The following is an example of convolving two signals; the convolution is done several different ways:
- Math... So much math.
- Using Convolution Shortcuts
- Geometrically, flipping and shifting \(h(t)\)
- Geometrically, flipping and shifting \(x(t)\)
Setup
The goal for this problem is to determine the output \(y(t)\) created by an input \(x(t)\) for a linear time invariant system if the system's impulse response \(h(t)\) is known.
In this particular example:
\(
\begin{align}
x(t)&=2u(t)-u(t-1)-u(t-3)\\
h(t)&=e^{-t}u(t)\\
y(t)&=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=\int_{-\infty}^{\infty}x(t-\tau)h(\tau)d\tau
\end{align}
\)
where the particular choice of which integral to use is up to the user.
Math
Given the relative complexity of \(x(t)\), it may make more sense to use the first form, where \(h(t)\) is flipped and shifted. That yields:
\(
\begin{align}
y(t)&=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\\
y(t)&=\int_{-\infty}^{\infty}\left(2u(\tau)-u(\tau-1)-u(\tau-3)\right)\left(e^{-(t-\tau)}u(t-\tau)\right)d\tau
\end{align}
\)
Distributing terms gives:
\(
\begin{align}
y(t)&=\int_{-\infty}^{\infty}2e^{-(t-\tau)}u(\tau)u(t-\tau)d\tau\dots\\
~&~-\int_{-\infty}^{\infty}e^{-(t-\tau)}u(\tau-1)u(t-\tau)d\tau\dots\\
~&~-\int_{-\infty}^{\infty}e^{-(t-\tau)}u(\tau-3)u(t-\tau)d\tau\\
\end{align}
\)
Taking the first part alone:
\(
\int_{-\infty}^{\infty}{\color{Brown}2e^{-(t-\tau)}}{\color{Red}u(\tau)}{\color{Blue}u(t-\tau)}d\tau
\)
note that the integrand is only non-zero when two conditions are met simultaneously:
\(
\begin{align}
{\color{Red}0}&{\color{Red}<\tau} & {\color{Blue}\tau}&{\color{Blue}<t}
\end{align}
\)
Furthermore, those conditions can only be met simultaneously if the outer conditions are met; that is, if:
\(
\begin{align}
{\color{Purple}0}&{\color{Purple}<t}
\end{align}
\)
These three conditions lead to three changes. The limits on \(\tau\) change the limits of the integral with respect to \(\tau\) while the condition on \(t\) can be represented by a unit step of \(t\). In other words:
\(
\begin{align}
\int_{-\infty}^{\infty}{\color{Brown}2e^{-(t-\tau)}}{\color{Red}u(\tau)}{\color{Blue}u(t-\tau)}d\tau &=
{\color{Purple}u(t)}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Brown}2e^{-(t-\tau)}}d\tau\\
{\color{Purple}u(t)}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Brown}2e^{-(t-\tau)}}d\tau&=
u(t)\left[2e^{-(t-\tau)} \right]_0^{t}=\left(2-2e^{-t}\right)u(t)
\end{align}
\)
Note with that last line that the integral is with respect to \(\tau\) and the exponent of \(\tau\) is +1. The other two components of the convolution integral can be processed in a similar way:
\(
\begin{align}
\int_{-\infty}^{\infty}{\color{Brown}-e^{-(t-\tau)}}{\color{Red}u(\tau-1)}{\color{Blue}u(t-\tau)}d\tau &=
{\color{Purple}u(t-1)}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau\\
{\color{Purple}u(t-1)}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau&=
-u(t-1)\left[e^{-(t-\tau)} \right]_1^{t}=-\left(1-1e^{-(t-1)}\right)u(t-1)
\end{align}
\)
and
\(
\begin{align}
\int_{-\infty}^{\infty}{\color{Brown}-e^{-(t-\tau)}}{\color{Red}u(\tau-3)}{\color{Blue}u(t-\tau)}d\tau &=
{\color{Purple}u(t-3)}\int_{{\color{Red}3}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau\\
{\color{Purple}u(t-3)}\int_{{\color{Red}3}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau&=
