### Lecture 16 - Transfer Functions

```Output ( s )
TF ( S ) 
Input ( s )
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Transfer Functions
Outline of Today’s Lecture
 A new way of representing systems
 Derivation of the gain transfer function
 Coordinate transformation effects
 hint: there are none!
 Development of the Transfer Function from an ODE
 Gain, Poles and Zeros
Observability
 Can we determine what are the states that produced a certain
output?
 Perhaps
 Consider the linear system
d
 x  Ax  Bu
 dt

 y  Cx  Du
We say the system is observable if for any time T>0 it is
possible to determine the state vector, x(T), through the
measurements of the output, y(t), as the result of input, u(t),
over the period between t=0 and t=T.
Observers / Estimators
Input u(t)
d
 x  Ax  Bu
 dt

 y  Cx  Du
Output y(t)
Noise
State xˆ
Observer/Estimator
Testing for Observability
y (t )   0 (t )CA0  1 (t )CA   2 (t )CA2  ...   n 1 (t )CAn 1  x(0)
 For x(0) to be uniquely determined, the material in the
parens must exist requiring
C

CA 


2
W0  CA   C T

 
...

CAn 1 


T
AC
T
2T
A C
T
 n 1T
... A
C 

T
T
to have full rank, thus also being invertible, the common test
 Wo is called the Observability Matrix
Testing for Observability
y (t )   0 (t )CA0  1 (t )CA   2 (t )CA2  ...   n 1 (t )CAn 1  x(0)
 For x(0) to be uniquely determined, the material in the
parens must exist requiring
C

CA 


2
W0  CA   C T

 
...

CAn 1 


T
AC
T
2T
A C
T
 n 1T
... A
C 

T
T
to have full rank, thus also being invertible, the common test
 Wo is called the Observability Matrix
Example: Inverted Pendulum
 Determine the observability pf the
Segway system with v as the output
0
0
A 
0

0
1
0
0 6.405
0
0
0 7.205
0
0

1

0
CA  0 0 6.4046 0
0


 0.01837 
 C = [0 1 0 0]
B
0




 0.008163
CA2  0 0 0 6.4046
CA3  0 0 46.1459 0
Using the model developed in Lecture 5 for the inverted pendulum
0


0


  x
2
2 2
J  ml 
ml g




0 
0  
  v   J ( M  m )  Mml 2 
J ( M  m )  Mml 2

F

0
0
1   
0

  



mlg ( M  m )
ml
0
0    

2
J ( M  m )  Mml
 J ( M  m )  Mml 2 

M  10kg
 x
v 
m  80kg
y  0 1 0 0   with
l  1m
 
 
J  100kg  m 2 / s 2
 
0
 x  
  0
d v 

dt   0
  
  0


1
0
0
0 
0 1
0. 0. 6.4046
0. 

Wo  
0.
6.4046
0. 0.


0. 
0. 0. 46.146
W0 is singular; therefore the system is not observable. We can not see
the state variable x as a result of the output!
Observable Canonical Form
 A system is in Observable Canonical Form if it can be put into the form
 a1 1 0 0 ...
 a
0 1 0 ...
2

d
z   ...
... ... ... ...
dt

 an 1 0 0 0 ...
 an 0 0 0 ...
y  1 0 0 ... 0 z  Du
0
 b1 
b 
0

 2 
... z   ...  u



1
b
 n 1 
 bn 
0 
Where ai are the
coefficients of the
characteristic equation
I  A
…
u
bn
S
an
bn-1

zn
S
b2

zn-1
an-1
…
S
a2
…
b1

z2
S
a1
D

z1
-1
S
y
Observable Canonical Form
 a1
 a
 2
d
z    a3
dt

  a4
 a5
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
 b1 
b 
0

 2
0 z   b3  u

 
1
b4 
 b5 
0
y  1 0 0 0 0 z  Du
The observable matrix has the form
1


a1

W0  
a12  a2

3

a
1  2a2 a1  a3

 a14  3a2 a12  2a3a1  a22  a4
0
1
a1
a12  a2
a13  2a2 a1  a3
0
0
1
a1
a12  a2
0
0
0
1
a1
0
0

0

0
1
Dual Canonical Forms
The Controllable/Reachable canonical form is
 z1   a1
z   1
2
d   
 z3    0
dt   
 ...   ...
 zn   0
 a2
0
1
...
0
 a3
0
0
...
...
0
1
an   z1  1 
0   z2   0 
   
0   z3    0  u
   
...   ...  0 
0   zn  0
Therefore for system
d
 Ax  Bu y  Cx
dt
Ac  AoT Bc  CoT and Cc  Bo T
T
 Ac z  Bc u
W r Wo
The Observable Canonical form is
T  W rWr 1
y  b1 b2
b3 ... bn  z  Du  Cc z  Du
 a1 1 0 0 ...
 a
0 1 0 ...
 2
d
z   ...
... ... ... ...
dt

 an 1 0 0 0 ...
 an 0 0 0 ...
