This document discusses the concept of BIBO (bounded-input bounded-output) stability through analyzing several examples of linear time-invariant systems. It addresses how the location of poles and zeros of a system's transfer function affects its stability and response. Specifically: 1) For a system to be BIBO stable, all poles must have negative real parts (lie in the left half plane), meaning the system is stable. 2) Zeros do not directly impact stability, but can affect the system's dynamic response. Stability depends solely on pole locations. 3) Repeating poles or non-repeating poles on the jω axis can lead to instability even if other poles are stable in the left half

1. BIBO STABILITY
EFFECT OF POLES

2. Problem 1:
• Let us take the first problem. This is the system, whose governing
equation is given by
y 𝑦̈(𝑡)+5ẏ(𝑡)+6𝑦(𝑡)=𝑢(𝑡)
• This is a second order system. If we take the Laplace transform on
both sides, we get
𝑠2𝑌(𝑠)−𝑠𝑦(0)−ẏ(0)+5[𝑠𝑌(𝑠)−𝑦(0)]+6𝑌(𝑠)=𝑈(𝑠)

3. We collect terms and we will get
[𝑠2
+5𝑠+6]𝑌(𝑠)=[𝑠𝑦(0)+5𝑦(0)+ y ̇(0)]+𝑈(𝑠)
So, now when we want to get the transfer function, we take all the initial
conditions as 0. we take 𝑦(0)=0 and ẏ(0)=0.

4. • Here we see that the order of the denominator polynomial of the transfer function 𝑛=2 and
the order of the numerator polynomial of the transfer function 𝑚=0.
• And we solve the denominator polynomial equal 0 to get the poles
• 𝑠2 +5𝑠+6=0
𝑠2
+3𝑠+ 2𝑠 +6=0
𝑠 𝑠 + 3 + 2 𝑠 + 3 =0
• The poles are -2 and -3 and there are no zeroes for this problem

5. • Now let us calculate the unit step response of the system.
• For a unit step response 𝑢(𝑡)=1 and 𝑈(𝑠)=1/𝑠.
• We can write
• A, B, C are called the residues.

6. If we take the inverse Laplace transform
This is the unit step response of the system.

7. what we observe from this unit step response.
1) As,𝑡→∞,𝑦(𝑡)→1/6, this is called as the steady state value.
2) We locate the poles -2 and -3 on the negative real axis.
3)What about BIBO stability here?
Step is a bounded input, and the corresponding 𝑦(𝑡) that we have
calculated is bounded.
The observation we can make here is that the system is bounded input
bounded output stable.

8. Problem 2:
𝑦 ̈(𝑡)+𝑦(𝑡)=𝑢(𝑡)
• take the Laplace transform on both sides
• Now in order to find the transfer function take all initial conditions to
be 0 = P(S)
• n is 2, second order system, m is 0. What about poles

9. solve for 𝑠2
+1=0
s = ±j
• we need to get the unit step response
• Find A ,B,C?
https://www.chilimath.com/lessons/advanced-algebra/partial-fraction-decomposition/
• A=1 ,B=-1 and C= 0

10. • Apply inverse Laplace transform
y(t)=1-cos(t)
• So, what happens as t→∞?
• the magnitude of y(t) remains bounded,
• we are getting a bounded output;
• so from this would we conclude that the system is BIBO stable.

11. Problem 3:
𝑦̈(𝑡)+𝑦̇(𝑡)=𝑢(𝑡)
take the Laplace transform on both sides,
Initial condition =0
n = 2, m = 0. What are the poles? The poles are at 0 and – 1
Apply unit step input 𝑈(𝑠)=1/𝑠 to find unit step response

12. A,B,C= ? Using partial fraction decomposition we get
A=-1, B=1,C= 1
Taking inverse Laplace transform
So the magnitude of y(t) as t tends to infinity, what happens to its magnitude
It tends to infinity, because we have a t term.
we have found a step input for which the output is unbounded

13. Problem 4:
𝑦̈(𝑡)+𝑦̇(𝑡)−2𝑦(𝑡)=𝑢(𝑡)
take the Laplace transform on both sides and putting initial condition
equal to zero we get
n = 2, m = 0, the poles are at -2 and +1.

