Lecture – 9
Conversion Between State Space and
Transfer Function Representations in
Linear Systems – I
Dr. Radhakant Padhi
Asst. Professor
Dept. of Aerospace Engineering
Indian Institute of Science - Bangalore
State Space Representation
(noise free linear systems)
z
z
State Space form
X = AX + BU
A
B
- System matrix- n x n
Y = CX + DU
C
- Output matrix- p x n
D
- Feed forward matrix – p x m
Transfer Function form
- Input matrix- n x m
Q: Is conversion between
the two forms possible?
A: Yes.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
2
Deriving Transfer Function Model
From Linear State Space Model
z
Known:
X = AX + BU
Y = CX + DU
z
Taking Laplace transform (with zero initial conditions)
sX ( s ) = AX ( s ) + BU ( s )
Y ( s ) = CX ( s ) + DU ( s )
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
3
Deriving Transfer functions from
State Space Description
z
The state equation can be placed in the form
( sI − A) X ( s ) = BU ( s )
z
Pre-multiplying both sides by
( sI − A) −1
−1
X ( s ) = ( sI − A) BU ( s )
z
Substituting for X (s) in the output equation,
Y ( s ) = ⎡⎣C ( sI − A) B + D ⎤⎦ U ( s )
−1
Transfer Function Matrix T ( s )
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
4
Example
z
State space model
⎡ x1 ⎤ ⎡ 0 1 ⎤ ⎡ x1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢ −2 −3⎥ ⎢ x ⎥ + ⎢1 ⎥ u
⎣ 2 ⎦ ⎣
⎦ ⎣ 2 ⎦ ⎣N⎦
A
B
⎡ x1 ⎤
y = [1 0] ⎢ ⎥ + [ 0] u
N ⎣ x2 ⎦ N
C
D
z
Using the expression for derived transfer function
Y (s)
1
1
= 2
=
U ( s ) s + 3s + 2 ( s + 1)( s + 2)
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
5
Example: Detail Algebra
T ( s ) = C ( sI − A ) B + D
−1
−1
⎛ ⎡ s 0 ⎤ ⎡ 0 1 ⎤ ⎞ ⎡0 ⎤
= [1 0] ⎜ ⎢
−⎢
⎟ ⎢ ⎥ + [ 0]
⎥
⎥
⎝ ⎣0 s ⎦ ⎣ −2 −3⎦ ⎠ ⎣1 ⎦
−1
⎡ s −1 ⎤ ⎡0 ⎤
= [1 0] ⎢
⎥ ⎢1 ⎥
+
2
3
s
⎣
⎦ ⎣ ⎦
⎛
⎞ ⎡ s + 3 1⎤ ⎡0 ⎤
1
= [1 0] ⎜⎜
⎟⎟ ⎢
⎥ ⎢1 ⎥
−
2
s
+
+
s
s
3
2
(
)
⎦⎣ ⎦
⎝
⎠⎣
⎡1⎤
1
1
= 2
=
1
0
[ ]⎢ ⎥ 2
s + 3s + 2
⎣ s ⎦ s + 3s + 2
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
6
Deriving State Space Model
From Transfer Function Model
z
The process of converting transfer function to state
space form is NOT unique. There are various
“realizations” possible.
z
All realizations are “equivalent” (i.e. properties do not
change). However, one representation may have some
advantages over others for a particular task.
z
Possible representations:
•
•
•
•
First companion form (controllable canonical form)
Jordan canonical form
Alternate first companion form (Toeplitz first companion form)
Second companion form (observable canonical form)
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
7
First Companion Form: SISO Case
(Controllable canonical form)
⎤
y(s) ⎡
1
= ⎢ n
H (s) =
⎥
n −1
+ " + a n −1 s + a n ⎦
u ( s ) ⎣ s + a1 s
y
(n )
+ a1 y
( n −1)
+ " + a n − 1 y + a n y = u
dny
d n −1 y
+ a1
+
n
n −1
dt
dt
"
+ an y = u
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
8
First Companion Form: SISO Case
(Controllable canonical form)
Choose output y(t ) and its (n − 1) derivatives as
