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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
• Dr. Radhakant Padhi, Asstt. Prof, IISC, Bangalore, through NPTEL
• 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
adding them
---(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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