# How to handle undefined functions in a convolution?

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.

[ • 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).
– Axel
Sep 14 at 18:09
• 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. $$u(n)=\begin{cases}1 & n\geq 0\\0&\text{otherwise}\end{cases}$$. Sep 14 at 18:20
• There being no "equation" does not make the step function undefined. By the way, it does have an explicit piecewise-constant equation.
– user67664
Sep 15 at 7:54

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}

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