CT Convolution
Impulse Response DT Convolution CT Convolution Properties of LTI Systems

Deriving the Convolution Integral

Consider an arbitrary input $x(t)$,

\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,

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

Notice that-

\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.


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

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