# Stability Analysis and 4 of the Significant Natures In Control System

Hello everyone, I hope that you all are doing good in your lives. The topic for today is stability analysis of control systems. However, we have gone through the basics and a few of the other important concepts of the control system before only. But if you want to have a review of them then you can check the previous blog links

The response of any linear time-invariant system depends on its poles. Every pole has its own contribution to the system response. However, the total response of the system is the summation of individual responses. Thus we can say that if any of the poles is unstable then it may lead to complete instability of the system. The nature of these poles can be either of these;

- Real Poles
- Single
- Multiple

- Complex conjugate pole pair
- Lastly repeated complex conjugate pair

## What is Stability?

A system is said to be stable if it produces bounded output for every bounded input It is mathematically expressed as;

There are basically three main types of stability;

### Absolute Stability

A system is absolutely stable if it is stable for all values of system parameters. For example, a unity feedback system with an open-loop transfer function;

is stable in a closed-loop for all values of k. Hence the system is absolutely stable.

### Conditional Stability

A system is conditionally stable if it is stable only upon satisfying certain conditions. For example, a unity feedback system with an open-loop transfer function;

is stable in closed loop for 0 < k < 30 and is unstable for k > 30.

### Relative Stability

It is a measure of stability between two or more systems. A system with less settling time is relatively more stable. In another way, the system with poles away from the imaginary axis (in the left half of the s-plane) is more stable.

## Stability Analysis of Single Real Pole

If the poles are s=a, then we get the transfer function G(s) = 1/(s-a). Thus the time response of the system is equal to **g(t) = e ^{at}**

### Case 1: a < 0

If a < 0 the pole lies in the left half of the s-plane.

For all values of ‘a’ less than zero, g(t) decreases with the increasing time and at t=∞, g(t)=0. This response is stable in nature.

### Case 2: a > 0

If a > 0 the pole lies in the right half of the s-plane.

For all values of ‘a’ greater than zero, g(t) increases with the increasing time, and at t=∞, g(t)=∞. However, this response is unstable in nature.

### Case 3: a = 0

For all values of ‘a’ equal zero, g(t) = 1. However, this g(t) is bounded but it does not approach’s to zero and its response is marginally stable.

## Stability Analysis of Multiple Real Pole

Assume that we are having two poles at s = a then, we get the transfer function as G(s) = 1/(s+a)^{2}. Thus the time response of the system is equal to g(t) = te^{at}.

It has the same conditionality as that in a single real pole analysis.

## Stability Analysis of Complex Conjugate Pole Pair

Let’s assume that we have a pair of conjugate poles i.e. s = a±jb then, we get the transfer function; G(s) = 1/(s-a+jb)(s-a-jb). Thus the time response of the system is equal to g(t) = 1/b e^{at}Sinbt.

### Case 1: a < 0

If the value of a < 0 then the poles lie in the left half of the s-plane.

For all values of ‘a’ less than zero, e^{at} decreases with the increasing time t. And hence the system response is a sine wave with decreasing amplitude. However, this response is stable in nature.

### Case 2: a > 0

If the value of a > 0 then the poles lie in the right half of the s-plane

For all values of ‘a’ greater than zero, e^{at} increases with the increasing time t. And hence the system response is a sine wave with increasing amplitude. However, this response is unstable in nature.

### Case 3: a = 0

For all values of ‘a’ equal to zero, g(t) = (1/b)Sinbt is a sine wave with constant amplitude. However, this response is marginally stable.

## Stability Analysis of Repeated Complex Conjugate Poles

If we assume that we are having multiple complex conjugate poles for example s= a ± jb then, we get the transfer function; G(s) = 1/(s-a+jb)^{2}(s-a-jb)^{2}

It has the same conditionality as that in a single complex conjugate pole pair analysis.

## Stability Analysis based on location of pole

- A system is said to be stable if all its poles are in the left half of the s-plane.
- A system is unstable if;
- Any of the pole is in the right half of the s-plane.
- Repeated poles are on imaginary axis.

