Spring 2008
Linear Systems and Signals
Lecture 4
Continuous-Time Convolution
Useful Functions
• Unit gate function (a.k.a. unit pulse function)
rect(x)
1
-1/2
0
1/2
x

0
 1
rect  x   
2
1

• What does rect(x / a) look like?
• Unit triangle function
(x)
1
-1/2
0
1/2
x

 0
 x   
1  2 x

1
2
1
x
2
1
x
2
x
1
2
1
x
2
x
4-2
Unit Impulse (Functional)
• Mathematical idealism for
an instantaneous event
• Dirac delta as generalized
function (a.k.a. functional)
Unit area:
Sifting






P (t ) 

provided g(t) is defined at t = 0

1
if a  0
Scaling:  d (at ) dt 
a
• Note that d(0) is undefined
1
2

t
d t   lim P t 
d (t ) dt  1
g (t )d (t ) dt  g (0)
1
 t 
rect 

2
 2 
 0
P (t ) 
1  t 
 
  2 
1



d t   lim P t 
 0
t
4-3
Unit Impulse (Functional)
• Generalized sifting
Assuming that a > 0
1 if  a  T  a

d
(
t

T
)
dt


a
0 if T  a or T  a
a
• By convention, plot Dirac delta as arrow at origin
Undefined amplitude at origin
Denote area at origin as (area)
Height of arrow is irrelevant
Direction of arrow indicates sign of area
d t 
(1)
t
0
• With d(t) = 0 for t  0, it is tempting to think
f(t) d(t) = f(0) d(t)
f(t) d(t-T) = f(T) d(t-T)
Simplify unit impulse
under integration only
4-4
Unit Impulse (Functional)
• We can simplify d(t)
under integration

  t d t dt   0

• What about?
1
  t d t dt  ?

Answer: 0
• What about?

  t d t  T dt  ?

By substitution of variables,



 t  T d t dt   T 
• Other examples

 j t


d
t
e
dt  1


 t 


d
t

2
cos
  dt  0





 4 
e 2 x t  d 2  t  dt  e 2 x 2 
• What about at origin?
0
 d t  dt  ?
0
 d t  dt  0

0
 d t  dt  1

4-5
Unit Impulse (Functional)
• Relationship between unit impulse and unit step

t

d   d

0

?

1

u t 
t0
t 0
t 0
du
 d t 
dt
• What happens at the origin for u(t)?
u(0-) = 0 and u(0+) = 1, but u(0) can take any value
Common values for u(0) are {0, ½, 1}
u(0) = ½ is used in impulse invariance filter design:
L. B. Jackson, “A correction to impulse invariance,” IEEE Signal
Processing Letters, vol. 7, no. 10, Oct. 2000, pp. 273-275.
4-6
LTI Systems
• Recall for LTI system with input f and output y
– Homogeneity: c f(t)  c y(t)
– Time-invariance: c f(t-T)  c y(t-T)
m
m
– Adding additivity: F t    ci f t  Ti    ci y t  Ti   Y t 
i 1
i 1
• If a signal ( F(t) ) can be expressed as a sum of
shifted ( t - Ti ) and weighted ( ci ) copies of a
simpler signal ( f(t) ), we easily find a system’s
output ( Y(t) ) to that signal if we only know
system’s output ( y(t) ) to that simpler signal
• A common choice for f(t) is the impulse
4-7
Impulse Response
• Impulse response of a system is response of the
system to an input that is a unit impulse (i.e., a
Dirac delta functional in continuous time)
• Linear constant coefficient differential equation
Q(D) y (t )  P( D) f (t )
Lathi (2.16a)
• When initial conditions are zero, this differential
equation is LTI and system has impulse response
h(t )  b0 d (t )  P( D) y0 (t ) u(t )
Lathi (2.23)
b0 is coefficient (could be 0) of b0 DN f(t) on right-hand side
N is highest order of derivative in differential equation
4-8
Impulse Response
• In following plug, where did bn come from?
h(t )  b0 d (t )  P( D) y0 (t ) u(t )
Lathi (2.23)
• In solving these differential equations for t  0,
f (t )  g (t ) u (t )
y (t )  m(t ) u (t )
• Funny things happen to y’(t) and y”(t)
y ' (t )  m' (t ) u (t )  m(t ) d (t )
y" (t )  m" (t ) u (t )  2 m' (t ) d (t )  m(t ) d ' (t )
• In differential equations class, solved for m(t)
– Likely ignored d(t) and d’(t) terms
– Solution for m(t) is really valid for t  0+
4-9
System Response
F(t)
• Signals as sum of impulses
 t  n 


