Question 8 of 8: Mason’s gain formula & second-order derivations

Question 8 — Mason’s gain formula & second-order derivations [5 + 5 + 10]

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.

Part A — Transfer function by Mason [5]

Approach. Each branch is a first-order feedback loop; its input–output gain is $\dfrac{g_{in}g_{out}\,(k/s)}{1-g_{fb}(k/s)}$. The three branches are in parallel, so add them.

1. Per-branch gains. Top ($g_{in}{=}1,k{=}1,g_{fb}{=}{-}2,g_{out}{=}3$): $\dfrac{3}{s+2}$; middle ($1,1,-1,2$): $\dfrac{2}{s+1}$; bottom ($2,3,-1,1$): $\dfrac{6}{s+3}$.
2. Sum. $$G(s)=\frac{3}{s+2}+\frac{2}{s+1}+\frac{6}{s+3} =\boxed{\frac{11s^2+40s+33}{(s+1)(s+2)(s+3)}}.$$

Part B — Complete the SFG & verify [5]

Approach. The target is third order, so a third integrator is needed; realise $G$ in phase-variable (controllable-canonical) form.

1. Factor. $s^3+9s^2+26s+24=(s+2)(s+3)(s+4)$; numerator $2s+20=2(s+10)$.
2. Phase-variable realisation. Forward chain $U\to1/s\to1/s\to1/s\to Y$; denominator coefficients $9,26,24$ become feedback gains $-9,-26,-24$ into the first node; numerator coefficients $2,20$ become output taps.
3. Mason check. Forward paths $P_1=\dfrac{20}{s^3}$, $P_2=\dfrac{2}{s^2}$; loops give $\Delta=1+\dfrac{9}{s}+\dfrac{26}{s^2}+\dfrac{24}{s^3}$. Then $$G=\frac{P_1+P_2}{\Delta}=\frac{(20+2s)/s^3}{(s^3+9s^2+26s+24)/s^3} =\boxed{\frac{2s+20}{s^3+9s^2+26s+24}}\ \checkmark$$

Part C — Derivation of $PO$ and $T_{settle}$ [10]

For $G(s)=\dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$ with a unit-step input, $C(s)=\dfrac{\omega_n^2}{s\,(s^2+2\zeta\omega_ns+\omega_n^2)}$. Using the Laplace-table pair for $\dfrac{\omega_n^2}{s(s^2+2\zeta\omega_ns+\omega_n^2)}$, the unit-step response of an under-damped system ($0\lt\zeta\lt1$) is

$$c(t)=1-\frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}\,\sin\!\bigl(\omega_dt+\phi\bigr),\qquad \omega_d=\omega_n\sqrt{1-\zeta^2},\ \ \phi=\cos^{-1}\zeta.$$

Percent overshoot. The peak occurs when $\dot c(t)=0$. Differentiating, the first maximum is at $t_p=\dfrac{\pi}{\omega_d}$. Substituting back, the sine term contributes $-\sin(\pi+\phi)=\sin\phi=\sqrt{1-\zeta^2}$, so

$$c(t_p)=1+e^{-\zeta\omega_n(\pi/\omega_d)}=1+e^{-\zeta\pi/\sqrt{1-\zeta^2}}.$$

Since the steady-state value is 1, the fractional overshoot is $c(t_p)-1$, giving

$$\boxed{PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}\ \%.}$$

Settling time. The response is bounded by the exponential envelope $1\pm\dfrac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}$. For the $\pm2\%$ band the decaying envelope must fall to $0.02$: $e^{-\zeta\omega_nt}\approx0.02$, i.e. $\zeta\omega_nT_s=\ln 50\approx3.91$. Rounding to the standard engineering value,

$$\boxed{T_{settle}(\pm2\%)=\frac{4}{\zeta\omega_n}.}$$