Question 2 of 6: Steady-state error with a pre-loop gain

Question 2: Steady-state error with a pre-loop gain (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 reference passes first through a gain block $K_{1}$; the difference between that scaled reference and the unity-feedback signal drives the plant

The error is defined at the reference, $E(s)=R(s)-Y(s)$, not at the summing junction, and $R(s)=1/s$.

Find. (a) $e_{ss}$ as a function of $K$ and $K_{1}$; (b) the value of $K_{1}$ that drives $e_{ss}$ to zero.

Approach. Reduce the inner unity-feedback loop, multiply by the pre-gain to get $Y/R$, subtract that from unity to form $E/R$, then apply the Final Value Theorem to a unit step.

1. Reduce the loop. The summing junction and the plant with unity feedback give $Y/(K_{1}R)=G/(1+G)$, so the overall closed-loop transfer function is$$\frac{Y(s)}{R(s)}=\frac{K_{1}G(s)}{1+G(s)}=\frac{K_{1}K}{(s+10)(s+12)+K} .$$The pre-gain $K_{1}$ multiplies the numerator only; it sits outside the loop and therefore cannot move a single closed-loop pole.
2. Form the error transfer function. With $E=R-Y$,$$\frac{E(s)}{R(s)}=1-\frac{K_{1}G(s)}{1+G(s)}=\frac{1+G(s)-K_{1}G(s)}{1+G(s)}=\frac{1+\left(1-K_{1}\right)G(s)}{1+G(s)} .$$This is not the classical $1/(1+G)$ shape, because the signal actually formed at the summing junction is $K_{1}R-Y$ rather than $R-Y$.
3. Apply the Final Value Theorem for a unit step. The DC gain of the plant is $G(0)=K/(10\times 12)=K/120$, and the closed-loop poles satisfy $s^{2}+22s+120+K=0$, which by inspection has positive coefficients for every $K>0$ — the loop is stable, so the limit is legitimate. With $R(s)=1/s$,$$e_{ss}=\lim_{s\to 0}s\,E(s)=\frac{1+\left(1-K_{1}\right)K/120}{1+K/120} .$$Multiplying numerator and denominator by 120 clears the fractions:$$\boxed{\,e_{ss}=\frac{120+K-K_{1}K}{120+K}=1-\frac{K_{1}K}{120+K}\,}$$Setting $K_{1}=1$ recovers the familiar type-0 result $e_{ss}=120/(120+K)=1/(1+K_{p})$ with $K_{p}=K/120$, which is a useful sanity check on the algebra.
4. Choose $K_{1}$ for zero error. The boxed expression vanishes when its numerator vanishes, $120+K-K_{1}K=0$, so$$\boxed{\,K_{1}=\frac{120+K}{K}=1+\frac{120}{K}\,}$$For example $K=240$ needs $K_{1}=1.5$, and $K=120$ needs $K_{1}=2$. The required pre-gain is exactly the reciprocal of the closed-loop DC gain, $K_{1}=1/\left[G(0)/(1+G(0))\right]$ — the block is a feed-forward reference prefilter that scales the command up by however much the loop attenuates it.

Because $K_{1}$ lies outside the loop it changes neither the closed-loop poles nor the disturbance or noise response; it only rescales the command. That makes the cancellation exact on paper but fragile in service: any drift in $K$ — a change in actuator gain, temperature, or supply voltage — reopens the error, because nothing in the loop is measuring and correcting it.