Signals and Systems
Lecture 4
Representation
Convolution
Examples
2006 Fall
of signals
Chapter 2
LTI Systems
Example 1
an LTI system
f1 t 
y1 t 
1
L
0
2
1
t
0
L
0
2
1
2006 Fall
4
t
2
t
y2 t   y1 t   y1 t  2
f 2 t   f1 t   f1 t  2
1
1
1
0
-1
2
4
t
Chapter 2
LTI Systems
§2.1 Discrete-time LTI Systems : The Convolution Sum
（卷积和）
§2.1.1 The Representation of Discrete-Time Signals
in Terms of impulses
Example 2
xn
3
1
1
2
01 2

n
xk  n  k 
xn    x 1 n  1  x0 n  x1 n  1  
xn 

 xk  n  k 
k  
2006 Fall
Representing DT Signals with
Sums of Unit Samples

Property of Unit Sample
x[n] [n  n0 ]  x[n0 ] [n  n0 ]
 Examples
 x[1] n  1
x[1] [n  1]  
n  1
 0
 x[0] n  0
x[0] [n]  
 0 n0
2006 Fall
 x[1] n  1
x[1] [n  1]  
 0 n 1
Written Analytically

x[n]   x[n  i ]
i  
   x[2] [n  2]  x[1] [n  1]  x[0] [n]  x[1] [n  1]  
Important to note
the “-” sign
x[n] 

 x[k ] [n  k ]
k  
Coefficients
2006 Fall
Basic Signals
Note the Sifting Property of the Unit Sample
Chapter 2
LTI Systems
§2.1.2 The Discrete-Time Unit Impulse Responses and the
Convolution-Sum Representation of LTI Systems
1. The Unit Impulse Responses
单位冲激响应


h  n  L 0 ,   n
2. Convolution-Sum （卷积和）
yn 

 xk  hn  k   xn hn
k  
k时刻的脉冲在n时刻的响应
系统在n时刻的输出包含所有时刻输入脉冲的影响
2006 Fall
Derivation of Superposition Sum

Now suppose the system is LTI, and
define the unit sample response h[n]:
 [n]  h[n]
– From Time-Invariance:
 [n  k ]  h[n  k ]
– From Linearity:
x[n] 
2006 Fall


k  
k  
 x[k ] [n  k ]  y[n]   x[k ]h[n  k ]  x[n] * h[n]
convolution sum
The Superposition Sum for DT Systems
Graphic View of Superposition Sum
2006 Fall
Hence a Very Important Property
of LTI Systems

The output of any DT LTI System is a
convolution of the input signal with the
unit-sample response, i.e.
Any DT LTI  y[n]  x[n] * h[n]


 x[k ]h[n  k ]
k  

2006 Fall
As a result, any DT LTI Systems are
completely characterized by its unit
sample response
Calculation of Convolution Sum

Choose the value of n and consider it fixed
y[ n ] 

 x[k ]h[n  k ]
k  
View as
functions of k
with n fixed
From x[n] and h[n]
to x[k] and h[n-k]
Note, h[n-k]–k is
the mirror image of
h[n]–n with the
origin shifted to n
2006 Fall
Calculating Successive Values:
Shift, Multiply, Sum








2006 Fall
y[n] = 0
y[-1] = 1
y[0] = 2
y[1] = -2
y[2] = -3
y[3] = 1
y[4] = 1
y[n] = 0
for n < -1
y[ n ] 

 x[k ]h[n  k ]
k  
for n > 4
Calculation of Convolution Sum
Use of Analytical Form

Suppose that x[n]  a nu[n] and h[n]  bnu[n]
then y[n]  x[n] * h[n]

 x[k ]h[n  k ]
  a u[k ]b u[n  k ]
 a b

k
 
k
k n 
k 0
n
2006 Fall
n k
k n k
 a (n  1)u[n] a  b
 n 1 n 1
 b  a
 b  a u[n] a  b
u[k ]  k  0,
u[n  k ]  n  k
n  0,
n  0时为0
Calculation of Convolution Sum
Use of Array Method
2006 Fall
Calculation of Convolution Sum
Example of Array Method
 If x[n]<0 for n<-1,x[-1]=1,x[0]=2,x[1]=3,
x[4]=5,…,and h[n]<0 for n<-2,h[-2]=-1,
h[-1]=5,h[0]=3,h[1]=-2,h[2]=1,….In this
case ,N=-1,M=-2,and the array is as follows
So,y[-3]=-1,
y[-2]=3, y[-1]=10,
y[0]=15,
y[1]=21,…, and
y[n]=0 for n<-3
2006 Fall
Calculation of Convolution Sum
Using Matlab

