19
EEE 303: Signals and Linear Systems
Convolve x(t ) with h(t ) as shown below where T = 3 sec.
y (t )  h(t )  x(t )  h(t ) 



n 
n 
n 
  (t  nT )   h(t )   (t  nT )   h(t  nT ) .
The resulting output is shown in Figure (b) above.
Periodic Convolution
If x1 (t ) and x2 (t ) are both periodic signals with common period T0 , the convolution of x1 (t ) and x2 (t ) does
not converge. In this case we define periodic convolution of x1 (t ) and x2 (t ) as,
T0
y(t )  x1 (t )  x2 (t )   x1 ( ) x2 (t   )d .
0
Frequency Response Function of a Linear System
Let, x(t )  e jt . The convolution integral gives the output as, y(t ) 
Or, y(t )  e jt






h( )e j (t  ) d .
h( )e j d  H ( j)e j ,
Where we define, H ( j ) 



h( )e j d  H ( j ) e jH ( j ) .
j (t H ( j ))
The output y (t ) is then expressed as, y(t )  H ( j) e
.
The output is a complex exponential of the same frequency as input multiplied by the complex constant
H ( j ) . This H ( j ) is called the frequency response of the system.
Similar things will happen for sin t and cos t input. The functions e jt ,sin t and cos t are called Eigen
function, as we get the same function in the output as in the input.
20
Example
1  t / RC
e
u (t ) . Find the expression for the frequency
RC
response. What will be the output of the system with, x(t )  cos 2000 t  cos 20000 t ?
The impulse response of a system is given as, h(t ) 
------------------------------------------------------------------------------------------------------------
1   / RC
1   ( j 1/ RC )
1/ RC
 j
e
u
(

).
e
d


e
d 



0
RC
RC
j  1/ RC
1
1/ RC
Or, H ( j ) 
e  j tan ( RC ) . H ( j1 )  0.707e j 45 ; H ( j2 )  0.995e j 84.29 .
2
 2  1/ RC 
H ( j ) 
Thus, y(t )  0.707 cos(2000 t  45 )  0.995cos(2000 t  84.29 ) . Ans
Block Diagram Representation
The impulse response and the differential equation descriptions represent only the input=output behavior of
a system. On the other hand block diagram representation describes a different set of internal computations
used to determine the system output.
A block diagram is an interconnection of elementary operations that act on the input signal. It is a more
detailed representation of the system as it describes how the system’s internal computations are ordered.
Block diagram representations consist of an interconnection of three elementary operations on signals:
1. Scalar multiplication: y (t )  cx(t ) ,

2. Addition: y (t )  x(t )  w(t ) , and
3. Integration: y(t ) 

t

x( )d .

t

d k y(t ) M d k x(t )
. In order to depict the
 bk
dt k
dt k
k 0
k 0
system in terms of integration operation, let us assume M  N , aN  1 and multiply both side of the above
N
An Nth order system is represented by the equation,
 ak
equation by D  N . After rearrangement,
y (t )  bN x(t )  D 1 (bN 1 x  aN 1 y )  D 2 (bN  2 x  aN  2 y ) 
 D  ( N 1) (b1 x  a1 y )  D  N (b0 x  a0 y )
(1)
Using the above equation we can draw the simulation diagram as,







This form of realization is called canonical realization. Differentiators are not used to simulate a system, as a
differentiator enhances noise. On the other hand integrators smooth or suppress noise present in an input.
21
Equation (1) in the previous page can be written in another form for N = 2 as,
y (t )  b2 x(t )  b1 D 1 x(t )  b0 D 2 x(t )  a1D 1 y (t )  a0 D 2 y (t )
(2)
Direct form-I and direct form-II implementation of the system is shown in figures (a) and (b) below.
Examples
Construct the block diagram representation of the system below.


Or,  D 2 
1
1
1
1
1 
1
, a0 
; b2  0, b1  , b0  0 .
D
v(t )  Di (t ) ; here, a1 

RC
LC
C
RC
LC 
C
The canonical block diagram representation of the system is shown below:
The system can be simulated using multiplier, summer and
integrator using OP-AMP.
State-variable Representation
The state-variable description of an LTI system consists of a series of coupled first-order differential
equations that describes how the state of the system evolves and an equation that relates the output of the
system to the current state variables and the input. These equations are written in matrix form. Statevariable analysis transforms an Nth order differential equation into N first-order differential equations of a
set of state variables.
The state of a system is defined as a minimal set of signals that represents the system’s entire memory of
the past. Given only the value of the state at a point in time t 0 , and the input for the times t  t0 we can
22
determine the output for all times (or, t  t0 ). The state of
a system is not unique and there are many possible statevariable descriptions corresponding to a system with a
given input-output characteristic.
Consider the electrical circuit shown in figure right. Derive
the state-variable description for this system if the input is
x(t ) and the output is the current through the resistor y (t ) . Choose the state variables as the voltage across
each capacitor.
x(t )  R1 y (t )  q1 (t )
or , y (t )  
1
1
q1 (t )  x(t )
R1
R1
This equation expresses the output as a function of the state variables and the input.
Let, i2 (t ) be the current through R2 . We get, i2 (t ) 
Thus, q2 (t ) 
1
1
q1 (t )  q2 (t ) . Also, i2 (t )  C2 q2 (t ) .
R2
R2
1
1
q1 (t ) 
q2 (t )
C2 R2
C2 R2
Again, current through C1 is, i1 (t )  C1q1 (t )  y (t )  i2 (t )  y (t ) 
1
1
q1 (t )  q2 (t )
R2
R2
Thus,
q1 (t )  y (t )  i2 (t ) 
1
1
1
1
1
1
1
y (t ) 
q1 (t ) 
q2 (t )  
q1 (t ) 
x(t ) 
q1 (t ) 
q2 (t )
C1
C1 R2
C1 R2
C1 R1
C1 R1
C1 R2
C1 R2
 1
1 
1
1
 

q2 (t ) 
x (t )
 q1 (t ) 
C1 R2
C1 R1
 C1 R1 C1 R2 
 q1 (t )    1
1 
1   q1 (t)   1 




