10 hours · 10 marks
The classical method finds transient response by writing integro-differential equations from KCL/KVL, converting them to ordinary differential equations (ODEs) with constant coefficients, and solving for natural + forced response. Switching, initial stored energy, and source type (DC / exponential / sinusoidal) determine constants via initial conditions from Chapter 1.
General solution of a linear ODE:
\[ y(t) = y_h(t) + y_p(t) \] where \(y_h\) is the complementary (homogeneous / natural / transient) solution and \(y_p\) is a particular (forced / steady-state) solution. For stable passive RLC networks, \(y_h\to 0\) as \(t\to\infty\).Standard form:
\[ \frac{dy}{dt} + ay = f(t) \quad\text{or}\quad \tau\frac{dy}{dt} + y = K\,f(t). \] Homogeneous solution: \(y_h = A e^{-at} = A e^{-t/\tau}\) with time constant \(\tau=1/a\).For DC forcing \(f(t)=F\) (constant), \(y_p = F/a\) (or final value \(y(\infty)\)). Complete response:
\[ y(t) = y(\infty) + \bigl[y(0^+)-y(\infty)\bigr]e^{-t/\tau},\quad t\ge 0. \]An \(n\)th-order linear DE with constant coefficients:
\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = f(t). \]Guess \(y_p\) matching the form of \(f(t)\), multiply by \(t^s\) if that form already appears in \(y_h\):
| Forcing \(f(t)\) | Trial \(y_p\) |
|---|---|
| Constant \(K\) | \(A\) |
| Polynomial \(a_0+a_1 t+\cdots+a_m t^m\) | \(A_0+A_1 t+\cdots+A_m t^m\) |
| \(Ke^{\alpha t}\) | \(Ae^{\alpha t}\) (or \(At^s e^{\alpha t}\) if \(\alpha\) is a root) |
| \(K\cos\omega t\) or \(K\sin\omega t\) | \(A\cos\omega t+B\sin\omega t\) |
| \(e^{\alpha t}\cos\omega t\) / \(\sin\) | \(e^{\alpha t}(A\cos\omega t+B\sin\omega t)\) |
Substitute into the DE, equate coefficients, solve for unknowns \(A,B,\ldots\).
Series RL with step \(V_s u(t)\), \(i(0^+)=I_0\):
\[ L\frac{di}{dt}+Ri=V_s \Rightarrow i(t)=\frac{V_s}{R}+\Bigl(I_0-\frac{V_s}{R}\Bigr)e^{-(R/L)t}. \]Series RC charging, \(v_C(0^+)=V_0\):
\[ RC\frac{dv_C}{dt}+v_C=V_s \Rightarrow v_C(t)=V_s+\bigl(V_0-V_s\bigr)e^{-t/(RC)}. \]Source \(v_s(t)=V_m e^{\alpha t}\) (\(\alpha\) usually negative for decaying sources in exams). For series RC:
\[ RC\,v_C' + v_C = V_m e^{\alpha t}. \] Homogeneous: \(v_{Ch}=A e^{-t/(RC)}\). Particular (undetermined coefficients): try \(v_{Cp}=B e^{\alpha t}\) if \(\alpha\neq -1/(RC)\): \[ RC(\alpha B)e^{\alpha t}+B e^{\alpha t}=V_m e^{\alpha t} \Rightarrow B=\frac{V_m}{1+\alpha RC}. \] \[ v_C(t)=A e^{-t/(RC)}+\frac{V_m}{1+\alpha RC}\,e^{\alpha t}. \] Apply \(v_C(0^+)\) to find \(A\).\(R=2\,\Omega\), \(C=1\,\text{F}\), \(v_s=10e^{-2t}\,\text{V}\), \(v_C(0^+)=0\). DE:
\[ 2\frac{dv_C}{dt}+v_C=10e^{-2t}. \] \(v_{Ch}=Ae^{-t/2}\). Try \(v_{Cp}=Be^{-2t}\): \[ 2(-2B)e^{-2t}+Be^{-2t}=10e^{-2t} \Rightarrow (-4B+B)=10\Rightarrow B=-10/3. \] \[ v_C(t)=Ae^{-t/2}-\frac{10}{3}e^{-2t}. \] \(v_C(0)=0\Rightarrow A-10/3=0\Rightarrow A=10/3\). \[ v_C(t)=\frac{10}{3}\bigl(e^{-t/2}-e^{-2t}\bigr)\,\text{V}. \]Final answer: \(v_C(t)=\dfrac{10}{3}\bigl(e^{-t/2}-e^{-2t}\bigr)\,\text{V}\).
