Question 5 of 6: Steady-state errors for step and ramp inputs

Question 5: Steady-state errors for step and ramp inputs (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 unity-feedback configuration of Figure 2, with four different forward-path transfer functions:

Find. For each case, the steady-state error $e_{ss}=\lim_{t\to\infty}e(t)$ for (i) a unit step $r(t)=u_s(t)$ and (ii) a unit ramp $r(t)=t$, the closed loop being assumed stable.

[Figure not reproduced: Figure 2 redrawn: the unity-feedback configuration for which the static error constants are defined. See the official exam paper.]

Approach. For unity feedback the error transform is $E(s)=R(s)/[1+G(s)]$, so the final-value theorem converts each case into a limit of $sG(s)^k$; classify each $G$ by its system type, evaluate the appropriate static error constant, and read the two errors from the standard table.

1. Set up the error expression. With unity feedback, $E(s)=R(s)-C(s)=\dfrac{R(s)}{1+G(s)}$, and provided the closed loop is stable the final-value theorem gives $$\boxed{\;e_{ss}=\lim_{s\to 0}\frac{sR(s)}{1+G(s)}\;}$$ The stability proviso matters: applying the theorem to an unstable loop returns a finite but meaningless number.
2. Define the static error constants. Substituting $R(s)=1/s$ for the unit step and $R(s)=1/s^{2}$ for the unit ramp gives, respectively, $e_{ss}=1/(1+K_p)$ with $K_p=\lim_{s\to 0}G(s)$, and $e_{ss}=1/K_v$ with $K_v=\lim_{s\to 0}sG(s)$. A type-0 system has finite $K_p$ and $K_v=0$; a type-1 system has $K_p=\infty$ and finite $K_v$; a type-2 system has both infinite, so both errors vanish and the acceleration constant $K_a=\lim_{s\to 0}s^{2}G(s)$ becomes the relevant measure.
3. Case (a) — type 0. The position constant is $K_p=\lim_{s\to 0}\dfrac{10}{(s+1)(s+2)}=\dfrac{10}{2}=5$, so the step error is $$\boxed{\;e_{ss,\text{step}}=\frac{1}{1+5}=0.1667\;}$$ while $K_v=\lim_{s\to0}sG(s)=0$, so the ramp error is infinite — a type-0 loop simply cannot follow a ramp. (The closed loop is stable here: $s^{2}+3s+12$ has both roots in the left half-plane.)
4. Case (b) — type 1. One free integrator drives $K_p\to\infty$, so the step is followed exactly. The velocity constant is $K_v=\lim_{s\to0}\dfrac{10}{(s+1)(s+5)}=\dfrac{10}{5}=2\ \text{s}^{-1}$, giving $$\boxed{\;e_{ss,\text{step}}=0,\qquad e_{ss,\text{ramp}}=\frac{1}{2}=0.5\;}$$
5. Case (c) — type 2. Two integrators make both $K_p$ and $K_v$ infinite, so $$\boxed{\;e_{ss,\text{step}}=0,\qquad e_{ss,\text{ramp}}=0\;}$$ The finite measure of tracking quality here is the acceleration constant $K_a=\lim_{s\to0}s^{2}G(s)=\dfrac{5\times 2}{6}=1.667$, which would give an error of $1/K_a = 0.6$ against a unit parabolic input.
6. Case (d) — type 1 with a resonant plant. The single $s$ in the denominator makes this type 1 despite the undamped pair at $s=\pm j\sqrt{3}$. Hence the step error is zero and $$K_v=\lim_{s\to0}s\cdot\frac{6s^{2}+2s+10}{s(s^{2}+3)}=\frac{10}{3}=3.333\ \text{s}^{-1} \quad\Longrightarrow\quad\boxed{\;e_{ss,\text{ramp}}=\frac{3}{10}=0.3\;}$$ The stability assumption is genuinely needed here, and it holds: the closed-loop polynomial is $s(s^{2}+3)+6s^{2}+2s+10=s^{3}+6s^{2}+5s+10$, and since $6\times 5=30>10$ the Routh test is satisfied.
7. Read the pattern. Error performance is governed entirely by the number of free integrators in the forward path and by the low-frequency gain, not by the detailed pole locations. Adding an integrator removes one order of steady-state error but costs $90^\circ$ of phase, which is exactly the trade-off Question 1 explores from the frequency-domain side.