1. Controller and Observer Design
(Design of Control System in State Space)
(Design of Control System in State Space by Pole placement)

2. References
Dr. Radhakant Padhi, Asstt. Prof, IISC, Bangalore, through NPTEL
Modern Control Engineering by Katsuhiko Ogata, PHI Pvt. Ltd New Delhi

3. Pole Placement Controller Design

4. 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

5. Pole Placement Technique
Assumptions are
The system is completely state controllable
The sate variable are measureable and available for feedback
Control input (u) is unconstrained and single
Note: For multi input system, the state feedback gain matrix is not unique

6. Pole Placement Technique
Objective:
The closed loop poles should lie 𝜇 1 , 𝜇 2 ,… 𝜇 𝑛 . 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

7. Philosophy of Pole placement control design
Let a system is represented by
𝑋 =𝐴𝑋+𝐵𝑢 ---(1)
Put input u as
𝑢=−𝐾𝑋, 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)
𝑋 =𝐴𝑋−𝐵𝐾𝑋
𝑋 = 𝐴−𝐵𝐾 𝑋= 𝐴 𝑋---(2)
𝐴 =(𝐴−𝐵𝐾) New closed loop state transition matrix
Its time response
𝑥 𝑡 = 𝑒 𝐴−𝐵𝐾 𝑥(0) ---(3)

8. Philosophy of Pole placement control design …B u
Fig 2: Closed loop Control system
With u=-KX -K X 𝑋 +
Philosophy: The matrix K is designed such a way that the two characterize equations are having same poles
𝑠𝐼− 𝐴 = 𝑠− 𝜇 1 𝑠− 𝜇 2 … 𝑠− 𝜇 𝑛 A B u
Fig 1: Open loop Control system X 𝑋 +

9. 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

10. Controller Design by method 1:
Let the system is 𝑋 =𝐴𝑋+𝐵𝑢 steps are
Step 1: Check controllability of the system
Step 2: Put u=-KX where 𝐾= 𝑘 1 𝑘 2 𝑘 3
So 𝑋 = 𝐴−𝐵𝐾 𝑋
Step 3: Write characteristic equations of above new system
𝑠𝐼−(𝐴−𝐵𝐾) =0
Step 4: Write Desired characteristic equation
𝑠− 𝜇 1 𝑠− 𝜇 2 𝑠− 𝜇 3 =0
Step 5: Compare above two characteristic equations and solve for k1, k2, k3 by equating the power of s on both sides

11. Controller Design by method 2:
Let the system is 𝑋 =𝐴𝑋+𝐵𝑢 steps are
Step 1: Check controllability of the system
Step 2: Put u=-KX where 𝐾= 𝑘 1 𝑘 2 … 𝑘 𝑛
Step 3: Let the system is in first companion form (Controllable canonical form) i.e

12. Controller Design by method 2…
Step 4: after putting the value of u in given system, now system will become 𝑋 = 𝐴−𝐵𝐾 𝑋= 𝐴 𝑋. So 𝐴
---(4)

13. Controller Design by method 2…
Step 5:
---(5)

14. Controller Design by method 2…
Step 6: Comparing equations (4) & (5) we have

15. What if the system is not given in first companion form?
Answer is to convert it into Companion Form as follows
Define a transform 𝑋=𝑇 𝑋
⇒ 𝑋 = 𝑇 −1 𝑋 put the value of 𝑋
⇒ 𝑋 = 𝑇 −1 (𝐴𝑋+𝐵𝑢)
⇒ 𝑋 = 𝑇 −1 𝐴𝑇 𝑋 + 𝑇 −1 𝐵 𝑢
Select the value of T such that 𝑇 −1 𝐴𝑇 is in first companion form
Put T=MW
Where 𝑀= 𝐵 𝐴𝐵… 𝐴 𝑛−1 𝐵 is the controllability matrix

16. What if the system is not given in first companion form?...

17. Controller Design using Method 2: Bass-Gura Approach
Step 1: Check controllability of the system
Step 2: Form the characteristic equation using matrix A
𝑠𝐼−𝐴 = 𝑠 𝑛 + 𝑎 1 𝑠 𝑛−1 + 𝑎 2 𝑠 𝑛−2 … 𝑎 𝑛−1 𝑠 1 + 𝑎 𝑛 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
𝑠− 𝜇 1 … 𝑠− 𝜇 𝑛 =𝑠 𝑛 + 𝛼 1 𝑠 𝑛−1 + 𝛼 2 𝑠 𝑛−2 … 𝛼 𝑛−1 𝑠 1 + 𝛼 𝑛 find𝛼i’s
Step 5: The required state feedback matrix is
𝐾= 𝛼 𝑛 − 𝑎 𝑛 𝛼 𝑛−1 − 𝑎 𝑛−1 … 𝛼 1 − 𝑎 𝑇 −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)

