Stability

Result: the system is unstable. It has 2 roots in the LHP, a repeated pair on the $j\omega$ axis at $\pm j1$, and no roots in the RHP.

The Routh–Hurwitz criterion states that the number of roots in the right half of the s-plane equals the number of sign changes in the first column of the Routh array. A row of all zeros means there are roots placed symmetrically about the origin; these are found from the auxiliary equation.

$$q(s) = s^6+3s^5+4s^4+6s^3+5s^2+3s+2 = 0$$

All coefficients are present and positive, so the necessary condition is satisfied.

Routh array

The $s^3$ row is all zeros. Auxiliary equation from the $s^4$ row:

$$A(s) = 2s^4+4s^2+2, \qquad \frac{dA}{ds} = 8s^3 + 8s$$

Replace the $s^3$ row by $8,\ 8$ and continue:

The $s^1$ row is zero again. New auxiliary equation $A_1(s) = 2s^2+2$, $\frac{dA_1}{ds} = 4s$:

First column: $1,\ 3,\ 2,\ 8,\ 2,\ 4,\ 2$. No sign change, so no root is in the right half plane.

Roots of the auxiliary equation

$$2s^4+4s^2+2 = 2(s^2+1)^2 = 0 \;\Rightarrow\; s = \pm j1,\ \pm j1$$

So $\pm j1$ is a repeated pair on the imaginary axis. Dividing $q(s)$ by $(s^2+1)^2$ leaves $s^2+3s+2 = (s+1)(s+2)$.

Location of roots

Comment on stability

There is no RHP root, but the imaginary-axis roots are repeated. Repeated poles on the $j\omega$ axis give terms like $t\sin t$ in the response, which grow without limit. Therefore the system is unstable. If the $j\omega$ roots were non-repeated, it would only be marginally stable.

Result: RHP = 0, LHP = 1 (at $s=-4$), $j\omega$ axis = 4 ($\pm j1$, each twice). The system is unstable.

The number of sign changes in the first column gives the RHP roots. A zero row gives the roots that are symmetric about the origin; these are found from the auxiliary polynomial.

$$s^5+4s^4+2s^3+8s^2+s+4=0$$

Routh array

The $s^3$ row is all zeros. Auxiliary polynomial from the $s^4$ row:

$$A(s) = 4s^4+8s^2+4, \qquad \frac{dA}{ds} = 16s^3+16s$$

New auxiliary polynomial $A_1(s)=4s^2+4$, $\frac{dA_1}{ds}=8s$:

First column: $1,\ 4,\ 16,\ 4,\ 8,\ 4$. No sign change, so there is no root in the RHP.

Roots from the auxiliary polynomial

$$4s^4+8s^2+4 = 4(s^2+1)^2 = 0 \Rightarrow s=\pm j1,\ \pm j1$$

These 4 roots lie on the $j\omega$ axis (repeated pair). The remaining root: $\frac{s^5+4s^4+2s^3+8s^2+s+4}{(s^2+1)^2} = s+4$, so $s=-4$.

Stability

No root is in the RHP, but the roots on the $j\omega$ axis are repeated. They produce a response term like $t\sin t$, which grows with time. So the system is unstable.

Relative stability tells how far the closed-loop poles lie to the left of the imaginary axis, i.e. how stable a stable system is. The basic Routh–Hurwitz test only checks for poles on or right of the $j\omega$ axis. To find out whether all roots lie to the left of a line $s=-\sigma$, the axis is shifted.

Method (axis shifting)

1. Take the characteristic equation $q(s)=0$.
2. Substitute $s = z - \sigma$, where $\sigma>0$ is the required margin. This moves the vertical line $s=-\sigma$ to the new imaginary axis ($z=0$).
3. Expand to get a polynomial $q(z)$.
4. Form the Routh array of $q(z)$.
   • Number of sign changes = number of roots to the right of $s=-\sigma$.
   • No sign change: all roots lie to the left of $s=-\sigma$, so every root has real part more negative than $-\sigma$. Each mode then decays at least as fast as $e^{-\sigma t}$.
   • A zero row: roots lie exactly on the line $s=-\sigma$.
5. Repeat for other $\sigma$ values, or keep $\sigma$ (or a gain $K$) as a variable to find the limit.

