Steady-state error of a cruise-control loop with a road grade: 22-Elec-B2 Advanced Control Systems, May 2013, Q1

Question 1: Steady-state error of a cruise-control loop with a road grade (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 cruise-control loop in which the road-grade disturbance is injected at the plant input with a negative sign, so that the signal reaching the vehicle dynamics is $C(s)e-d$.

Given data
Controller	$C(s)=\dfrac{K}{10s+1}$
Plant (vehicle)	$P(s)=\dfrac{4}{(3s+1)^{2}}$
Controller gain, part (a) and (b)	$K=2$
Set-point (constant)	$r=4$
Grade disturbance	$d=0$ in (a), $d=1$ in (b)
Target error in (c)	25 % of the part-(b) value

Find. The steady-state tracking error $e_{ss}=\lim_{t\to\infty}\bigl[r(t)-y(t)\bigr]$ for the level grade and for the loaded grade, and whether raising $K$ alone can cut the loaded-grade error to one quarter of its value.

Figure Q1.1 - Cruise-control loop. The grade disturbance d is injected with a negative sign at the plant input, so the signal driving P(s) is C(s)e − d.

Approach. Write the error as the superposition of a reference term and a disturbance term, evaluate both at DC with the final-value theorem, then test the gain that part (c) demands against the Routh stability limit of the same loop.

Write the error in terms of the two inputs. The plant sees $C(s)E(s)-D(s)$, and the feedback is unity and negative, so$$Y(s)=P(s)\bigl[C(s)E(s)-D(s)\bigr],\qquad E(s)=R(s)-Y(s).$$Substituting the second relation into the first and collecting $E$,$$\boxed{\,E(s)=\frac{R(s)}{1+P(s)C(s)}+\frac{P(s)\,D(s)}{1+P(s)C(s)}\,}$$The reference and the grade therefore share one denominator, the return difference $1+PC$, but the grade is weighted by the extra factor $P(s)$ because it must pass through the vehicle dynamics to reach the output.
Evaluate the loop gain at DC. Both transfer functions are finite at the origin, so$$P(0)=\frac{4}{(3\cdot 0+1)^{2}}=4,\qquad C(0)=\frac{K}{10\cdot 0+1}=K,\qquad L(0)=P(0)C(0)=4K.$$With $K=2$ this gives $L(0)=8$. The loop contains no integrator, so it is a Type-0 loop and a finite steady-state error to a constant reference is unavoidable.
Part (a): level grade. With $R(s)=4/s$ and $D(s)=0$, the final-value theorem gives$$e_{ss}=\lim_{s\to 0}sE(s)=\frac{r}{1+L(0)}=\frac{4}{1+8}=\frac{4}{9}$$$$\boxed{\,e_{ss}\big|_{d=0}=\tfrac{4}{9}=0.4444\,}$$The vehicle settles about 0.44 units below the demanded speed even on a flat road.
Part (b): the grade appears. Now $D(s)=1/s$ as well, and the two contributions simply add:$$e_{ss}=\frac{r}{1+L(0)}+\frac{P(0)\,d}{1+L(0)}=\frac{4}{9}+\frac{4\times 1}{9}=\frac{8}{9}$$$$\boxed{\,e_{ss}\big|_{d=1}=\tfrac{8}{9}=0.8889\,}$$Because $P(0)=4$ happens to equal the set-point, the unit grade doubles the error. Physically the car slows on the hill and the proportional-plus-lag controller has no integral action with which to recover.
Part (c): the gain the specification demands. The target is one quarter of the part-(b) value, $\tfrac14\cdot\tfrac89=\tfrac29=0.2222$. Since $K$ enters only through $L(0)=4K$,$$\frac{r+P(0)d}{1+4K}=\frac{8}{1+4K}=\frac{2}{9}\;\Longrightarrow\;1+4K=36\;\Longrightarrow\;\boxed{\,K_{\text{req}}=8.75\,}$$That is the gain the accuracy specification alone would require.
Test that gain against stability. The closed-loop characteristic polynomial is $1+P(s)C(s)=0$, i.e.$$(10s+1)(3s+1)^{2}+4K=90s^{3}+69s^{2}+16s+(1+4K)=0.$$The Routh array for $a_{3}s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ requires $a_{2}a_{1}\gt a_{3}a_{0}$, so$$69\times 16\gt 90\,(1+4K)\;\Longrightarrow\;1104-90\gt 360K\;\Longrightarrow\;\boxed{\,K\lt K_{\text{crit}}=\tfrac{1014}{360}=2.8167\,}$$At $K=K_{\text{crit}}$ the auxiliary polynomial $69s^{2}+(1+4K)=0$ places a pair of roots on the imaginary axis at $\omega_{\text{osc}}=\sqrt{16/90}=0.4216$ rad/s, and a numerical root solve confirms it.
Verdict and the best that gain alone can do. The required gain 8.75 is more than three times the stability limit 2.8167, so the specification cannot be met by raising $K$. Pushing $K$ as far as the loop will tolerate gives$$e_{ss}\big|_{K\to K_{\text{crit}}}=\frac{8}{1+4(2.8167)}=0.6522,$$$$\boxed{\,\text{Not possible: }K_{\text{req}}=8.75\gg K_{\text{crit}}=2.8167;\ \text{best attainable }e_{ss}=0.6522\,}$$which is 73.4 % of the part-(b) error — a reduction of only 26.6 %, and even that only at zero damping margin. The proper fix is structural, not a gain change: add integral action (a PI controller), which drives the Type-0 loop to Type 1 and makes the steady-state error zero for both the constant reference and the constant grade.

