**Stability**

By Prof. Seungchul Lee

http://iai.postech.ac.kr/

Industrial AI Lab at POSTECH

http://iai.postech.ac.kr/

Industrial AI Lab at POSTECH

Table of Contents

- In order for a system $G(s)={N(s) \over D(s)}$ to be stable all of the roots of the characteristic polynomial need to lie in the left-half plane (LHP).
- The characteristic equation is the denominator of the transfer function.
- The roots of the characteristic equation are the exact same as the poles of the transfer function.
- The eigenvalues of matrix $A$ in the equivalent state space representation are the same as the roots of the characteristic polynomial.
- In order to have a stable system, roots of $G(s)$ must be in LHP.

- When a pole is negative
- This root exists in the left half plane
- Transfer function will ultimately die out
- The system will eventually be at rest (stable)

- When a pole is positive
- This root exists in the right half plane
- Transfer function will blow up into infinity
- The system is unstable

- Transfer function of multiple poles

- The last one blows up to infinity to make the whole transfer function unstable

- Conclusion: a single root in the right half plane makes the whole system unstable

Calculating the roots of the system for larger than the second-order polynomial becomes time-consuming and possibly even impossible in a closed-form

- How can we determine the stability of a higher order polynomial without solving for the roots directly?
- The great thing about the Routh-Hurwitz criterion is that we do not have to solve for the roots of the characteristic equation

$$G(s) = {1 \over s^4 + 3s^3-5s^2+s+2}$$

- If all of the signs are not the same, the system is unstable
- If you build up a transfer function with a series of poles, then the only way to get a negative coefficient is to have at least one pole exists in right-half plane

- The great thing about the Routh-Hurwitz criterion is that we do not have to solve for the roots of the characteristic equation

- However, we cannot claim that all positive coefficients are still either stable or unstable

$$ \begin{align*} G(s) &= \frac{1}{s^4 + 2s^3 + 3s^2 + 10s + 8} \\\\ &= {1 \over s^2-s+4} \centerdot {1 \over s+2} \centerdot {1 \over s+1} \end{align*} $$

Routh array is a table that can be populated with the coefficients of the polynomial with a few simple rules

- The number of RHP roots of $D(s)$ is equal to the number of sign changes in the first column of the Routh array

- Determine the number of roots in RHP by counting the number of sign changes
- We can determine the number of roots in the right-half plane by looking at this first column
- For example, it changes sign twice which means that there are two roots in the right half plane

- A zero in a row with at least one non-zero appearing later in that same row
- If you are attempting to access stability of the system, you do not need to complete the rest of the table at this point
- The system is always unstable because completing Routh array will always result in a sign change of the first column

- If you are interested in the number of roots located in the right half plane, you can complete the table like below
- You replace that zero with the Greek symbol epsilon $\epsilon >0$

- When you finish completing the table, you can take the limit as epsilon $\epsilon$ goes to zero

- You can see that we still have two unstable roots or two roots in the right half plane

- You replace that zero with the Greek symbol epsilon $\epsilon >0$

- The second special case is when there is an entire row of zeros, not just a single zero in the row

- Auxiliary polynomial $P(s)$: the row directly above the row of zeros

- Then $P(s)$ is a factor of the original polynomial $D(s)$

$$
\begin{align*}
D(s) &= P(s) \cdot R(s) \\\\\\
s^5 + 2s^4 + 6s^3 + 10s^2 + 8s +12 & = \underbrace{(s^2 + 2)}_{\text{marginally stable}}\;\;\underbrace{(s^3 + 2s^2 + 4s + 6)}_{\text{stable}}
\end{align*}
$$

- Apply the Routh-Hurwitz criterion again to $R(s)$

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- It is useful to start with scalar systems to get some intuition about what is going on

$$\begin{cases}
a>0: \text{unstable}\\
a<0: \text{asymptotically stable}\\
a=0: \text{critically stable}
\end{cases}$$

- From scalars to matrices?

- We cannot say that $A>0$, but we can do the next best thing - eigenvalues !

- The eigenvalues tell us how the matrix $A$ 'acts' in different directions (eigenvectors)

- We are interested in the stability of a closed loop system from an open loop system.
- Stability of a
**closed loop**system can also be determined from the**open loop frequency response**

- Stability of a

- The closed-loop system is

$$H(s) = \frac{KG}{1+KG}$$

- A pole exists when the characteristic polynomial in the denominator becomes zero.

$$1+KG(s) = 0 \implies KG(s) = -1 = 1 \angle (2k+1) \pi, \quad k = 0, \pm1, \pm 2, \cdots$$

- A value of $s^*$ is a closed loop pole if

Closed-loop poles in the LHP indicate stability

- The closeness of the poles to the RHP indicate how near to instability the system is

- Suppose the Bode plot of the open-loop transfer function is given.

- Question: tell the stability of a closed-loop system from the open-loop frequency response

- At 180$^o$ of phase lag of the loop, the reference and feedback signal are added.
- If the magnitude of the loop is greater than 1 the error grows exponentially (unstable)

**Relative stability**is indicated by how close the open-loop frequency response is to the point of 180$^o$ of phase lag and a magnitude of 1

- More specifically,
- Gain margin is the distance from a magnitude of 1 (0 dB) at the frequency where $\phi = -180^o$ (phase crossover frequency)
- Phase margin is the distance from a phase of -180$^o$ at the frequency where $M$ = 0 dB (gain crossover frequency)

- In order to be stable, both gain and phase margin must be positive

- Gain and phase margins tell how stable the system would be in closed-loop
- These quantities can be read from the open-loop data

- What if $K$ proportional controller is implemented?