Step Response from a Pole–Zero Description, and an Inverse Transform: 22-Mec-A3 System Analysis and Control, December 2018, Q6

Question 6: Step Response from a Pole–Zero Description, and an Inverse Transform (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) a first-order system specified only by its zero at $s=-1$, its pole at $s=-2$ and a gain factor of $2$, driven by a unit step; (b) the transformed response $Y(s)=\dfrac{s+4}{s(s+1)(s+2)}$.

Find. (a) $y(t)$ for the unit-step input; (b) $y(t)$ by inverse transformation.

Time responses for Question 6. Curve (a) is the step response of $2(s+1)/(s+2)$, which jumps immediately to $2$ because the transfer function is proper but not strictly proper. Curve (b) is the inverse transform of $(s+4)/[s(s+1)(s+2)]$, which starts at zero and climbs to its final value of $2$.

Approach. Assemble the transfer function from the stated pole, zero and gain; multiply by $1/s$ and expand in partial fractions for part (a). For part (b) the transform is already given, so the work is a three-term residue calculation, checked against the initial- and final-value theorems.

(a) Build the transfer function. “Gain factor of 2” means the multiplying constant in pole–zero form, so

$$G(s)=\frac{2(s+1)}{s+2}$$

This is proper but not strictly proper: numerator and denominator have the same degree. Its DC gain is $G(0)=2(1)/2=1$ and its high-frequency gain is $G(\infty)=2$, and that difference is what produces the jump seen below.

(a) Form the step response and expand. With $R(s)=1/s$,

$$Y(s)=\frac{2(s+1)}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2},\qquad A=\left.\frac{2(s+1)}{s+2}\right|_{s=0}=1,\qquad B=\left.\frac{2(s+1)}{s}\right|_{s=-2}=\frac{-2}{-2}=1$$

Inverting with pairs (1) and (15) of the supplied table,

$$\boxed{y(t)=1+e^{-2t},\qquad t\ge0}$$

(a) Interpret and check the two endpoints. The Initial Value Theorem gives

$$y(0^{+})=\lim_{s\to\infty}sY(s)=\lim_{s\to\infty}\frac{2(s+1)}{s+2}=2,\qquad y(\infty)=\lim_{s\to0}sY(s)=\frac{2(1)}{2}=1$$

both of which the closed form reproduces. The response therefore decays from $2$ down to $1$ rather than rising: the zero at $s=-1$ feeds the input through directly, so the output jumps instantaneously to twice the input and then relaxes to the DC gain of unity with time constant $\tau=0.5$ s. A first-order system with no finite zero could never behave this way — it would start at zero and rise monotonically.

(b) Expand the given transform into residues. Writing $Y(s)=\dfrac{A}{s}+\dfrac{B}{s+1}+\dfrac{C}{s+2}$ and using the cover-up rule at each simple pole,

$$A=\left.\frac{s+4}{(s+1)(s+2)}\right|_{s=0}=\frac{4}{2}=2,\qquad B=\left.\frac{s+4}{s(s+2)}\right|_{s=-1}=\frac{3}{-1}=-3,\qquad C=\left.\frac{s+4}{s(s+1)}\right|_{s=-2}=\frac{2}{2}=1$$

A free check comes before any inversion: $Y(s)$ is strictly proper, so the residues must sum to $y(0)=0$, and indeed $2-3+1=0$.

(b) Invert term by term. Using pairs (1) and (15) once more,

$$\boxed{y(t)=2-3e^{-t}+e^{-2t},\qquad t\ge0}$$

The final value is $2$, matching $\lim_{s\to0}sY(s)=4/2=2$, and both exponentials decay, so the Final Value Theorem is legitimately applicable here. The slower mode $e^{-t}$ carries the larger residue and therefore dominates the approach to steady state: the response is within 2% of $2$ after $t=4.31$ s — the $4\tau=4$ s rule of thumb is slightly optimistic here, because the residue $-3$ is larger than the final value itself. The curve rises monotonically from zero with an initial slope of $\dot{y}(0)=3-2=1$, and it never overshoots: $\dot{y}(t)=3e^{-t}-2e^{-2t}>0$ for every $t\ge0$, since $3>2e^{-t}$ always.

Question 6 — final results
Item	Result
(a) Transfer function	$G(s)=2(s+1)/(s+2)$
(a) Step response	$y(t)=1+e^{-2t}$
(a) Initial / final value	$y(0^{+})=2$; $y(\infty)=1$ (decaying, not rising)
(a) Time constant	$\tau=0.5$ s
(b) Residues	$A=2$, $B=-3$, $C=1$ (sum $=0=y(0)$)
(b) Time response	$y(t)=2-3e^{-t}+e^{-2t}$
(b) Final value	$y(\infty)=2$

Concept & Theory

Topic. Inverse Laplace transformation by partial fractions, and the effect of a transfer-function zero on the initial response.

Key relations. Cover-up (residue) rule $R_i=\left[(s-p_i)Y(s)\right]_{s=p_i}$ for simple poles; Initial Value Theorem $y(0^{+})=\lim_{s\to\infty}sY(s)$; Final Value Theorem $y(\infty)=\lim_{s\to0}sY(s)$ when all poles of $sY(s)$ are in the LHP; residues of a strictly proper $Y(s)$ sum to $y(0^{+})$.

Why this works. Each simple pole of $Y(s)$ contributes one exponential mode whose amplitude is the residue there, so the inverse transform is fully determined by evaluating the remaining factors at each pole. The relative degree of $Y(s)$ controls the start of the response: when numerator and denominator degrees are equal, as in part (a), there is a direct feed-through term and the output steps instantly. The residue-sum identity is the fastest sanity check available — it catches an arithmetic slip in any one residue without recomputing the others.

Common pitfall. Assuming every step response starts at zero and rises. Part (a) does the opposite because its zero makes the transfer function only proper, and a candidate who writes $y(t)=1-e^{-2t}$ by reflex has the transient exactly backwards. The other frequent error is misreading “gain factor of 2” as a DC gain of 2, which would give $G(s)=4(s+1)/(s+2)$ — in pole–zero form the gain factor is the leading constant, not $G(0)$.

Source. Ogata, Modern Control Engineering, 5th ed., §2.4 (inverse Laplace transformation by partial fractions) and §5.3; Nise, Control Systems Engineering, 8th ed., §2.2 and §4.3 (effect of zeros on the transient response).

Question 6 of 7: Step Response from a Pole–Zero Description, and an Inverse Transform

Question 6: Step Response from a Pole–Zero Description, and an Inverse Transform (25 marks)