Second-Order Circuits

1. Overdamped \(\left(\alpha>\omega_0\right)\), i.e. \(L>4R^{2}C\). The roots are real, distinct and negative:

2. Critically damped \(\left(\alpha=\omega_0\right)\), i.e. \(L=4R^{2}C\). The roots are real and equal:

3. Underdamped \(\left(\alpha<\omega_0\right)\), i.e. \(L<4R^{2}C\). The roots are complex, \(s_{1,2}=-\alpha\pm j\omega_d\) with \(\omega_d=\sqrt{\omega_0^{2}-\alpha^{2}}\):

The constants \(A_1\) and \(A_2\) are found from the initial conditions \(v(0)\) and \(\dfrac{dv(0)}{dt}\).

Series RLC:

Parallel RLC:

General solution procedure: (1) find \(x(0)\) and \(\tfrac{dx(0)}{dt}\) from the \(t=0^{-}\) circuit, using continuity of inductor current and capacitor voltage; (2) find the final value \(x(\infty)\) from the \(t\to\infty\) circuit; (3) compute \(\alpha,\ \omega_0\) and identify the case; (4) write \(x(t)=x(\infty)+\text{transient}\) and solve for the two constants.

For \(t<0\) the switch is open and the circuit splits into two independent sub-circuits. The 4 A source current flows through the inductor, so

Since \(30u(-t)=30\) for \(t<0\), the voltage source is active. The capacitor behaves as an open circuit, so its voltage equals the voltage across the parallel 20 Ω resistor. By voltage division:

For \(t>0\) the switch is closed and the voltage source (now \(30u(-t)=0\)) acts as a short. This leaves a parallel \(RLC\) circuit fed by the 4 A source, with the two 20 Ω resistors in parallel: \(R=20\parallel20=10\ \Omega\). The characteristic roots are:

Since \(\alpha>\omega_0\) the response is overdamped. With final value \(I_s=4\):

Applying \(i(0)=4\) gives \(A_2=-A_1\). Differentiating (1) and using \(L\dfrac{di(0)}{dt}=v(0)=15\):

Hence the inductor current, and then \(i_R=\dfrac{v}{20}=\dfrac{L}{20}\dfrac{di}{dt}\):

Initial conditions from the \(t=0^{-}\) circuit (the 2 A source is active, the 50 V step is off):

For \(t>0\) we have a series \(RLC\) circuit with a step input (\(R=6\ \Omega,\ L=1\ \text{H},\ C=0.04\ \text{F}\)):

The response is underdamped with final capacitor voltage \(V_f=50\):

From \(v(0)=-12=50+A\) we get \(A=-62\). Since \(i(0)=C\dfrac{dv(0)}{dt}=0\):

Convert the current source to a voltage source. The initial conditions are:

For \(t>0\) we have a series \(RLC\) circuit (\(R=5\ \Omega,\ L=1\ \text{H},\ C=4\ \text{F}\)). Note \(\alpha\) is a positive quantity:

The roots and general solution (final value \(V_s=20\)):

Apply \(v(0)=10\) and \(\dfrac{dv(0)}{dt}=\dfrac{i(0)}{C}=\dfrac{2}{4}=0.5\):

At \(t=0^{-}\), mesh analysis on the resistive network gives:

Solving (1) and (2): \(i_1=5,\ i_2=\tfrac{5}{3}\), so the inductor and capacitor initial values are:

For \(t>0\) we have a series \(RLC\) circuit with \(R=6\parallel12=4\ \Omega\):

With final value \(V_s=10\) and \(v(0)=20\): \(A=10\). Using \(i_C(0)=-5=C(-40+B)\) with \(C=\tfrac18\) gives \(B=0\). Then \(i_C(t)=-\tfrac12(10)e^{-4t}\), and the inductor current is \(i=-i_C\):

At \(t=0^{-}\) (switch at \(a\)):

For \(t>0\) (switch at \(b\)) we have a series \(RLC\) circuit:

With final value \(V_s=24-12=12\) and \(v(0)=0\): \(A=-12\). From \(i(0)=3=C(-5A+B)=0.02(60+B)\) we get \(B=90\):

At \(t=0^{-}\):

For \(t>0\) the source is short-circuited, leaving a source-free parallel \(RLC\) circuit (\(R=1,\ L=1,\ C=0.25\)):

Since \(\alpha=\omega_0\) the response is critically damped, \(s_{1,2}=-2\):

Using \(v=L\dfrac{di}{dt}\) at \(t=0\): \(v(0)=2=B+4\Rightarrow B=-2\). Hence:

For \(t=0^{-}\), \(u(t)=0\), so \(v(0)=0\) and \(i(0)=0\).

