Steady-State Error of a Type-2 Loop: 22-Mec-A3 System Analysis and Control, December 2019, Q3

Question 3: Steady-State Error of a Type-2 Loop (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.

Figure 2.0 — unity-negative-feedback loop with forward transfer function 10/[s²(4 + s)].

Given. Unity negative feedback with forward transfer function $G(s) = \dfrac{10}{s^{2}\left(4+s\right)}$ (Fig 2.0), driven by (a) the ramp $r(t)=10t$ and (b) the combined input $r(t)=4+6t+3t^{2}$.

Find. The steady-state error $e_{ss} = \lim_{t\to\infty}\left[r(t)-c(t)\right]$ for each input.

Approach. Classify the loop by type, evaluate the three static error constants, apply the final-value theorem term by term — and then test the closed-loop characteristic equation with the Routh criterion, because the final-value theorem is only valid for a stable loop.

Write the error transform and the loop type. For unity feedback $E(s) = \dfrac{R(s)}{1+G(s)}$. Since $G(s)=10/\left[s^{2}(4+s)\right]$ has a double pole at the origin, the loop is type 2.
Evaluate the static error constants. $$K_p=\lim_{s\to0}G(s)=\infty, \qquad K_v=\lim_{s\to0}sG(s)=\lim_{s\to0}\frac{10}{s(4+s)}=\infty, \qquad K_a=\lim_{s\to0}s^{2}G(s)=\frac{10}{4}=2.5\ \text{s}^{-2}.$$ A type-2 loop therefore tracks steps and ramps exactly and follows a parabola with a finite offset.
(a) Ramp input. With $r(t)=10t$, $R(s)=10/s^{2}$, and $$e_{ss}=\lim_{s\to0}\frac{sR(s)}{1+G(s)}=\frac{10}{K_v}=\frac{10}{\infty}=\boxed{\;0\;}$$ formally — the two integrations in the forward path absorb the constant velocity demand.
(b) Combined input, term by term. Superposition applies because the loop is linear. The step $4$ gives $4/(1+K_p)=0$; the ramp $6t$ gives $6/K_v=0$; the parabola $3t^{2}$ must be written in the standard form $\tfrac{1}{2}At^{2}$, so $A = 6$ and $$e_{ss}=\frac{A}{K_a}=\frac{6}{2.5}=\boxed{\;2.4\;}$$ Only the parabolic term contributes.
Test the closed loop for stability — the step that governs the answer. The characteristic equation is $$1+G(s)=0 \;\Longrightarrow\; s^{2}(4+s)+10 = 0 \;\Longrightarrow\; s^{3}+4s^{2}+0\cdot s+10 = 0.$$ The coefficient of $s$ is missing, which already violates the necessary condition that all coefficients of a stable polynomial be present and of one sign.
Confirm with the Routh array. $$\begin{array}{c|cc} s^{3} & 1 & 0\\ s^{2} & 4 & 10\\ s^{1} & \dfrac{4(0)-1(10)}{4}=-2.5 & 0\\ s^{0} & 10 & \end{array}$$ The first column reads $1,\;4,\;-2.5,\;10$: two sign changes, so $\boxed{\;\text{two closed-loop poles lie in the right half plane}\;}$ (the roots are $s=-4.495$ and $s=+0.247\pm j1.471$).

The two results must be read together. The error constants are perfectly well defined and give the textbook answers $e_{ss}=0$ for the ramp and $e_{ss}=2.4$ for the parabolic input, and those are the values the marking scheme expects. But the final-value theorem may only be applied when $sE(s)$ has all its poles in the left half plane, and here it does not: the physical response diverges, oscillating with a growing envelope at about 1.47 rad/s. The complete answer therefore states both the formal error figures and the fact that this particular loop can never reach them.

Check: the printed loop is unstable as drawn. In an examination answer, quote the error constants, then note explicitly that the steady-state error is only realisable if the loop is first stabilised — for instance by cascading a phase-lead network such as $D(s)=(1+\alpha s)$, which supplies the missing $s$ coefficient. Adding a single zero at $s=-1$ makes the characteristic polynomial $s^{3}+4s^{2}+10s+10$, whose Routh first column is all positive.

Question 3 — error analysis of the type-2 loop
Quantity	Value	Remark
System type	2	double pole at the origin
$K_p$	$\infty$	zero step error
$K_v$	$\infty$	zero ramp error
$K_a$	$2.5\ \text{s}^{-2}$	finite parabolic error
(a) $e_{ss}$ for $r=10t$	$0$	formal value
(b) $e_{ss}$ for $r=4+6t+3t^{2}$	$2.4$	from the $3t^{2}$ term only
Closed-loop stability	unstable (2 RHP poles)	$s^{3}+4s^{2}+10$; errors not physically realised

Topic: System type, static error constants, and the stability precondition for the final-value theorem.

Key relations: $E(s)=R(s)/[1+G(s)]$; $K_p=\lim_{s\to0}G$, $K_v=\lim_{s\to0}sG$, $K_a=\lim_{s\to0}s^{2}G$; $e_{ss}=A/(1+K_p),\;A/K_v,\;A/K_a$ for step, ramp and parabola $\tfrac{1}{2}At^{2}$.

Why this works: Each integration in the forward path adds one power of $s$ that cancels against the input transform in the limit, which is why a type-$n$ loop tracks the first $n$ polynomial inputs exactly and shows a constant offset on the $(n+1)$th. The final-value theorem is a limit statement about a Laplace transform, and it is only meaningful when the time function actually converges — formally, when every pole of $sE(s)$ lies strictly in the left half plane.

Common pitfall: Two, both present here. First, reading $3t^{2}$ as $A=3$: the standard parabolic input is $\tfrac{1}{2}At^{2}$, so $A=6$ and the error is 2.4, not 1.2. Second, and more serious, quoting a steady-state error for an unstable loop — always run the Routh test on $1+G(s)$ before believing any final-value result.

Source: Ogata, Modern Control Engineering, 5th ed., Ch. 5.8 (steady-state errors in unity-feedback systems) and Ch. 5.6 (Routh stability); Nise, Control Systems Engineering, 8th ed., Ch. 7.

Question 3 of 6: Steady-State Error of a Type-2 Loop

Question 3: Steady-State Error of a Type-2 Loop (25 marks)