- However, a system is marginally stable if non repeated poles are on the imaginary axis and remaining poles are in the left half of the s-plane.

### Routh Hurwitz Stability Analysis

The necessary condition for the routh criteria is that if the system with closed-loop characteristic equation Q(s) = a_{0}s^{n} + a_{1}s^{n-1} + a_{2}s^{n-2} +————–+a_{n-1}s +a_{n} = 0 may be stable if all coefficients in the characteristic equation have the same sign and no coefficient is zero, otherwise, the system is not stable.

Sufficiency Condition – The system is stable if all elements in the first column of routh array have the same sign. The routh is formed as;

The first two rows are formed from the characteristic equation where;

In this way all elements in the third row can be calculated thus;

However, this procedure is continued till the n+1^{th} row completes the routh array. If there are sign changes in the first column of routh array then, the system is unstable and the number of poles is equal to the number of sign changes in the right half of the s-plane.

#### Special Case 1

When a zero appears in the first column of routh array. In this particular case, the calculation of the next row of routh array is not possible. To overcome this problem there are two possible solutions;

1. Thus to avoid such error the 0 is replaced by ε where ε is such that ε -> 0 and the investigation about sign change in the first column of routh array the system is unstable and if there is no sign change in the first column of routh array the system is marginally stable.

##### For example:- Investigate the stability of system with characteristic equation Q(s) = s^{4} + 3s^{3} + 2s^{2} + 6s + 4 = 0

Solution:- The characteristic equation is Q(s) = s^{4} + 3s^{3} + 2s^{2} + 6s + 4 = 0 Thus we get the routh array as;

As there are two sign changes, the system is unstable.

2. Another way is to replace s in the characteristic equation with 1/z and obtain a new characteristic equation. However, this method fails when the same problem occurs with the newly formed equation.

##### For example:- Investigate the stability of system with characteristic equation Q(s) = s^{4} + 2s^{3} + 2s^{2} + 4s + 6 = 0

Solution:- The characteristic equation is Q(s) = s^{4} + 2s^{3} + 2s^{2} + 4s + 6 = 0. therefore we get the routh array as;

Replacing s by 1/z we get the characteristic equation as Q(s) = 1/z^{4} + 2/z^{3} + 2/z^{2} + 4/z + 6 = 0 i.e. Q(s) = 6z^{4} + 4z^{3} + 2z^{2} + 2z + 1 = 0. Thus we get the routh array as;

#### Special Case 2

When all elements in a row of routh array become zero. In this case, further calculation becomes impossible to complete. Thus in such a case, the row just above the row with all zero elements is considered to form an auxiliary equation. This auxiliary equation is differentiated with respect to s to get a new auxiliary equation of the row with all zeros and further calculations are performed to form routh array. Then if there is no sign change, the system is marginally stable otherwise the system is unstable.

##### For example:- Investigate the stability of system with characteristic equation Q(s) = s^{6} + 3s^{5} + 5s^{4} + 9s^{3} + 8s^{2} + 6s + 4 = 0

Solution:- The characteristic equation is Q(s) = s^{6} + 3s^{5} + 5s^{4} + 9s^{3} + 8s^{2} + 6s + 4 = 0 thus we get the routh array as;

As all the elements of the row s^{3} are 0, consider the equation of row s^{4} we get the auxiliary equation as; A(s) = 2s4 + 6s2 +4 after differentiating the auxiliary equation we get dA(s)/ds = 8s^{3} +12s

Replace 0s in the s^{3} row by 8 and 12

There is no sign change in the first column of the routh array and the system is marginally stable.

## Conclusion

However here we are at the end of the blog. I hope that all your doubts are clear and have got a clear idea about the topic. If you do like the topic then please make sure to share it with others. And also if you do have any doubt regarding the topic then please feel free to mention it down below.

Regards,

Have a nice day 🙂

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