⎡ x1 = y ⎤
⎢
⎥
dy
⎢ x2 =
⎥
dt ⎥
⎢
⎢
⎥
#
⎢
⎥
d n −1 y ⎥
⎢
⎢⎣ xn = dt n −1 ⎥⎦
dy ⎤
⎡
⎢ x1 = dt ⎥
⎢
⎥
2
⎢ x = d y ⎥
⎯⎯→ differentiating ⎯⎯→ ⎢ 2 dt 2 ⎥
⎢
⎥
#
⎢
⎥
⎢
dn y⎥
⎢ xn = n ⎥
dt ⎦
⎣
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
9
First Companion Form: SISO Case
(Controllable canonical form)
⎡ x1 ⎤ ⎡ 0
⎢ x ⎥ ⎢ 0
⎢ 2 ⎥ ⎢
⎢ x3 ⎥ ⎢ 0
⎢
⎥=⎢
⎢ # ⎥ ⎢ #
⎢ xn −1 ⎥ ⎢ 0
⎢
⎥ ⎢
⎢⎣ xn ⎥⎦ ⎢⎣ − an
1
0
0
#
0
− an −1
0
0 0
1
0 0
0
1 0
#
#
#
0
0 0
− an − 2 " "
0 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤
" 0
0 ⎥⎥ ⎢⎢ x2 ⎥⎥ ⎢⎢ 0 ⎥⎥
" 0
0 ⎥ ⎢ x3 ⎥ ⎢ 0 ⎥
" 0
⎥⎢
⎥ + ⎢ ⎥u
#
#
# ⎥ ⎢ # ⎥ ⎢0 ⎥
0 0
1 ⎥ ⎢ xn −1 ⎥ ⎢ 0 ⎥
⎥⎢
⎥ ⎢ ⎥
" " − a1 ⎥⎦ ⎢⎣ xn ⎥⎦ ⎢⎣ 1 ⎥⎦
⎡ x1 ⎤
⎢x ⎥
⎢ 2 ⎥
⎢ x3 ⎥
y = [1 0 0 " " 0 ] ⎢
⎥ + [0]u
#
⎢
⎥
⎢ xn −1 ⎥
⎢
⎥
⎣⎢ xn ⎦⎥
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
10
First Companion Form: SISO Case
(Controllable canonical form): Block Diagram
n integrators
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
11
Example:
(Constant in the Numerator)
3 integrators
C (s)
s + 7s + 2
T (s) =
= 3
2
R ( s ) s + 9 s + 26 s + 24
2
( s 3 + 9 s 2 + 2 6 s + 2 4 )C ( s ) = 2 4 R ( s )
w ith z e ro in itia l c o n d itio n s
c + 9 c + 2 6 c + 2 4 c = 2 4 r
L e t x1 c ,
x 1 = x 2
x 2 = x 3
x 2 c ,
x 3 c
x 3 = − 2 4 x 1 − 2 6 x 2 − 9 x 3 + 2 4 r
y = c = x1
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
12
Example:
(Polynomial in the Numerator)
For the block containing denominator
( s 3 + 9 s 2 + 26 s + 24) X 1 ( s ) = 24 R ( s )
x1 = x2
x2 = x3
x3 = −24 x1 − 26 x2 − 9 x3 + r
For Numerator Block
C ( s ) = ( s 2 + 7 s + 2) X 1 ( s )
Taking inverse Laplace transform
y = x3 + 7 x2 + 2 x1
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
13
Example With A Polynomial In
The Numerator
3 integrators
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
14
Systems Having Single Input but
Multiple Outputs
Generalization of the concepts discussed to
this case is straight forward:
™ A and B matrices remain same
™ C and D matrices get modified
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
15
Example
y1 ( s )
2s + 3
and
= H1 ( s ) = 2
u ( s)
3s + 4 s + 5
y2 ( s )
3s + 2
= H 2 (s) = 2
u (s)
3s + 4 s + 5
y1 ( s ) ⎛ z ( s ) ⎞⎛ y1 ( s ) ⎞ ⎛
1
⎞
=⎜
H1 ( s ) =
⎟⎜
⎟= ⎜ 2
⎟ ( 2 s + 3)
u ( s ) ⎝ u ( s ) ⎠⎝ z ( s ) ⎠ ⎝ 3s + 4 s + 5 ⎠
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
16
Example – contd.
4
5
1
z + z + z = u
3
3
3
Define x1 z , x2 z
1
⎡ x ⎤ ⎡ 0
⎢ x ⎥ = ⎢− 5 3 − 4
⎣ ⎦ ⎣
1
2
⎤⎡ x ⎤ ⎡ 0 ⎤
+⎢ ⎥u
⎥
⎢
⎥
3⎦ ⎣ x ⎦ ⎣1 3⎦
1
2
y = 2 z + 3 z = 2 x + 3 x
1
2
1
⎡x ⎤
y = [3 2] ⎢ ⎥ + [0] u
⎣x ⎦
1
1
2
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
17
Example – contd.
⎛ z ( s ) ⎞ ⎛ y2 ( s ) ⎞ ⎛
1
⎞
H 2 (s) = ⎜
⎟⎜
⎟= ⎜ 2
⎟ ( 3s + 2 )
⎝ u ( s ) ⎠ ⎝ z ( s ) ⎠ ⎝ 3s + 4 s + 5 ⎠
A and B are the same
y2 = 3 z + 2 z = 3 x2 + 2 x1
⎡ x1 ⎤
y2 = [ 2 3] ⎢ ⎥ + [ 0] u
⎣ x2 ⎦
Note: Block diagram representation is fairly straight forward.
The realization requires n integrators.
