Closed-loop transfer function, poles and zeros, step response: 22-Mec-A3 System Analysis and Control, May 2018, Q1

Question 1: Closed-loop transfer function, poles and zeros, step response (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. A single-loop system with unity negative feedback around the forward-path plant

$$G(s)=\frac{14{,}000}{s^{3}+45s^{2}+3100s+500}, \qquad H(s)=1 .$$

Find. (a) the closed-loop transfer function $T(s)$, (b) its poles and finite zeros, (c) the partial-fraction expansion of $Y(s)$ for a unit step, and (d) the final value $y(\infty)$.

[Figure not reproduced: Question 1 — unity-negative-feedback loop as printed on page 2 of the examination paper. See the official exam paper.]

Approach. Reduce the single loop with the standard formula $T=G/(1+GH)$, factor the resulting cubic to expose the poles, expand $Y(s)=T(s)/s$ into partial fractions, and read the final value from the Final Value Theorem.

Close the loop. For a single loop with negative feedback the closed-loop transfer function is $T=G/(1+GH)$. With $H=1$,$$T(s)=\frac{\dfrac{14{,}000}{s^{3}+45s^{2}+3100s+500}}{1+\dfrac{14{,}000}{s^{3}+45s^{2}+3100s+500}}=\frac{14{,}000}{s^{3}+45s^{2}+3100s+500+14{,}000}.$$Adding the numerator constant to the plant constant term gives$$\boxed{\,T(s)=\frac{Y(s)}{R(s)}=\frac{14{,}000}{s^{3}+45s^{2}+3100s+14{,}500}\,}$$Only the constant term moves — the loop gain adds to $a_{0}$ and leaves every other coefficient of the plant untouched.
Factor the closed-loop cubic. Testing the divisors of 14,500 shows $s=-5$ is a root: $(-5)^{3}+45(-5)^{2}+3100(-5)+14{,}500=-125+1125-15{,}500+14{,}500=0$. Synthetic division then gives$$s^{3}+45s^{2}+3100s+14{,}500=(s+5)\left(s^{2}+40s+2900\right).$$Completing the square on the quadratic, $s^{2}+40s+2900=(s+20)^{2}+50^{2}$, so its roots are $-20\pm j50$.
List the poles and zeros. The three closed-loop poles are$$\boxed{\,s_{1}=-5,\qquad s_{2,3}=-20\pm j50\,}$$and because the numerator of $T(s)$ is the constant 14,000 there are no finite zeros (all three zeros lie at infinity). The complex pair has natural frequency $\omega_{n}=\sqrt{2900}=53.85$ rad/s and damping ratio $\zeta=40/(2\times 53.85)=0.371$, while the real pole contributes a $0.2$ s time constant.
Expand $Y(s)=T(s)/s$ in partial fractions. Write$$Y(s)=\frac{14{,}000}{s(s+5)\left(s^{2}+40s+2900\right)}=\frac{A}{s}+\frac{B}{s+5}+\frac{Cs+D}{s^{2}+40s+2900}.$$Covering up each simple pole gives the two real residues directly:$$A=\left.\frac{14{,}000}{(s+5)(s^{2}+40s+2900)}\right|_{s=0}=\frac{14{,}000}{5(2900)}=\frac{28}{29}=0.96552,$$$$B=\left.\frac{14{,}000}{s(s^{2}+40s+2900)}\right|_{s=-5}=\frac{14{,}000}{(-5)(25-200+2900)}=\frac{14{,}000}{(-5)(2725)}=-\frac{112}{109}=-1.02752 .$$Matching the $s^{2}$ coefficient of the recombined numerator gives $A+B+C=0$, hence $C=112/109-28/29=196/3161=0.06201$, and matching the $s^{1}$ coefficient ($3100A+2900B+5D=0$) gives $D=-8400/3161=-2.65739$.
State the expansion and invert it. Substituting the residues,$$\boxed{\,Y(s)=\frac{0.96552}{s}-\frac{1.02752}{s+5}+\frac{0.06201\,s-2.65739}{(s+20)^{2}+50^{2}}\,}$$Rewriting the numerator of the quadratic term about $(s+20)$ splits it into a cosine and a sine transform, so inverting term by term with pairs (1), (15), (22) and (23) of the appended Laplace table gives$$y(t)=0.96552-1.02752e^{-5t}+e^{-20t}\left[0.06201\cos 50t-0.07795\sin 50t\right].$$A free check: at $t=0$ the three coefficients sum to $0.96552-1.02752+0.06201=0$, which is required because $T(s)$ has relative degree three.
Predict the final value. All closed-loop poles are in the open left half-plane, so the Final Value Theorem is legitimate:$$y(\infty)=\lim_{s\to 0}sY(s)=\lim_{s\to 0}\frac{14{,}000}{s^{3}+45s^{2}+3100s+14{,}500}=\frac{14{,}000}{14{,}500}.$$$$\boxed{\,y(\infty)=0.9655\,}$$Equivalently the position error constant is $K_{p}=G(0)=14{,}000/500=28$, giving a steady-state error $e_{ss}=1/(1+K_{p})=1/29=0.0345$ — the same 3.45 % shortfall seen in $y(\infty)$.