14. unit step response: 𝑌(𝑠)=𝑃(𝑠)𝑈(𝑠)=
t→∞, the magnitude tends to infinity because we have 𝑒𝑡, the magnitude of 𝑒𝑡 tends to infinity
Taking inverse Laplace transform

15. Problem 5:
𝑦̈(𝑡)+𝑦(𝑡)=𝑢(𝑡)
• take the Laplace transform on both sides
• Now in order to find the transfer function take all initial conditions to
be 0 = P(S)
• n is 2, second order system, m is 0.
𝑠2
+1=0
s = ±j

16. • And we saw that in problem 2 when it was subjected to a unit step
input, the output was bounded in magnitude
• Now subject it to a cosine input of a very specific angular frequency
i.e 𝜔=1 rad/s. If 𝑢(𝑡)=cos𝑡, then 𝑈(𝑠)=
𝑠
𝑠2+1
.
•
• Now, we use a property of Laplace transform, the complex
differentiation theorem.

17. Taking inverse Laplace transform
We observe that, as 𝑡→∞,𝑦(𝑡)→∞.

18. Problem 6 :
same as problem 3 but with a different input
𝑦̈(𝑡)+𝑦̇(𝑡)=𝑢(𝑡) , u(t) = cos(t),
poles at 0,-1
• 𝜔=1 rad/s. If 𝑢(𝑡)=cos𝑡, then 𝑈(𝑠)=
𝑠
𝑠2+1

19. • A,B,C=?
• Solving the partial fractions, we get 𝐴=1/2, 𝐵=−1/2 and 𝐶=1/2.
• Putting the values we get
• Taking inverse Laplace, we have
• Earlier we saw that, for the same system when we gave the unit step
response, 𝑦(𝑡)→∞.
• Now we gave another bounded input cos𝑡, we are getting a bounded
output.

20. BIBO Stability
• A system is said to be BIBO stable, if the output remains bounded in
magnitude for all time given any bounded input
• We will look into the location of poles and its influence on the
stability of the system

21. Conclusion
Problem # System Poles Input output
Problem No 2 𝑦̈(𝑡)+𝑦(𝑡)=𝑢(𝑡) ±𝑗 Unit Bounded as t→∞
Problem No5 cost Unbounded as t→∞
Problem No 3 𝑦̈(𝑡)+𝑦̇(𝑡)=𝑢(𝑡) O , -1 unit Unbounded as
Problem No 6 cost Bounded as
if we use the definition of BIBO stability which states that a system is BIBO stable if its
output is bounded for all possible bounded inputs, we should brand these two systems as
unstable
t→∞
t→∞

22. conclusion
1. For BIBO stability, all poles of the plant transfer function must lie in the LHP or
in other words having negative real parts.
2. If there exists even one pole of the plant transfer function in the right half
plane, that is have has a positive real part, then the plant or the system is
not BIBO stable.
3. If there are repeating poles of the plant transfer function on the imaginary
axis (j omega axis) with all remaining poles in the LHP, then system is not
BIBO stable.
4. If there are non repeating poles of the plant transfer function on the imaginary
axis with all remaining poles in the LHP then the system is unstable or critically
stable.

23. Effect of zeros:
• We identified the fact that the location of the poles influences the
stability of the system.
• And for BIBO stability all poles of the system transfer function should
lie in left half complex plane.
• In other words, they should have negative real parts.
• Now to find out the impact of zeros on the system response, Let us
consider two systems:

24. Problem 1:
The first plant transfer function is
The order of these systems is 2 because the order of the denominator polynomial is
always going to be the order of the system for the class of systems that we study.
For the first system, n=2 and m=0 .
We want to figure out the unit step response.

25. • Evaluating the partial fractions, we get
• take the inverse Laplace transform, we get
• We observe that the two poles are in the left of complex plane. So the
system is BIBO stable
• As t→∞, y (t)→ 1/10

26. Problem 2
• n=2 and m=1
• The poles are at -1 and -10 there is a zero at -2.
• The difference between system 1 and system 2 is, we have introduced a zero at -2.
• We want to figure out the unit step response. U (s)=1/s
• Evaluating the partial fractions, we get

27. Evaluating the partial fractions, we get
If we take the inverse Laplace transform, we get
we can see that as t→∞, y (t)→1/5 So, it is bounded, the system is
BIBO stable because the poles are in the left half plane.

29. • We observe that the zero is going to affect the dynamic response of
the system but it is not going to have a direct impact on the stability.
Because stability is dependent on the location of the poles.
• Zeros will affect the coefficients of the response function

30. Problem 4: