Stability

By Prof. Seungchul Lee
http://iai.postech.ac.kr/
Industrial AI Lab at POSTECH

# 1. Stability of Open Loop SystemÂ¶

• 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.

$$G(s) = {1 \over s+a}$$

$$\mathscr{L}^{-1}(G(s)) = e^{-at} u(t)$$

• 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

\begin{align*} G(s) &= {1 \over s+1} \centerdot {1 \over s+3} \centerdot {1 \over s-2}\\\\ G(s) &= {A \over s+1} + {B \over s+3} + {C \over s-2} \\\\ \mathscr{L}^{-1}(G(s)) &= Ae^{-t} + Be^{-3t} + Ce^{2t} \end{align*}

• 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

# 2. Routh-Hurwitz CriterionÂ¶

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
• 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*}

## 2.1. Normal CaseÂ¶

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

$$As^6 + Bs^5 + Cs^4 + Ds^3 + Es^2 + Fs + G$$

• 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

$$G(s) = {1 \over s^4 + 2s^3 + 3s^2 + 10s + 8}$$

## 2.2. Special Case 1Â¶

• 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

## 2.3. Special Case 2Â¶

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

$$D(s) = s^5 + 2s^4 + 6s^3 + 10s^2 + 8s +12$$

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

$$6s^2 + 12s^0 \implies P(s) = s^2 + 2$$

• 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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# 3. Stability with State Space RepresentationÂ¶

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

$$\dot{x} = ax \quad \implies x(t) = e^{at} x(0)$$

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

• From scalars to matrices?

$$\dot{x} = Ax + Bu \\ y = Cx + Du$$

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

$$Av = \lambda v$$

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

$\begin{cases} \text{Re}\,(\lambda)>0: \text{unstable}\\ \text{Re}\,(\lambda)<0: \text{asymptotically stable}\\ \text{Re}\,(\lambda) \leq 0: \text{critically stable} \end{cases}$

# 4. Stability of Closed Loop SystemÂ¶

• 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

## 4.1. Root Locus (Stability in Time)Â¶

• 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

$$\begin{cases} \lvert KG(s^*) \rvert = 1 \quad \implies \quad K = \frac{1}{\lvert G(s^*)\rvert}\\\\ \angle KG(s^*) = (2k+1)\pi \end{cases}$$

• Closed-loop poles in the LHP indicate stability

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

## 4.2. Relative Stability (Stability in Frequency)Â¶

• 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?