22-Mec-A3 System Analysis and Control · December 2016

Question 5 of 6: Routh–Hurwitz Stability, Step Response and Gain Tolerance

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Routh–Hurwitz Stability, Step Response and Gain Tolerance (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a): stability of a fifth-order characteristic equation

Given. $s^{5}+s^{4}+2s^{3}+s^{2}+s+K=0$, with $K$ an adjustable parameter.

Find. The range of $K$ (if any) for which every root lies in the left half plane, and the number of right-half-plane roots otherwise.

Approach. Build the Routh array symbolically in $K$ and require every first-column entry to be strictly positive.

1. Lay out the first two rows from the coefficients. Odd powers go in the $s^{5}$ row, even powers in the $s^{4}$ row:

2. Verify the row computations. Each entry is the usual $2\times2$ determinant divided by the pivot above:$$s^{3}\text{ row}:\quad\frac{(1)(2)-(1)(1)}{1}=1,\qquad\frac{(1)(1)-(1)(K)}{1}=1-K$$$$s^{2}\text{ row}:\quad\frac{(1)(1)-(1)(1-K)}{1}=K,\qquad\frac{(1)(K)-(1)(0)}{1}=K$$$$s^{1}\text{ row}:\quad\frac{(K)(1-K)-(1)(K)}{K}=(1-K)-1=-K$$
3. Apply the criterion. Stability requires every first-column entry to be positive. The $s^{2}$ and $s^{0}$ entries demand $K>0$, but the $s^{1}$ entry is identically $-K$, which demands $K<0$. These cannot both hold:$$\boxed{\;\text{No value of }K\text{ makes this system stable.}\;}$$
4. Count the unstable roots. For any $K>0$ the first column reads $1,\,1,\,1,\,+K,\,-K,\,+K$ — two sign changes — so there are two roots in the right half plane for every positive gain. At $K=0$ a root sits at the origin (the constant term vanishes) and for $K<0$ the $s^{0}$ entry turns negative, so the system is unstable there too. Direct numerical root-finding at $K=0.5$, $2$ and $100$ confirms exactly two right-half-plane roots in each case.
5. Interpret. The failure is structural, not a matter of tuning. The coefficients are all positive, which is necessary for stability but nowhere near sufficient above second order — this equation is a textbook demonstration of that gap. No gain adjustment can rescue the loop; it needs compensation that changes the coefficient pattern itself.

Part (b): unit step response

Given. $P(s)=(s+2)/[(s+0.5)(s+4)]$ driven by a unit step, $R(s)=1/s$, from rest.

Find. $c(t)$ for $t\ge0$.

Approach. Form $C(s)=P(s)/s$, expand in partial fractions over the three simple poles, and invert term by term.

1. Form the response transform and set up the expansion. $$C(s)=\frac{s+2}{s(s+0.5)(s+4)}=\frac{A_{0}}{s}+\frac{A_{1}}{s+0.5}+\frac{A_{2}}{s+4}$$
2. Evaluate the residues by cover-up. $$A_{0}=\left.\frac{s+2}{(s+0.5)(s+4)}\right|_{s=0}=\frac{2}{(0.5)(4)}=1$$$$A_{1}=\left.\frac{s+2}{s(s+4)}\right|_{s=-0.5}=\frac{1.5}{(-0.5)(3.5)}=-\frac{6}{7}=-0.857$$$$A_{2}=\left.\frac{s+2}{s(s+0.5)}\right|_{s=-4}=\frac{-2}{(-4)(-3.5)}=-\frac{1}{7}=-0.143$$
3. Check before inverting. Because $C(s)$ is strictly proper and the system starts from rest, $c(0)=0$, so the residues must sum to zero: $1-\tfrac{6}{7}-\tfrac{1}{7}=0$. They do.
4. Invert. $$\boxed{\;c(t)=1-\frac{6}{7}e^{-0.5t}-\frac{1}{7}e^{-4t}=1-0.857e^{-0.5t}-0.143e^{-4t}\qquad(t\ge0)\;}$$
5. Describe the response. The final value is $c(\infty)=1=P(0)$, the DC gain. Both poles are real and both residues are negative, so the rise is monotonic with no overshoot. The fast pole at $-4$ carries only one seventh of the response and is gone within about a second; the slow pole at $-0.5$ dominates, giving 2 % settling at $t=\ln(0.857/0.02)/0.5=7.52\ \text{s}$. The zero at $-2$ does not add a mode — it only redistributes the residues, here pulling weight onto the slow pole.

Part (c): permissible variation of the amplifier gain

Given. $s^{3}+(4+K)s^{2}+6s+(16+8K)=0$, nominal $K=2$.

Find. How far $K$ may drift from 2 before the closed loop becomes unstable.

Approach. Build the Routh array in $K$, impose positivity on the whole first column, and compare the resulting interval with the nominal value.

1. Construct the array. For a cubic $a_{3}s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ the array is short:

2. Impose positivity on every first-column entry. $$4+K>0\;\Rightarrow\;K>-4;\qquad 8-2K>0\;\Rightarrow\;K<4;\qquad 16+8K>0\;\Rightarrow\;K>-2$$The binding lower limit is $K>-2$ (it is tighter than $K>-4$), so$$\boxed{\;-2<K<4\;}$$
3. Express the tolerance about the nominal gain. With the design value $K=2$:$$\Delta K_{\text{up}}=4-2=+2,\qquad\Delta K_{\text{down}}=2-(-2)=-4$$so the gain may increase by 2 (a factor of 2, i.e. $20\log_{10}2=6.02\ \text{dB}$ of gain margin) or decrease by 4 before stability is lost. Since a physical amplifier gain cannot go negative, the practical statement is that $K$ may take any value in $0<K<4$ — it may double, but must not more than double.
4. Identify the mode at the boundary. At $K=4$ the $s^{1}$ row vanishes and the auxiliary polynomial $(4+K)s^{2}+(16+8K)=8s^{2}+48$ gives roots $s=\pm j\sqrt{6}=\pm j2.449$: the system breaks into a sustained oscillation at $2.449\ \text{rad/s}$. At the nominal $K=2$ the roots are $-5.102$ and $-0.449\pm j2.503$ — already lightly damped ($\zeta\approx0.18$), which is why so little upward gain headroom remains.