7 hours · 7 marks
The Laplace transform converts linear constant-coefficient ODEs in time \(t\) into algebraic equations in complex frequency \(s\). Initial conditions enter automatically through differentiation theorems, so switched RLC problems become circuit algebra with impedances \(sL\), \(1/(sC)\) plus IC sources.
| \(f(t)\) for \(t\ge 0\) | \(F(s)\) |
|---|---|
| \(u(t)\) (unit step) | \(1/s\) |
| \(\delta(t)\) | \(1\) |
| \(t\) | \(1/s^2\) |
| \(e^{-at}\) | \(1/(s+a)\) |
| \(t e^{-at}\) | \(1/(s+a)^2\) |
| \(\sin\omega t\) | \(\omega/(s^2+\omega^2)\) |
| \(\cos\omega t\) | \(s/(s^2+\omega^2)\) |
| \(e^{-at}\sin\omega t\) | \(\omega/((s+a)^2+\omega^2)\) |
| \(e^{-at}\cos\omega t\) | \((s+a)/((s+a)^2+\omega^2)\) |
Procedure: (1) find \(i_L(0^-)\), \(v_C(0^-)\); (2) replace circuit by s-domain impedances + IC sources; (3) solve for \(I(s)\) or \(V(s)\) by nodal/mesh/Ohm; (4) partial-fraction expand; (5) inverse Laplace to \(t\)-domain for \(t>0\).
Series RL, \(v_s=V_s u(t)\), \(i(0^-)=I_0\):
\[ \bigl(R+sL\bigr)I(s)= \frac{V_s}{s}+L I_0 \Rightarrow I(s)=\frac{V_s/L}{s(s+R/L)}+\frac{I_0}{s+R/L}. \] Inverse (or recognize standard form): \[ i(t)=\frac{V_s}{R}\bigl(1-e^{-(R/L)t}\bigr)+I_0 e^{-(R/L)t},\quad t>0. \]\(R=4\,\Omega\), \(L=2\,\text{H}\), \(V_s=12\,\text{V}\), \(i(0^-)=0\).
\[ I(s)=\frac{12/s}{4+2s}=\frac{12}{s(2s+4)}=\frac{6}{s(s+2)} =\frac{3}{s}-\frac{3}{s+2}. \] \[ i(t)=3\bigl(1-e^{-2t}\bigr)\,\text{A}. \]Final answer: \(i(t)=3(1-e^{-2t})\,\text{A}\).
\(v_s=V_m e^{-\alpha t}u(t)\Rightarrow V_s(s)=V_m/(s+\alpha)\). Series RC with zero IC:
\[ I(s)=\frac{V_m/(s+\alpha)}{R+1/(sC)}=\frac{V_m\,sC}{(s+\alpha)(RCs+1)}, \] \[ V_C(s)=\frac{1}{sC}I(s)=\frac{V_m}{(s+\alpha)(s+1/(RC))}. \] Partial fractions → \(v_C(t)\) same as classical undetermined-coefficient result.\(R=2\,\Omega\), \(C=1\,\text{F}\), \(v_s=10e^{-2t}\), \(v_C(0^-)=0\). Voltage divider in the \(s\)-domain:
\[ V_C(s)=\frac{1/(sC)}{R+1/(sC)}\cdot\frac{10}{s+2} =\frac{1}{RCs+1}\cdot\frac{10}{s+2} =\frac{10}{(2s+1)(s+2)} =\frac{5}{(s+1/2)(s+2)}. \] Partial fractions \(V_C(s)=\dfrac{A}{s+1/2}+\dfrac{B}{s+2}\): \[ A=\frac{5}{-1/2+2}=\frac{5}{3/2}=\frac{10}{3},\qquad B=\frac{5}{-2+1/2}=\frac{5}{-3/2}=-\frac{10}{3}. \] \[ v_C(t)=\frac{10}{3}\bigl(e^{-t/2}-e^{-2t}\bigr)\,\text{V}. \]This matches the classical undetermined-coefficient result in Chapter 2.
Final answer: \(v_C(t)=\dfrac{10}{3}\bigl(e^{-t/2}-e^{-2t}\bigr)\,\text{V}\).