Source \(v_s=V_m\cos\omega t\) (or \(\sin\)). For series RL:
\[ L i'+Ri = V_m\cos\omega t. \] Particular solution: \(i_p=A\cos\omega t+B\sin\omega t\). Substitute: \[ L(-\omega A\sin\omega t+\omega B\cos\omega t)+R(A\cos\omega t+B\sin\omega t)=V_m\cos\omega t. \] Equate \(\cos\) / \(\sin\) coefficients: \[ RA + L\omega B = V_m,\qquad -L\omega A + RB = 0 \] \[ A=\frac{R V_m}{R^2+(\omega L)^2},\quad B=\frac{\omega L V_m}{R^2+(\omega L)^2}. \] Complete: \(i(t)=Ce^{-(R/L)t}+A\cos\omega t+B\sin\omega t\); fix \(C\) from \(i(0^+)\).Phasor shortcut for \(i_p\): \(\mathbf{I}=V_m\angle 0^\circ / (R+j\omega L)\), then convert to time domain and still add the transient term.
\(R=5\,\Omega\), \(L=2\,\text{H}\), \(V_s=20\,\text{V}\), \(i(0^+)=0\):
\[ i(t)=\frac{20}{5}\bigl(1-e^{-(5/2)t}\bigr)=4\bigl(1-e^{-2.5t}\bigr)\,\text{A}. \]Final answer: \(i(t)=4(1-e^{-2.5t})\,\text{A}\).
KVL: \(L\dfrac{di}{dt}+Ri+\dfrac{1}{C}\int i\,dt = v_s(t)\). Differentiate:
\[ L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{1}{C}i = \frac{dv_s}{dt}. \] Or in terms of \(v_C\): \[ LC\,v_C''+RC\,v_C'+v_C=v_s(t). \] Characteristic equation (natural response): \[ s^2+\frac{R}{L}s+\frac{1}{LC}=0. \] Define \(\alpha = R/(2L)\) (neper frequency), \(\omega_0=1/\sqrt{LC}\) (resonant radian frequency). \[ s_{1,2}=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}. \]| Case | Condition | Natural response form |
|---|---|---|
| Overdamped | \(\alpha>\omega_0\) (\(R>2\sqrt{L/C}\)) | \(A_1 e^{s_1 t}+A_2 e^{s_2 t}\) (two real roots) |
| Critically damped | \(\alpha=\omega_0\) | \((A_1+A_2 t)e^{-\alpha t}\) |
| Underdamped | \(\alpha<\omega_0\) | \(e^{-\alpha t}(A_1\cos\omega_d t+A_2\sin\omega_d t)\), \(\omega_d=\sqrt{\omega_0^2-\alpha^2}\) |
For \(v_s=V_s u(t)\), right-hand side of the \(i\)-equation is \(0\) for \(t>0\), so \(i(t)\) is purely natural with \(i(\infty)=0\). Capacitor voltage:
\[ v_C(t)=V_s + v_{Ch}(t),\quad v_C(\infty)=V_s. \] Constants from \(v_C(0^+)\) and \(i(0^+)=C v_C'(0^+)\).Common data: \(L=1\,\text{H}\), \(C=1\,\text{F}\), \(V_s=10\,\text{V}\), zero IC. Then \(\omega_0=1\). Critical \(R=2\,\Omega\).
(a) Overdamped \(R=3\,\Omega\): \(\alpha=1.5\), \(s_{1,2}=-1.5\pm\sqrt{2.25-1}=-1.5\pm\sqrt{1.25}=-1.5\pm 1.118\).
\[ s_1=-0.382,\quad s_2=-2.618. \] \(v_C=10+A_1 e^{s_1 t}+A_2 e^{s_2 t}\). \(v_C(0)=0\Rightarrow A_1+A_2=-10\). \(i=C v_C'\), \(i(0)=0\Rightarrow s_1 A_1+s_2 A_2=0\). Solving: \(A_1\approx 11.18\), \(A_2\approx -21.18\) (exact via \(\sqrt{5}/2\) form often left symbolic in exams).(b) Critical \(R=2\,\Omega\): \(v_C=10+(A_1+A_2 t)e^{-t}\). \(v_C(0)=0\Rightarrow A_1=-10\). \(v_C'(0)=0\Rightarrow -A_1+A_2=0\Rightarrow A_2=-10\).
