Type-1 system — gain for a ramp-error spec, then step/ramp errors: 22-Mec-A3 System Analysis and Control, May 2013, Q2

Question 2: Type-1 system — gain for a ramp-error spec, then step/ramp errors (equal value)

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. Unity-feedback loop with

$$G(s)=\frac{K}{s\,(s+25)},\qquad H(s)=1.$$

Design target: a steady-state ramp error of $5\%$, i.e. velocity error constant $K_v=20$ (the paper labels this constant $K_r$).

Find. the gain $K$; then (a) the steady-state error to a unit step, and the full transient error responses to (b) a unit step and (c) a unit ramp.

Error transients for the designed loop ($K=500$). The unit-step error starts at $1$ and decays to $0$ (type-1 loop rejects a step exactly); the unit-ramp error starts at $0$ and settles to the design value $e_{ss}=0.05$.

Approach. A single free integrator makes this a type-1 loop, so $K_v$ fixes the ramp error and $K$; the transient error $E(s)=R(s)/(1+G)$ is then inverted for each input.

Gain from the ramp-error spec. For a type-1 loop the velocity error constant and ramp error are $$K_v=\lim_{s\to0}sG(s)=\frac{K}{25},\qquad e_{ss,\text{ramp}}=\frac{1}{K_v}.$$ Setting $e_{ss,\text{ramp}}=0.05$ gives $K_v=20$ and $$\boxed{\,K=25\,K_v=500\,}.$$ The closed-loop characteristic equation is then $s^{2}+25s+500=0$, with poles $-12.5\pm j18.54$ ($\omega_n=22.36$, $\zeta=0.559$).
(a) Steady-state error to a unit step. Because the loop contains an integrator, the static position constant is infinite: $$K_p=\lim_{s\to0}G(s)=\infty\;\Longrightarrow\;\boxed{\,e_{ss,\text{step}}=\frac{1}{1+K_p}=0\,}.$$ The integrator drives the step error to zero in the steady state.
(b) Transient error to a unit step. The error transfer function is $\dfrac{E}{R}=\dfrac{1}{1+G}=\dfrac{s(s+25)}{s^{2}+25s+500}$, so for $R=1/s$ $$E(s)=\frac{s+25}{s^{2}+25s+500}=\frac{(s+12.5)+12.5}{(s+12.5)^{2}+18.54^{2}}.$$ Inverting, $$\boxed{\,e(t)=e^{-12.5t}\!\left[\cos 18.54t+0.674\sin 18.54t\right]\,}.$$ The error jumps to $e(0^{+})=1$ (the plant cannot respond instantly) and rings down to zero with time constant $1/12.5=0.08$ s — consistent with $e_{ss,\text{step}}=0$.
(c) Transient error to a unit ramp. For $R=1/s^{2}$, $$E(s)=\frac{s+25}{s\,(s^{2}+25s+500)}=\frac{0.05}{s}+\frac{-0.05(s+12.5)+0.375}{(s+12.5)^{2}+18.54^{2}}.$$ Inverting, $$\boxed{\,e(t)=0.05-e^{-12.5t}\!\left[0.05\cos 18.54t-0.0202\sin 18.54t\right]\,}.$$ The ramp error starts at $e(0)=0$ and climbs to the steady value $e(\infty)=0.05$ — exactly the $5\%$ the gain was designed to deliver.

Quantity	Value
Loop gain	$K=500$ ($K_v=20$)
Closed-loop poles	$-12.5\pm j18.54$ ($\zeta=0.559$)
(a) steady-state step error	$0$ (type-1)
(b) step-error transient	$e^{-12.5t}(\cos18.54t+0.674\sin18.54t)$
(c) ramp-error transient	$0.05-e^{-12.5t}(0.05\cos18.54t-0.0202\sin18.54t)$

Topic: System type and static error constants; transient error response from the error transfer function.

Key relations: type-1 $\Rightarrow K_p=\infty,\ e_{ss,\text{step}}=0$; $K_v=\lim_{s\to0}sG$, $e_{ss,\text{ramp}}=1/K_v$; $E(s)=R(s)\big/(1+G(s))$.

Why this works: Each pure integrator in the forward path raises the system type by one, killing the steady-state error to the correspondingly lower-order input and leaving a finite error to the next-higher input. Designing to a $K_v$ target therefore sets the gain directly; the transient is the natural closed-loop mode $e^{-\zeta\omega_n t}$ superposed on the steady error.

Common pitfall: confusing the steady-state error (a single number, from the final-value theorem) with the transient error $e(t)$ (the whole time function). Also, applying the final-value theorem to a ramp on a type-0 loop, where the error diverges and the theorem is invalid.

Question 2 of 6: Type-1 system — gain for a ramp-error spec, then step/ramp errors

Question 2: Type-1 system — gain for a ramp-error spec, then step/ramp errors (equal value)