18. Controller Design using Method 3:Ackermann’s Formula
Let 𝐴 =𝐴−𝐵𝐾
Desired characteristic equation
𝑠𝐼− 𝐴 = 𝑠− 𝜇 1 … 𝑠− 𝜇 𝑛 =𝑠 𝑛 + 𝛼 1 𝑠 𝑛−1 + 𝛼 2 𝑠 𝑛−2 … 𝛼 𝑛−1 𝑠 1 + 𝛼 𝑛
Caley-Hamilton theorem states that every matrix A satisfies it own characteristic equation. So
Φ 𝐴 = 𝐴 𝑛 + 𝛼 1 𝐴 𝑛−1 + 𝛼 2 𝐴 𝑛−2 … 𝛼 𝑛−1 𝐴 1 + 𝛼 𝑛 𝐼=0
For case n=3 consider the following identities

19. Controller Design using Method 3:Ackermann’s Formula …
Multiplying the above identities with 𝛼 3 , 𝛼 2 & 𝛼 1 respectively and adding them
---(6)

20. Controller Design using Method 3:Ackermann’s Formula …
From Caley-Hemilton theorem for 𝐴
𝛼 3 𝐼+ 𝛼 2 𝐴 +𝛼 1 𝐴 2 =ϕ 𝐴 =0
Also we have for A
𝛼 3 𝐼+ 𝛼 2 𝐴 +𝛼 1 𝐴 2 =ϕ 𝐴 ≠0
Putting the values ϕ 𝐴 & ϕ 𝐴 of in equation (6)

21. Controller Design using Method 3:Ackermann’s Formula …
=>
Since system is completely controllable inverse of the controllability matrix exists we obtain
=> (7)

22. Controller Design using Method 3:Ackermann’s Formula …
Pre multiplying both sides of the equation (2) with [0 0 1]

23. 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
𝛼 𝑖 ′ 𝑠 are the coefficients of desired characteristic polynomial

24. Example Example 1: Consider the system defined by 𝑋 =𝐴𝑋+𝐵𝑢 where
𝐴= −1 −5 −6 𝐵=
By using the state feedback control u=-KX, it is desired to the closed loop poles at 𝑠=−2±𝑗4 and s=-10. Determine the sate feedback gain matrix K.
Solution:
First check the controllability of above system

25. Example .. Controllability matrix 𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 = 0 0 1 0 1 −6 1 −6 31
𝑀 =−1 so rank of M =3. Hence system is completely state controllable.
Now we will solve this problem with previous three methods

26. Example .. Method 1: Direct substitution method
Put u=-KX where 𝐾= 𝑘 1 𝑘 2 𝑘 3
So 𝑋 = 𝐴−𝐵𝐾 𝑋
Write characteristic equations of above new system
𝑠𝐼−(𝐴−𝐵𝐾) =0
𝑠 𝑠 𝑠 − −1 −5 − 𝑘 1 𝑘 2 𝑘 3 =0
⇒ 𝑠𝐼−(𝐴−𝐵𝐾) = 𝑠 𝑘 3 𝑠 𝑘 2 𝑠+ 1+ 𝑘 1

27. Example… Desired Characteristic equation
Φ 𝐴 = 𝑠+2−𝑗4 𝑠+2+𝑗4 𝑠+10 = 𝑠 𝑠 2 +60𝑠+200
comparing above two characteristic equations
k1 = 199, k2 = 55, k3 = 8
So 𝐾=

28. Example Method 2: Characteristic equation of the given system
𝑠𝐼−𝐴 = 𝑠 −1 0 0 𝑠 −1 1 5 𝑠+6
Φ 𝐴 = 𝑠 3 +6 𝑠 2 +5𝑠+1 Comparing with
Φ 𝐴 = 𝑠 3 + 𝑎 1 𝑠 2 + 𝑎 2 𝑠+ 𝑎 3
a1 = 6, a2 = 5, a3 = 1

29. Example… Desired Characteristic equation
Φ 𝐴 = 𝑠+2−𝑗4 𝑠+2+𝑗4 𝑠+10 = 𝑠 𝑠 2 +60𝑠+200 Comparing with
Φ 𝐴 = 𝑠 3 + 𝛼 1 𝑠 2 + 𝛼 2 𝑠+ 𝛼 3
𝛼 1 =14, 𝛼 2 =60, 𝛼 3 =200,
Sate feedback gain matrix K is
𝐾= 𝛼 3 − 𝑎 𝛼 2 − 𝑎 𝛼 1 − 𝑎 𝑇 −1
Where T= I (identity matrix as system is in controllable canonical form) 𝐾=

30. Example… Method 3: Ackermann’s Formula 𝐾= 0 0 1 𝐴 𝐴𝐵 𝐴 2 𝐵 −1 Φ(𝐴)
𝐾= 𝐴 𝐴𝐵 𝐴 2 𝐵 −1 Φ(𝐴)
Φ 𝐴 = 𝐴 𝐴 2 +60𝐴+200𝐼
Φ 𝐴 = −1 −5 − −1 −5 − −1 −5 −
⇒Φ 𝐴 = − −7 −43 117
𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 = −6 1 −6 31

31. Example… So
𝐾= −6 1 − − − −7 −43 117
⇒𝐾=

32. 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.

33. 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

34. Summary wise Define a linear combination of
control variables as new control
cariable. i.e
𝑣=𝜂 𝑢 1 +(1−𝜂) 𝑢 2
Figure Reference:

35. Next: Observer Design
Thanks