# EE 550 Lecture no. 7

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```					Feedback Connection
State Space Model

u                     S1                       y1=y
u1

S2
u2
y2
                                 
x1 (t )  A1 x1 (t )  B1u1 (t )     x 2 (t )  A2 x 2 (t )  B2 u2 (t )
y1 (t )  C1 x1 (t )  D1u1 (t )     y 2 ( t )  C 2 x 2 ( t )  D2 u 2 ( t )

but
u1 = u – y2 = u – C2x2 - D2y1
= u – C2x2 – D2(C1x1+D1u1)
or
(I+D2D1)u1 = u – D2C1x1 - C2x2

solving for u1 gives

u1 = (I+D2D1)-1[u – D2C1x1 - C2x2]

.
x1 =[A1-B1(I+D2D1)-1D2C1]x1- B1(I+D2D1)-1C2x2
+ B1(I+D2D1)-1u

y = y1=C1x1 +D1u1 = C1x1 +D1(u-y2)

= C1x1+ D1u – D1(C2x2 + D2y)

y = (I+D1D2)-1[C1x1+ D1u – D1C2x2]
.
x2 = A2x2+ B2 y

 x   A  B ( I  D D ) 1 D C                B1 ( I  D2 D1 ) 1 C2    x1 
 1    1 1       2 1      2 1

 x2   B2 ( I  D1 D2 ) C1
1
A2  B2 ( I  D1 D2 ) 1 D1C2   x2 
 
 

 B1 ( I  D2 D1 ) 1 
                     u
 B2 ( I  D1 D2 ) D1 
1

x 
y  ( I  D1 D2 ) 1 C1    ( I  D1 D2 ) 1 D1C2  1   ( I  D1 D2 ) 1 D1u
 x2 
Feedback Connection
y1

S1                  y
u    -

S2
y2

y(s) = G1(s)u1(s) = G1(s)(u(s)- G2(s)y(s))
(I+ G1(s)G2(s))y = G1(s)u(s)

We assume (I+ G1(s)G2(s))-1 exist
(i.e. det ((I+ G1(s)G2(s)) ≠ 0)

y(s) = (I+ G1(s)G2(s))-1G1(s)u(s)

or
G(s) = (I+ G1(s)G2(s))-1G1(s) = G1(I+ G2(s)G1(s))-1

Note that :
Making the assumption that det ((I+ G1(s)G2(s)) ≠ 0 was
essential for the closed loop mathematical formulation to
make sense. To see this
Consider the example:

1  s        1 
           s  1
1 0
G2 ( s)  
0 1
G1 ( s )   s
1         s 

 s

s  1
    

1      1 
     s  1
I  G1 ( s )G2 ( s )   s
1     1 
          
s    s  1

And the det (I + G1(s)G2(s)) = 0
From the block diagram, (I + G1(s)G2(s))y(s)=G1(s)u(s)
So

1       1                  1  1      1 
      s  1  y1 ( s )    s       s  1  u1 ( s ) 
 s
1      1   y2 ( s)  1              s  u 2 ( s )
                                                  
s     s  1                 s       s  1
If

1             1                 1  s 
1                 
u ( s)   s , then we have  s          s  1  y1 ( s)    s 2 
0                   1            1   y2 ( s )  1 
                                                      
s           s  1                s 
2

For which there is no solution
We have seen that det ((I+ G1(s)G2(s)) ≠ 0 is absolutely
essential.
Even when ((I+ G1(s)G2(s))-1 exists the transfer function
from u(s) to another point in the loop may not be proper.
Example:

s                  y(s)
u(s)          -        s 1

Here det (I+ G1(s)G2(s)) = 1+G(s) = 1+ -s/(s+1) =
1/(s+1)≠0

However
s
s
G (s)  s  1           s
s   s 1 s
1
s 1
Improper System

Improper transfer functions do not correspond to good
systems.

Problem??

NOISE
Well-posedness
Definition: Let every subsystem of a composite system be
described by a rational transfer function. Than
the composite system is said to be well posed if
the transfer function of every subsystem is
proper and the closed transfer function from any
point chosen as an input terminal to every other
point along the directed path is well defined and
proper.

Theorem: Consider the feedback system

u                   G1                      y
-

G2

Let G1(s) and G2(s) be q    ×   p and p   ×   q proper rational
transfer matrices. Than the overall transfer function
G(s) = G1(s)(I+G2(s)G1(s))-1
is proper if and only if I+G2(∞)G1(∞) is nonsingular.
Discrete Time Systems
Inputs and outputs of discrete-time systems are defined
only at discrete instants of time, to, t1, … . The discrete
instants of time are assumed to be an integral multiplies of
some basic unit T, say
to = 0 , t1 = T, t2 = 2T, …
in which case T is often not explicitly shown and assumed
that the time parameter, denoted by k, takes integral
values,
k = 0, ±1, ±2, …
so we define {y(k) = y(kT)} and {u(k) = u(kT)}
as the discrete output and input sequences.

For a linear relaxed discrete time system, we have


y(k )   g (k , m)u(m)
m
where g(k, m) is called the weighting sequence or

the impulse response. It is the response to the input

1 n  m
 (n  m)  
0 n  m
If the system is causal, and relaxed at ko then we have

k
y(k )   g (k , m)u (m)
mk
o
If the system is time invariant and if we take ko = 0, then

we have

k
y(k )   g (k  m)u(m)                          *
m0

Z – Transform

The Z – Transform of the sequence {u(k), k = 0, 1, 2, … }

is defined as


u ( z )  {u (k )}   u (k ) z  k
k 0

If the Z – transform is applied to * then

y(z) = G(z) u(z)
State Space Model

Time Varying

x(k+1) = A(k)x(k) + B(k)u(k)

y(k) = C(k)x(k) + D(k)u(k)

Time Invariant Systems

x(k+1) = Ax(k) + Bu(k)

y(k) = Cx(k) + Du(k)                          **

Transfer Function from State Space

Let X(z) be the Z-transform of x(k)


X ( z )  {x(k )}   x(k ) z k
k 0

Let x(0) = xo then


{x(k  1)}   x(k  1) z k
k 0

Let m=k+1
                         
{x(k  1)}   x(m) z    ( m1)
 z  x ( m) z  m
m1                       m1

 z{ x(m) z m  x(0)}
m 0

 zX(z) - xo

Applying z-transform to (**), gives

zX(z) –z xo = Ax(z) + Bu(z)
y(z)=Cx(z)+Du(z)
x(z)= (zI - A)-1xo + (zI - A)-1Bu(z)
y(z)= C[(zI - A)-1xo + (zI - A)-1Bu(z)] + Du(z)

If xo = 0, then
y(z) = (C(zI - A)-1B + D) u(z)

G(z) = C(zI - A)-1B + D

```
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 views: 10 posted: 8/12/2011 language: English pages: 10