Question 3 of 6: Compensator Constraints for Type 1 Behaviour and Robust Stability

Question 3: Compensator Constraints for Type 1 Behaviour and Robust Stability (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.

Given. Plant $G(s)=1/(s^{2}+2\zeta s+1)$ with damping ratio $\zeta>0$ and undamped natural frequency $\omega_{n}=1\ \text{rad/s}$; series compensator $D(s)=K(s+a)/(s+b)$; negative unity feedback around the cascade. The loop transfer function is therefore$$L(s)=D(s)G(s)=\frac{K(s+a)}{(s+b)\left(s^{2}+2\zeta s+1\right)}$$

Find. The constraints on the three compensator parameters that make the loop (a) type 1, (b) type 1 and stable, and (c) type 1 and stable for every positive $K$.

Approach. Read the system type directly off the number of poles $L(s)$ has at the origin — which pins $b$ immediately — then form the closed-loop characteristic polynomial and apply Routh’s criterion to convert stability into inequalities on $K$ and $a$; finally take the worst case of those inequalities over all $K>0$.

Part (a) — the type 1 condition

1. Recall the definition of system type. For a negative unity-feedback loop the system type is the number $N$ of poles of $L(s)$ at $s=0$, because the steady-state error constants are $K_{p}=\lim_{s\to0}L(s)$, $K_{v}=\lim_{s\to0}sL(s)$ and $K_{a}=\lim_{s\to0}s^{2}L(s)$. Type 1 means exactly one free integrator, giving a finite non-zero $K_{v}$ and hence zero steady-state error to a step but a finite error to a ramp.
2. Locate the only available integrator. The plant contributes no pole at the origin: $G(0)=1$ is finite. The compensator’s only pole is at $s=-b$, so the loop can acquire a pole at the origin only by setting$$\boxed{\;b=0\;}$$which turns the compensator into $D(s)=K(s+a)/s$ — a proportional-plus-integral (PI) controller in factored form.
3. Guard against cancellation of the new integrator. With $b=0$ the loop transfer function is $L(s)=K(s+a)/\left[s\left(s^{2}+2\zeta s+1\right)\right]$, and$$K_{v}=\lim_{s\to0}sL(s)=\frac{Ka}{1}=Ka$$For the system to be genuinely type 1, $K_{v}$ must be finite and non-zero, which requires the numerator not to vanish at the origin. Hence$$\boxed{\;b=0,\qquad K\neq0,\qquad a\neq0\;}$$If $a$ were zero the compensator zero would cancel its own pole and the loop would collapse back to type 0.

Part (b) — adding stability

1. Form the closed-loop characteristic polynomial. With $b=0$, clearing $1+L(s)=0$ gives$$s\left(s^{2}+2\zeta s+1\right)+K(s+a)=0\quad\Longrightarrow\quad s^{3}+2\zeta s^{2}+(1+K)s+Ka=0$$
2. Build the Routh array. For the cubic $s^{3}+\alpha_{2}s^{2}+\alpha_{1}s+\alpha_{0}$ the array is$$\begin{array}{c|cc} s^{3} & 1 & 1+K\\ s^{2} & 2\zeta & Ka\\ s^{1} & \dfrac{2\zeta(1+K)-Ka}{2\zeta} & 0\\ s^{0} & Ka & \end{array}$$
3. Impose positivity of the whole first column. Since $\zeta>0$ the $s^{2}$ entry is positive automatically. The remaining two conditions are$$Ka>0\qquad\text{and}\qquad 2\zeta(1+K)>Ka$$The first requires $K$ and $a$ to share a sign; taking the physically sensible positive-gain branch, $K>0$ and $a>0$. For a cubic the classical Routh–Hurwitz statement is equivalent to $\alpha_{2}\alpha_{1}>\alpha_{0}$, which is exactly the second inequality.
4. Express the constraint as a bound on the compensator zero. Dividing by $K>0$,$$\boxed{\;b=0,\quad K>0,\quad a>0,\quad a<\frac{2\zeta(1+K)}{K}\;}$$In words: the PI zero must be placed to the right of a limit that depends on both the plant damping and the loop gain. Pushing the zero too far out into the left half-plane makes $K_{v}=Ka$ large but destabilises the loop, which is the familiar integral-windup trade-off in PI tuning.

Part (c) — stability for every positive gain

1. Take the worst case over all admissible gains. The bound of part (b) must hold simultaneously for every $K>0$, so $a$ must be smaller than the infimum of the right-hand side:$$a<\inf_{K>0}\frac{2\zeta(1+K)}{K}=\inf_{K>0}\,2\zeta\left(1+\frac{1}{K}\right)$$
2. Evaluate the infimum. The function $2\zeta(1+1/K)$ decreases monotonically in $K$ and tends to $2\zeta$ as $K\to\infty$, so the infimum is $2\zeta$ and is not attained. The constraint therefore becomes$$\boxed{\;b=0,\qquad 0<a<2\zeta\;}$$with $K$ then free to take any positive value.
3. Interpret the result and check the boundary. The quantity $2\zeta$ is the coefficient of $s^{2}$ in the plant denominator, i.e. the sum of the magnitudes of the real parts of the two plant poles. The compensator zero must sit closer to the origin than that total damping. At exactly $a=2\zeta$ the loop is stable for every finite $K$ but degenerates as $K\to\infty$; for $a>2\zeta$ there is always a finite gain beyond which the loop goes unstable, namely $K>2\zeta/(a-2\zeta)$.
4. Verify with a worked case. Take $\zeta=0.5$, so the bound is $0<a<1$. Choosing $a=0.9$ and testing the extreme gain $K=100$, the characteristic polynomial is $s^{3}+s^{2}+101s+90$, and $\alpha_{2}\alpha_{1}=101>90=\alpha_{0}$ — stable, as predicted. Choosing instead $a=1.2$ (outside the bound) with $K=100$ gives $s^{3}+s^{2}+101s+120$, where $101<120$ and the loop is unstable.