Signals and Systems 1
Lecture 9
Dr. Ali. A. Jalali
September 9, 2002
Signals and Systems 1
Lecture # 9
Convolution
EE 327 fall 2002
Example for total response of system
Total response = ZIR + ZSR
+
R
f(t)
C
y(t)
-
1 t
y (t )  Rf (t )   f ( )d
C 
1 0
1 t
y (t )  Rf (t )   f ( )d   f ( )d
C 
C 0
1 t
y (t )  Rf (t )  vc (t0 )   f ( )d
C 0
IC response, Force response and Steady state response
EE 327 fall 2002
Signals and Systems 1
The Unit Impulse Response Model
1.
The unit impale function is defined implicitly by
its sifting property:



 (t  t0 ) f (t )dt  f (t0 )
where f(t) is assumed to be continuous at t  t0 .
pt  .
Approximation to impulse function: t  t 0   lim
0
EE 327 fall 2002
Signals and Systems 1
The Unit Impulse Response Model
1.
Using the sifting property lead the
following product function.
Approximation of the sifting property


The value of integral is: f t 0   .   f t 0 
EE 327 fall 2002
Signals and Systems 1
Unit Impulse Response
The response of an LTI system to an input
of unit impulse function is called the unit
impulse response.
x(t)=(t)
LTI
y(t)=h(t)
Important: When determining the unit impulse
response h(t) of an LTI system, it is
necessary to make all initial conditions
zero. (output due to input, not energy stored in system)
EE 327 fall 2002
Signals and Systems 1
Convolution
If the unit impulse response h(t) of a linear
continuous system is known, the system
output y(t) can be found for any input x(t).
Approximation by pulse

Sifting property of pulse
EE 327 fall 2002
x(t )   x( ) f (t   )d

Signals and Systems 1
Convolution Integral
1. The convolution integral is one of the most important
results used in the study of the response of linear
systems.
2. If we know the unit impulse response h(t) for a linear
system, by using the convolution integral we can
compute the system output for any known input x(t).
3. In the following integration integral h(t) is the
system’s unit impulse response.

x(t )   x( )h(t   )d

EE 327 fall 2002
Signals and Systems 1
Convolution Evaluation
1. The convolution integral can be evaluated in three
distinct ways.
a) Analytical method,
b) Graphical method,
c) Numerical convolution
We will discuss about these and about convolution
properties in class. (see class notes)
EE 327 fall 2002
Signals and Systems 1
Convolution
1. The main convolution theorem
states that the response of a
system at rest (zero initial
condition) due to any input is
the convolution of that input
and the system impulse
response.
The Properties of Convolution
1. Commutatively
f1 (t) * f 2 (t)  f 2 (t) * f1 (t)
2. Distributivity
f1 (t) *[f 2 (t)  f 3 (t)]  f1 (t) * f 2 (t)  f1 (t) * f 3 (t)
3. Associatively
f1 (t) *{f 2 (t) * f 3 (t)}  {f1 (t) * f 2 (t)} * f 3 (t)
The Properties of Convolution
1. Duration: If duration of f1 (t) and f 2 (t)
respectively are [t1 , T1 ] and [t2, T2 ]
then,
f (t )  f1 (t ) * f 2 (t )
 0, t  t1  t 2
T1 T2

   f1 ( ) f 2 (t   )d , t1  t 2  t  T1  T2
 t1 t2
 0, t  T1  T2
The Properties of Convolution
1. Time shifting: If
f(t)  f 2 (t) * f1 (t)
Then, convolution of shifted signals are:
f1 (t -  1 ) * f 2 (t)  f(t -  1 )
f1 (t) * f 2 (t -  2 )  f(t -  2 )
f1 (t -  1 ) * f 2 (t -  2 )  f(t -  1 -  2 )
2. Continuity:
This property simply states\that the convolution is a
continuous function of the parameter t.
Graphical Convolution
1. This is a good method for understanding every step
in the convolution procedure.

f(t)  f1 (t) * f 2 (t)   f1 ( )f 2 (t -  )d ,    t  
Example:
-
f 2 (t)
2
f1 (t)
1
t
0
3
t
0
1
Graphical Convolution
Step 1: Duration property {f1 (t )
[t1, T1]=[0, 3]
and f 2 (t )
[t2, T2]=[0, 1]}
Convolution of these two signals is zero in the
following intervals.
f1 (t) * f 2 (t)  0, t  t1  t 2  0  0  0
f1 (t) * f 2 (t)  0, t  T1  T2  1  3  4
Therefore we need to evaluate the integral in the
interval of
0  t  4.
Graphical Convolution
Step 2: flip the signal that has the simpler shape
about the vertical axis. Convolution for t = 0 is:
f 2 (- )
f1 ( )
2
1

0
3

-1
0
Graphical Convolution
Step 3: shift signal f 2 (- ) to the left and to the right,
we form the signal f 2 (t - ) for t  (, 0)
and t  (0, )
f 2 (t - )
2
f1 ( )
1
0

3

-1+t
t
Start shifting the signal f 2 (t -  ) to the right (t>0).
0
Graphical Convolution
f1 ( )
1
0
3

f 2 (t - )
2

-1+t 0 t
t
f1 (t) * f 2 (t)   1 2d  2t, 0  t  1
0
Graphical Convolution
f1 ( )
1
0
-1+t
t 3

f 2 (t - )
2

0
t
-1+t
t
3
f1 (t) * f 2 (t)   1 2d  2, 0  t  3
t -1
Graphical Convolution
f1 ( )
1

0
2
-1+t 3 t
f 2 (t - )

0
3
-1+t 3 t
4
f1 (t) * f 2 (t)   1 2d  8 - 2t, 3  t  4
t -1
Graphical Convolution
f1 ( )
1
0
-1+t 3 t

f 2 (t - )
2

0
3 -1+t
3
t
f1 (t) * f 2 (t)   1 0d  0, t  4
t -1
Summary of solution
t0
 0,
 2t ,
0  t 1

f1 (t) * f 2 (t)   2,
1 t  3
8  2t , 3  t  4

 0,
t4
f(t)
2
t
0
1
2
3
4