Convolution

The convolution of two signals in the time domain is equivalent to the multiplication of their representation in frequency domain. Mathematically, we can write the convolution of two signals as

y(t)=x1(t)∗x2(t)$$y(t)=x_1(t)\ast x_2(t)$$

=∫∞−∞x1(p).x2(t−p)dp$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}_1(p).x_2(t-p)dp$$

Steps for convolution

• Take signal x₁(t) and put t = p there so that it will be x₁(p).
• Take the signal x₂(t) and do the step 1 and make it x₂(p).
• Make the folding of the signal i.e. x₂(-p).
• Do the time shifting of the above signal x₂[-(p-t)]
• Then do the multiplication of both the signals. i.e. x1(p).x2[−(p−t)]$x_1(p).x_2[-(p-t)]$

Example

Let us do the convolution of a step signal u(t) with its own kind.

y(t)=u(t)∗u(t)$y(t)=u(t)\ast u(t)$

=∫∞−∞[u(p).u[−(p−t)]dp$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}[u(p).u[-(p-t)]dp$

Now this t can be greater than or less than zero, which are shown in below figures

So, with the above case, the result arises with following possibilities

y(t)={0,∫t01dt,ift<0fort>0$y(t)=\{\begin{array}{ll}0,& ift<0\\ {\int }_0^{t}1dt,& fort>0\end{array}$

={0,t,ift<0t>0=r(t)$=\{\begin{array}{ll}0,& ift<0\\ t,& t>0\end{array}=r(t)$

Properties of Convolution

Commutative

It states that order of convolution does not matter, which can be shown mathematically as

x1(t)∗x2(t)=x2(t)∗x1(t)$$x_1(t)\ast x_2(t)=x_2(t)\ast x_1(t)$$

Associative

It states that order of convolution involving three signals, can be anything. Mathematically, it can be shown as;

x1(t)∗[x2(t)∗x3(t)]=[x1(t)∗x2(t)]∗x3(t)$$x_1(t)\ast [x_2(t)\ast x_3(t)]=[x_1(t)\ast x_2(t)]\ast x_3(t)$$

Distributive

Two signals can be added first, and then their convolution can be made to the third signal. This is equivalent to convolution of two signals individually with the third signal and added finally. Mathematically, this can be written as;

x1(t)∗[x2(t)+x3(t)]=[x1(t)∗x2(t)+x1(t)∗x3(t)]$$x_1(t)\ast [x_2(t)+x_3(t)]=[x_1(t)\ast x_2(t)+x_1(t)\ast x_3(t)]$$

Area

If a signal is the result of convolution of two signals then the area of the signal is the multiplication of those individual signals. Mathematically this can be written

If y(t)=x1∗x2(t)$y(t)=x_1\ast x_2(t)$

Then, Area of y(t) = Area of x₁(t) X Area of x₂(t)

Scaling

If two signals are scaled to some unknown constant “a” and convolution is done then resultant signal will also be convoluted to same constant “a” and will be divided by that quantity as shown below.

If, x1(t)∗x2(t)=y(t)$x_1(t)\ast x_2(t)=y(t)$

Then, x1(at)∗x2(at)=y(at)a,a≠0$x_1(at)\ast x_2(at)=\frac{y(at)}{a},a\ne 0$

Delay

Suppose a signal y(t) is a result from the convolution of two signals x1(t) and x2(t). If the two signals are delayed by time t1 and t2 respectively, then the resultant signal y(t) will be delayed by (t1+t2). Mathematically, it can be written as −

If, x1(t)∗x2(t)=y(t)$x_1(t)\ast x_2(t)=y(t)$

Then, x1(t−t1)∗x2(t−t2)=y[t−(t1+t2)]