Difference between revisions of "Convolution Shortcuts"

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(Exponential and Shifted Step)
(Convolution Between Exponentials)
 
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\end{align}</math></center>
 
\end{align}</math></center>
  
==Convolution with Other Singularities==
+
==Convolution with Steps and Ramps==
 
<center><math>\begin{align}
 
<center><math>\begin{align}
 
u(t)*f(t)&=\int_{-\infty}^{t}f(\tau)~d\tau\\
 
u(t)*f(t)&=\int_{-\infty}^{t}f(\tau)~d\tau\\
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u(t)*q(t)=r(t)*r(t)=u(t)*u(t)*u(t)*u(t)&=\frac{1}{6}t^3u(t)\\
 
u(t)*q(t)=r(t)*r(t)=u(t)*u(t)*u(t)*u(t)&=\frac{1}{6}t^3u(t)\\
 
\mbox{equivalent of }n\mbox{ steps convolved together}&=\frac{1}{(n-1)!}t^{n-1}u(t)
 
\mbox{equivalent of }n\mbox{ steps convolved together}&=\frac{1}{(n-1)!}t^{n-1}u(t)
 +
\end{align}</math></center>
 +
 +
==Convolution Between Exponentials==
 +
Note - the following work if $$a$$ and/or $$b$$ is 0. 
 +
* Same exponent
 +
<center><math>\begin{align}
 +
(e^{-at}\,u(t))*(e^{-at}\,u(t))&=\int_{-\infty}^{\infty} e^{-a\tau}\,u(\tau)\,e^{-a(t-\tau)}\,u(t-\tau)\,d\tau\\
 +
&=u(t)\int_{0}^t e^{-a\tau}\,e^{-a(t-\tau)}\,d\tau=u(t)\int_{0}^t e^{-a\tau}\,e^{-at}\,e^{a\tau}\,d\tau\\
 +
&=e^{-at}u(t)\int_{0}^t e^{-a\tau}\,e^{a\tau}\,d\tau=e^{-at}u(t)\int_{0}^t d\tau\\
 +
&=e^{-at}u(t)\left[ \tau \right]_0^t=e^{-at}u(t)\left[t-0\right]\\
 +
&=t\,e^{-at}\,u(t)
 +
\end{align}</math></center>
 +
 +
* Different exponents
 +
<center><math>\begin{align}
 +
(e^{-at}\,u(t))*(e^{-bt}\,u(t))&=\int_{-\infty}^{\infty} e^{-a\tau}\,u(\tau)\,e^{-b(t-\tau)}\,u(t-\tau)\,d\tau\\
 +
&=u(t)\int_{0}^t e^{-a\tau}\,e^{-b(t-\tau)}\,d\tau=u(t)\int_{0}^t e^{-a\tau}\,e^{-bt}\,e^{b\tau}\,d\tau\\
 +
&=e^{-bt}u(t)\int_{0}^t e^{-a\tau}\,e^{b\tau}\,d\tau=e^{-bt}u(t)\int_{0}^t e^{(b-a)\tau}\,d\tau\\
 +
&=e^{-bt}u(t)\left[ \frac{e^{(b-a)\tau}}{b-a}\right]_0^t=e^{-bt}u(t)\left[\frac{e^{(b-a)t}}{b-a}-\frac{1}{b-a}\right]\\
 +
&=\left[\frac{e^{-at}-e^{-bt}}{b-a}\right]\,u(t)
 +
\end{align}</math></center>
 +
 +
* "Single" exponent
 +
<center><math>\begin{align}
 +
(e^{-at}\,u(t))*(u(t))&=(e^{-at}\,u(t))*(e^{-0t}\,u(t))\\
 +
&=\left[\frac{e^{-at}-e^{-0t}}{0-a}\right]\,u(t)=\left[\frac{1-e^{-at}}{a}\right]\,u(t)
 
\end{align}</math></center>
 
\end{align}</math></center>
  

Latest revision as of 01:43, 18 September 2023

The following is a list of convolutions that are good to know. In each case, \(f(t)\) represents an arbitrary function while \(a\) and \(b\) represent constants.

