CONTROL SYSTEM

1. 26-Aug-16 1
SEMINAR PRESENTATION
ON
RH CRITERION
Presented By:
LUCKY CHAUHAN
B.TECH (I.T) – 1st year
(Roll No. 20190947)
DEPARTMENT OF INFORMATION AND TECHNOLOGY ENGINEERING
GRAPHIC ERA UNIVERSITY, DEHRADUN

2. CONTENTS
• STABILITY
• WHY INVESTIVAGATE STABILITY?
• EXAMPLE OF UNSTABLE SYSTEM
• ROUTH STABILITY CRITERION
• RULES
• SPECIAL CASES
• PRACTICAL USE
• REFRENCES

3. STABILITY
A process to find out if a system is stable or
not.
1. A system is stable if every bounded input
yields a bounded output.
2. A system is unstable if any bounded input
yields an unbounded output.

4. Why investigate stability?
• Most important issue to design a control
system.
• An unstable feedback system is of no practical
value.
• A stable system should exhibit a bounded
output if the corresponding input is bounded.

5. 5
Dramatic photographs showing the collapse of
the Tacoma Narrows suspension bridge on
November 7, 1940.
(a) Photograph showing the twisting motion
of the bridge’s center span just before failure.
(b) A few minutes after the first piece of
concrete fell, this second photograph shows a
600-ft section of the bridge breaking out of
the suspension span and turning upside down
as it crashed in Puget Sound, Washington.
(Taken from:-1)
One famous example of an
unstable system:

6. 6
stability region:
according to the
roots obtained
by the
characteristic
equation.
Taken from:- 2

7. • Using these concepts, we can say:-
1. stable system: if the natural response
approaches zero as time approaches
infinity.
2. unstable system: if the natural response
grows without bound as time approaches
infinity.
3. marginally stable: if the natural response
neither decays nor grows but remains
constant or oscillates as time approaches
infinity.

8. 8
Contributions of
characteristic
equation roots to
closed-loop
response.
𝑡𝑎𝑘𝑒𝑛 𝑓𝑟𝑜𝑚: 2

9. Routh-Hurwitz Stability Criterion
• The characteristic equation of the nth order continuous system can be write as:
• The stability criterion is applied using a Routh table which is defined as;
• Where are coefficients of the characteristic equation.
{𝑡aken from:3}

10. Routh-Hurwitz Criterion

11. RULES
• There should be no missing term.
3 𝑠4
+ 4𝑠3
+ 5𝑠 + 1 = 0
𝑠2
missing.
• All variables should be of same sign.
5𝑠3
− 3𝑠2
+ 5𝑠 + 1 = 0
-ve sign before the variable 3

12. Closed-loop control system with
T(s) = Y(s)/R(s) = 1/(s3 + s2+ 2s+24).
EXAMPLE:

13. Routh array for the closed-loop control system
with T(s) = Y(s)/R(s) = 1/(s3+ s2 +2s+24)

14. Routh-stability Criterion: Special cases
• Two special cases can occur:
– Routh table has zero only in the first column of a
row
– Routh table has an entire row that consists of
zeros.
s3 1 3 0
s2 3 4 0
s1 0 1 2
s3 1 3 0
s2 3 4 0
s1 0 0 0

15. Example of special case:1

16. Example of Special case:2
Determine the number of right-half-lane poles in
the closed-loop transfer function
  5 4 3 2
10
7 6 42 8 56
T s
s s s s s

    
𝒔 𝟓
1 6 8
𝑠4 7 42 56
𝑠3
0 0 0
𝑠1
𝑠0

17. Special case:2
• To solve it we will form polynomial
• P=7𝑠4+42𝑠2+56
• Then take its derivative
𝒔 𝟓
1 6 8
𝑠4 7 42 56
𝑠3
0 0 0
𝑠1
𝑠0

18. • Solving for the remainder of the Routh table
• There are no sign changes, so there are no
poles on the right half plane. The system is
stable.

19. 19
PRACTICAL USE:
Find the values of controller gain Kc that make the feedback
control system of following equation stable.
The characteristic equation is:-
All coefficients are positive provided that 1 + Kc > 0 or Kc < -1.
The Routh array is
10 8
17 1 + Kc
b1 b2
c1
10𝑆3+17𝑆2+8S+(1+𝐾𝐶)=0

20. 20
To have a stable system, each element in the left column of the
Routh array must be positive. Element b1 will be positive if
Kc < 7.41/0.588 = 12.6. Similarly, c1 will be positive if Kc > -1.
Thus, we conclude that the system will be stable if
-1<𝐾𝐶 < 12.6

21. REFRENCES
1. www.calvin.edu/../chap-6
2. site.ntvc.edu.cn/..automation_7.ppt
3. www.slideshare.com
4.BOOKS:-
• B.S MANKE
• I.J NAGRATHA

22. THANK YOU