1. **Problem Statement:** You are given a passive RLC circuit with an inductor (1 H), capacitor (2 F), resistor (4 Ω), a voltage source $v_s(t)$, and a current source $i_s(t) = 3v_s(t)$. The goal is to find the transfer function $$G(s) = \frac{V_o(s)}{V_s(s)}$$ and the differential equation describing the system. 2. **Circuit Analysis Setup:** - Inductor impedance: $$Z_L = sL = s \times 1 = s$$ - Capacitor impedance: $$Z_C = \frac{1}{sC} = \frac{1}{2s}$$ - Resistor impedance: $$Z_R = 4$$ - The resistor and capacitor are in parallel, so their combined impedance is: $$Z_{RC} = \left( \frac{1}{Z_R} + \frac{1}{Z_C} \right)^{-1} = \left( \frac{1}{4} + 2s \right)^{-1} = \frac{1}{\frac{1}{4} + 2s} = \frac{4}{1 + 8s}$$ 3. **Total impedance seen by the voltage source:** The inductor is in series with the parallel combination, so: $$Z_{total} = Z_L + Z_{RC} = s + \frac{4}{1 + 8s}$$ 4. **Expressing output voltage $V_o(s)$:** $V_o(s)$ is the voltage across the parallel resistor and capacitor, so: $$V_o(s) = I(s) \times Z_{RC}$$ where $I(s)$ is the current through the series branch. 5. **Current source relation:** Given $i_s(t) = 3v_s(t)$, in Laplace domain: $$I_s(s) = 3V_s(s)$$ 6. **Applying Kirchhoff's Voltage Law (KVL):** The voltage source $V_s(s)$ drives the series combination of $Z_L$ and $Z_{RC}$ with current $I(s)$, and the current source $I_s(s)$ is connected at the node of $V_o(s)$. 7. **Node voltage relation:** At the node where $V_o(s)$ is measured, the current from the voltage source branch $I(s)$ and the current source $I_s(s)$ combine: $$I(s) = I_s(s) + I_{Z_{RC}}(s)$$ But since $V_o(s) = I_{Z_{RC}}(s) \times Z_{RC}$, and $I_{Z_{RC}}(s) = \frac{V_o(s)}{Z_{RC}}$, we have: $$I(s) = 3V_s(s) + \frac{V_o(s)}{Z_{RC}}$$ 8. **Express $V_s(s)$ in terms of $I(s)$ and $Z_{total}$:** $$V_s(s) = I(s) Z_{total} = I(s) \left(s + \frac{4}{1 + 8s}\right)$$ 9. **Substitute $I(s)$ from step 7 into step 8:** $$V_s(s) = \left(3V_s(s) + \frac{V_o(s)}{Z_{RC}}\right) \left(s + \frac{4}{1 + 8s}\right)$$ 10. **Rearranging to find $G(s) = \frac{V_o(s)}{V_s(s)}$:** Divide both sides by $V_s(s)$: $$1 = \left(3 + \frac{G(s)}{Z_{RC}}\right) \left(s + \frac{4}{1 + 8s}\right)$$ 11. **Solve for $G(s)$:** $$1 = 3 \left(s + \frac{4}{1 + 8s}\right) + G(s) \frac{s + \frac{4}{1 + 8s}}{Z_{RC}}$$ Recall $Z_{RC} = \frac{4}{1 + 8s}$, so: $$\frac{s + \frac{4}{1 + 8s}}{Z_{RC}} = \left(s + \frac{4}{1 + 8s}\right) \times \frac{1 + 8s}{4} = \frac{(s(1 + 8s) + 4)(1 + 8s)}{4(1 + 8s)} = \frac{s(1 + 8s) + 4}{4}$$ Simplify numerator: $$s(1 + 8s) + 4 = s + 8s^2 + 4$$ So: $$\frac{s + \frac{4}{1 + 8s}}{Z_{RC}} = \frac{8s^2 + s + 4}{4}$$ 12. **Rewrite equation:** $$1 = 3 \left(s + \frac{4}{1 + 8s}\right) + G(s) \times \frac{8s^2 + s + 4}{4}$$ Calculate $3 \left(s + \frac{4}{1 + 8s}\right)$: $$3s + \frac{12}{1 + 8s}$$ 13. **Isolate $G(s)$:** $$1 - 3s - \frac{12}{1 + 8s} = G(s) \times \frac{8s^2 + s + 4}{4}$$ Multiply both sides by 4: $$4 - 12s - \frac{48}{1 + 8s} = G(s)(8s^2 + s + 4)$$ 14. **Express $G(s)$ explicitly:** $$G(s) = \frac{4 - 12s - \frac{48}{1 + 8s}}{8s^2 + s + 4} = \frac{(4 - 12s)(1 + 8s) - 48}{(8s^2 + s + 4)(1 + 8s)}$$ Expand numerator: $$(4 - 12s)(1 + 8s) = 4 + 32s - 12s - 96s^2 = 4 + 20s - 96s^2$$ So numerator: $$4 + 20s - 96s^2 - 48 = -44 + 20s - 96s^2$$ 15. **Final transfer function:** $$G(s) = \frac{-44 + 20s - 96s^2}{(8s^2 + s + 4)(1 + 8s)}$$ 16. **Differential Equation:** Recall $G(s) = \frac{V_o(s)}{V_s(s)}$, so: $$(8s^2 + s + 4)(1 + 8s) V_o(s) = (-44 + 20s - 96s^2) V_s(s)$$ Expand left side: $$(8s^2 + s + 4)(1 + 8s) = 8s^2 + 64s^3 + s + 8s^2 + 4 + 32s = 64s^3 + 16s^2 + 33s + 4$$ Rewrite: $$\left(64s^3 + 16s^2 + 33s + 4\right) V_o(s) = (-96s^2 + 20s - 44) V_s(s)$$ Convert to time domain by replacing $s^n$ with $\frac{d^n}{dt^n}$: $$64 \frac{d^3 v_o}{dt^3} + 16 \frac{d^2 v_o}{dt^2} + 33 \frac{d v_o}{dt} + 4 v_o = -96 \frac{d^2 v_s}{dt^2} + 20 \frac{d v_s}{dt} - 44 v_s$$ This is the differential equation describing the system behavior. **Final answers:** $$G(s) = \frac{-44 + 20s - 96s^2}{(8s^2 + s + 4)(1 + 8s)}$$ $$64 \frac{d^3 v_o}{dt^3} + 16 \frac{d^2 v_o}{dt^2} + 33 \frac{d v_o}{dt} + 4 v_o = -96 \frac{d^2 v_s}{dt^2} + 20 \frac{d v_s}{dt} - 44 v_s$$