Question 2 of 6: Unit Step Responses by Partial Fractions

Question 2: Unit Step Responses by Partial Fractions (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) $G_a(s) = \dfrac{1}{(s+2)^2(s+1)}$ — a repeated real pole at $s=-2$ together with a simple pole at $s=-1$. (b) $G_b(s) = \dfrac{54}{(2s+6)(s^2+3s+9)}$ — a real factor with a leading coefficient of 2 and an underdamped quadratic. Input in both cases is the unit step, $R(s)=1/s$.

Find. The time functions $c(t)$ for $t \ge 0$, in closed form.

Approach. Form $C(s)=G(s)/s$, expand in partial fractions (using the derivative rule for the repeated pole and completing the square for the quadratic), then invert term by term with the supplied Laplace table. Verify each result against $c(0)=0$ and the final-value theorem.

1. Set up the expansion for (a). With a unit-step input, $$C_a(s)=\frac{1}{s(s+2)^2(s+1)} =\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}+\frac{D}{(s+2)^2}$$ The repeated pole needs two terms, and $C$ must be found by differentiation rather than by simple residue evaluation.
2. Evaluate the simple residues. Multiplying through and setting the appropriate pole value, $$A=\left.\frac{1}{(s+2)^2(s+1)}\right|_{s=0}=\frac{1}{4\cdot 1}=\frac14,\qquad B=\left.\frac{1}{s(s+2)^2}\right|_{s=-1}=\frac{1}{(-1)(1)}=-1$$
3. Evaluate the repeated-pole residues. The coefficient of the squared term is a direct residue, while the coefficient of the simple term requires one derivative: $$D=\left.\frac{1}{s(s+1)}\right|_{s=-2}=\frac{1}{(-2)(-1)}=\frac12$$ $$C=\left.\frac{d}{ds}\!\left[\frac{1}{s(s+1)}\right]\right|_{s=-2} =\left.\frac{-(2s+1)}{(s^2+s)^2}\right|_{s=-2}=\frac{3}{4}$$
4. Invert term by term. Using $1/s \to 1$, $1/(s+\alpha) \to e^{-\alpha t}$ and $1/(s+\alpha)^2 \to te^{-\alpha t}$ from the supplied table, $$\boxed{c_a(t)=\tfrac14-e^{-t}+\tfrac34 e^{-2t}+\tfrac12\,t\,e^{-2t}}$$ Two independent checks confirm this: at $t=0$ the coefficients sum to $0.25-1+0.75+0=0$ as required for a strictly proper system, and as $t\to\infty$ the response tends to $1/4$, which equals the DC gain $G_a(0)=1/(4\cdot 1)$. The response is monotone — all poles are real, so no overshoot is possible.

Part (b) looks like a different problem but is really a test of the same discipline applied one step earlier: the real factor is not monic, and until it is normalised the pole is easy to misread.

5. Normalise the leading coefficient of (b) first. Since $2s+6 = 2(s+3)$, dividing numerator and denominator by 2 gives $$G_b(s)=\frac{54}{2(s+3)(s^2+3s+9)}=\frac{27}{(s+3)(s^2+3s+9)}$$ The real pole is at $s=-3$, not at $s=-6$; halving the numerator at the same time is essential, and the DC gain is $27/(3 \times 9)=1$.
6. Expand with the quadratic kept intact. $$C_b(s)=\frac{27}{s(s+3)(s^2+3s+9)} =\frac{A}{s}+\frac{B}{s+3}+\frac{Cs+D}{s^2+3s+9}$$ The residues at the two real poles are $A=27/(3\cdot 9)=1$ and $B=27/[(-3)(9-9+9)]=-1$. Matching the coefficient of $s^3$ gives $A+B+C=0$, hence $C=0$, and matching $s^2$ gives $3A+3B+D=0$, hence $D=-3$. The vanishing of $C$ is what makes the answer a pure sine rather than a sine-plus-cosine combination.
7. Complete the square and invert. $$s^2+3s+9=\left(s+\tfrac32\right)^2+\tfrac{27}{4},\qquad \omega_d=\frac{3\sqrt3}{2}=2.598\ \text{rad/s}$$ so the third term is $-3/[(s+1.5)^2+\omega_d^2]$, whose inverse is $-(3/\omega_d)e^{-1.5t}\sin\omega_d t$. Collecting, $$\boxed{c_b(t)=1-e^{-3t}-\frac{2}{\sqrt3}\,e^{-1.5t}\sin\!\left(\frac{3\sqrt3}{2}t\right)}$$ Again $c_b(0)=1-1-0=0$ and $c_b(\infty)=1$, matching the unity DC gain. The quadratic factor has $\zeta = 3/(2\times 3) = 0.5$; a numerical sweep of the closed form gives a peak of 1.0815 at $t=1.641\ \text{s}$, an overshoot of 8.15 per cent — markedly less than the 16.3 per cent a $\zeta=0.5$ pair alone would give, because the real pole at $s=-3$ slows the leading edge.