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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
Steps for convolution
Example
Let us do the convolution of a step signal u(t) with its own kind.
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
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 individually 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 multiplication of those individual signals. Mathematically this can be written
If
Then, Area of y(t) = Area of x_{1}(t) X Area of x_{2}(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,
Then,
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,
Then,
Solved Examples
Example 1 − Find the convolution of the signals u(t1) and u(t2).
Solution − Given signals are u(t1) and u(t2). Their convolution can be done as shown below −
Example 2 − Find the convolution of two signals given by
Solution −
x_{2}(n) can be decoded as x_{2}(n) = {2,2,2,2,2} Original first
x_{1}(n) is previously given = {3,−2,3} = 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)
= {3−2Z^{−1} + 2Z^{−2}} × {2+2Z^{−1} + 2Z^{−2} + 2Z^{−3} + 2Z^{−4}}
= 6+2Z^{−1} + 6Z^{−2} + 6Z^{−3} + 6Z^{−4} + 6Z^{−5}
Taking inverse Ztransformation of the above, we will get the resultant signal as
x(n) = {6,2,6,6,6,0,4} Origin at the first
Example 3 − Determine the convolution of following 2 signals 
x(n) = {2,1,0,1}
h(n) = {1,2,3,1}
Solution −
Taking the Ztransformation 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 multiplication of their Ztransformations
That is
Y(Z) = X(Z) × h(Z)
= {2 + 2Z^{−1} + 2Z^{−3}} × {1 + 2Z^{−1} + 3Z^{−2} + Z^{−3}}
= {2 + 5Z^{−1} + 8Z^{−2} + 6Z^{−3} + 3Z^{−4} + 3Z^{−5} + Z^{−6}}
Taking the inverse Ztransformation, the resultant signal can be written as;
y(n) = {2,5,8,6,6,1} Original first
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