          jw   |  jw' (new axis)
               |   |
     x         |   |
        x      |   |        Shift: s = z - sigma
   ------------+---+-------- sigma
     x       -sigma 0
               |   |

Example

$q(s)=s^3+6s^2+11s+6$ (roots $-1, -2, -3$). Check whether all roots lie left of $s=-0.5$. Put $s=z-0.5$:

$$q(z) = z^3 + 4.5z^2 + 5.75z + 1.875$$

No sign change, so all roots lie to the left of $s=-0.5$. With $s=z-1$, $q(z) = z^3+3z^2+2z$; the constant term becomes zero, showing a root exactly at $s=-1$.

Uses

• Ensures a minimum decay rate, i.e. settling time $t_s \approx 4/\sigma$.
• Lets the designer find the range of gain $K$ that keeps all poles left of a chosen line.

Apply the Routh array with $K$ in it. For stability, every element of the first column must be positive. Marginal stability occurs when the $s^1$ row becomes zero.

$$s^4+20s^3+15s^2+2s+K = 0$$

Routh array

(i) Range of K for stability

• From $s^1$ row: $29.8 - 20K > 0 \Rightarrow K < 1.49$
• From $s^0$ row: $K > 0$

$$0 < K < 1.49$$

(ii) Marginal stability

Yes. At $K = 1.49$ the $s^1$ row becomes all zero, and the system has a pair of roots on the $j\omega$ axis (marginally stable, sustained oscillation).

Auxiliary equation from the $s^2$ row:

$$\begin{aligned} 14.9s^2 + K &= 0 \\ 14.9s^2 + 1.49 &= 0 \\ s^2 &= -0.1 \Rightarrow s = \pm j0.316 \end{aligned}$$

So the frequency of sustained oscillation is $\omega = \sqrt{0.1} = 0.316$ rad/s.

Answer: stable for $0 < K < 1.49$; marginally stable at $K_{mar} = 1.49$, oscillating at $\omega = 0.316$ rad/s.

Result: stable for $K>-1$ and $b(3+K)>1+K$. For $K>0$ this means any $K$ when $b\ge1$, and $0<K<\frac{3b-1}{1-b}$ when $\frac13<b<1$. It just oscillates at $K=\frac{3b-1}{1-b}$, with $\omega=\sqrt{3+K}=\sqrt{\frac{2}{1-b}}$ rad/s.

The characteristic equation is $1+G(s)=0$. The Routh array gives the stability conditions in terms of $K$ and $b$. Sustained oscillation occurs when the $s^1$ row becomes zero.

Characteristic equation

$$\begin{aligned} s^3+bs^2+3s+1 + K(s+1) &= 0 \\ s^3 + bs^2 + (3+K)s + (1+K) &= 0 \end{aligned}$$

Routh array

Conditions for stability

1. $b > 0$
2. $1+K>0 \Rightarrow K>-1$
3. $b(3+K) - (1+K) > 0 \Rightarrow K(b-1) > 1-3b$

Interpreting condition 3 (for $K>0$, $b>0$):

• If $b \ge 1$: condition 3 holds for every $K>0$, so the system is stable for all $K>0$.
• If $b < 1$: $K < \dfrac{3b-1}{1-b}$. A positive range exists only if $b > \frac{1}{3}$.

$$\text{Range: } 0 < K < \frac{3b-1}{1-b}\quad\left(\tfrac13<b<1\right)$$

Sustained oscillation

The system just oscillates when the $s^1$ row is zero:

$$b(3+K) = 1+K \quad\Rightarrow\quad K_{mar} = \frac{3b-1}{1-b}$$

Auxiliary equation from the $s^2$ row:

$$\begin{aligned} bs^2 + (1+K) &= 0 \\ \omega^2 &= \frac{1+K}{b} = 3+K = \frac{2}{1-b} \\ \omega &= \sqrt{\frac{2}{1-b}}\ \text{rad/s} \end{aligned}$$

Example: for $b = 0.5$: $K_{mar} = \frac{0.5}{0.5}=1$ and $\omega = \sqrt{4} = 2$ rad/s. Check: $s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5)$, which has roots $\pm j2$.

The closed-loop characteristic equation $1+G(s)=0$ is tested with the Routh array. The system just oscillates (is marginally stable) when the $s^1$ row becomes zero.

Characteristic equation

$$\begin{aligned} s(s+3)(s+7) + K(s+13) &= 0 \\ s^3 + 10s^2 + (21+K)s + 13K &= 0 \end{aligned}$$

Routh array

Range of K

• $s^1$ row: $210-3K>0 \Rightarrow K<70$
• $s^0$ row: $13K>0 \Rightarrow K>0$

$$0 < K < 70$$

Sustained oscillation

At $K = 70$ the $s^1$ row is zero, so the system just oscillates. Auxiliary equation from the $s^2$ row:

$$\begin{aligned} 10s^2 + 13K &= 0 \\ 10s^2 + 910 &= 0 \\ s^2 &= -91 \Rightarrow s = \pm j9.54 \end{aligned}$$

Answer: stable for $0<K<70$; it just oscillates at $K=70$, with frequency $\omega=\sqrt{91}=9.54$ rad/s.

For unity feedback, the characteristic equation is $1+G(s)=0$. All first-column Routh elements must be positive.

Characteristic equation

$$\begin{aligned} s[s(s+10)+T] + K &= 0 \\ s^3 + 10s^2 + Ts + K &= 0 \end{aligned}$$

Routh array

Conditions

• $s^0$ row: $K > 0$
• $s^1$ row: $10T - K > 0 \Rightarrow K < 10T$
• From these, $T > 0$ (also needed so that all coefficients are positive).

$$0 < K < 10T$$

Interpretation: for a given $T$, the gain can be raised up to $10T$. At $K = 10T$ the system is marginally stable and oscillates at $\omega = \sqrt{K/10} = \sqrt{T}$ rad/s (from $10s^2+K=0$). For example, with $T = 5$ the system is stable for $0<K<50$.

Answer: relation for stability: $0 < K < 10T$ (with $T>0$).

To make all roots lie to the left of $s=-a$, shift the imaginary axis to $s=-a$ by putting $s = z - a$. Then apply the Routh criterion to the new polynomial in $z$ (relative stability).

Characteristic equation

$$\begin{aligned} 1 + \frac{K}{s(1+Ts)} &= 0 \\ Ts^2 + s + K &= 0 \end{aligned}$$

Shift the axis: $s = z - a$

$$\begin{aligned} T(z-a)^2 + (z-a) + K &= 0 \\ Tz^2 + (1-2aT)z + (Ta^2 - a + K) &= 0 \end{aligned}$$

Routh array in z

For no sign change (all roots left of $z=0$, i.e. left of $s=-a$):

1. $T > 0$
2. $1-2aT > 0 \Rightarrow T < \dfrac{1}{2a}$
3. $Ta^2 - a + K > 0 \Rightarrow K > a - a^2T = a(1-aT)$

$$0 < T < \frac{1}{2a}, \qquad K > a(1-aT)$$

Physical meaning: the sum of the closed-loop roots is $-\frac{1}{T}$, so both roots can be left of $-a$ only if $\frac{1}{T} > 2a$. The gain must be large enough that the slower real root moves past $-a$.

Example: for $a=1$ and $T=0.25$: $K > 1(1-0.25)=0.75$. With $K=1$, $0.25s^2+s+1=0$ gives $s = -2, -2$, both left of $-1$.

Result: the system is marginally stable. It has roots at $\pm j\sqrt2$ and $\pm j2$, one root at $s=-7$, and none in the RHP.

For unity negative feedback, the characteristic equation is $1+G(s)=0$. Its Routh array is formed and the first column is examined.

Characteristic equation

$$\begin{aligned} s(s^4+7s^3+6s^2+42s+8) + 56 &= 0 \\ s^5 + 7s^4 + 6s^3 + 42s^2 + 8s + 56 &= 0 \end{aligned}$$

Routh array

The $s^3$ row is all zeros. Auxiliary equation from the $s^4$ row:

$$A(s) = 7s^4 + 42s^2 + 56, \qquad \frac{dA}{ds} = 28s^3 + 84s$$

First column: $1,\ 7,\ 28,\ 21,\ 9.33,\ 56$. No sign change, so no root is in the RHP.

