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.
$ \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*} $
- "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
Prescript: While $ u(t)*u(t)=t\,u(t)=r(t) $, the same does not hold for discrete. Instead, $ u[n]*u[n]=(n+1)\,u[n]=r[n]+u[n] $. Given that, the $ r[n] $ function is not used in the shortcuts below.
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 - if $ \alpha $ and/or $ \beta $ is 0, you might need to have a philosophical conversation with a Math person to decide what $ 0^0 $ is... If it is assumed that $ 0^0=1 $, then $ 0^nu[n]=\delta[n] $ and the formulas below work when $ \alpha $ and/or $ \beta $ is 0.
$ \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}=\beta^n\,u[n]\sum_{k=0}^{n} \left(\frac{\alpha}{\beta}\right)^k\\ &=\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)\,u[n]\mbox{ probably best form if $\alpha$<1, or}\\ &=\left(\frac{\alpha^{n+1}-1}{\alpha-1}\right)\,u[n]\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*} $
Geometric Series and Shifted Geometric Series
Find $ y[n] $ if:
$ \begin{align*} x[n]&=\left(\frac{1}{2}\right)^{n-2}u[n-2] & h[n]&=\left(\frac{1}{3}\right)^nu[n]\end{align*} $
The first step for this would be to pull the shift out of $ x[n] $ and rewriting it as:
$ \begin{align*} x[n]&=\delta[n-2]*\left(\frac{1}{2}\right)^{n}u[n]\end{align*} $
Then proceed as follows:
$ \begin{align*} y[n]&=x[n]*h[n]\\ ~&=\left(\delta[n-2]*\left(\frac{1}{2}\right)^{n}u[n]\right) * \left(\left(\frac{1}{3}\right)^nu[n]\right) & ~& \text{replace variables with formulas}\\ ~&=\delta[n-2]*\left(\left(\frac{1}{2}\right)^{n}u[n] * \left(\frac{1}{3}\right)^nu[n]\right) & ~& \text{convolution is associative}\\ ~&=\delta[n-2]*\left(\left(\frac{\left(\frac{1}{2}\right)^{n+1}-\left(\frac{1}{3}\right)^{n+1}}{\frac{1}{2}-\frac{1}{3}}\right)u[n]\right) & ~& \text{formula for convolution of two geometric series is known}\\ ~&=\left(\frac{\left(\frac{1}{2}\right)^{n-1}-\left(\frac{1}{3}\right)^{n-1}}{\frac{1}{2}-\frac{1}{3}}\right)u[n-2]& ~& \text{convolution of a shifted impulse with a function gives the shifted function}\end{align*} $
Shifting Issues
When dealing with shifts, in order to use the trick of replacing a shifted function with the convolution of a shifted impulse and an unshifted function, you must make sure that the variable has the same shift everywhere in the shifted function. For example, $ x(t)=e^{-2(t-3)}u(t-3) $ has $ t $ shifted by 3 everywhere, so you can write $ x(t)=\delta(t-3)*\left(e^{-2t}u(t)\right) $. If instead you had something like $ x(t)=e^{-2t}u(t-3) $, you would need to figure out how to make the shifts on $ t $ match the shift in the argument of the step:
$ \begin{align*} x(t)&=e^{-2t}u(t-3)\\ ~&=e^{-2(t{\color{red}-3}{\color{blue}+3})}u(t-3)\\ ~&=e^{\color{blue}-6}e^{-2(t{\color{red}-3})}u(t-3)\\ ~&=e^{\color{blue}-6}\left(\delta(t-3)*e^{-2t}u(t)\right) \end{align*} $
Questions
Post your questions by editing the discussion page of this article. Edit the page, then scroll to the bottom and add a question by putting in the characters *{{Q}}, followed by your question and finally your signature (with four tildes, i.e. ~~~~). Using the {{Q}} will automatically put the page in the category of pages with questions - other editors hoping to help out can then go to that category page to see where the questions are. See the page for Template:Q for details and examples.
External Links
References
