Question 3 of 6: Algebraic design of a proper, stable controller from a specified sensitivity

Question 3: Algebraic design of a proper, stable controller from a specified sensitivity (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. A unity-feedback loop around a two-lag plant, with the transfer function from reference to error prescribed in advance up to four free numerator coefficients.

Find. A set of coefficients $b_{i}$ and the corresponding controller $C(s)$ that is both proper and stable, and then the closed-loop transfer function $Y(s)/R(s)$.

Approach. Recognise the specified ratio as the sensitivity function, invert it to get $C(s)$, impose the high-frequency and properness constraints to fix $b_{3}$ and $b_{2}$, then match coefficients for the lowest-order controller that places the closed-loop poles at the prescribed triple root.

1. Identify what has been specified. With $d=0$ the loop gives $E=R-Y$ and $Y=P C E$, so$$\frac{E(s)}{R(s)}=\frac{1}{1+P(s)C(s)}\equiv S(s),$$the sensitivity function. The question therefore hands us the sensitivity and asks for the controller that realises it — an algebraic (model-matching) design rather than a root-locus or loop-shaping one.
2. Invert the relation. Solving $S=1/(1+PC)$ for the controller,$$C(s)=\frac{1-S(s)}{S(s)\,P(s)}=\frac{\bigl[(s+1)^{3}-n(s)\bigr](4s+1)(5s+1)}{n(s)}.$$Everything now hinges on the degree of the bracket, because the plant denominator contributes two extra powers of $s$ to the controller numerator.
3. Constraint 1: the high-frequency value of $S$. The controller is at best proper and the plant is strictly proper of relative degree two, so $L=PC$ is strictly proper and $L(\infty)=0$. Hence $S(\infty)=1$, which forces the leading coefficients of $n(s)$ and $(s+1)^{3}$ to agree:$$\boxed{\,b_{3}=1\,}$$A design that ignores this ships an improper controller no matter what else is chosen.
4. Constraint 2: properness. With $b_{3}=1$,$$(s+1)^{3}-n(s)=(3-b_{2})s^{2}+(3-b_{1})s+(1-b_{0}),$$so the controller numerator has degree $\deg\bigl[(s+1)^{3}-n\bigr]+2$ while its denominator has degree 3. Properness demands $\deg\bigl[(s+1)^{3}-n\bigr]\le 1$, i.e. the $s^{2}$ term must vanish:$$\boxed{\,b_{2}=3\,}$$Both leading coefficients are thus fixed by structure alone; only $b_{1}$ and $b_{0}$ remain as design freedom.
5. Use that freedom for the lowest-order controller. Take the general bi-proper first-order form $C(s)=\dfrac{n_{1}s+n_{0}}{s+p}$ and impose the required closed-loop characteristic polynomial. Since $1+PC=\dfrac{(s+p)(20s^{2}+9s+1)+n_{1}s+n_{0}}{(s+p)(20s^{2}+9s+1)}$, the numerator must be proportional to $(s+1)^{3}$, and the leading coefficient 20 fixes the constant of proportionality:$$(s+p)(20s^{2}+9s+1)+n_{1}s+n_{0}=20(s+1)^{3}=20s^{3}+60s^{2}+60s+20.$$
6. Match coefficients. Expanding the left side gives $20s^{3}+(9+20p)s^{2}+(1+9p+n_{1})s+(p+n_{0})$, so$$9+20p=60\;\Rightarrow\;p=2.55,\qquad 1+9(2.55)+n_{1}=60\;\Rightarrow\;n_{1}=36.05,\qquad 2.55+n_{0}=20\;\Rightarrow\;n_{0}=17.45.$$$$\boxed{\,C(s)=\frac{36.05\,s+17.45}{s+2.55}\,}$$This controller is proper (degree 1 over degree 1) and stable (its only pole is at $s=-2.55$), with a zero at $s=-0.4840$. It performs no pole–zero cancellation with the plant, so the loop is internally stable with all three closed-loop poles at $s=-1$.
7. Read off the coefficients $b_{i}$. Substituting back, $n(s)=\dfrac{(s+p)(20s^{2}+9s+1)}{20}=(s+2.55)(s^{2}+0.45s+0.05)$, so$$\boxed{\,n(s)=s^{3}+3s^{2}+1.1975\,s+0.1275\,}$$i.e. $b_{3}=1$, $b_{2}=3$, $b_{1}=1.1975$, $b_{0}=0.1275$ — and the first two agree with the structural constraints derived above, which is the consistency check the design should be judged on. Note $n(s)$ is Hurwitz, as it must be, since its roots are the controller poles together with the plant poles that the sensitivity cancels.
8. Interpret the design. Evaluating the sensitivity at DC, $S(0)=b_{0}=0.1275$, so the loop leaves 12.75 % of a step reference as steady-state error. That is the price of insisting on a stable controller: driving the error to zero would require $b_{0}=0$, which puts a pole of $C(s)$ at the origin — integral action, which is marginally stable rather than stable. The figure below shows the shape of the realised sensitivity.
9. (b) Closed-loop transfer function. Complementary sensitivity and sensitivity always sum to one, $T+S=1$, so$$T(s)=\frac{Y(s)}{R(s)}=1-S(s)=\frac{(s+1)^{3}-n(s)}{(s+1)^{3}}=\frac{(3-b_{1})s+(1-b_{0})}{(s+1)^{3}}$$$$\boxed{\,\frac{Y(s)}{R(s)}=\frac{1.8025\,s+0.8725}{(s+1)^{3}}\,}$$Its DC gain is $0.8725=1-b_{0}$, consistent with the 12.75 % steady-state error found above, and a direct evaluation of $PC/(1+PC)$ gives $\dfrac{36.05s+17.45}{20(s+1)^{3}}$, the same expression.