Question 3.1 d from Chapter 3 of Oppenheim's Signals & Systems I have

$$x[n]=(-1)^n{u[-n]-u[-n-8]}$$ $$h[n]=u[n]-u[n-8]$$

and the question asks for $y[n]$ which is the convolution of $x[n]$ and $h[n]$ namely

$$y[n]=x[n]\star h[n]$$

I'm stuck with the limits on the sum


Should the limit be '$0$' or '$n-k$' or '$n-8$'?

  • 1
    $\begingroup$ The limits are always the same: $-\infty$ to $\infty$. However, in many instances, the signals have value $0$ for infinitely many arguments, e.g. for all $k < 0$ or all $k > 35$, say, in which case, instead of wasting time adding infinitely many $0$s in the sum, you can reduce the limits on the sum and add only those numbers which might possibly be nonzero .nonzero $\endgroup$ Nov 16, 2013 at 14:17

1 Answer 1


Here is another example of how incredibly poor choice of notation misleads students everywhere. I do not possess a copy of any of the various tomes on signal processing that seem to be revered as the fifth Gospel on this site, and so I do not know if the book cited by the OP actually has written $(1)$, but I assert that it is against mathematical common sense to write things like $$y[n] = x[n] \star h[n].\tag{1}$$ It is generally accepted mathematical convention that is a symbol has the same meaning everywhere it appears in an equation or expression. For example, when we write $$y(t) = x(t)h(t),~ -\infty < t < \infty, \tag{2}$$ we mean by this that for every choice of real number $t$, the value of $y(t)$ is the same as the product of the values of $x(t)$ and $h(t)$; for example, $y(3) = x(3)h(3)$, and $y(313.012) = x(313.012)h(313.012)$, ans so on and so forth. So how are we supposed to make sense of this monstrosity $(1)$? Clearly $y[3]$ is not $x[3]\star h[3]$ whatever meaning we might ascribe to the latter quantity; the value of $y[3]$ depends on a lot of other values of $x$ and $h$. The poor notation misleads students into writing things like


which is not the correct expression for the convolution (or the correlation, for that matter) of $$x[n]=(-1)^n{u[-n]-u[-n-8]}\quad \text{and} \quad h[n]=u[n]-u[n-8]\tag{3}$$

The correct way to do this is write $$y[n] = (x\star h)[n] = \sum_{k=-\infty}^\infty x[k]h[n-k]$$ and then substitute $n-k$ for $n$ in $h[n]=u[n]-u[n-8]$ to get $h[n-k] = u[n-k]- u[n-k-8]$ and not $u[-n-k]-u[(-n-8)-k]$ the way that the OP has it.

Some initial thought might have shown that (assuming that $u[n]$ is the unit step), that $h[n]$ is $0$ for $n < 0$ (since then both $u[n]$ and $u[n-8]$ are $0$, and that $h[n] = 0$ for $n \geq 8$ (since then both $u[n]$ and $u[n-8]$ are $1$. Consequently for any given fixed integer $n$, $h[n-k]$ is nonzero only for $$0 \leq n-k \leq 7 \Rightarrow n-7 \leq k \leq n$$ and so the limits on the sum could be reduced from $-\infty$ and $\infty$ to $n-7$ and $n$ (cf. my comment on the OP's question). I will leave it to the OP to figure out whether the fact that $x[n]$ is also time-limited can be used to further reduce the limits on the sum. Hint: the answer to this might depend on the specific value of $n$, and you may need to try several different values of $n$, e.g. $0, 3, 8, 10, -40$ etc to see if you can discern a pattern.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.