For \(t>0\) we have a parallel \(RLC\) circuit with a step input:

Since \(\alpha<\omega_0\) the response is underdamped, \(s_{1,2}\cong-50\pm j5000\), with final value \(I_s=6\ \text{mA}\):

From \(i(0)=0=6+A\Rightarrow A=-6\ \text{mA}\). From \(v(0)=L\dfrac{di(0)}{dt}=0\): \(5000B-50A=0\Rightarrow B=-0.06\ \text{mA}\). Hence:

For \(t=0^{-}\): \(i(0)=3+\dfrac{12}{4}=6\) and \(v(0)=0\).

For \(t>0\) we have a parallel \(RLC\) circuit with a step input (\(R=5,\ L=5,\ C=0.05\)):

Since \(\alpha=\omega_0\) the response is critically damped, \(s_{1,2}=-2\), with final value \(I_s=3\):

From \(\dfrac{v(0)}{L}=\dfrac{di(0)}{dt}=B-2(3)=0\Rightarrow B=6\):

For \(t<0\): \(i(0)=0,\ v(0)=0\). Writing KVL for the two mesh currents (\(i_0\) in the larger loop, \(i\) in the smaller loop):

Differentiating (2): \(4\tfrac{di}{dt}+25(i+i_0)=0\Rightarrow i_0=-0.16\tfrac{di}{dt}-i\)  (3), and \(\tfrac{di_0}{dt}=0.16\tfrac{d^2i}{dt^2}-\tfrac{di}{dt}\)  (4). Substituting into (1):

The characteristic equation \(s^2+30.25s+250=0\) gives underdamped roots:

So \(i(t)=I_s+e^{-15.125t}\left(A_1\cos4.608t+A_2\sin4.608t\right)\). As \(t\to\infty\), \(i_0(\infty)=\tfrac{20}{10}=2=-i(\infty)\), so \(I_s=i(\infty)=-2\) and \(A_1=-I_s=2\). From \(\tfrac{di(0)}{dt}=0\):

Let \(v\) = capacitor voltage and \(i\) = inductor current. At \(t=0^{-}\) the circuit is in steady state:

By continuity, at \(t=0^{+}\): \(v(0^{+})=16\) and \(i(0^{+})=2\). For \(t>0\) the source is disconnected, leaving a source-free series \(RLC\) circuit with \(R=8+12=20\ \Omega,\ L=1\ \text{H},\ C=\tfrac{1}{36}\ \text{F}\):

Since \(\alpha>\omega_0\) the response is overdamped, \(s_{1,2}=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}=-2,\,-18\), so:

For \(\tfrac{di(0)}{dt}\), KVL around the loop at \(t=0^{+}\) gives \(-v(0)+20i(0)+v_L(0)=0\Rightarrow v_L(0)=16-40=-24\), so \(\tfrac{di(0)}{dt}=\tfrac{v_L(0)}{L}=-24\):

Solving (1) and (2): \(B=1.25,\ A=0.75\). Since the labelled \(i_x=-i\) and \(v_R=8i\):

Compute the damping parameters for the series circuit:

The real, distinct roots and general solution:

Apply the initial conditions. From \(i(0)=0\): \(A_1+A_2=0\). The inductor slope follows from \(L\tfrac{di(0)}{dt}=v_L(0)=-Ri(0)-v_C(0)=-12\), so \(\tfrac{di(0)}{dt}=-12\):

Compute the damping parameters for the parallel circuit:

The damped frequency and general solution:

Apply the initial conditions. From \(v(0)=10\): \(B_1=10\). KCL gives \(C\tfrac{dv(0)}{dt}=-\tfrac{v(0)}{R}-i_L(0)\), so \(\tfrac{dv(0)}{dt}=\tfrac{1}{0.125}\left(-\tfrac{10}{2}-0\right)=-40\):