-u(t-3)\left[e^{-(t-\tau)} \right]_3^{t}=-\left(1-1e^{-(t-3)}\right)u(t-3)
\end{align}
\)
meaning (finally) that:
\(
\begin{align}
y(t)=x(t)*h(t)=\left(2-2e^{-t}\right)u(t)-\left(1-1e^{-(t-1)}\right)u(t-1)-\left(1-1e^{-(t-3)}\right)u(t-3)
\end{align}
\)
Shortcuts
Full color version:
To use the shortcuts, first distribute the various components of the convolution:
\(
\begin{align}
x(t)&={\color{RedOrange}2u(t)-u(t-1)-u(t-3)}\\
h(t)&={\color{Blue}e^{-t}u(t)}\\
x(t)*h(t)&={\color{RedOrange}2u(t)}*{\color{Blue}e^{-t}u(t)}{\color{RedOrange}-u(t-1)}*{\color{Blue}e^{-t}u(t)}{\color{RedOrange}-u(t-3)}*{\color{Blue}e^{-t}u(t)}
\end{align}
\)
These can be rearranged, pulling out any shifts, as:
\(
\begin{align}
x(t)*h(t)&={\color{Orange}2}{\color{Blue}e^{-t}u(t)}*{\color{Red}u(t)}{\color{Orange}-}{\color{Blue}e^{-t}u(t)}*{\color{Red}u(t)}{\color{Orange}*\delta(t-1)}{\color{Orange}-}{\color{Blue}e^{-t}u(t)}*{\color{Red}u(t)}{\color{Orange}*\delta(t-3)}
\end{align}
\)
Note that each component now has a convolution of
\(
\begin{align}
{\color{Blue}e^{-t}u(t)}*{\color{Red}u(t)}
\end{align}
\)
in it. Convolution with a step generates an integral, and integrating the right-sided exponential yields:
\(
\begin{align}
{\color{Red}\int_{-\infty}^{t}}{\color{Blue}e^{-\tau}u(\tau)}d\tau=u(t)\left[-e^{-\tau}\right]_0^t={\color{Fuchsia}(1-e^{-t})u(t)}
\end{align}
\)
which means the rearranged functions can be re-written as:
\(
\begin{align}
x(t)*h(t)&={\color{Orange}2}{\color{Fuchsia}(1-e^{-t})u(t)}{\color{Orange}-}{\color{Fuchsia}(1-e^{-t})u(t)}{\color{Orange}*\delta(t-1)}{\color{Orange}-}{\color{Fuchsia}(1-e^{-t})u(t)}{\color{Orange}*\delta(t-3)}
\end{align}
\)
Now the shifts can be applied to give:
\(
\begin{align}
x(t)*h(t)&=2(1-e^{-t})u(t)-(1-e^{-(t-1)})u(t-1)-(1-e^{-(t-3)})u(t-3)
\end{align}
\)
Black and white only:
To use the shortcuts, first distribute the various components of the convolution:
\(
\begin{align}
x(t)&={\color{Black}2u(t)-u(t-1)-u(t-3)}\\
h(t)&={\color{Black}e^{-t}u(t)}\\
x(t)*h(t)&={\color{Black}2u(t)}*{\color{Black}e^{-t}u(t)}{\color{Black}-u(t-1)}*{\color{Black}e^{-t}u(t)}{\color{Black}-u(t-3)}*{\color{Black}e^{-t}u(t)}
\end{align}
\)
These can be rearranged, pulling out any shifts, as:
\(
\begin{align}
x(t)*h(t)&={\color{Black}2}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}-}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}*\delta(t-1)}{\color{Black}-}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}*\delta(t-3)}
\end{align}
\)
Note that each component now has a convolution of
\(
\begin{align}
{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}
\end{align}
\)
in it. Convolution with a step generates an integral, and integrating the right-sided exponential yields:
\(
\begin{align}
{\color{Black}\int_{-\infty}^{t}}{\color{Black}e^{-\tau}u(\tau)}d\tau=u(t)\left[-e^{-\tau}\right]_0^t={\color{Black}(1-e^{-t})u(t)}
\end{align}
\)
which means the rearranged functions can be re-written as:
\(
\begin{align}
x(t)*h(t)&={\color{Black}2}{\color{Black}(1-e^{-t})u(t)}{\color{Black}-}{\color{Black}(1-e^{-t})u(t)}{\color{Black}*\delta(t-1)}{\color{Black}-}{\color{Black}(1-e^{-t})u(t)}{\color{Black}*\delta(t-3)}
\end{align}
\)
Now the shifts can be applied to give:
\(
\begin{align}
x(t)*h(t)&=2(1-e^{-t})u(t)-(1-e^{-(t-1)})u(t-1)-(1-e^{-(t-3)})u(t-3)
\end{align}
\)