Convolution with Impulses

\(\begin{align} \delta(t)*f(t)&=f(t)\\ \delta(t-a)*f(t)&=f(t-a)\\ \delta(t)*f(t-b)&=f(t-b)\\ \delta(t-a)*f(t-b)&=f(t-a-b)\\ \end{align}\)

Convolution with Steps and Ramps

\(\begin{align} u(t)*f(t)&=\int_{-\infty}^{t}f(\tau)~d\tau\\ r(t)*f(t)=u(t)*u(t)*f(t)&=\int_{-\infty}^{t}\int_{-\infty}^{\gamma}f(\tau)~d\tau~d\gamma\\ \end{align}\)

Convolution Between Singularity Functions

\(\begin{align} u(t)*u(t)&=r(t)=tu(t)\\ u(t)*r(t)=u(t)*u(t)*u(t)&=q(t)=\frac{1}{2}t^2u(t)\\ u(t)*q(t)=r(t)*r(t)=u(t)*u(t)*u(t)*u(t)&=\frac{1}{6}t^3u(t)\\ \mbox{equivalent of }n\mbox{ steps convolved together}&=\frac{1}{(n-1)!}t^{n-1}u(t) \end{align}\)

Convolution Between Exponentials

Note - the following work if $$a$$ and/or $$b$$ is 0.

  • Same exponent
\(\begin{align} (e^{-at}\,u(t))*(e^{-at}\,u(t))&=\int_{-\infty}^{\infty} e^{-a\tau}\,u(\tau)\,e^{-a(t-\tau)}\,u(t-\tau)\,d\tau\\ &=u(t)\int_{0}^t e^{-a\tau}\,e^{-a(t-\tau)}\,d\tau=u(t)\int_{0}^t e^{-a\tau}\,e^{-at}\,e^{a\tau}\,d\tau\\ &=e^{-at}u(t)\int_{0}^t e^{-a\tau}\,e^{a\tau}\,d\tau=e^{-at}u(t)\int_{0}^t d\tau\\ &=e^{-at}u(t)\left[ \tau \right]_0^t=e^{-at}u(t)\left[t-0\right]\\ &=t\,e^{-at}\,u(t) \end{align}\)
  • Different exponents
\(\begin{align} (e^{-at}\,u(t))*(e^{-bt}\,u(t))&=\int_{-\infty}^{\infty} e^{-a\tau}\,u(\tau)\,e^{-b(t-\tau)}\,u(t-\tau)\,d\tau\\ &=u(t)\int_{0}^t e^{-a\tau}\,e^{-b(t-\tau)}\,d\tau=u(t)\int_{0}^t e^{-a\tau}\,e^{-bt}\,e^{b\tau}\,d\tau\\ &=e^{-bt}u(t)\int_{0}^t e^{-a\tau}\,e^{b\tau}\,d\tau=e^{-bt}u(t)\int_{0}^t e^{(b-a)\tau}\,d\tau\\ &=e^{-bt}u(t)\left[ \frac{e^{(b-a)\tau}}{b-a}\right]_0^t=e^{-bt}u(t)\left[\frac{e^{(b-a)t}}{b-a}-\frac{1}{b-a}\right]\\ &=\left[\frac{e^{-at}-e^{-bt}}{b-a}\right]\,u(t) \end{align}\)
  • "Single" exponent
\(\begin{align} (e^{-at}\,u(t))*(u(t))&=(e^{-at}\,u(t))*(e^{-0t}\,u(t))\\ &=\left[\frac{e^{-at}-e^{-0t}}{0-a}\right]\,u(t)=\left[\frac{1-e^{-at}}{a}\right]\,u(t) \end{align}\)

Examples

Exponential and Shifted Step

Find \(y(t)\) if \(x(t)=u(t-a)\) and \(h(t)=e^{-2t}u(t)\):

\(\begin{align} y(t)&=x(t)*h(t)\\ ~&=(u(t-a)) * (e^{-2t}u(t))\\ ~&=\delta(t-a) * u(t) * e^{-2t}u(t)\\ ~&=\delta(t-a) * \int_{-\infty}^{t} e^{-2\tau} u(\tau)~d\tau = \delta(t-a) * u(t)\int_{0}^{t} e^{-2\tau} ~d\tau\\ ~&=\delta(t-a) * \left( \frac{1-e^{-2t}}{2} \right)u(t)\\ ~&=\left( \frac{1-e^{-2(t-a)}}{2}\right) u(t-a) \end{align}\)

Questions

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External Links

References