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
18
Jordan Canonical Form
(Non-repeated roots)
H (s) =
r
r
r
y(s)
= d0 + 1 + 2 + " + n
u (s)
s − λ1 s − λ2
s − λn
All the poles of the transfer function are distinct, i.e. no repeated poles
ri 's are called "residues" of the reduced transfer function [ H ( s ) − b0 ]
y (s) = d 0u (s) +
r u (s)
r1u ( s ) r2 u ( s )
+
+ " + n
s − λ1 s − λ2
s − λn
Let
r1u ( s ) ⎤
⎡
x
(
s
)
=
⎢ 1
s − λ1 ⎥
⎢
⎥
⎢
⎥
#
⎢
⎥
r
u
(
s
)
⎢ x (s) = n
⎥
n
⎢⎣
s − λn ⎥⎦
→
⎡ x1 − λ1 x1 = r1u ⎤
⎢
⎥
#
⎢
⎥
⎢
⎥
#
⎢
⎥
⎣ xn − λn xn = rn u ⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
19
Jordan Canonical Form
(Non-repeated roots)
⎡ x1 ⎤ ⎡λ1
⎢ x ⎥ ⎢ 0
⎢ 2⎥ = ⎢
⎢ # ⎥ ⎢0
⎢ ⎥ ⎢
⎣ xn ⎦ ⎣ 0
0
λ2
0
0
y = [1 1 "
" 0 ⎤ ⎡ x1 ⎤ ⎡ r1 ⎤
⎥
⎢
⎥
⎢
⎥
" 0 ⎥ ⎢ x2 ⎥ ⎢ r2 ⎥
+
u
% # ⎥⎢ # ⎥ ⎢# ⎥
⎥⎢ ⎥ ⎢ ⎥
" λn ⎦ ⎣ xn ⎦ ⎣ rn ⎦
⎡ x1 ⎤
⎢x ⎥
1] ⎢ 2 ⎥ + [ d 0 ] u
⎢#⎥
⎢ ⎥
⎣ xn ⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Jordan Canonical Form
(Non-repeated roots): Block Diagram
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Jordan Canonical Form: Example
(Non-repeated roots)
Given
By partial fraction,
y( s)
1
=
u ( s ) ( s + 1)( s + 2)
1 ⎤
⎡ 1
y(s) = ⎢
−
u (s)
⎥
⎣ s +1 s + 2 ⎦
Define two transfer functions
1
x1 ( s ) =
u (s) ,
s +1
1
x2 ( s ) =
u ( s)
s+2
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
22
Jordan Canonical Form: Example
(Non-repeated roots)
This leads to
u ( s ) = x1 ( s )( s + 1), u ( s ) = x2 ( s )( s + 2)
Differential equations corresponding to the x1 , x2
x1 = u − x1
x2 = u − 2 x2
Output Equation
y = x1 − x2
⎡ x1 ⎤ ⎡ −1 0 ⎤ ⎡ x1 ⎤ ⎡1⎤
⎢ x ⎥ = ⎢ 0 −2 ⎥ ⎢ x ⎥ + ⎢1⎥ u
⎣ 2 ⎦ ⎣
⎦ ⎣ 2 ⎦ ⎣N⎦
A
⎡ x1 ⎤
y = [1 −1] ⎢ ⎥
⎣ x2 ⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
B
23
Jordan Canonical Form: Example
Repeated Roots
Following the same procedure
H (s) =
y(s)
2
=
u ( s ) ( s − 3 )3
Let
x2 ( s )
x1 ( s ) =
s −3
x3 ( s )
x2 ( s ) =
s −3
u (s)
x3 ( s ) =
s −3
⇒
x1 − 3 x1 = x2
⇒
x2 − 3 x2 = x3
⇒
x3 − 3 x3 = u
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
24
Jordan Canonical Form: Example
Repeated Roots
z
1
1 ⎡ u (s) ⎤
y(s) =
u (s) = 2
⎢
⎥
3
−
−
−
s
3
s
3
s
3
( ) ( ) ⎢⎣(
)⎥
( s − 3)
⎦
2
Output Equation
x3 ( s )
=2
z
Finally
x3 ( s )
x (s)
1
=2 2
= 2 x1 ( s )
( s − 3) ( s − 3) ( s − 3)
⎡ x1 ⎤ ⎡ 3 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0 3 1 ⎥ ⎢ x ⎥ + ⎢0 ⎥ u
⎢ 2⎥ ⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ x3 ⎥⎦ ⎢⎣0 0 3⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣1 ⎥⎦
N N N
X
A
X
B
y = [ 2 0 0] X + [ 0] u
N
C
D
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
25
Jordan Canonical Form:
What if complex conjugate roots?
Roots always exist in complex conjugate pairs!
H1,2 =
=
λ + jυ
λ − jυ
+
s + (σ − jω ) s + (σ + jω )
2 ⎡⎣λ s + ( λσ − ωυ ) ⎤⎦
The system can be realized
partially in other forms
(like First companion form)
s 2 + 2σ s + (σ 2 + ω 2 )
First companion form:
0
1 ⎤ ⎡ x1 ⎤ ⎡0 ⎤
⎡ x1 ⎤ ⎡
⎥ ⎢ ⎥ + ⎢ ⎥u
⎢ ⎥ = ⎢
2
2
⎣ x2 ⎦ ⎢⎣ − (σ + ω ) −2σ ⎥⎦ ⎣ x2 ⎦ ⎣1 ⎦
y1,2 = ⎡⎣ 2 ( λσ − ωυ )
⎡ x1 ⎤
2λ ⎤⎦ ⎢ ⎥
⎣ x2 ⎦
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
27
ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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