Question 1 — final results
Quantity	Value
Loop DC gain, $L(0)=4K$ at $K=2$	8
(a) $e_{ss}$, level grade	$4/9=0.4444$
(b) $e_{ss}$, unit grade	$8/9=0.8889$
(c) Gain required for $0.25\,e_{ss}$	$K_{\text{req}}=8.75$
(c) Routh stability limit	$K_{\text{crit}}=2.8167$
(c) Oscillation frequency at $K_{\text{crit}}$	0.4216 rad/s
(c) Verdict	Not possible; best attainable $e_{ss}=0.6522$ (73.4 % of (b))

Topic. Steady-state error of a Type-0 loop driven simultaneously by a reference and an input disturbance.

Key relations. Sensitivity $S(s)=\dfrac{1}{1+P C}$; input-disturbance transfer $\dfrac{P}{1+PC}$; final-value theorem $e_{ss}=\lim_{s\to 0}sE(s)$; Routh–Hurwitz condition for a cubic, $a_{2}a_{1}\gt a_{3}a_{0}$.

Why this works. The two inputs enter the same loop, so superposition applies and each gets the same return difference in the denominator. What differs is the numerator: the reference is passed straight to the comparator, while the disturbance must first traverse the plant, so it is scaled by $P(0)$ at DC. Because neither $C$ nor $P$ has a pole at the origin, $L(0)$ is finite and the error is inversely proportional to $1+L(0)$ — accuracy is bought entirely with loop gain. The Routh test then puts a hard ceiling on that gain, because a triple-lag plant contributes 270 degrees of phase and the loop must cross unity gain before it reaches 180 degrees of lag.

Common pitfall. Two mistakes recur. The first is to treat the grade as an output disturbance and forget the $P(0)$ weighting, which halves the part-(b) answer. The second is to answer part (c) with the arithmetic alone: 8.75 looks like a perfectly ordinary gain until the Routh array is written, and quoting it as the answer is exactly the error the question is testing for.

Source. Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 7th ed., §3.6 (Routh–Hurwitz) and §4.2–4.3 (steady-state tracking and disturbance rejection).

Question 1 of 6: Steady-state error of a cruise-control loop with a road grade

Question 1: Steady-state error of a cruise-control loop with a road grade (25 marks)