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<math>y(t)=u(-t)\frac{e^{3\tau}}{3} |^t </math><br />
+
<math>y(t)=\left.  u(-t)\frac{e^{3\tau}}{3}\right |_t^0 </math><br />
  
 
<math>y(t)=\frac{u(-t)}{3}(1 - e^{3t})</math><br />
 
<math>y(t)=\frac{u(-t)}{3}(1 - e^{3t})</math><br />
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<math>h(t) = e^{-2t} u(t)</math><br />
 
<math>h(t) = e^{-2t} u(t)</math><br />
  
<math>y(t) = h(t)*x(t)</math><br />
 
  
<math>y(t) =  \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau</math><br />
+
<math>
 
+
\begin{align}
<math>y(t) =  \int_{-\infty}^{\infty} e^{-2\tau} u(\tau)u(-(t - \tau)) d\tau</math><br />
+
y(t) &= h(t)*x(t)\\
 
+
&=  \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau\\
<math>y(t) =  \int_{0}^{\infty} e^{-2\tau} u(-t + \tau) d\tau</math><br />
+
& =  \int_{-\infty}^{\infty} e^{-2\tau} u(\tau)u(-(t - \tau)) dt\\
 +
&=  \int_{0}^{\infty} e^{-2\tau} u(-t + \tau) d\tau
 +
\end{align}
 +
</math>
  
 
Since <math>u(-t + \tau) = 1</math><br />
 
Since <math>u(-t + \tau) = 1</math><br />
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<math>h[n] = u[-n]</math><br />
 
<math>h[n] = u[-n]</math><br />
  
<math>x[n] = 3^{n}u[-n]</math><br />
+
<math>x[n] = 5^{n}u[n]</math><br />
  
 
<math>y[n] = h[n]*x[n]</math><br />
 
<math>y[n] = h[n]*x[n]</math><br />
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<math>y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k]</math><br />
 
<math>y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n - k]</math><br />
  
<math>y[n] = \sum_{k=-\infty}^{\infty}3^{k}u[-k]u[-n + k]</math><br />
+
<math>y[n] = \sum_{k=-\infty}^{\infty}5^{k}u[k]u[n - k]</math><br />
  
<math>y[n] = \sum_{k=-\infty}^{0}3^{k}u[-n + k]</math><br />
+
<math>y[n] = \sum_{k=0}^{\infty}5^{k}u[n - k]</math><br />
  
since <math>u[-n + k] = 1</math><br />
+
since <math>u[n - k] = 1</math><br />
  
<math>k \geq n</math><br />
+
<math>n \geq k</math><br />
  
 
<math>u[k]=\begin{cases}  
 
<math>u[k]=\begin{cases}  
\sum_{k=n}^{0}3^{k},  & \mbox{if }n \leq 0 \\
+
\sum_{k=0}^{n}n^{k},  & \mbox{if }n \geq 0 \\
0,  & \mbox{if }n > 0
+
0,  & \mbox else
 
\end{cases}</math><br />
 
\end{cases}</math><br />
  
Substitute <math>m = -k</math><br />
+
<math>y[n] = u[n]\frac{1 - 5^{n + 1}}{1 - 5}</math><br />
 
+
<math>y[n] = u[-n]\sum_{m=-n}^{0}3^{-m}</math><br />
+
 
+
<math>y[n] = u[-n]\sum_{m=0}^{-n}(\frac{1}{3})^{m}</math><br />
+
 
+
<math>y[n] = u[-n]\frac{1 - (\frac{1}{3})^{-n + 1}}{1-\frac{1}{3}}</math><br />
+
 
+
<math>y[n] = u[-n]\frac{3 - 3^{-n}}{2}</math><br />
+
  
 
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Latest revision as of 12:38, 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) $


$ \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)) dt\\ &= \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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Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal