When convolving with undefined functions, do we just treat them like variables? (eg. How do I handle $u \left( k \right)$ and $u \left( n - k \right)$ since they are just step functions with no equation listed?)

I treated them like variables and got this answer but it doesn't look right. Am I supposed to put them in the limits, strike them down, remove them, or what?

Original problem:

  1. Convolve $h \left( n \right)$ and $x \left( n \right)$ to get $y \left( n \right)$ below. Put $y \left( n \right)$ in closed form when possible.

a. $h \left( n \right) = (\frac{1}{2})^n\, u \left( n \right)$ and $x ( n ) = 3^n u ( n )$.

I'm sure that this is very easy and the solution is obvious as my professor usually does a few steps and just arrives at a solution. However, I don't know what to do with basic functions like this within a convolution.

[Convolution with unit step functions][1]

  • $\begingroup$ Forgot to add the original problem. Convolve h(n) and x(n) to get y(n) below. Put y(n) in closed form when possible. (a) h(n) = (1/2) nu(n) and x(n) = 3 nu(n). $\endgroup$
    – Axel
    Sep 14 at 18:09
  • 1
    $\begingroup$ Hello @Axel, in this case the signals are discrete and you are supposed to perform a summation. $u(n)$ is just the step function, i.e. $$u(n)=\begin{cases}1 & n\geq 0\\0&\text{otherwise}\end{cases}$$. $\endgroup$
    – cjferes
    Sep 14 at 18:20
  • $\begingroup$ There being no "equation" does not make the step function undefined. By the way, it does have an explicit piecewise-constant equation. $\endgroup$
    – user67664
    Sep 15 at 7:54

1 Answer 1


The function $u[n]$ is not unknown, but instead is the common notation for a step function. In discrete time $n$, it corresponds to $$u[n]=\begin{cases}1&n\geq0\\0&n<0\end{cases}\,. $$

We also have that both $h[n]$ and $x[n]$ are discrete-time signals, and hence we have to use the definition of the discrete-time convolution, i.e. $$\begin{align} y[k]&=\sum_{n=-\infty}^{\infty}h[n]x[k-n]=\sum_{n=-\infty}^{\infty}\frac{1}{2^n}u[n]\cdot 3^{k-n}u[k-n]\\ &=3^k\sum_{n=-\infty}^{\infty}\frac{1}{6^n}\big(u[n]\cdot u[k-n]\big).%\qquad \text{ where }u[n]=0\text{ for }n<0\\ %&=3^k\sum_{n=0}^{\infty}\frac{1}{6^n}u[k-n]\qquad \text{ where }u[k-n]=u[k]-u[n]\\ \end{align}$$

Note that $u[n]\cdot u[k-n]=1$ only if $n\geq 0$ AND $k-n\geq0$, i.e. only if $k\geq n\geq 0$, where the free variable is $k$. Hence,

  • we can reduce the infinite summation terms to only the range $n\in\{0,1,\ldots,k\}$; and
  • we have that $y[k]=0$ for $k<0$ as the summation terms would be all zero.

Therefore, we can rewrite $y[k]$ with a step function $u[k]$ and derive $$\begin{align} y[k]&=3^ku[k]\sum_{n=0}^k\frac{1}{6^n}=3^k\frac{1-(\frac{1}{6})^{k+1}}{1-\frac{1}{6}}u[k]=3^k\left(\frac{6^{k+1}-1}{6^{k+1}}\right)\left(\frac{6}{5}\right)u[k]=\frac{6^{k+1}-1}{5\cdot 2^k}u[k]\,. \end{align}$$

  • $\begingroup$ I'm glad you're using the bracket notation for discrete-time functions, e.g. $x[n]$. $\endgroup$ Sep 14 at 21:37
  • 1
    $\begingroup$ @robertbristow-johnson Formality in notation helps quite a bit in understanding/learning :) $\endgroup$
    – cjferes
    Sep 14 at 21:41
  • 1
    $\begingroup$ I know. That's why I'm sorta anal about it. $\endgroup$ Sep 14 at 21:45

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