Revision as of 11:37, 30 November 2018

CT and DT Convolution Examples

In this course, it is important to know how to do convolutions in both the CT and DT world. Sometimes there may be some confusion about how to deal with certain positive or negative input combinations. Here are some examples for how to deal with them.

CT Examples

Example 1: t is positive for both h(t) and x(t)

$ x(t) = u(t) $

$ h(t) = e^{-2t} u(t) $

$ y(t) = h(t)*x(t) $

$ y(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau $

$ y(t) = \int_{-\infty}^{\infty} e^{-2\tau} u(\tau)u(t - \tau) d\tau $

$ y(t) = \int_{0}^{\infty} e^{-2\tau} u(t - \tau) d\tau $

Since $ u(t - \tau) = 1 $

$ \tau \leq t $

$ y(t)=\begin{cases} \int_{0}^{t} e^{-2\tau}d\tau, & \mbox{if }t \geq 0 \\ 0, & \mbox else \end{cases} $

$ y(t)=\begin{cases} \frac{e^{-2t}-1}{-2} , & \mbox{if }t \geq 0 \\ 0, & \mbox else \end{cases} $

$ y(t)=\frac{u(t)}{2}(1-e^{-2t}) $

Example 2: t is negative for both h(t) and x(t)

$ x(t) = u(-t) $

$ h(t) = e^{3t} u(-t) $

$ y(t) = h(t)*x(t) $

$ y(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau $

$ y(t) = \int_{-\infty}^{\infty} e^{3\tau} u(-\tau)u(-(t - \tau)) d\tau $

$ y(t) = \int_{-\infty}^{0} e^{3\tau} u(-t + \tau) d\tau $

Since $ u(-t + \tau) = 1 $

$ \tau \geq t $

$ y(t)=\begin{cases} \int_{t}^{0} e^{3\tau}d\tau, & \mbox{if }t \leq 0 \\ 0, & \mbox else \end{cases} $

$ y(t)=\left. u(-t)\frac{e^{3\tau}}{3}\right |_t^0 $

$ y(t)=\frac{u(-t)}{3}(1 - e^{3t}) $

Example 3: t is negative for x(t) and positive for h(t)

$ x(t) = u(-t) $

$ h(t) = e^{-2t} u(t) $

$ y(t) = h(t)*x(t) $

$ y(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau $

$ y(t) = \int_{-\infty}^{\infty} e^{-2\tau} u(\tau)u(-(t - \tau)) d\tau $

$ y(t) = \int_{0}^{\infty} e^{-2\tau} u(-t + \tau) d\tau $

$ \begin{align} y(t) &= h(t)*x(t)\\ &= \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau\\ & = \int_{-\infty}^{\infty} e^{-2\tau} u(\tau)u(-(t - \tau)) d\ta\\ &= \int_{0}^{\infty} e^{-2\tau} u(-t + \tau) d\tau \end{align} $

Since $ u(-t + \tau) = 1 $

$ \tau \geq t $

$ y(t)=\begin{cases} \int_{t}^{\infty} e^{-2\tau}d\tau, & \mbox{if }t \geq 0 \\ \int_{0}^{\infty} e^{-2\tau}d\tau, & \mbox{if }t < 0 \end{cases} $

$ y(t)=\begin{cases} \frac{e^{-2t}}{2}, & \mbox{if }t \geq 0 \\ \frac{1}{2}, & \mbox{if }t < 0 \end{cases} $

DT Examples

Example 1: n is positive for both h[n] and x[n]

$ h[n] = u[n] $

$ x[n] = 4^{-n}u[n] $

$ y[n] = x[n]*h[n] $

$ y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k] $

$ y[n] = \sum_{k=-\infty}^{\infty}4^{-k}u[k]u[n - k] $

$ u[k]=\begin{cases} 1, & \mbox{if }k \geq 0 \\ 0, & \mbox{if }k < 0 \end{cases} $

$ y[n] = \sum_{k=0}^{\infty}4^{-k}u[n - k] $

$ u[n-k]=\begin{cases} 1, & \mbox{if }k \leq n \\ 0, & \mbox else \end{cases} $

$ y[n]=\begin{cases} \sum_{k=0}^{n}4^{-k}, & \mbox{if }n \geq 0 \\ 0, & \mbox{if }n < 0 \end{cases} $

$ y[n]=\begin{cases} \frac{1-(\frac{1}{4})^{n+1}}{1-\frac{1}{4}}, & \mbox{if }n \geq 0 \\ 0, & \mbox else \end{cases} $

$ y[n]=\begin{cases} \frac{4-(\frac{1}{4})^{n}}{3}, & \mbox{if }n \geq 0 \\ 0, & \mbox else \end{cases} $

$ y[n] = \frac{4-(\frac{1}{4})^{n}}{3}u[n] $

Example 2: n is negative for both h[n] and x[n]

$ h[n] = u[-n] $

$ x[n] = 3^{n}u[-n] $

$ y[n] = h[n]*x[n] $

$ y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k] $

$ y[n] = \sum_{k=-\infty}^{\infty}3^{k}u[-k]u[-n + k] $

$ y[n] = \sum_{k=-\infty}^{0}3^{k}u[-n + k] $

since $ u[-n + k] = 1 $

$ k \geq n $

$ u[k]=\begin{cases} \sum_{k=n}^{0}3^{k}, & \mbox{if }n \leq 0 \\ 0, & \mbox{if }n > 0 \end{cases} $

Substitute $ m = -k $

$ y[n] = u[-n]\sum_{m=-n}^{0}3^{-m} $

$ y[n] = u[-n]\sum_{m=0}^{-n}(\frac{1}{3})^{m} $

$ y[n] = u[-n]\frac{1 - (\frac{1}{3})^{-n + 1}}{1-\frac{1}{3}} $

$ y[n] = u[-n]\frac{3 - 3^{-n}}{2} $

Example 3: n is negative for x[n] and positive for h[n]

$ h[n] = u[-n] $

$ x[n] = 5^{n}u[n] $

$ y[n] = h[n]*x[n] $

$ y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k] $

$ y[n] = \sum_{k=-\infty}^{\infty}5^{k}u[k]u[n - k] $

$ y[n] = \sum_{k=0}^{\infty}5^{k}u[n - k] $

since $ u[n - k] = 1 $

$ n \geq k $

$ u[k]=\begin{cases} \sum_{k=0}^{n}n^{k}, & \mbox{if }n \geq 0 \\ 0, & \mbox else \end{cases} $

$ y[n] = u[n]\frac{1 - 5^{n + 1}}{1 - 5} $

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