Roots on the imaginary axis

$$7s^4+42s^2+56 = 7(s^2+2)(s^2+4) = 0 \Rightarrow s = \pm j1.414,\ \pm j2$$

The fifth root: $\frac{q(s)}{(s^2+2)(s^2+4)} = s+7$, so $s=-7$.

Conclusion

The poles on the $j\omega$ axis are simple (non-repeated), and none are in the RHP. The system is marginally stable: its response contains sustained oscillations at 1.414 rad/s and 2 rad/s.

Result: RHP = 2, LHP = 2, $j\omega$ axis = 0. The system is unstable.

For unity negative feedback the characteristic equation is $1+G(s)=0$. The number of sign changes in the first column of the Routh array gives the number of roots in the right half plane.

Characteristic equation

$$\begin{aligned} s(s^3+6s^2+11s+6) + 200 &= 0 \\ s^4 + 6s^3 + 11s^2 + 6s + 200 &= 0 \end{aligned}$$

Routh array

First column: $1,\ 6,\ 10,\ -114,\ 200$.

Sign changes: $+10 \to -114$ (one) and $-114 \to +200$ (two), so there are 2 sign changes.

Root location

The numerical roots ($s \approx 1.28 \pm j2.54$ and $s \approx -4.28 \pm j2.54$) agree with this.

Answer: 2 roots in the RHP, 2 in the LHP, none on the imaginary axis. The closed-loop system is unstable; the gain of 200 is too high.

A system "just oscillates" when it is marginally stable. In the Routh array, the $s^1$ row then becomes zero, and the auxiliary equation from the $s^2$ row gives the frequency of oscillation.

Characteristic equation

$$\begin{aligned} 1 + \frac{K(s+1)}{s^3+as^2+2s+1} &= 0 \\ s^3 + as^2 + (2+K)s + (1+K) &= 0 \end{aligned}$$

Routh array

Condition for sustained oscillation

$s^1$ row $=0$:

$$a(2+K) = 1+K \qquad \text{...(1)}$$

Auxiliary equation from the $s^2$ row, with $s = j\omega$:

$$\begin{aligned} as^2 + (1+K) &= 0 \\ \omega^2 &= \frac{1+K}{a} \qquad \text{...(2)} \end{aligned}$$

From (1), $\frac{1+K}{a} = 2+K$, so (2) becomes $\omega^2 = 2+K$.

Substitute $\omega = 2$ rad/s

$$\begin{aligned} 4 &= 2+K \Rightarrow K = 2 \\ a &= \frac{1+K}{2+K} = \frac{3}{4} = 0.75 \end{aligned}$$

Check

With $K=2$, $a=0.75$: $s^3+0.75s^2+4s+3 = (s^2+4)(s+0.75)$. The roots are $\pm j2$ and $-0.75$, so the system oscillates at 2 rad/s.

Answer: $K = 2$, $a = 0.75$.

The stability of a system depends only on where the poles of its closed-loop transfer function (the roots of the characteristic equation) lie in the s-plane. Each pole $s = \sigma \pm j\omega$ adds a term $e^{\sigma t}\sin(\omega t)$ to the response.

• The further left the poles are, the faster the response decays. This means greater relative stability.
• Poles close to the $j\omega$ axis (dominant poles) decay slowly and decide the transient response.
• Imaginary part $\omega$ sets the frequency of oscillation; real poles give no oscillation.

Result: 2 sign changes, so the closed-loop system has 2 roots in the RHP and is unstable.

The closed-loop stability is decided by the characteristic equation $1+G(s)H(s)=0$.

$$\begin{aligned} 2s^5+3s^4+2s^3+s^2+2s + 2 &= 0 \end{aligned}$$

Routh array

First column: $2,\ 3,\ 1.33,\ -0.5,\ 6,\ 2$.

There are 2 sign changes ($1.33 \to -0.5$ and $-0.5 \to 6$), so 2 closed-loop poles lie in the right half s-plane.

Answer: the closed-loop system is unstable, with 2 poles in the RHP and 3 in the LHP. The roots are approximately $0.51\pm j0.75$, $-0.76\pm j0.80$ and $-1$.

