Controller and Observer Design (Design of Control System in State Space)

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Controller and Observer Design (Design of Control System in State Space)
```Controller and Observer
Design
(Design of Control System in State Space)
(Design of Control System in State Space by Pole
placement)
References
• Modern Control Engineering by Katsuhiko Ogata, PHI Pvt. Ltd New
Delhi
Pole Placement
Controller Design
Pole Placement Technique
• Poles of a control system (stable/unstable) can be place at desired
location by pole placement technique. This is done to
•
•
•
•
•
Improve the performance of the system
Make the system stable
Increase the damping
Increase the response time
Etc
Pole Placement Technique
• Assumptions are
1. The system is completely state controllable
2. The sate variable are measureable and available for feedback
3. Control input (u) is unconstrained and single
Note: For multi input system, the state feedback gain matrix is not unique
Pole Placement Technique
• Objective:
• The closed loop poles should lie 1 ,  2 ,…  . Which are their “desired
locations”.
• Difference from classical approach:
• Not only the dominants poles, but “all poles” are forced to lie at specified
desired locations.
• In classical approach only dominants poles are placed at desired location
• Necessary and Sufficient condition:
• The system is completely state controllable
Philosophy of Pole placement control design
• Let a system is represented by
•  =  +  ---(1)
• Put input u as
•  = −, put in equation (1)
• K is called state feedback gain matrix (1xn) and X is state vector (nx1)
• So KX will be scalar (=> single input)
•
•
•
•
•
=  −
=  −   = ---(2)
= ( − )
New closed loop state transition matrix
Its time response
=  − (0) ---(3)
Philosophy of Pole placement control design …

u
B
X
++
B
+
+
A
Fig 1: Open loop Control system
• Philosophy: The matrix K is designed such
equations are having same poles
•  −  =  − 1  − 2 …  −
X
u
A
-K
Fig 2: Closed loop Control system
a way thatWith
theu=-KX
two characterize
Placement control design (Controller Design)
• There are three method:
• Method 1: Direct substitution method (when order of system n≤3)
• Method 2: Bass-Gura Approach
• Method 3: Ackermann’s formula
Controller Design by method 1:
• Let the system is  =  +  steps are
• Step 1: Check controllability of the system
• Step 2: Put u=-KX where  = 1 2 3
• So  =  −
• Step 3: Write characteristic equations of above new system
•  − ( − ) = 0
• Step 4: Write Desired characteristic equation
•  − 1  − 2  − 3 = 0
• Step 5: Compare above two characteristic equations and solve
for k1, k2, k3 by equating the power of s on both sides
Controller Design by method 2:
• Let the system is  =  +  steps are
• Step 1: Check controllability of the system
• Step 2: Put u=-KX where  = 1 2 …
• Step 3: Let the system is in first companion form (Controllable
canonical form) i.e
Controller Design by method 2…
• Step 4: after putting the value of u in given system, now system will
become  =  −   = . So
---(4)
Controller Design by method 2…
• Step 5:
---(5)
Controller Design by method 2…
• Step 6: Comparing equations (4) & (5) we have
What if the system is not given in first
companion form?
• Answer is to convert it into Companion Form as follows
• Define a transform
=
• ⇒  =  −1
put the value of
• ⇒  =  −1 ( + )
• ⇒  =  −1   +  −1
• Select the value of T such that  −1  is in first companion form
• Put
T=MW
• Where
=   … −1  is the controllability matrix
What if the system is not given in first
companion form?...
