Unity-feedback system — real-pole range, step response, steady-state error: 22-Mec-A3 System Analysis and Control, May 2013, Q1

Question 1: Unity-feedback system — real-pole range, step response, steady-state error (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 open-loop transfer function

$$G(s)=\frac{K(s+12)}{s^{2}+25},\qquad H(s)=1,\qquad K>0\ \text{a scalar gain.}$$

The plant is an undamped second-order term ($s^{2}+25$, poles at $\pm j5$) with a real open-loop zero at $s=-12$.

Find. (i) the closed-loop denominator; (ii) the range of $K$ for which the closed-loop poles are real; (iii) with $K=24$, the closed-loop unit-step response $c(t)$ and the steady-state error $e_{ss}$ following a step.

Closed-loop unit-step response for $K=24$. The response is under-damped ($\zeta\approx0.68$), overshoots, and settles to $c(\infty)=0.920$; the gap to the input value of $1$ is the steady-state error $e_{ss}=0.080$.

Approach. Form $1+G(s)H(s)=0$ to get the characteristic polynomial, apply the discriminant test for real roots, then invert $C(s)=T(s)/s$ by partial fractions and read $e_{ss}$ from the position error constant.

Closed-loop denominator. For unity feedback the closed-loop transfer function is $T(s)=G/(1+G)$, whose denominator is the numerator of $1+G$: $$1+G(s)=\frac{(s^{2}+25)+K(s+12)}{s^{2}+25}\;\Longrightarrow\; \boxed{\,D(s)=s^{2}+K\,s+(25+12K)\,}.$$ The loop turns the marginally-stable plant into a genuine second-order system with adjustable damping $Ks$ and stiffness $25+12K$.
Range of $K$ for real poles. The two poles are real when the discriminant of $D(s)$ is non-negative: $$K^{2}-4(25+12K)\ge 0\;\Longrightarrow\;K^{2}-48K-100\ge0\;\Longrightarrow\;(K-50)(K+2)\ge0.$$ Hence $K\le-2$ or $K\ge 50$. For a physical positive gain the poles are real (and equal at the boundary) for $$\boxed{\,K\ge 50\,}\qquad(\text{repeated real pole exactly at }K=50).$$ For $0<K<50$ the closed-loop poles are complex conjugates.
Pole location at $K=24$. With $K=24$, $D(s)=s^{2}+24s+313$, so $$s=\frac{-24\pm\sqrt{24^{2}-4(313)}}{2}=-12\pm j13,\qquad \omega_n=\sqrt{313}=17.69\ \text{rad/s},\ \ \zeta=\frac{12}{\omega_n}=0.678.$$ Since $24<50$ the poles are complex, as expected — the response will be under-damped.
Unit-step response. The closed-loop transfer function is $T(s)=\dfrac{24(s+12)}{s^{2}+24s+313}$; for a unit step $R(s)=1/s$, $$C(s)=\frac{24(s+12)}{s\,(s^{2}+24s+313)}=\frac{A}{s}+\frac{B\,s+D}{s^{2}+24s+313}.$$ Matching coefficients gives $A=\dfrac{24\cdot12}{313}=0.9201$, $B=-A=-0.9201$, $D=24(1-A)=1.917$. Completing the square $s^{2}+24s+313=(s+12)^{2}+13^{2}$ and rewriting the numerator about $(s+12)$, $$\boxed{\,c(t)=0.920+e^{-12t}\!\left[-0.920\cos 13t+0.997\sin 13t\right]\,},\qquad t\ge0.$$ The transient decays with time constant $1/12\approx0.083$ s; the ringing frequency is $13$ rad/s.
Steady-state error to a step. This is a type-0 loop (no free integrator), so the static position error constant is finite: $$K_p=\lim_{s\to0}G(s)=\frac{24\cdot12}{25}=11.52,\qquad e_{ss}=\frac{1}{1+K_p}=\frac{1}{12.52}=\boxed{0.0799}.$$ Equivalently the response settles to $c(\infty)=T(0)=288/313=0.920$, and $e_{ss}=1-c(\infty)=0.080$ — the two routes agree.

Quantity	Value
Closed-loop denominator	$s^{2}+Ks+(25+12K)$
Real closed-loop poles (positive $K$)	$K\ge 50$
Poles at $K=24$	$-12\pm j13$ ($\zeta=0.678$, $\omega_n=17.7$)
Step response $c(t)$	$0.920+e^{-12t}(-0.920\cos13t+0.997\sin13t)$
Steady-state step error	$e_{ss}=0.080$ ($K_p=11.52$)

Topic: Second-order closed-loop analysis — characteristic equation, discriminant test for real vs complex poles, time response and static error constants.

Key relations: $1+GH=0$; real roots $\Leftrightarrow$ discriminant $\ge0$; $\omega_n=\sqrt{a_0}$, $\zeta=a_1/2\omega_n$; type-0 position constant $K_p=\lim_{s\to0}G$, $e_{ss}=1/(1+K_p)$; final-value theorem $c(\infty)=\lim_{s\to0}sC(s)$.

Why this works: The closed-loop poles are the roots of $D(s)$, so their character (real/complex) is fixed by the sign of the quadratic discriminant. Because the forward path contains no $1/s$ factor, a step leaves a finite residual error governed by the DC loop gain $K_p$; the same number appears as the deficit $1-c(\infty)$ because the closed-loop DC gain is $K_p/(1+K_p)$.

Common pitfall: reading $e_{ss}$ off the output as $0.92$ instead of the error $0.08$, or assuming the zero at $-12$ removes the steady-state error — a forward-path zero shapes the transient and DC gain but does not add an integrator, so the loop stays type-0.

Question 1 of 6: Unity-feedback system — real-pole range, step response, steady-state error

Question 1: Unity-feedback system — real-pole range, step response, steady-state error (equal value)