Question 3 of 8: Second-order RLC two-port: transfer function & step specs

Question 3 — Second-order RLC two-port: transfer function & step specs [20]

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.

Find. $G(s)$, filter type, DC gain, $PO$, and $R_2$ for $PO=15\%$.

[Figure not reproduced: Figure Q3.2 (redrawn) — series $L$ into node A, $R_1$ to ground, series $R_2$ to the output node, $C$ to ground; $v_o$ taken across $C$. See the official exam paper.]

Approach. Node equations at A and B (output) in the $s$-domain, eliminate the internal node, then read $\omega_n,\zeta$ from the standard second-order denominator.

1. Node equations. With node voltages $V_A,V_B$ ($V_B=v_o$): $$\frac{V_A-v_i}{sL}+\frac{V_A}{R_1}+\frac{V_A-V_B}{R_2}=0,\qquad \frac{V_B-V_A}{R_2}+V_B\,sC=0.$$
2. Transfer function. Eliminating $V_A$ and substituting the values gives a clean second order (two storage elements $L,C$): $$G(s)=\frac{v_o}{v_i}=\frac{2\times10^{8}}{11\,s^2+3.0\times10^{4}\,s+2\times10^{8}}=\frac{1.818\times10^{7}}{s^2+2727\,s+1.818\times10^{7}}.$$
3. Filter type & DC gain. As $s\to0$ the inductor is a short and the capacitor an open, so $v_o=v_i$: the numerator equals the constant term of the denominator. It is a $\boxed{\text{2nd-order low-pass filter, DC gain }=1\ (0\text{ dB})}$.
4. Natural frequency and damping. Matching $s^2+2\zeta\omega_n s+\omega_n^2$: $$\omega_n=\sqrt{1.818\times10^{7}}=\boxed{4264\ \text{rad/s}},\qquad \zeta=\frac{2727}{2\omega_n}=\boxed{0.320}.$$ Under-damped ($\zeta\lt1$).
5. Percent overshoot. $$PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.320\pi/\sqrt{1-0.320^2}}=\boxed{34.6\%}.$$
6. Adjust $R_2$ for $PO=15\%$. First invert the $PO$ formula: $\zeta=\dfrac{-\ln(0.15)}{\sqrt{\pi^2+\ln^2(0.15)}}=0.517$. Re-deriving the denominator with $R_2$ symbolic gives $\zeta(R_2)=\dfrac{10^{4}R_2+5\times10^{4}}{2\sqrt{2\times10^{9}R_2+2\times10^{11}}}$; setting $\zeta=0.517$ and solving the positive root: $$\boxed{R_2\approx20.4\ \Omega}.$$ (Raising $R_2$ adds damping, lowering the overshoot from 34.6% to 15%.)