The response therefore settles about 3.4 % below the commanded unit step in roughly $4/5=0.8$ s, the slow real pole at $-5$ dominating the tail while the lightly damped pair at $-20\pm j50$ produces a fast 50 rad/s ripple that has decayed within about 0.2 s.

Quantity	Result
Closed-loop transfer function $T(s)$	$14{,}000/(s^{3}+45s^{2}+3100s+14{,}500)$
Poles	$-5$, $-20\pm j50$
Finite zeros	none (three zeros at infinity)
Partial-fraction residues $A,B$	$0.96552$, $-1.02752$
Quadratic numerator $Cs+D$	$0.06201s-2.65739$
Final value $y(\infty)$	$0.9655$
Steady-state error $e_{ss}$	$0.0345$ (3.45 %)

Topic: Single-loop reduction and partial-fraction inversion of a third-order step response.

Key relations: $T=G/(1+GH)$; residue (cover-up) rule $A_{i}=[(s-p_{i})Y(s)]_{s=p_{i}}$; $(s+\sigma)^{2}+\omega_{d}^{2}$ inverts to $e^{-\sigma t}(\cos\omega_{d}t,\ \sin\omega_{d}t)$; Final Value Theorem $y(\infty)=\lim_{s\to0}sY(s)$; $e_{ss}=1/(1+K_{p})$ with $K_{p}=G(0)$ for a type-0 loop.

Why this works: Negative feedback adds the loop numerator to the plant denominator, so for a plant with a constant numerator only the $a_{0}$ coefficient changes; every other coefficient, and therefore the high-frequency behaviour, is untouched. Because the closed-loop numerator carries no $s$ terms, $T(s)$ has no finite zeros and the residues must satisfy $A+B+C=0$ — a relative degree of three forces $y(0^{+})=0$. The Final Value Theorem is valid only because all three poles were first confirmed to be in the left half-plane; the resulting 3.45 % offset is exactly the type-0 steady-state error $1/(1+K_{p})$.

Common pitfall: Students frequently add the loop gain to the wrong coefficient (or to the numerator) when closing the loop, and then apply the Final Value Theorem without checking stability — on an unstable loop the limit still returns a finite number that is completely meaningless.

Source: Dorf & Bishop, Modern Control Systems, 14th ed. §2.7 (block-diagram reduction), §5.6–5.7 (steady-state error); Nise, Control Systems Engineering, 8th ed. Ch. 4 (time response of higher-order systems).

Question 1 of 6: Closed-loop transfer function, poles and zeros, step response

Question 1: Closed-loop transfer function, poles and zeros, step response (25 marks)