I'm currently working through some convolution examples and I'm unsure of something.

The question is given as:

Consider the input \$\ x(n) = u(n) \$ and the impulse response \$\ h(n) = (0.5)^n u(n) \$ for a certain system. What is the output of the system?

From the convolution equation, I understand that the first step produces: \$\ y(n) = \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m) \$

The question goes on to say:

But since \$\ u(m) = 1 \$ for \$\ n \ge 0 \$ and both \$\ x(n) \$ and \$\ h(n) \$ start at \$\ n = 0 \$, we have \$\ y(n) = \sum\limits_{m=0}^{m=n} = (.5)^{n-m} \$

I understand that \$\ u(m) \$ disappears as it is a constant, but how is \$\ u(n-m) \$ removed?

Thanks in advance for any help on this.

EDIT: Additionally, why do the limits change from +/- infinity to \$\ m=n \$ and \$\ n = 0 \$?

  • 1
    \$\begingroup\$ Hint: \$u(n-m) = 1 \$ for \$ m \le n\$ \$\endgroup\$ Nov 9, 2013 at 23:22

4 Answers 4


Since others have provided answers, I'll try to answer with an intuitive approach.

The original equation is:

\$ y(n) = \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m) \$

Now, due the to the presence of \$u(m)\$, the terms for \$m < 0 \$ are zero. Thus, the sum is unchanged by changing the lower limit from \$- \infty \$ to zero. This change of lower limit makes the presence of \$u(m)\$ redundant and so it can be removed.

\$ y(n) = \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m) = \sum\limits_{m=0}^\infty (.5)^{n-m} u(n-m) \$

Similarly, due to the presence of \$u(n-m)\$, the terms for \$m > n \$ are zero. Thus, the sum is unchanged by changing the upper limit from \$\infty \$ to \$m=n\$. This change of upper limit makes the presence of \$u(n - m)\$ redundant and so it can be removed.

\$ y(n) = \sum\limits_{m=0}^\infty (.5)^{n-m} u(n-m) = \sum\limits_{m=0}^{m=n} (.5)^{n-m} \$

  • \$\begingroup\$ Thanks. Why does the presence of \$\ u(n-m) \$ makes the terms for \$\ m>n \$ zero? \$\endgroup\$
    – user32485
    Nov 11, 2013 at 16:02
  • \$\begingroup\$ @user32485, (1) the unit step function equals zero when the argument is negative and (2) (n-m) is negative when m > n. \$\endgroup\$ Nov 11, 2013 at 17:23

Think about what is happening when you do convolution. You are taking h(m) and reflecting it about m = 0, and offsetting it by n. You can demonstrate this:
On one piece of paper, draw x(m) = u(m). On a second piece, draw h(-m). Line them up at zero. This is n = 0. The two overlap only at zero. Now shift the second piece to the right by one. Now you have n = 1. They overlap at 0 and 1. Shift it right again and you have n = 2. And so on.
You should be able to see why the summation is zero for all m < 0, and for all m > n


It's simple. You've already answered the question with your another question:

"EDIT: Additionally, why do the limits change from +/- infinity to m=n and n=0?"

Instead of writing your equation with u(..), you just analyze where your step response begins and then you can change your summation limits to the step response limits. This way your equation is valid just when u(...) = 1 so you can remove it from there. Therefore you know that your final answer has a limitation in time:

y(t) = .... with (t > t0) or (t < t0) depending on how is your u(t)

And you can do that because when step response is zero, nothing will happen and you equation goes to zero. This is why you change your summation/integral limits so your equation only cares when u(t) is not zero. And it's the same reason you can remove u(t) from it.


\$u(n) \$ is the unit step and \$u(n-m) \$ is still a unit step delayed by \$m \$ samples. Irrespective of any delay, the unit step remains the same, with each sample \$S_n ,n \gt m\$ having a value of one; which is equivalent to :

\$\ \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m)= \sum\limits_{m=0}^{m=n} (.5)^{n-m} \cdot (1) =\sum\limits_{m=0}^{m=n} (.5)^{n-m}\$

  • \$\begingroup\$ \$u(n)\$ is the unit step. \$\endgroup\$ Nov 9, 2013 at 23:22
  • \$\begingroup\$ my bad, edited. :P \$\endgroup\$
    – K. Rmth
    Nov 9, 2013 at 23:27

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