DSP - Operations on Signals Convolution

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

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 multippcation of both the signals. i.e. $x_{1}(p).x_{2}[−(p−t)]$

Example

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

$y(t) = u(t)*u(t)$

$= int_{-infty}^{infty}[u(p).u[-(p-t)]dp$

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 possibipties

$y(t) = egin{cases}0, & ifquad t<0\int_{0}^{t}1dt, & forquad t>0end{cases}$

$= egin{cases}0, & ifquad t<0\t, & t>0end{cases} = r(t)$

Properties of Convolution

Commutative

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

Associative

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

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 inspanidually with the third signal and added finally. Mathematically, this can be written as;

Area

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

If $y(t) = x_{1}*x_{2}(t)$

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

Scapng

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 spanided by that quantity as shown below.

If, $x_{1}(t)*x_{2}(t) = y(t)$

Then, $x_{1}(at)*x_{2}(at) = frac{y(at)}{a}, a e 0$

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, $x_{1}(t)*x_{2}(t) = y(t)$

Then, $x_{1}(t-t_{1})*x_{2}(t-t_{2}) = y[t-(t_{1}+t_{2})]$

Solved Examples

Example 1 − Find the convolution of the signals u(t-1) and u(t-2).

Solution − Given signals are u(t-1) and u(t-2). Their convolution can be done as shown below −

$y(t) = u(t-1)*u(t-2)$

$y(t) = int_{-infty}^{+infty}[u(t-1).u(t-2)]dt$

$= r(t-1)+r(t-2)$

$= r(t-3)$

Example 2 − Find the convolution of two signals given by

$x_{1}(n) = lbrace 3,-2, 2 brace $

$x_{2}(n) = egin{cases}2, & 0leq nleq 4\0, & x > elsewhereend{cases}$

Solution −

x₂(n) can be decoded as $x_{2}(n) = lbrace 2,2,2,2,2 brace Originalfirst$

x₁(n) is previously given $= lbrace 3,-2,3 brace = 3-2Z^{-1}+2Z^{-2}$

Similarly, $x_{2}(z) = 2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}$

Resultant signal,

$X(Z) = X_{1}(Z)X_{2}(z)$

$= lbrace 3-2Z^{-1}+2Z^{-2} brace imes lbrace 2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4} brace$

$= 6+2Z^{-1}+6Z^{-2}+6Z^{-3}+6Z^{-4}+6Z^{-5}$

Taking inverse Z-transformation of the above, we will get the resultant signal as

$x(n) = lbrace 6,2,6,6,6,0,4 brace$ Origin at the first

Example 3 − Determine the convolution of following 2 signals −

$x(n) = lbrace 2,1,0,1 brace$

$h(n) = lbrace 1,2,3,1 brace$

Solution −

Taking the Z-transformation of the signals, we get,

$x(z) = 2+2Z^{-1}+2Z^{-3}$

And $h(n) = 1+2Z^{-1}+3Z^{-2}+Z^{-3}$

Now convolution of two signal means multippcation of their Z-transformations

That is $Y(Z) = X(Z) imes h(Z)$

$= lbrace 2+2Z^{-1}+2Z^{-3} brace imes lbrace 1+2Z^{-1}+3Z^{-2}+Z^{-3} brace$

$= lbrace 2+5Z^{-1}+8Z^{-2}+6Z^{-3}+3Z^{-4}+3Z^{-5}+Z^{-6} brace$

Taking the inverse Z-transformation, the resultant signal can be written as;

$y(n) = lbrace 2,5,8,6,6,1 brace Originalfirst$