Controller Design using Method 2: Bass-Gura
Approach
•
•
•
•
•
•
•
•
•
Step 1: Check controllability of the system
Step 2: Form the characteristic equation using matrix A
−  =   + 1  −1 + 2  −2 …−1 1 + find ai’s
Step 3: find the transformation matrix T if system is not in first companion
T=MW
Step 4: Write the desired characteristic equation
− 1 …  −  =   +1  −1 + 2  −2 …−1 1 + findi’s
Step 5: The required state feedback matrix is
−1 − −1 … 1 − 1  −1
=  −
Note: Above approach is for any system (controllable canonical form or
not). If system is in controllable canonical form put T=I (identity matrix)
Controller Design using Method 3:Ackermann’s
Formula
• Let  =  −
• Desired characteristic equation
•  −  =  − 1 …  −  =   +1  −1 + 2  −2 …−1 1 +
• Caley-Hamilton theorem states that every matrix A satisfies it own characteristic
equation. So
• Φ  =  +1 −1 + 2 −2 …−1 1 +  = 0
• For case n=3 consider the following identities
Controller Design using Method 3:Ackermann’s
Formula …
• Multiplying the above identities with 3 , 2 & 1 respectively and
---(6)
Controller Design using Method 3:Ackermann’s
Formula …
• From Caley-Hemilton theorem for
• 3  + 2  +1 2 = ϕ  = 0
• Also we have for A
• 3  + 2  +1 2 = ϕ  ≠ 0
• Putting the values ϕ  & ϕ  of in equation (6)
0
Controller Design using Method 3:Ackermann’s
Formula …
• =>
• Since system is completely controllable inverse of the controllability
matrix exists we obtain
=>
---(7)
Controller Design using Method 3:Ackermann’s
Formula …
• Pre multiplying both sides of the equation (2) with [0 0 1]
Controller Design using Method 3:Ackermann’s
Formula …
• Hence
• For an arbitrary positive integer n ( number of states) Ackermann’s
formula for the state feedback gain matrix K is given by
′  are the coefficients of desired characteristic polynomial
Example
• Example 1: Consider the system defined by  =  +  where
0
1
0
0
•= 0
= 0
0
1
−1 −5 −6
1
• By using the state feedback control u=-KX, it is desired to the closed
loop poles at  = −2 ± 4 and s=-10. Determine the sate feedback
gain matrix K.
• Solution:
• First check the controllability of above system
Example ..
0 0
1
• Controllability matrix  =   2  = 0 1 −6
1 −6 31
•  = −1 so rank of M =3. Hence system is completely state
controllable.
• Now we will solve this problem with previous three methods
Example ..
• Method 1: Direct substitution method
• Put u=-KX where  = 1 2 3
• So  =  −
• Write characteristic equations of above new system
•  − ( − ) = 0
0 0
0
1
0
0
• 0  0 − 0
0
1 + 0 1 2 3
0 0
−1 −5 −6
1
=0
• ⇒  − ( − ) =  3 + 6 + 3  2 + 5 + 2  + 1 + 1
Example…
• Desired Characteristic equation
• Φ  =  + 2 − 4  + 2 + 4  + 10 =  3 + 14 2 + 60 + 200
• comparing above two characteristic equations
• k1 = 199, k2 = 55, k3 = 8
• So  = 199
55
8
Example
• Method 2: Characteristic equation of the given system
•
•
•
•
−1
0
−  = 0
−1
1 5 +6
3
Φ  =  + 6 2 + 5 + 1 Comparing with
Φ  =  3 + 1  2 + 2  + 3
a1 = 6, a2 = 5, a3 = 1
Example…
• Desired Characteristic equation
• Φ  =  + 2 − 4  + 2 + 4  + 10 =  3 + 14 2 + 60 + 200
Comparing with
• Φ  =  3 + 1  2 + 2  + 3
• 1 = 14, 2 = 60, 3 = 200,
• Sate feedback gain matrix K is
2 − 2
1 − 1  −1
•  = 3 − 3
• Where T= I (identity matrix as system is in controllable canonical form)
•  = 199 55 8
Example…
• Method 3: Ackermann’s Formula
•  = 0 0 1   2  −1 Φ()
• Φ  = 3 + 142 + 60 + 200
0
1
0 3
• Φ  = 0
0
1 + 14
−1 −5 −6
199 55
8
• ⇒ Φ  = −8 159
7
−7 −43 117
0 0
•  =   2  = 0 1
1 −6
0
1
0
0
−1 −5
1
−6
31
0
1
−6
2
0
1
0
1 0
+ 60 0
0
1 + 200 0 1
−1 −5 −6
0 0
0
0
1
Example…
• So
0 0
•= 0 0 1 0 1
1 −6
• ⇒  = 199 55 8
1
−6
31
−1
199
−8
−7
55
159
−43
8
7
117
Choice of closed loop poles:
• Don’t choose the closed loop poles far away from the open loop
poles, otherwise it will damage high control effort.
• Don’t choose the closed loop poles very negative, otherwise the
system will be fast reacting (i.e it will have a small time constant)
• In frequency domain it will lead to large bandwidth and hence noise get
amplified.
Controller for multi input system
• The state feedback gain matrix (K) becomes a matrix of mxn (Not
vector of 1xn unlike single input system)
• m = no of inputs and n = no of states
• The state feedback gain matrix (K) is not unique
Summary wise
• Define a linear combination of
control variables as new control
cariable. i.e
•  = 1 + (1 − )2
• Figure Reference:
• www.optisyn.com
Next: Observer Design
Thanks
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