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1. Convolution 1
SignalsandSystem Lecture06
Convolution
❖Definition
❖Properties
❖Applications
❖Operation

2. Convolution 2
SignalsandSystem Lecture06
Definition
The convolution of one function , i.e. the input signal, x(t)
with a second function, i.e. the impulse response, h(t) gives
the output, y(t) of a linear time-invariant system (LTI).
SYSTEM LTI
h(t)
Input signal
x(t)
Output signal
y(t)

3. Convolution 3
SignalsandSystem Lecture06
Definition
1.If the input signal is unit impulse function, δ(t). Then,
the output signal is equal with the impulse response
function of the system
y(t) = h(t)
2. Iftheunitimpulsefunctionhastimeshiftofδ(t-t0).
Then,thesystem’simpulseresponsealsohasatime
shiftt0.
y(t)=h(t-t0)

4. Convolution 4
SignalsandSystem Lecture06
Definition
3. The output for any input signal can be obtained from an
integration between the input signals and the system’s
impulse response.
Mathematically, convolution can be written as:
where * indicates convolution operation.
y(t) = x(t)  h(t) = −
x( )h(t − )d

5. Convolution 5
SignalsandSystem Lecture06
Convolution Properties

6. Convolution 6
SignalsandSystem Lecture06
STEP OPERATION
1Express each function in terms of a dummy variable τ.
2Reflect one of the functions: h(τ) →h(-τ).
3Add a time-offset, t, which allows h(t-τ) to slide along the τ-axis.
4Start t at -∞ and slide it all the way to +∞. Wherever the two
functions intersect, find the integral of their product.
5The resulting waveform is the convolution of functions f and h.
Convolution Operation

7. Convolution 7
SignalsandSystem Lecture06
Convolution Operation
Example 1:
Let x(t) be the input signal for an LTI system with the impulse response h(t), where:
x(t) = e
− at
u(t), a  0
and
h(t) = u(t)
Determine the output signal y(t) for the system.

8. Convolution 8
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
1Express each
function in terms of
a dummy variable τ.
x(τ)
τ
h(τ)
τ

9. Convolution 9
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
2Apply time reversal
one of the functions:
h(τ) →h(-τ).
h(τ)
τ

10. Convolution 10
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
3 Apply time shifting
operation, t, which
allows h(t-τ) to slide
along the τ-axis.

11. Convolution 11
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
4aStart t at -∞ and slide
it all the way to +∞.
Wherever the two
functions overlap,
find the integral of
their product.
At t < 0,
y(t) = x(t)  h(t)

= −
x( )h(t − )d
= 0

12. Convolution 12
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
4bStart t at -∞ and slide
it all the way to +∞.
Wherever the two
functions overlap,
find the integral of
their product.
At t > 0,

y(t) = 
−
x( )h(t − )d
t
= 
e
− a
(1)d
0
= -
1
e
− a
t
=
1
(1 − e
− at
)
a
0 a

13. Convolution 13
SignalsandSystem Lecture06
Convolution Operation
Solution 1:
STEP OPERATION VISUALIZATION
5The resulting
waveform is the
convolution of
functions f and h.
Therefore, for all t, output signal is:
y(t) =
1
(1 − e
− at
)u(t)
a
1/a)

14. Convolution 14
SignalsandSystem Lecture06
Convolution Operation
Example 2:
Given two following signals:
r(t) = u(t) − u(t −1)
s(t) = 2u(t + 2) − 2u(t − 2)
Signal y(t) is the product of convolution between the two
signals. Plot y(t).

15. Convolution 15
SignalsandSystem Lecture06
Convolution Operation
STEP OPERATION VISUALIZATION
1Express each
function in terms of
a dummy variable τ.
s(τ)
τ
r(τ)
τ
Solution 2:

16. Convolution 16
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
STEP OPERATION VISUALIZATION
2Flip one of the
functions: r(τ) →r(-τ)
r(τ) r(-τ)
τ
τ

17. Convolution 17
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
STEP OPERATION VISUALIZATION
3Add a time-offset, t,
which allows r(t-τ) to
slide along the τ-axis.
r(-τ) r(t-τ)
τ τ
t-1 t

18. Convolution 18
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
STEPOPERATION VISUALIZATION
4aStart t at -∞ and
slide it all the
way to +∞.
Wherever the
two functions
intersect, find
the integral of
their product.
At t <-2,
s( )
r(t − ) 2 y(t) = r(t)* s(t)

1
= 
−
s( )r(t − )d

t -1 t -2 0 2 = 
−
0d

= 0
t < -2

19. Convolution 19
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
STEPOPERATION VISUALIZATION
4bStart t at -∞ and
slide it all the
way to +∞.
Wherever the
two functions
intersect, find
the integral of
their product.
At -2 < t <-1,

2
y(t) = −
s( )r(t − )d
t
1
= −2
2(1)d
= 2
t
= 2(t + 2)
t -1 -2 t 0 2
−2
-2<t, t-1< -2 ; -2 < t < -1
t < -1

20. Convolution 20
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
VISUALIZATION
4c Start t at -∞ and
slide it all the
way to +∞.
Wherever the
two functions
intersect, find
the integral of
their product.
At – 1 < t <2,
-2 0 2-1
1
2
t-1 t
2(1)d

t−1
t−1
y(t) =s( )r(t − )d
=
= 2
t
= 2(t − t + 1)
= 2

−
t


21. Convolution 21
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
VISUALIZATIONVISUALIZATION
4d Start t at -∞ and
slide it all the
way to +∞.
Wherever the
two functions
intersect, find
the integral of
their product.
At 2 < t < 3,
-2 0
1
r(t-)
s()
2
1 t -1 2 t
2
= 2(3 − t)
=
y(t) =
t −1
t −1

−


= 2
2
= 2(2 − t +1)
2(1)d
s( )r(t − )d

22. Convolution 22
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
VISUALIZATIONSTEPOPERATION VISUALIZATION
4eStart t at -∞ and
slide it all the
way to +∞.
Wherever the
two functions
intersect, find
the integral of
their product.
At t > 3,
s()
2
y(t) = r(t)* s(t)

r(t-)
= 
s( )r(t − )d
1
−
= 0
-2 0 2 t -1 t

23. Convolution 23
SignalsandSystem Lecture06
Convolution Operation
Solution 2:
VISUALIZATIONSTEPOPERATION VISUALIZATION
5The resulting
waveform is
the

convolution of
functions f and
h.
0,

2(t + 2),

2,

 2(3 − t),
t  −2 & t  3
−1  t  2
2  t  3 y(t)
2
-2-10123
t

24. Convolution 24
SignalsandSystem Lecture06
The End
VISUALIZATION