2.7 Convolution

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2.7 Convolution Convolution means folding. Convolution helps to determine the effect a system has on an input signal. It provides a means of viewing and characterizing physical systems. A convolution is an integral that expresses the amount of overlap of one

function g as it is shifted over another function f

The convolution of two signals consists of time-reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product.

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It is used in finding the response y(t) of a system to an excitation x(t), knowing the system impulse response h(t). This can be defined using :

or simply

dtxhdthxty )()()()()(

)(*)()(*)()( txththtxty

2.8.2 Convolution Integral

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We can establish link between LT and convolution integral. Eg: Given two functions f1(t) and f2(t) with LT of F1(s) and F2(s). The convolution will be:

dtfftftftft

)()()(*)()( 20

121 )()()](*)([)( 2121 sFsFtftfLsF

Eg: Given that and . Thus, the convolution will be :tetx 4)( teth 25)(

0)(202

201

20

14

25

)]()([)(*)(

21

1

1

tforeess

L

ssL

sXsHLtxth

tt

Example 2.8

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2.7.3 Steps to evaluate the convolution integral

1. Folding : take the mirror image of h(λ) about the ordinate axis to obtain h(-λ).

2. Displacement : shift or delay h(-λ) by t to obtain h(t-λ).

3. Multiplication : find the product of h (t-λ) and x(λ).

4. Integration : for a given time t, calculate the area under the product h (t-λ) x(λ) for 0<λ<t to get

y(t) at t.

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CONVOLUTION PROCESS

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Example 2.10 Find the convolution of the two signals [ x1(t) and x2(t) ]:

)(2 tx

1 2 3 t

1

)(1 tx

t

2

1

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Example 2.10 (..cont’d) 1st, we fold signal x1(t) or reflect

it at y-axis Then shift it by t

λ

)(1 x

2

0-1

(a)

λ

)(1 tx

2

0t -1 t

(b)

Now, multiply the two functions and integrate to determine the area of the overlapping region

For 0 < t < 1, there is no overlap of the two function.

Hence, y(t)=x1(t)*x2(t) = 0

)(1 tx

1 2 3 λ

1

)(2 x

tt-1

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Example 2.10 (..cont’d)

)1(22)1)(2()(1

1 tdtyt

t

For 1 < t < 2, the two signals overlap between t and 1.

For 2 < t < 3, the two signals overlap between (t-1) and t.

22)1)(2()(1

1

t

t

ttdty

)(1 tx

1 2 3 λ

1

)(2 x

t-1

2

t

)(1 tx

1 2 3 λ

1

)(2 x

t-1

2

t

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Example 2.10 (..cont’d)

)13(22)1)(2()(3

1

31

tdty

tt

tty 28)(

For 3 < t < 4, the two signals overlap between (t-1) and 3.

1 2 3 λ

1

t-1

)(1 tx

2 )(2 x

t 4

For t > 4, the two signals does not overlap.

1 2 3 λ

1

t-1

)(1 tx

2 )(2 x

t4

0)( ty

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Example 2.10 (..cont’d)

0,28

,2,22

,0

)(t

tty

Combining all the equations, we obtain:

443322110

ttttt )(ty

t

2

1 2 3 4

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Exercise 1Graphically, convolve the two functions.

,0

,26,

)( tt

ty

x1(t)

t

1

1

x2(t)

t

1

1 2

2

Answer:

otherwisett

3220

y(t)

t

1

1 2

2

3

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Exercise 2Given g(t) and f(t). Graphically find y(t) = g(t)*

f(t). g(t)

t

1

10

f(t)

t

3

0

te3

,0,)1(3

,1(3

)( t

t

ee

e

ty

Answer:

0≤ t ≤ 1

t ≥ 0

elsewhere