CT Convolution

Deriving the Convolution Integral

Consider an arbitrary input $x(t)$,

(1)

\begin{aligned} x(t)& \approx ...+x(-T_p)rect\bigg(\frac{t+T_p}{T_p}\bigg)+x(0)rect(t)+x(T_p)rect\bigg(\frac{t-T_p}{T_p}\bigg)+...\nonumber \\ & = \sum_{n=-\infty}^{\infty}x(nT_p)rect\bigg(\frac{t-nT_p}{T_p}\bigg)\nonumber \\ & = \sum T_px(nT_p)\bigg[\frac{1}{T_p}rect\bigg(\frac{t-nT_p}{T_p}\bigg) \bigg]\nonumber\\ \end{aligned}

Suppose $h_p$ is the response of the system to an input $\frac{1}{T_p}rect\bigg(\frac{t-nT_p}{T_p}\bigg)$.Then $y(t)$ corresponding to $x(t)$ is $y(t)=\sum T_px(nT_p)h_p(t-nT_p)$.
Taking limit $T_p$ $\rightarrow$ 0,then $\frac{1}{T_p}rect(\frac{t}{T_p})$ $\rightarrow$ $\delta(t)$.Let the impulse response be $h(t)$.Now, look at $y(t)=\sum T_px(nT_p)h_p(t-nT_p)$ and let $\tau=nT_p$.
Then, $y(t)=\int d\tau x(\tau)h(t-\tau)$.Therefore,

(2)

\begin{align} y(t)=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \end{align}

Notice that-

(3)

\begin{align} x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau \end{align}

We already know the above equation from the shifting property.

Examples

CT convolution can be best understood by the the following examples:-

Example 1

Find the convolution of $x(t)=u(t+3)$ and $h(t)=e^{-t}u(t)$.

Solution:

(4)

\begin{align} y(t)=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \end{align}

Now,

(5)

\begin{align} x(\tau)=u(\tau+3) \nonumber \end{align}

(6)

\begin{align} h(\tau)=e^{-\tau}u(\tau) \nonumber \end{align}

Let $w_t(\tau)=x(\tau)h(t-\tau)$

From the above figure, for $t<-3$,$w_t(\tau)=0$ when $-\infty < \tau < \infty$.

From the above figure, for $t>-3$,

(7)

\begin{cases} w_t(\tau)= e^{\tau-t} & \text{if}\ -3 \leq \tau \leq t \\ \nonumber 0 & \text{otherwise} \nonumber \end{cases}

Therefore,

(8)

\begin{cases} y(t)= 0 & \text{if}\ t<-3 \\ \nonumber \int_{-3}^{t}e^{\tau-t}d\tau & \text{if}\ t \geq -3 \\ \nonumber \end{cases}

So, for $t \geq -3$,

(9)

\begin{align} y(t)=e^{\tau-t}\big|_{-3}^{t}=1-e^{-3-t} \nonumber \end{align}

Example 2

Find the convolution of $x(t)=u(t+1)-u(t-1)$ and $h(t)=e^{-t} [u(t)-u(t-1)]$

Solution:

For $t<-1$, $w_t(\tau)=0 \Rightarrow y(t)=0$
For $-1 \leq t <0$ and $-1 \leq \tau \leq t$, $w_t(\tau)=e^{\tau-t} \Rightarrow y(t)=\int_{-1}^{t}e^{\tau-t}d\tau=e^{\tau-t}\big|_{-1}^{t}=1-e^{-1-t}$
For $0 \leq t <1$ and $-1+t \leq \tau \leq t$, $w_t(\tau)=e^{\tau-t}\Rightarrow y(t)=\int_{-1+t}^{t}e^{\tau-t}d\tau=e^{\tau-t}\big|_{-1+t}^{t}=1-e^{-1}$

For $1 \leq t <2$ and $-1+t \leq \tau \leq 1$, $w_t(\tau)=e^{\tau-t}\Rightarrow y(t)=\int_{-1+t}^{1}e^{\tau-t}d\tau=e^{\tau-t}\big|_{-1+t}^{1}=e^{-1-t}-e^{-1}$
For $2 \leq t$ and $-\infty \leq \tau \leq \infty$, $w_t(\tau)=0\Rightarrow y(t)=0$
So,

(10)

\begin{cases} y(t)= 0 & \text{if}\ t<-1 \\ \nonumber 1-e^{-1-t}& \text{if}\ -1\leq t \leq 0 \\ \nonumber 1-e^{-1}& \text{if}\ 0\leq t \leq 1 \\ \nonumber e^{-1-t}-e^{-1} & \text{if}\ 1\leq t \leq 2 \\ \nonumber 0 & \text{if}\ t > 2 \\ \nonumber \end{cases}

Example 3