 Ao z  Bo u
0
 b1 
b 
0

 2 
... z   ...  u



1
b
n

1


 bn 
0 
y  1 0 0 ... 0 z  Du  Co z  Du
Ac  TAT 1 Bc  TB
Cc  CT 1
A0  Ac T   TAT 1   T 1 AT T T
T
T
B0  Cc T   CT 1   T 1 C T
T
C0  Bc T   TB   B T T T
T
Input u(t)
Output y(t)
d
 x  Ax  Bu
 dt

 y  Cx  Du
Observers / Estimators
Noise
Observer/Estimator
State
B
+
+

y
C
A
u
x
B
++
+
+

A
L
C
y
_ +
xˆ
Alternative Method of Analysis
 Up to this point in the course, we have been concerned about the




structure of the system and discribed that structure with a state
space formulation
Now we are going to analyze the system by an alternative method
that focuses on the inputs, the outputs and the linkages between
system components.
The starting point are the system differential equations or
difference equations.
However this method will characterize the process of a system
block by its gain, G(s), and the ratio of the block output to its
input.
Formally, the transfer function is defined as the ratio of the
Laplace transforms of the Input to the Output:
TF ( S ) 
Output ( s )
for s a complex number and the Laplace Variable
Input ( s )
System Response From Lecture 11
 We derived for u(t )  est where s  i
t
y (t )  Ce x (0)   Ce A( t  ) Be s d  De st
At
0
 Ce At x (0)  Ce At  e  sI  A Be s d  De st
t
0
 Ce At x (0)  Ce At ( sI  A) 1  e 

sI  A t

 I  B  De st


 Ce At  x (0)  ( sI  A) 1 B   C  sI  A  B  D e st
1
}
}
Transient
State
Transfer function is defined as
G ( S )  C  sI  A B  D
1
where s is a complex number
Derivation of Gain
 Consider an input of
u(t )  e st  e i  t  e t eit  e t  cos t  i sin t 
  0 has the effect of damping and determines the damping rate
with s  i , the eigenvalues of dynamics matrix, A, our solution was

x (t )  e At x (0)   e A( t  ) Be s d  e At x (0)   e
t

0
 e At x (0)   sI  A

1
 e
sI  A t
 
t
 
0
sI  A

Be s d


 1 B  e At x (0)   sI  A  B   sI  A  Be st
1

y (t )  Ce At x (0)   sI  A B  C  sI  A  B e st
1
1
The first term is the transient and dies away if A is stable.
If x(0)  ( sI  A) 1 B then
x (t )  ( sI  A) 1 Be st  ( sI  A) 1 Bu(t )
y (t )  (C ( sI  A) 1 B  D )e st   C ( sI  A) 1 B  D  u (t )
thus
G ( s )   C ( sI  A) 1 B  D  is the output y in response to input u
which we call the transfer function from u to y
1
Example
G ( s )  C ( sI  A) 1 B  D
1
b
k
m
x
u(t)
mx  bx  kx  u (t )
 0
d  x 

k


dt  x   
 m
 x
y  1 0  
 x
1 
 x   0

b      u (t )
   x  1
m
1 

 0
s
0


  0
 k

G ( s )  1 0  

b
 0 s   
   1
m
 m

1
G( s) 
b
k
s2  s 
m
m
The general form of the input is u(t )  e st
if s  0 then u(t )  e0t  1 a step function
m
m
G ( s ) s 0 
for a steady state response y  G ( s )u(t ) 
k
k
G( s) 
Example
1
m
 2
b
k ms  bs  k
s2  s 
m
m
From a set of Mathematics Tables, sin t 
i  it iwt
(e  e )
2
i  it
i
e
and u2 (t )  eit then u(t )  u1 (t )  u1 (t )
2
2
y (t )  y1 (t )  y2 (t )  G ( s )u1 (t ) s  i  G ( s )u2 (t ) s i
Let u1 (t ) 
b
k
G (i ) 
m
u(t)
x
G (i ) 
m

m  i   b  i   k
2
k
k 2  m 2  ib
2
 m
   b 
2 2
2
m
k 2  m 2  ib
which is a complex number
and can be written in exponential form: G (i )  Mei
1
2
2 2 2
with M 
km

m


mb





2
 k  m 2    b 2
 b 
and   tan 1 
2 
 k  m 
G ( i ) can be treated similarly so that
M  it  
y (t ) 
(ie
 ieit   )  M sin t   
2
Coordination Transformations
x1
d
d
 x  Ax  Bu  z  Az  Bu
  dt
 dt

 y  Cz  Du
 y  Cx  Du

z2
x2
for A  TAT 1, B  TB and C  CT 1
Since the ouput y is unchanged by the transformation
z1
y  C ( sI  A) 1 B  D
and
y  C ( sI  A) 1 B  D
y  G ( s )u(t )  G ( s )u(t )  G ( s )  G  s 
 Thus the Transfer function is invariant under coordinate
transformation
Linear System Transfer Functions
General form of linear time invariant (LTI) system is expressed:
(n)
 n1
 n 2 
 m
 m1
 m2 
y  a1 y  a2 y  ...  an2 y  an1 y  an y  b0 u  b1 u  b2 u  ...  bn2u  bn1u  bnu
For an input of u(t)=est such that the output is y(t)=y(0)est
s
n
 a1s n 1  a2 s n 2  ...  an 2 s 2  an 1 2  an  y0e st   b0 s m  b1s m 1  b2 s m 2  ...  bn 2 s 2  bn 1s  bn  e st
b s bs


s  a s
m
y (t )  y (0)e
st
0
m 1
1
n 1
n
1
 b2 s m 2  ...  bn 2 s 2  bn 1s  bn 
 a2 s n 2  ...  an 2 s 2  an 1 2  an 
e st
The transfer function form is then
b0 s m  b1s m1  b2 s m2  ...  bn 2 s 2  bn 1s  bn b( s )
y ( t )  G ( s )u ( t )  G ( s )  n

s  a1s n 1  a2 s n 2  ...  an 2 s 2  an 1 2  an
a(s)
Note that the transfer function for a simple ODE can be written as the ratio
of the coefficients between the left and right sides multiplied by powers of s
The order of the system is the highest exponent of s in the denominator.
