to solve problems in cyclic (periodic) shift.
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