Question 6 of 6: Laplace-domain transfer function and step response
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — nodal/mesh analysis, first- and second-order transients, AC steady-state phasors, complex power and power-factor correction, Thévenin’s theorem and maximum-power transfer, and Laplace-domain analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the source-free RLC circuit and s-domain transfer functions.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms and a common cosine time reference (ω = 377 rad/s) is used throughout Q2.
Question 6: Laplace-domain transfer function and step response [5 + 5 + 10]
Given. $V_{in}=5\text{ V}$ (applied as a step when the switch closes), a series $2\,\Omega$ into node $X$, a $2\,\Omega$ from $X$ to ground, a series inductor $L=2\text{ H}$ from $X$ to the output node, and a $2\,\Omega$ output resistor across which $v_o$ is measured; $i_L(0)=0$.
Find. The s-domain model, $H(s)$, and $v_o(t)$.
[Figure not reproduced: Figure 6 — redrawn. Closing the switch applies the $5\text{ V}$ step; the inductor (impedance $2s$, zero initial current) feeds the output $2\,\Omega$. See the official exam paper.]
Approach. Replace the inductor by its impedance $sL=2s$ (zero initial current means no added source), reduce the resistive ladder to get $H(s)$, then apply the $5/s$ step input and invert by partial fractions.
Laplace model, part (i). With $i_L(0)=0$ the inductor becomes a pure impedance $Z_L=sL=2s$; the source is a step $V_{in}(s)=5/s$. Every resistor keeps its value, giving a $2\,\Omega$ input resistor to node $X$, a $2\,\Omega$ from $X$ to ground, and the branch $2s+2\,\Omega$ carrying the output.
Ladder reduction for $H(s)$, part (ii). Let $V_X$ be the node voltage. The output branch is $2s$ in series with the $2\,\Omega$ output resistor, so $V_o=\dfrac{2}{2s+2}V_X=\dfrac{V_X}{s+1}$. KCL at $X$:$$\frac{V_{in}-V_X}{2}=\frac{V_X}{2}+\frac{V_X}{2s+2}.$$ Solving, $V_X=\dfrac{s+1}{2s+3}V_{in}$, hence$$H(s)=\frac{V_o}{V_{in}}=\frac{V_X/(s+1)}{V_{in}}=\boxed{\,\frac{1}{2s+3}\,}$$
Apply the step input, part (iii). With $V_{in}(s)=5/s$:$$V_o(s)=\frac{5}{s(2s+3)}=\frac{2.5}{s\,(s+1.5)}.$$
Partial fractions and inversion.$$\frac{2.5}{s(s+1.5)}=\frac{1.667}{s}-\frac{1.667}{s+1.5}\quad\Longrightarrow\quad \boxed{\,v_o(t)=\tfrac{5}{3}\left(1-e^{-1.5t}\right)\text{ V},\ t\ge0\,}$$
Sanity check. At $t=0^+$, $i_L=0$ forces $v_o(0)=0$ — matches. As $t\to\infty$ the inductor is a short and the two $2\,\Omega$ resistors from $X$ appear in parallel ($1\,\Omega$), giving $v_o(\infty)=5\cdot\frac{1}{2+1}=\frac{5}{3}=1.67\text{ V}$ — matches the steady state of the exponential.