Simple Transfer Functions
Differential
Equation
Transfer
Function
Name
yu
s
Differentiator
yu
1
s
1
s2
1
sa
Integrator
yu
y  ay  u
y  2n y  n 2 y  u
y  k pu  kd u  ki  udt
2nd order Integrator
1st order system
1
s  2n s  n
2
k p  kd s 
ki
s
2
Damped Oscillator
PID Controller
A Different
Method
 Design a controller that will
control the angular position to a
given angle, 0
If Va (t )  e i t  V0eit  V0e st ,
then from the Convolution Equation
 (t )  0e st
replacing the variables of the
differential equations with this result, we get
 b KK b  d
d2
KV0e st
st
st
0e    J  JR  dt 0e   JR
dt 2
a 
a

In Lect. 5, we formed the state space rep. from
K b d Va

ia   R dt  R
 b KK b  d KVa
d 2
a
a

  

 2

dt
J
JR
dt
JRa
d

Ki
b
d

a



 a
 dt
J
J dt
d 2  b KK b  d KVa
 


dt  J JRa  dt
JRa
 2  b KK b   st KV0e st
s   
 s   0e 
J
JR
JRa
a  


K
JRa
0e st 
V0e st 

 b KK b 
s2   
s
J
JR
a 

Output ( s )  (t )

Input ( s ) u (t ) u ( t )V0est
K
JRa

 b KK b 
s2   
s
J
JR
a 

A Different
Method
 Design a controller that will control the angular position to a given angle, 0
d 2  b KK b  d KVa
 


dt  J JRa  dt
JRa
K
JRa
G( s) 
 b KK b 
s2   
s
 J JRa 
A Different
Method
 Design a controller that will
control the angular position to a
given angle, 0
R=0.2
J= 10^-5
K=6*10^-5
Kb=5.5*10^-2
b=4*10^-2
gs=(K/(J*R))/(b*s^2/J+K*Kb*s/(J*R))
step(gs)
Output ( s )  (t )

Input ( s ) u (t ) u ( t )V0est
K
JRa

 b KK b 
s2   
s
J
JR
a 

We will refer to this as the transfer of the
input to the output, or transfer function
K
JRa
G( s) 
 b KK b 
s2   
s
 J JRa 
Output ( s )  G ( s )  Input ( s )
Which was the same for the state space
Later, we will learn how to control it
Gain, Poles and Zeros
G ( s )  C ( sI  A) 1 B  D 
G (0)  D  CA1B 
b( s )
b( s )
K
a( s)
a( s )
yo bm

K
uo am
 The roots of the polynomial in the denominator, a(s), are called
the “poles” of the system
 The poles are associated with the modes of the system and these are
the eigenvalues of the dynamics matrix in a state space representation
 The roots of the polynomial in the numerator, b(s) are called the
“zeros” of the system
 The zeros counteract the effect of a pole at a location
 The value of G(s) is the zero frequency or steady state gain of the
system
Summary
 A new way of representing systems
 The transfer function
G( s)  C( sI  A)1 B  D 
 Derivation of the gain transfer function
bs
a s
 Coordinate transformation effects
 hint: there are none!
 Development of the Transfer Function from an ODE
(n)
 n1
 n 2 
 m
 m1
 m2 
y  a1 y  a2 y  ...  an2 y  an1 y  an y  b0 u  b1 u  b2 u  ...  bn2u  bn1u  bnu
b0s m  b1s m1  b2 s m2  ...  bn2 s 2  bn1s  bn b( s)
G( s )  n

n 1
n 2
2
s  a1s  a2 s  ...  an2 s  an1 2  an a(s)
 Gain, Poles and Zeros
G( s)  C ( sI  A) 1 B  D 
b( s)
b( s)
K
a( s)
a( s)
Next: Block Diagrams
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