The convolution of two discrete-time signals can
be carried out using the Matlab M-file conv.
 Example:
–
–
–
–
–
–
–
2006 Fall
p=[0 ones(1,10) zeros(1,5)];
x=p; h=p;
y=conv(x,h);
n=-1:14;
subplot(2,1,1),Stem(n, x(1:length(n)))
n=-2:24;
subplot(2,1,2),Stem(n, y(1:length(n)))
Calculation of Convolution Sum
Using Matlab Result
2006 Fall
Conclusion

Any DT LTI Systems are completely
characterized by its unit sample response.
y[n]  x[n] * h[n]

Calculation of convolution sum:
– Step1:plot x and h vs k, since the convolution sum is
–
–
–
–
–
2006 Fall
on k;
Step2:Flip h[k] around vertical axis to obtain h[-k];
Step3:Shift h[-k] by n to obtain h[n-k] ;
Step4:Multiply to obtain x[k]h[n-k];
Step5:Sum on k to compute
x[k ]h[n  k ]
k
Step6:Index n and repeat step 3 to 6.

Conclusion

Calculation Methods of Convolution Sum
– Using graphical representations;
– Compute analytically;
– Using an array;
– Using Matlab.
2006 Fall
Chapter 2
LTI Systems
§2.2 Continuous-Time LTI Systems : The Convolution Integral
（卷积积分）
§2.2.1 The Representation of Continuous-Time Signals
in Terms of impulses
xt   


x  t   d
——Sifting Property
§2.2.2 The Continuous-Time Unit Impulse Response and the
Convolution Integral Representation of LTI Systems
y  t   x  t   h t   


2006 Fall
x   h  t    d
Representation of CT Signals

Approximate any input x(t) as a sum of
shifted, scaled pulses (in fact, that is
how we do integration)

x (t )  x(kt ), kt  t  (k  1)t
2006 Fall
Representation of CT Signals (cont.)
  (t) has a unit area
 x(k)  (t  k)


x (t ) 

 x(k)
k  

( t  k ) 
limit as 0

2006 Fall
x(t )   x( ) (t   )d

Sifting
property
of the
unit
impulse
Response of a CT LTI System

Now suppose the system is LTI, and
define the unit impulse response h(t):
(t)
 h(t)
– From Time-Invariance:
(t  )

h(t  )
– From Linearity:

x(t )   x( ) (t   )d


2006 Fall
 y (t )   x( )h(t   )d  x (t ) * h(t )

Superposition Integral for CT Systems
Graphic View of Staircase Approximation
2006 Fall
CT Convolution Mechanics

To compute superposition integral

y (t )  x(t ) * h(t )   x( )h(t   )d

– Step1:plot x and h vs , since the convolution
–
–
–
–
–
2006 Fall
integral is on ;
Step2:Flip h( around vertical axis to obtain h(-;
Step3:Shift h(-) by n to obtain h(n-) ;
Step4:Multiply to obtain x(h(n-;

Step5:Integral on  to compute
x( )h(t   )d

Step6:Increase t and repeat step 3 to 6.

Basic Properties of Convolution
Commutativity: x(t) h(t) h(t)  x(t)
 Distributivity :

x(t ) * [h1 (t )  h2 (t )]  x(t ) * h1 (t )  x(t ) * h2 (t )

Associativity:
x(t ) * [h1 (t ) * h2 (t )]  [ x(t ) * h1 (t )] * h2 (t )
x(t) (t  to ) x(t  to )
(Sifting property: x(t)  (t) x(t))
 An integrator:

t
x(t ) * u(t )   x( )d

2006 Fall
Convolution with Singularity Functions
f (t ) *  (t   )  f (t   )
f (t ) *  (t )  f (t )
t
f (t ) * u(t )   f ( )d

f (t ) * 
(k )
(t )  f
(k )
(t )
f (t ) *  ( k ) (t   )  f ( k ) (t   )
2006 Fall
More about Response of LTI Systems

How to get h(t) or h[n]:
– By experiment;
– May be computable from some known
mathematical representation of the given
system.

Step response:
t
s(t )   h( )d
s[n ] 

n
 h[k ]
k  
2006 Fall
Summary

What we have learned ?
– The representation of DT and CT signals;
– Convolution sum and convolution integral
Definition; Mechanics;
– Basic properties of convolution;

What was the most important point in the lecture?
 What was the muddiest point?
 What would you like to hear more about?
2006 Fall
Readlist

Signals and Systems:
– 2.3,2.4
– P103~126

Question: The solution of LCCDE
(Linear Constant Coefficient Differential or
Difference Equations)
2006 Fall
Problem Set

2.21(a),(c),(d)
 2.22(a),(b),(c)
2006 Fall