Consider the following sequences:

$x_{1}(n) = A$ (a constant), $x_{2}(n) = u(n)$, $x_{3}(n) =\delta(n)-\delta(n-1)$.

($\circledast$ stands for linear convolution)

If I perform the operation $x_{2}\circledast (x_{3}\circledast x_{1})$, the value I am getting is $0$, where as if I perform $(x_{2}\circledast x_{3})\circledast x_{1}$ the value I am getting is A.

Since Convolution is associative, why are the answers different?

My approach: In general, assuming $x_{1}(n)$, $x_{2}(n)$, $x_{3}(n)$ are of infinite lengths,

$$x_{2}\circledast (x_{3}\circledast x_{1})=\sum_{k=-\infty}^{\infty}x_2(k).x_{3}\circledast x_{1}(n-k)=\sum_{k=-\infty}^{\infty}x_2(k)\sum_{l=-\infty}^{\infty}x_3(l)x_1(n-k-l)$$

Let $m=n-k-l$,


$$=\sum_{m=-\infty}^{\infty}x_{2}\circledast x_{3}(n-m)x_1(m)=(x_{2}\circledast x_{3})\circledast x_{1}$$.

Hence I feel Convolution is associative even if the sequences are of infinite lengths. Where did I go wrong?

  • $\begingroup$ waitasecond, are we doing discrete time signals or continuous-time signals? $\endgroup$ Nov 2 '18 at 10:25
  • $\begingroup$ discrete time signals $\endgroup$
    – Narendra
    Nov 2 '18 at 10:56

The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences $x_1[n]$ and $x_2[n]$. Note that the convolution sum $x_1\star x_2$ does not converge, i.e., $x_3\star (x_1\star x_2)$ gives yet another (infinite) result.

In continuous time you have the same problem. Associativity of continuous convolution relies on Fubini's theorem for switching the order of integration. If the assumptions of Fubini's theorem are not satisfied for the given functions, convolution is not associative.

  • 2
    $\begingroup$ Sometimes, math is useful! $\endgroup$ Nov 2 '18 at 13:20

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