What is Convolution in Signals and Systems?

Signals and SystemsElectronics & ElectricalDigital Electronics

What is Convolution?

Convolution is a mathematical tool to combining two signals to form a third signal. Therefore, in signals and systems, the convolution is very important because it relates the input signal and the impulse response of the system to produce the output signal from the system. In other words, the convolution is used to express the input and output relationship of an LTI system.

Explanation

Consider a continuous-time LTI system which is relaxed at t = 0, i.e., initially,no input is applied to it. Now, if the impulse signal [δ(t)] is input to the system, then output of the system is called the impulse response h(t) of the system and is given by,

$$\mathrm{h(t)=T[\delta(t)]}$$

As any arbitrary signal x(t) can be represented as −

$$\mathrm{x(t)=\int_{-\infty}^{\infty}x(\tau)\:\delta (t-\tau)d\tau}$$

Then, the output of the system corresponding to x(t) is given by,

$$\mathrm{y(t)=T[x(t)]}$$

$$\mathrm{\Rightarrow y(t)=T\left [\int_{-\infty}^{\infty}x(\tau) \:\delta (t-\tau )d\tau\right ]}$$

For a continuous-time linear system, the output is given by,

$$\mathrm{y(t)=\int_{-\infty}^{\infty}x(\tau)\:T\left [\delta( t-\tau )\right ]d\tau\:\:\:\:\:\:...(1)}$$

Now, if the response of the system due to impulse signal δ(t) is h(t), then the response of the linear system due to delayed impulse signal is given by,

$$\mathrm{h(t,\tau)=T[\delta(t-\tau)]}$$

By substituting the value of $T[\delta(t-\tau)]$ in the equation (1), we get,

$$\mathrm{y(t)=\int_{-\infty}^{\infty}x(\tau)\:h(t,\tau)d\tau\:\:\:\:\:\:...(2)}$$

Again, for a time-invariant system, the output corresponding to the input delayed by τ units is equal to the output delayed by τ units, i.e.,

$$\mathrm{h(t,\tau)=h(t-\tau)}$$

On putting the value of $h(t,\tau)$ in the equation (2), we have,

$$\mathrm{y(t)=\int_{-\infty}^{\infty}x(\tau)\:h(t-\tau)d\tau\:\:\:\:\:\:...(3)}$$

The expression in equation(3) is called the convolution integral or simply convolution.

Therefore, the convolution of two continuous-time signals x(t) and h(t) is represented as,

$$\mathrm{y(t)=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)\:h(t-\tau)d\tau}$$

The limits of the integration in the convolution integral depends on whether the arbitrary signal x(t) and the impulse response h(t) are causal or not. Therefore,

  • If both x(t)and h(t) are non-causal, then,

$$\mathrm{y(t)=\int_{-\infty}^{\infty}x(\tau)\:h(t-\tau)d\tau}$$

  • If the signal x(t) is non-causal and the impulse response h(t) is causal, then,

$$\mathrm{y(t)=\int_{-\infty}^{t}x(\tau)\:h(t-\tau)d\tau}$$

  • If the signal x(t) is causal and h(t) is non-causal, then

$$\mathrm{y(t)=\int_{0}^{\infty}x(\tau)\:h(t-\tau)d\tau}$$

  • If both x(t) and h(t) are causal, then,

$$\mathrm{y(t)=\int_{0}^{t}x(\tau)\:h(t-\tau)d\tau}$$

raja
Updated on 15-Dec-2021 06:48:37

Advertisements