F
n


rect







n  

 t  n
F t   lim  F n  rect 
 0
 
n  
F t  

t

t=n


   F   d t    d

• But we know how to calculate the impulse
response ( h(t) ) of a system expressed as a
differential equation




F t    F   d t    d   F   ht    d  Y (t )
• Therefore, we know how to calculate the system
output for any input, F(t)
4 - 10
Convolution Integral
• Commonly used in engineering, science, math
f1 t   f 2 t    f1   f 2 t    d

• Convolution properties
–
–
–
–
Commutative: f1(t) * f2(t) = f2(t) * f1(t)
Distributive: f1(t) * [f2(t) + f3(t)] = f1(t) * f2(t) + f1(t) * f3(t)
Associative: f1(t) * [f2(t) * f3(t)] = [f1(t) * f2(t)] * f3(t)
Shift: If f1(t) * f2(t) = c(t), then
f1(t) * f2(t - T) = f1(t - T) * f2(t) = c(t - T).
– Convolution with impulse, f(t) * d(t) = f(t)
– Convolution with shifted impulse, f(t) * d(t-T) = f(t-T)
4 - 11
important later in modulation
Graphical Convolution Methods
• From the convolution integral, convolution is
equivalent to
f 1 t   f 2 t  



f 1   f 2 t   d
– Rotating one of the functions about the y axis
– Shifting it by t
– Multiplying this flipped, shifted function with the other
function
– Calculating the area under this product
– Assigning this value to f1(t) * f2(t) at t
4 - 12
Graphical Convolution Example
• Convolve the following two functions:
f(t)
g(t)
3
2
*
t
t
2
-2
2
• Replace t with  in f(t) and g(t)
• Choose to flip and slide g() since it is simpler
and symmetric
3
g(t-)
• Functions overlap like this:
2
f()
-2 + t
2+t
2

4 - 13
Graphical Convolution Example
•
Convolution can be divided into 5 parts
I.
3
t < -2
•
•
Two functions do not overlap
Area under the product of the
functions is zero
2
-2 + t
•
•
f()
3
Part of g(t) overlaps part of f(t)
Area under the product of the
functions is
2t

2
2+t
-2  t < 0
II.
g(t-)
g(t-)
2
-2 + t
f()
2+t
2
2
 2

32  t 
3t 2
0 3(  2)d  3  2  2    2  62  t    2  6
0
2t
4 - 14

Graphical Convolution Example
III. 0  t < 2
•
•
3
Here, g(t) completely overlaps f(t)
Area under the product is just
 





3



2
d


3


2

0
 2


2
2
2
6
•
•
2
2+t
-2 + t

0
Part of g(t) and f(t) overlap
Calculated similarly to -2  t < 0
t4
V.
f()
2
IV. 2  t < 4
•
•
g(t-)
3
g(t-)
2
f()
-2 + t
2

2+t
g(t) and f(t) do not overlap
Area under their product is zero
4 - 15
Graphical Convolution Example
• Result of convolution (5 intervals of interest):
0
 3
 t 2  6
 2
y (t )  f (t ) * g (t )  6
3 2
 t  12 t  24
2
0
for t  2
for  2  t  0
for 0  t  2
for 2  t  4
for t  4
y(t)
6
t
-2
0
2
4
4 - 16