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