Question 5 of 6: Transient Terms of a Two-Input Servo with Rate Feedback

Question 5: Transient Terms of a Two-Input Servo with Rate Feedback (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. The two-input block diagram of Figure 3, read directly from the examination drawing:

All initial conditions are zero.

Find. (a) the transient term of $c(t)$ driven by $R$ alone, as a function of $K$; (b) the transient term driven by $D$ alone, as a function of $K$; (c) the gain that makes the dominant closed-loop time constant exactly $1$ s.

[Figure not reproduced: Figure 3 — redrawn from page 4. The rate signal $sC$ is subtracted at the inner junction (ahead of the amplifier) while the output $C$ is subtracted at the outer junction; the disturbance enters through $0.4/(s+1)$ and is summed at the output node. See the official exam paper.]

Approach. Reduce the diagram once, algebraically and with both inputs retained, to obtain a single characteristic polynomial; superposition then gives each transfer function by setting the other input to zero. The transient terms are the inverse Laplace transforms of the partial-fraction terms associated with the system poles, which is why one reduction serves all three parts.

1. Reduce the diagram with both inputs present. Let $E=R-C$ be the outer error and let $u=KE-sC$ be the signal entering the amplifier. The disturbance branch is summed at the output node, downstream of the plant block, so the plant output and the filtered disturbance simply add there: $$C \;=\; \frac{10\bigl(K(R-C)-sC\bigr)}{s+1} \;+\; \frac{0.4}{s+1}\,D .$$ Multiplying through by $(s+1)$ and collecting the terms in $C$: $$C\bigl[(s+1)+10K+10s\bigr] \;=\; 10K\,R \;+\; 0.4\,D,$$ $$\boxed{\;C(s)\,\bigl[\,11s + 1 + 10K\,\bigr] \;=\; 10K\,R(s) \;+\; 0.4\,D(s). \;}$$ The closed loop is first order in both channels, with the single system pole $$s_{1} = -\frac{1+10K}{11},\qquad \tau = \frac{11}{1+10K}.$$
2. Note why the rate feedback is so powerful here. The tachometer term contributes $10s$ to the characteristic polynomial — ten times the plant's own $s$ coefficient. It is this term that turns the pole into $-(1+10K)/11$ rather than $-(1+10K)$, and it is the reason a modest gain suffices in part (c).

(a) Reference acting alone ($D=0$)

1. Form the reference transfer function. Setting $D=0$, $$\frac{C(s)}{R(s)} = \frac{10K}{11s+1+10K} = \frac{10K/11}{s+a},\qquad a \equiv \frac{1+10K}{11}.$$ The system contributes exactly one natural mode, $e^{-at}$.
2. Invert for a unit-step reference. With $R(s)=1/s$, $$C(s) = \frac{10K/11}{s(s+a)} \;\Longrightarrow\; c(t) = \frac{10K}{1+10K}\Bigl(1-e^{-at}\Bigr).$$ Separating the forced and natural parts, $$\boxed{\;c_{\text{transient}}(t) \;=\; -\,\frac{10K}{1+10K}\, \exp\!\left(-\frac{1+10K}{11}\,t\right),\qquad c_{\text{ss}} = \frac{10K}{1+10K}. \;}$$ For a general reference the transient term is any multiple of $\exp[-(1+10K)t/11]$: the mode is what depends on $K$, and it decays faster as $K$ rises. At $K=1$, for example, the steady-state output is $0.9091$ and the transient is $-0.9091e^{-t}$.

(b) Disturbance acting alone ($R=0$)

1. Form the disturbance transfer function. Setting $R=0$ in the reduction of step 1, $$\boxed{\;\frac{C(s)}{D(s)} = \frac{0.4}{11s+1+10K} = \frac{0.4/11}{s+a},\qquad a=\frac{1+10K}{11}. \;}$$ It is worth seeing why the filter pole at $s=-1$ does not survive. Writing $Y$ for the plant output, $C=Y+0.4D/(s+1)$ and $Y=-10(K+s)C/(s+1)$, so $C\bigl[(s+1)+10K+10s\bigr]/(s+1) = 0.4D/(s+1)$: the plant's own $(s+1)$ cancels the disturbance filter's, because the two have the same pole and the disturbance is injected downstream of the plant. The disturbance channel therefore carries exactly one natural mode, $e^{-at}$, the same mode as the reference channel, and differs from it only in the size of its residue.
2. Invert for a unit-step disturbance. With $D(s)=1/s$, $$C(s) = \frac{0.4/11}{s(s+a)} \;\Longrightarrow\; c(t) = \frac{0.4}{1+10K}\Bigl(1-e^{-at}\Bigr),$$ so the transient part is $$\boxed{\;c_{\text{transient}}(t) = -\,\frac{0.4}{1+10K}\, \exp\!\left(-\frac{1+10K}{11}\,t\right). \;}$$ Setting $t=0$ gives $0.4/(1+10K)-0.4/(1+10K)=0$, confirming the zero initial condition. As in part (a), it is the mode that depends on $K$ for a general disturbance; the coefficient above is the one belonging to a unit step.
3. Interpret the steady-state offset. The disturbance leaves a residual output $0.4/(1+10K)$, which the loop suppresses in proportion to $1+10K$. Raising $K$ from $0.3$ to $3$ cuts the offset from $0.100$ to $0.0129$. Because the loop contains no integrator, the offset can be made small but never zero by proportional action alone.
4. Evaluate at the design gain of part (c). At $K=1$ the pole is $a=1$ and $$c(t)\big|_{K=1} = \frac{0.4}{11}\Bigl(1-e^{-t}\Bigr) = 0.03636\Bigl(1-e^{-t}\Bigr),$$ so the transient term is $-0.03636\,e^{-t}$ and the residual output settles at $0.03636$. Contrast the reference channel at the same gain, $-0.9091e^{-t}$: identical mode, residue larger by the factor $10K/0.4 = 25$.

(c) Gain for a 1 s closed-loop time constant

1. Impose the specification on the closed-loop pole. From step 1 the closed-loop time constant is $\tau=11/(1+10K)$. Setting $\tau=1$ s, $$\frac{11}{1+10K}=1 \;\Longrightarrow\; 1+10K = 11 \;\Longrightarrow\; \boxed{\;K = 1 . \;}$$ The closed-loop pole then sits at $s=-1$, the step response of part (a) becomes $c(t)=0.9091(1-e^{-t})$, and the disturbance rejection improves to a steady offset of $0.4/11=0.03636$ — a factor of eleven better than the open-loop value $0.4$.
2. Sanity-check the design. Substituting $K=1$ into the characteristic polynomial gives $11s+11=11(s+1)$, whose root is $s=-1$ and whose time constant is $1/1=1$ s, as required. Note that $K=1$ happens to place the closed-loop pole exactly where the plant and disturbance-filter poles already sit, at $s=-1$; nothing degenerates, because those two open-loop factors cancel out of the closed-loop response anyway (step 1 of part (b)), and the loop remains first order.