Form the Routh array with $K$ and make every first-column element positive.

$$s^4 + 2s^3 + (4+K)s^2 + 9s + 25 = 0$$

Routh array

Conditions

1. $s^2$ row: $\frac{2K-1}{2}>0 \Rightarrow K>0.5$
2. $s^1$ row: with $2K-1>0$, we need $18K-109>0 \Rightarrow K > \frac{109}{18} = 6.056$
3. $s^0$ row: $25>0$ always true.

Condition 2 is stricter, so:

$$K > 6.056$$

Marginal value

At $K = 109/18 = 6.056$, the $s^1$ row is zero. The auxiliary equation $\frac{50}{9}s^2 + 25 = 0$ gives $s = \pm j2.12$, so the system would oscillate at 2.12 rad/s.

Answer: the system is stable for $K > 6.056$ (i.e. $K>109/18$). There is no upper limit.

The PI controller $K_p + \frac{K_I}{s}$ adds a pole at the origin. Find the closed-loop characteristic equation and apply Routh's criterion.

Characteristic equation

$$\begin{aligned} G(s) &= \left(K_p + \frac{K_I}{s}\right)\frac{1}{(s+1)(s+2)} = \frac{K_ps + K_I}{s(s+1)(s+2)} \\ 1+G(s) = 0 &\Rightarrow s(s^2+3s+2) + K_ps + K_I = 0 \\ &\Rightarrow s^3 + 3s^2 + (2+K_p)s + K_I = 0 \end{aligned}$$

Routh array

Conditions for stability

• $s^0$ row: $K_I > 0$
• $s^1$ row: $3(2+K_p) - K_I > 0 \Rightarrow K_I < 3(K_p+2)$
• This also requires $K_p > -2$.

$$0 < K_I < 3(K_p + 2)$$

Example: with $K_p = 1$, the integral gain must satisfy $0 < K_I < 9$; with $K_p = 4$, $0 < K_I < 18$. A larger $K_p$ allows a larger $K_I$.

For unity feedback, the characteristic equation is $1+G(s)=0$, with $G(s)=\frac{K}{s(s^2+s+1)(s+2)}$.

Characteristic equation

$$\begin{aligned} s(s^2+s+1)(s+2) + K &= 0 \\ s^4 + 3s^3 + 3s^2 + 2s + K &= 0 \end{aligned}$$

Routh array

Conditions

• $s^1$ row: $2 - \frac{9K}{7} > 0 \Rightarrow K < \frac{14}{9} = 1.556$
• $s^0$ row: $K > 0$

$$0 < K < 1.556$$

At $K = 14/9$ the system is marginally stable. The auxiliary equation $\frac73 s^2 + \frac{14}{9} = 0$ gives $\omega = \sqrt{2/3} = 0.816$ rad/s.

Answer: stable for $0 < K < \frac{14}{9}$ (= 1.556).

Result (assuming oscillation at 2 rad/s): $K = 1$, $b = 0.5$. In general, $K = \frac{3b-1}{1-b}$ and $\omega=\sqrt{\frac{2}{1-b}}$.

Assumption: the end of the question is missing. It is taken in its usual form: "...so that the system oscillates at a frequency of 2 rad/s". The general relations are also derived, so any other given frequency can be substituted.

Characteristic equation

$$\begin{aligned} s^3+bs^2+3s+1 + K(s+1) &= 0 \\ s^3 + bs^2 + (3+K)s + (1+K) &= 0 \end{aligned}$$

Routh array

Conditions for sustained oscillation

The $s^1$ row must be zero:

$$b(3+K) = 1+K \qquad \text{...(1)}$$

Auxiliary equation (from the $s^2$ row) with $s=j\omega$:

$$bs^2 + (1+K) = 0 \Rightarrow \omega^2 = \frac{1+K}{b} \qquad \text{...(2)}$$

From (1) and (2): $\omega^2 = 3 + K$.

Substitute $\omega = 2$ rad/s

$$\begin{aligned} 4 &= 3 + K \Rightarrow K = 1 \\ b &= \frac{1+K}{\omega^2} = \frac{2}{4} = 0.5 \end{aligned}$$

Check: $s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5)$. The roots are $\pm j2$ and $-0.5$, so the system oscillates at 2 rad/s.

Answer: $K = 1$, $b = 0.5$. In general: $K = \frac{3b-1}{1-b}$ and $\omega = \sqrt{\frac{2}{1-b}}$. The system is stable when $b(3+K) > 1+K$.