Convolution Shortcuts

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.

Continuous Functions

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

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

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" exponential with a step

\(\begin{align} (e^{-at}\,u(t))*(u(t))&=(e^{-at}\,u(t))*(e^{-0t}\,u(t))\\ &=\left(\frac{e^{-at}-e^{-0t}}{0-a}\right)\,u(t)=\left(\frac{1-e^{-at}}{a}\right)\,u(t) \end{align}\)

Discrete Functions

Convolution with Impulses

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

Convolution with Multiple Steps

\(\begin{align} u[n]*f[n]&=\sum_{k=-\infty}^{n}f[k]\\ u[n]*u[n]*f[n]&=\sum_{k=-\infty}^{n}\sum_{l=-\infty}^{k}f[l]\\ \end{align}\)

Convolution Between Singularity Functions

\(\begin{align} u[n]*u[n]&=(n+1)\,u[n]\\ u[n]*u[n]*u[n]&=\frac{(n+1)(n+2)}{2}\,u[n]\\ u[n]*u[n]*u[n]*u[n]&=\frac{(n+1)(n+2)(n+3)}{6}\,u[n]\\ \mbox{equivalent of }M\mbox{ steps convolved together}&=\left(\frac{1}{(M-1)!}\prod_{k=1}^{M-1}(n+k)\right)\,u[n] \end{align}\)

Convolution Between Geometric Series

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

• Same geometric constants

\(\begin{align} \left(\alpha^n\,u[n]\right)*\left(\alpha^n\,u[n]\right)&=\sum_{k=-\infty}^{\infty}\alpha^k\,u[k]\,\alpha^{n-k}\,u[n-k]\\ &=u[n]\sum_{k=0}^{n}\alpha^k\,\alpha^{n-k}=u[n]\sum_{k=0}^{n}\alpha^k\,\alpha^n\,\alpha^{-k}\\ &=\alpha^n u[n]\sum_{k=0}^{n} \alpha^k\,\alpha^{-k}=\alpha^n\,u[n]\sum_{k=0}^{n} 1\\ &=(n+1)\,\alpha^n\,u[n]\\ \end{align}\)

• Different geometric constants

\(\begin{align} \left(\alpha^n\,u[n]\right)*\left(\beta^n\,u[n]\right)&=\sum_{k=-\infty}^{\infty}\alpha^k\,u[k]\,\beta^{n-k}\,u[n-k]\\ &=u[n]\sum_{k=0}^{n}\alpha^k\,\beta^{n-k}=u[n]\sum_{k=0}^{n}\alpha^k\,\beta^n\,\beta^{-k}\\ &=\beta^n u[n]\sum_{k=0}^{n} \alpha^k\,\beta^{-k}=\alpha^n\,u[n]\sum_{k=0}^{n} \left(\frac{\alpha}{\beta}\right)^n\\ &=\beta^n\,u[n]\left(\frac{1-\left(\frac{\alpha}{\beta}\right)^{n+1}}{1-\frac{\alpha}{\beta}}\right)\\ &=\beta^{n+1}\,u[n]\left(\frac{1-\left(\frac{\alpha}{\beta}\right)^{n+1}}{\beta-\alpha}\right)\\ &=\left(\frac{\beta^{n+1}-\alpha^{n+1}}{\beta-\alpha}\right)\,u[n]=\left(\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}\right)\,u[n]\mbox{ Use larger value between $\alpha$ and $\beta$ to decide who goes first}\\ \end{align}\)

• Geometric series with a step

\(\begin{align} (\alpha^n\,u[n])*(u[n])&=(\alpha^n\,u[n])*(1^n\,u[n])\\ &=\left(\frac{1^{n+1}-\alpha^{n+1}}{1-\alpha}\right)\,u[n]\\ &=\left(\frac{1-\alpha^{n+1}}{1-\alpha}\right)\mbox{ probably best form if $\alpha$<1, or}\\ &=\left(\frac{\alpha^{n+1}-1}{\alpha-1}\right)\mbox{ if $\alpha$>1}\\ \end{align}\)

Examples

Exponential and Shifted Step

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

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

Questions

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