\[ v_C(t)=10-10(1+t)e^{-t}\,\text{V}. \](c) Underdamped \(R=1\,\Omega\): \(\alpha=0.5\), \(\omega_d=\sqrt{1-0.25}=\sqrt{0.75}=\sqrt{3}/2\).
\[ v_C=10+e^{-0.5t}(A_1\cos\omega_d t+A_2\sin\omega_d t). \] \(v_C(0)=0\Rightarrow A_1=-10\). \(v_C'(0)=0\Rightarrow -0.5 A_1+\omega_d A_2=0\Rightarrow A_2=-5/\omega_d\).Final answer (critical case boxed): \(v_C(t)=10\bigl[1-(1+t)e^{-t}\bigr]\,\text{V}\).
\(v_s=V_m e^{\alpha t}\). Particular solution \(v_{Cp}=B e^{\alpha t}\) (if \(\alpha\) is not a natural root):
\[ B=\frac{V_m}{LC\alpha^2+RC\alpha+1}. \] Add the homogeneous solution of the appropriate damping type; apply ICs.\(v_s=V_m\cos\omega t\). Particular solution via undetermined coefficients or phasors:
\[ \mathbf{V}_C = \frac{1/(j\omega C)}{R+j\omega L+1/(j\omega C)}\,V_m\angle 0^\circ. \] Convert to \(A\cos\omega t+B\sin\omega t\), then add natural transient matching damping case; constants from \(v_C(0^+)\), \(i(0^+)\).Parallel R, L, C driven by a current source \(i_s(t)\). Node voltage \(v\) satisfies:
\[ C\frac{dv}{dt}+\frac{v}{R}+\frac{1}{L}\int v\,dt = i_s(t). \] Differentiate: \[ C v'' + \frac{1}{R}v' + \frac{1}{L}v = i_s'(t). \] Characteristic: \(s^2 + \dfrac{1}{RC}s + \dfrac{1}{LC}=0\). Now \(\alpha = 1/(2RC)\), \(\omega_0=1/\sqrt{LC}\). Same over / critical / under-damped forms for \(v_h(t)\).Current step \(I_s u(t)\): for \(t>0\), \(i_s'=0\), so \(v(t)\) is natural + constant particular. In DC steady state inductor shorts the node \(\Rightarrow v(\infty)=0\), and \(i_L(\infty)=I_s\). Thus \(v(t)=v_h(t)\) only, with ICs \(v(0^+)=v_C(0^+)\) and \(i_L(0^+)\).
\(R=1\,\Omega\), \(L=1\,\text{H}\), \(C=1\,\text{F}\), \(i_s=2u(t)\,\text{A}\), zero IC. \(\alpha=1/(2RC)=0.5\), \(\omega_0=1\) → underdamped, \(\omega_d=\sqrt{3}/2\).
\[ v(t)=e^{-0.5t}(A_1\cos\omega_d t+A_2\sin\omega_d t). \] \(v(0^+)=0\Rightarrow A_1=0\). Capacitor current \(C v'(0^+)=i_s(0^+)-v(0)/R-i_L(0)=2\), so \(v'(0^+)=2\). \[ v'(0)= -0.5 A_1 + \omega_d A_2 =2 \Rightarrow A_2=2/\omega_d. \] \[ v(t)=\frac{2}{\omega_d}e^{-0.5t}\sin\omega_d t\,\text{V},\quad \omega_d=\sqrt{3}/2. \]Final answer: \(v(t)=\dfrac{4}{\sqrt{3}}\,e^{-t/2}\sin\bigl(\dfrac{\sqrt{3}}{2}t\bigr)\,\text{V}\).
\(i_s=I_m e^{\alpha t}\). Particular \(v_p=Be^{\alpha t}\) with
\[ B=\frac{I_m}{C\alpha + 1/R + 1/(L\alpha)} \] (from substituting into the integro-differential form, or equivalently from the impedance of the parallel tank at “frequency” \(\alpha\)). Add \(v_h\); apply ICs.