  
 C R 
C
R
C
R
C
R

1
1
1
2

1
2




   1 1  x(t ) ;


 

 

1
1


 q2 (t )  
 

C2 R2
C2 R2  q 2 (t)   0 

 
 1   q1 (t )  

 
1




y (t )  R1 
   R  x(t ) ;


 0   q2 (t )   1 
Or,
State-equation
Output equation
q(t) = Aq(t) + Bx(t)
, where the matrices A, B, C and D describe the internal structure of the system.
y(t) = Cq(t) + Dx(t)
Here, A is called the system matrix, B is called the input matrix, C is called the output matrix and D is called
the transmission matrix.
Find the state-variable description of the circuit depicted in figure below.
(Ans)
23
In block diagram representation, the state variable corresponds to the outputs of the integrators. Consider
the following example:
The block diagram above indicates that,
For an Nth order system with m-input and p-output, the dimensions of A, B, C and D are as follows:
A  ( N  N ), B  ( N  m), C  ( p  N ) and D  ( p  m) .
State-space representation of a differential equation
The general form of an Nth-order differential equation is,
Let us define N state variables q1 (t ), q1 (t ),
, qN (t ) as,
q1 (t )  q2 (t )
q1 (t )  y (t )
q2 (t )  y (t )
qN (t )  y N 1 (t )
Then,
q2 (t )  q3 (t )
qN (t )  y N (t )  aN q1 (t )  aN 1q2 (t ) 
y(t )  q1 (t ) .
And,
In matrix form the above two equations can be expressed as,
 a1qN (t )  x(t )
24
Where,
Solution of state equations
Let us determine the solution of the state equation, q(t) = Aq(t) + Bx(t); x(0) = x0 .
Rewriting the equation, q(t)  Aq(t)  Bx(t)
Multiplying both sides by e  At we get, e  At [q(t)  Aq(t)] 
d  At
[e q(t)]  e  At Bx(t)
dt
Integrating both sides between the limits 0 and t , we get
t
t
0
0
e At q(t)   e A Bx( )d
Or,
t
e q(t)  q(0)   e A Bx( )d
 At
or, q(t)  eAt q(0) 
0
t
e
A ( t  )
0
Bx( )d  eAt q(0)  eAt  Bx(t )
If the initial state is known at t  t0 , the above equation becomes,
t
q(t)  e A ( t t0 )q(t0 )   e A ( t  ) Bx( )d .
t0
At
The matrix function e is known as the state transition matrix of the system.
The output equation is given by,
t
y(t)  Ce At q( 0)   Ce A (t  ) Bx( )d  Dx(t) .
t0
The zero-state response of the system is(when q(0)  0 ),
y(t)  CeAt B  x(t )  Dx(t)  CeAt B  x(t )  Dδ(t)  x(t)
Or,
y(t)  [CeAt B  Dδ(t)]  x(t)  h(t)  x(t)
The matrix h(t) is known as impulse response matrix.
Evaluation of e At
Let, A be an ( N  N ) matrix and I be an ( N  N ) identity matrix. The Eigen values i , i  1, 2,
, N of A
are the roots of the Nth-order polynomial, det[ A   I]  0 .
Evaluation of the state transition matrix e At is based on the Cayley-Hamilton theorem. This theorem states
that the matrix A satisfies its own characteristic equation, i.e.,
Q( )   N  aN 1 N 1   a1  a0  0
If,
Then,
Now,
Q( A)  A N  aN 1A N 1 
e At  I  At 
2 2
At

2!

 a1A  a0 I  0  A N  aN 1A N 1 
 a1A  a0I
(1)
N N
A t

2!
Applying equation (1) , e At may be expressed as,
e At   0 I  1A   2 A 2 
  N 1A N 1
If all i s are distinct we may write,  0  1i   2 i 
as,
2
(2)
  N 1i
N 1
i t
 e from where we can found  s
We can also calculate e At using Laplace
transform.
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Example
Find a state-space representation of the system as shown in figure below.
The system is of order 3. We assume three state variables q1 (t ), q2 (t ) and q3 (t ) . From figure above we
obtain,
(Ans)
Example
Find a state-space representation of the circuit shown in figure below assuming that the outputs are the
currents flowing in R1 and R2 .
We choose the state variables q1 (t )  iL (t ) and q2 (t )  vC (t ) . Let, x1 (t )  v1 (t ) and x2 (t )  v2 (t ) . Also,
y1 (t )  i1 (t ) and y2 (t )  i2 (t ) .
Applying KVL to the two loops we get,
26
Rearranging and writing in matrix form we get,
(Ans)
0
1
Example: Find e At for A  
 using Cayley-Hamilton theorem.
 6 5
The characteristic polynomial of A is,

1
  2  5  6  (  2)(  3) .
6  5
Thus the Eigen values of A are, 1  2 and 2  3 .
q ( )   I  A 
 0
 61
Hence we have, e At  0I  1A  
1

, where  s are obtained as,
0  51 
 0  2 1  e 2t 
 0  31  e
3t
2 t
3t
2 t
3t
 solving we get,  0  3e  2e and 1  e  e .

(Ans)