\(v_s=V_m\cos\omega t\,u(t)\Rightarrow V_s(s)=V_m s/(s^2+\omega^2)\). For series RL, zero IC:
\[ I(s)=\frac{V_m s}{(s^2+\omega^2)(R+sL)}. \] Expand into terms \(As+B\) over \(s^2+\omega^2\) and \(C/(s+R/L)\); invert to transient exponential + steady AC.\(R=5\,\Omega\), \(C=0.2\,\text{F}\), \(V_s=10\,\text{V}\), \(v_C(0^-)=0\). \(\tau=RC=1\,\text{s}\).
\[ V_C(s)=\frac{10/s}{5+1/(0.2s)}\cdot\frac{1}{0.2s} =\frac{10}{s(1+s)}=\frac{10}{s(s+1)} =\frac{10}{s}-\frac{10}{s+1}. \] (Using voltage divider \(V_C=\frac{1/(sC)}{R+1/(sC)}\cdot V_s/s=\frac{1}{RCs+1}\cdot 10/s\).) \[ v_C(t)=10\bigl(1-e^{-t}\bigr)\,\text{V}. \]Final answer: \(v_C(t)=10(1-e^{-t})\,\text{V}\).
Series impedance \(Z(s)=R+sL+1/(sC)\). With source \(V_s(s)\) and ICs:
\[ I(s)=\frac{V_s(s)+L i(0^-)-v_C(0^-)/s}{R+sL+1/(sC)}. \] \[ V_C(s)=\frac{v_C(0^-)}{s}+\frac{1}{sC}I(s). \]\(V_s(s)=V_0/s\), often zero IC \(\Rightarrow\)
\[ I(s)=\frac{V_0/s}{R+sL+1/(sC)}=\frac{V_0}{L}\cdot\frac{1}{s^2+(R/L)s+1/(LC)}. \] Complete the square / factor depending on damping; invert using tables.\(R=2\,\Omega\), \(L=1\,\text{H}\), \(C=1/5\,\text{F}\), \(V_0=10\,\text{V}\), zero IC. \(\alpha=R/(2L)=1\), \(\omega_0=\sqrt{5}\), \(\omega_d=2\).
\[ I(s)=\frac{10}{s^2+2s+5}=\frac{10}{(s+1)^2+4}. \] \[ i(t)=10\cdot\frac{1}{2}e^{-t}\sin 2t=5e^{-t}\sin 2t\,\text{A}. \] \[ V_C(s)=\frac{1}{sC}I(s)=\frac{5}{s}\cdot\frac{10}{(s+1)^2+4}=\frac{50}{s\bigl((s+1)^2+4\bigr)}. \] Partial fractions → \(v_C(t)=10+e^{-t}(A\cos 2t+B\sin 2t)\); with \(v_C(0)=0\), \(i(0)=0\): \(A=-10\), and from \(i=C v_C'\) get \(B\). Standard result: \[ v_C(t)=10-e^{-t}(10\cos 2t+5\sin 2t)\,\text{V}. \]Final answer: \(i(t)=5e^{-t}\sin 2t\,\text{A}\); \(v_C(t)=10-e^{-t}(10\cos 2t+5\sin 2t)\,\text{V}\).
\(V_s(s)=V_m/(s+\alpha)\). Then
\[ I(s)=\frac{V_m/(s+\alpha)}{R+sL+1/(sC)} \] (zero IC). Factor denominator; expand including the \((s+\alpha)\) pole unless cancelled.\(V_s(s)=V_m s/(s^2+\omega^2)\) or \(\omega V_m/(s^2+\omega^2)\). \(I(s)\) is a rational function; invert to damped natural terms + undamped steady sinusoid (forced response).
\(R=3\,\Omega\), \(L=1\,\text{H}\), \(C=1/2\,\text{F}\), source shorted for \(t>0\) (source-free), \(i(0^-)=2\,\text{A}\), \(v_C(0^-)=4\,\text{V}\).
\[ I(s)=\frac{L i(0^-)-v_C(0^-)/s}{R+sL+1/(sC)} =\frac{2-4/s}{3+s+2/s} =\frac{(2s-4)/s}{(s^2+3s+2)/s} =\frac{2(s-2)}{(s+1)(s+2)}. \] Partial fractions \(I(s)=\dfrac{A}{s+1}+\dfrac{B}{s+2}\): \[ A=\frac{2(-1-2)}{-1+2}=-6,\qquad B=\frac{2(-2-2)}{-2+1}=8. \] \[ i(t)=\bigl(-6e^{-t}+8e^{-2t}\bigr)\,\text{A}. \] Check: \(i(0)=-6+8=2\,\text{A}\) matches \(i(0^-)\).Final answer: \(i(t)=-6e^{-t}+8e^{-2t}\,\text{A}\).
Parallel admittance \(Y(s)=1/R + 1/(sL) + sC\). With current source \(I_s(s)\) and ICs (using parallel IC current sources \(C v(0^-)\) out of the positive node and \(i_L(0^-)/s\) into/out per sign convention):
\[ V(s)=\frac{I_s(s)+C v(0^-)-i_L(0^-)/s}{sC+1/R+1/(sL)}. \]\(i_s=I_0 u(t)\Rightarrow I_s=I_0/s\). For zero IC:
\[ V(s)=\frac{I_0/s}{sC+G+1/(sL)}=\frac{I_0}{C}\cdot\frac{1}{s^2+(G/C)s+1/(LC)},\quad G=1/R. \] Invert according to damping (\(\alpha=G/(2C)=1/(2RC)\)).\(R=1\,\Omega\), \(L=1\,\text{H}\), \(C=1\,\text{F}\), \(I_0=2\,\text{A}\), zero IC. \(\alpha=0.5\), \(\omega_d=\sqrt{3}/2\).
\[ V(s)=\frac{2}{s^2+s+1}=\frac{2}{(s+1/2)^2+3/4}. \] \[ v(t)=2\cdot\frac{1}{\omega_d}e^{-t/2}\sin\omega_d t =\frac{4}{\sqrt{3}}e^{-t/2}\sin\Bigl(\frac{\sqrt{3}}{2}t\Bigr)\,\text{V}. \]Final answer: \(v(t)=\dfrac{4}{\sqrt{3}}\,e^{-t/2}\sin\bigl(\dfrac{\sqrt{3}}{2}t\bigr)\,\text{V}\) (matches Ch2 classical Example 4).
\(i_s=I_m e^{-\alpha t}\Rightarrow I_s(s)=I_m/(s+\alpha)\).
\[ V(s)=\frac{I_m/(s+\alpha)}{sC+G+1/(sL)}. \] Clear the denominator; partial-fraction including pole at \(-\alpha\); invert.Same \(R=L=C=1\), \(i_s=3e^{-2t}u(t)\), zero IC.
\[ V(s)=\frac{3/(s+2)}{s+1+1/s} =\frac{3s}{(s+2)(s^2+s+1)}. \] Expand \(V(s)=\dfrac{A}{s+2}+\dfrac{Bs+C}{s^2+s+1}\). Cover-up at \(s=-2\): \[ A=\frac{3(-2)}{(-2)^2+(-2)+1}=\frac{-6}{3}=-2. \] Clear denominators: \(A(s^2+s+1)+(Bs+C)(s+2)=3s\). With \(A=-2\): \[ -2s^2-2s-2 + Bs(s+2)+C(s+2)=3s \Rightarrow (-2+B)s^2+( -2+2B+C)s+(-2+2C)=3s. \] Equate: \(B=2\), \(-2+2C=0\Rightarrow C=1\), and check linear: \(-2+4+1=3\) ✓. So \[ V(s)=\frac{-2}{s+2}+\frac{2s+1}{(s+1/2)^2+(\sqrt{3}/2)^2}. \] Rewrite \(2s+1=2(s+1/2)+0\): \[ v(t)=-2e^{-2t}+2e^{-t/2}\cos\Bigl(\frac{\sqrt{3}}{2}t\Bigr)\,\text{V}. \]Final answer: \(v(t)=-2e^{-2t}+2e^{-t/2}\cos\bigl(\dfrac{\sqrt{3}}{2}t\bigr)\,\text{V}\).
| Aspect | Classical | Laplace |
|---|---|---|
| Setup | ODE + find \(y_h\), \(y_p\) | Algebraic \(s\)-domain circuit |
| Initial conditions | Applied at the end | Built into L/C models |
| Best for | Simple 1st/2nd order insight | Switched networks, nonzero IC, simultaneous equations |
| Inversion | Not needed | Partial fractions + transform table |