System type, loop gain and steady-state errors: 22-Mec-A3 System Analysis and Control, December 2013, Q3

Question 3: System type, loop gain and steady-state errors

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 Fig. 2 with loop gain function $G(s)=(s+1)(s+3)/[s(s+2)(s+4)]$.

Find. (a) the type number; (b) the gain constant of the loop gain function; (c) the steady-state errors $e_{ss}$ for a unit step and for a unit ramp input.

Figure 3.1 — The unity-feedback system of Fig. 2. The error signal is $E(s)=R(s)/[1+G(s)]$, so the steady-state error follows directly from the low-frequency behaviour of $G$.

Approach. Rewrite $G(s)$ in Bode (time-constant) form, whose leading factor $K/s^N$ exposes both the type number $N$ and the gain constant $K$ at a glance; then evaluate the static error constants $K_p$ and $K_v$ and apply the final-value theorem.

Convert to time-constant form. Divide each factor by its own constant term so every binomial reads $(\tau s+1)$: $$G(s)=\frac{(s+1)(s+3)}{s(s+2)(s+4)} =\frac{1\times 3}{2\times 4}\cdot\frac{(s+1)\left(\tfrac{s}{3}+1\right)} {s\left(\tfrac{s}{2}+1\right)\left(\tfrac{s}{4}+1\right)}$$ which collapses to $$G(s)=\frac{0.375\,(1+s)(1+s/3)}{s\,(1+s/2)(1+s/4)}$$
Read the type number. Type is the number of pure integrators, i.e. the multiplicity of the pole at the origin. There is exactly one factor of $s$ in the denominator, so $$\boxed{\;N=1\quad(\text{type-1 system})\;}$$
Read the gain of the loop gain function. With the time-constant form in place, the gain constant is the coefficient multiplying $1/s^N$: $$K=\frac{(1)(3)}{(2)(4)}=\frac{3}{8}=0.375$$ Equivalently, $K=\lim_{s\to 0} s\,G(s)$, which for a type-1 system is the velocity error constant itself, so $K=K_v=0.375\ \text{s}^{-1}$.
Evaluate the static error constants. For unity feedback, $$K_p=\lim_{s\to 0}G(s)=\lim_{s\to 0}\frac{(1)(3)}{s(2)(4)}\to\infty, \qquad K_v=\lim_{s\to 0}sG(s)=\frac{(1)(3)}{(2)(4)}=0.375$$ The integrator makes $K_p$ unbounded, which is the defining property of a type-1 loop.
Apply the final-value theorem to each input. With $E(s)=R(s)/[1+G(s)]$ and $e_{ss}=\lim_{s\to 0}sE(s)$: $$e_{ss}\big|_{\text{step}}=\frac{1}{1+K_p}=\frac{1}{\infty}=0, \qquad e_{ss}\big|_{\text{ramp}}=\frac{1}{K_v}=\frac{1}{0.375}=\frac{8}{3}$$ so $$\boxed{\;e_{ss}(\text{unit step})=0,\qquad e_{ss}(\text{unit ramp})=\tfrac{8}{3}=2.667\;}$$
Interpret the numbers. The integrator drives the step error to zero exactly — the output eventually reaches the commanded position no matter how small the gain. A ramp command, however, is tracked with a permanent lag of 2.667 units, which is large because the loop gain constant is well below unity. Raising the numerator gain by a factor of ten would cut the ramp error to 0.267 at the cost of reduced damping, which is the classic accuracy-versus-stability trade-off that Question 4 explores explicitly.

Quantity	Result
(a) Type number	$N=1$
(b) Gain of the loop gain function	$K=3/8=0.375$
Position error constant	$K_p=\infty$
Velocity error constant	$K_v=0.375\ \text{s}^{-1}$
(c) $e_{ss}$, unit step	0
(c) $e_{ss}$, unit ramp	$8/3=2.667$

Topic. System type, the Bode gain constant and the static error constants of a unity-feedback loop.

Key relations. Bode form $G(s)=\dfrac{K\prod(1+s/z_i)}{s^N\prod(1+s/p_j)}$; $K_p=\lim_{s\to0}G$, $K_v=\lim_{s\to0}sG$, $K_a=\lim_{s\to0}s^2G$; and $e_{ss}=1/(1+K_p)$, $1/K_v$, $1/K_a$ for step, ramp and parabolic inputs respectively.

Why this works. The final-value theorem converts a limit in time into a limit in $s$ near the origin, so only the low-frequency structure of the loop matters for steady-state accuracy. Writing $G$ in time-constant form isolates that structure: everything except $K/s^N$ tends to unity as $s\to0$, which is why the type number and the Bode gain between them determine every steady-state error the system can exhibit.

Common pitfall. Quoting the "gain" as the coefficient of the pole-zero (root-locus) form — here that coefficient is 1, not 0.375 — and then computing a ramp error of 1 instead of 2.667. The gain of the loop gain function in this context is the Bode constant, which must be extracted after normalising every factor to $(\tau s+1)$. A second trap is applying the error-constant formulas to a non-unity-feedback loop, where they do not hold without modification.

Source. Nise, Control Systems Engineering, 8th ed., §7.2–7.4 (steady-state error, system type, static error constants); Ogata, Modern Control Engineering, 5th ed., §5.8; Dorf and Bishop, §5.6.

Question 3 of 6: System type, loop gain and steady-state errors

Question 3: System type, loop gain and steady-state errors