AC Circuit Analysis II (Complete Response)

If we need to find out the transient response of an AC circuit to a certain input applied at time moment $t=0$ (e.g., after a switch is closed), then the phasor method discussed above is no longer sufficient. Now the DE describing the circuit will have to be solved to obtain the complete solution including the homogeneous (transient) solution as well as the particular (steady state) solution.

As a simple example, the RC and RL circuits shown below are composed of a resistor $R$ and capacitor $C$ or an inductor $L$ in series with an external voltage input $v(t)$, which is turned on at $t=0$, wither by a switch or a step voltage $v(t)$, such as a square wave input. Any of the variables $v_R(t)$, $v_C(t)$, $v_L(t)$ and $i(t)$ can be considered as the circuit's response to this input.

• The RC circuit can be described by KVL:
  
  $$v_R(t)+v_C(t)=R i(t)+v_C(t)=RC\frac{d}{dt} v_c(t)+v_c(t)=\tau \frac{d}{dt} v_C(t)+v_C(t)=v(t)$$
  
  
  where $\tau=RC$ is the time constant of the system with the dimension of time:
  
  $$[RC]=\frac{[V]}{[I]}\frac{[Q]}{[V]}=[T]$$
  
  

• Similarly, for the RL circuit we have:
  
  $$v_R(t)+v_L(t)=R i(t)+L\frac{d}{dt} i(t)=v(t)\;\;\;\;\mbox{or}\;\;\;\;\; \tau \frac{d}{dt} i(t)+i(t)=\frac{v(t)}{R}$$
  
  
  where $\tau=L/R$ is the time constant with the dimension of time: $L$ is $[L]=[V][T]/[I]=[R][T]$ and the dimension of $\tau$ is:
  
  $$[L/R]=\frac{[V][T]}{[I]}\frac{1}{[R]}=\frac{[R][T]}{[R]}=[T]$$
  
  

Homogeneous Solution

When the external input is zero $v_C(t)=0$, the DE is homogeneous (zero on right hand side):

$$RC\frac{d}{dt} v_c(t)+v_c(t)=\tau \frac{d}{dt} v_C(t)+v_C(t)=0$$

To find the transient solution $v_C(t)$ due to the non-zero initial condition $v_C(0)=V_0$, we need to solve this DE by substituting the general solution $v_C=A e^{st}$ and its derivative $dv_C(t)/dt=s A e^{st}$ into the DE, and get

$$\tau s+1=0,\;\;\;\;\;s=-\frac{1}{\tau}$$

i.e., $v_C(t)=Ae^{-t/\tau}$ for ($t\ge 0$). Assume the initial voltage on $C$ is $v_C(0)=V_0$ (initial condition), we can find the unknown constant $A$:

$$v_C(0)=A e^{-t/\tau}\vert _{t=0}=A e^{0}=V_0,\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;A=V_0$$

and the solution is

$$v_C(t)=V_0 e^{-t/\tau}$$

The current through $R$ and $C$ is

$$i(t)=C\frac{d\;v_C(t)}{dt}=C\frac{d}{dt}[V_0 e^{-t/\tau}] =-V_0 \frac{C}{\tau} e^{-t/\tau}=-\frac{V_0}{R} e^{-t/\tau}$$

The voltage across $R$ is

$$v_R(t)=i(t)R=-V_0 e^{-t/\tau}=-v_C(t)$$

This result can be verified: $v_C(t)+v_R(t)=0$.

Complete Solution 1: Constant Input

When a non-zero external input $v(t)\ne 0$ is applied to the circuit, the DE is no longer homogeneous:

$$v_R(t)+v_C(t)=\tau\frac{d}{dt} v_c(t)+v_c(t)=v(t)$$

We want to solve the DE for $v_C(t)$ as the circuit's complete response to the input as well as the initial condition $v_C(0)=V_0$, after a switch is closed at time moment $t=0$.

First we consider a constant (DC) input $v(t)=V_s$ applied to the circuit at $t=0$, i.e., a step input $v(t)=V_s u(t)$. The solution of this inhomogeneous DE is composed of two parts,

• particular solution (forced response) $v'_C(t)=v_C(t)\vert _{t\rightarrow \infty}=V_s$ due to the DC input $V_s$;
• homogeneous solution (natural response) : $v''_C(t)=A e^{-t/\tau}$ due to the initial condition $v_C(0)=V_0$.

Therefore the overall solution is:

$$v_C(t)=v'_C(t)+v''_C(t)=V_s+A e^{-t/\tau}$$

From the initial condition $v_C(t)\vert _{t=0}=v_C(0)=V_0$, we have

$$v_C(0)=V_0=V_s+A,\;\;\;\;\;\mbox{i.e.,}\;\;\;\;A=V_0-V_s$$

and the solution becomes

$$v_C(t)=V_s+(V_0-V_s) e^{-t/\tau}$$

In particular, for zero initial condition $v_C(0)=0$, the solution is

$$v_C(t)=V_s+(-V_s) e^{-t/\tau}=V_s(1-e^{-t/\tau})$$

The voltage across $R$ is

$$v_R(t)=V_s-v_C(t)=V_s-[V_s+(V_0-V_s) e^{-t/\tau}]=(V_s-V_0) e^{-t/\tau}$$

and the current through the circuit is

$$i(t)=\frac{V_R(t)}{R}=\frac{V_s-V_0}{R} e^{-t/\tau}$$

The two plots below show $v_C(t)$ (red) and $v_R(t)$ (green) under different initial conditions $V_0$ (purple, 2, 1) and inputs $V_s$ (blue, 1, 2). Note that when $t<0$ $v_R(t)=0$ and when $t\ge 0$ $v_R(t)<V_s-v_c(0)=V_s-V_0$, which can be a negative value if $v_C(0)=V_0>V_s$. ($v_R(t)$ measured with respective to the wire at the bottom treated as the ground), while the voltage source $V_s>0$ is positive. This is because right after the switch is closed, the voltage on the left side of $C$ drops from $V_0=2V$ to $V_s=1V$, causing the voltage on its right side to also drop from $0$ to $-1V$, lower than the ground level of $0V$.

A Shortcut Method:

All first-order systems can be described by the canonical form of a first-order DE which can be solved once for all. The responses of all first order systems take the same form:

$$f(t)=f(\infty)+[f(0)-f(\infty)] e^{-t/\tau}$$

in terms of three essential components of the system's response:

1. $f(0)$ is the initial value;
2. $f(\infty)$ is the steady-state response;
3. $\tau$ is the time constant of the system.

In particular, note that

• when $t=0$, $f(t)=f(\infty)+f(0)-f(\infty)=f(0)$ the initial condition;
• when $t\rightarrow \infty$, $f(t)=f(\infty)$ the steady state response;
• when $0 < t < \infty$, the difference $f(0)-f(\infty)$ between the initial value and the steady-state value of the response decays exponentially. This term is the transient response of the system.

Here is how to find the three components specifically:

• Denote the value of $f(t)$ immediately before and after the moment $t=0$ by $f(0_-)$ and $f(0_+)$, respectively. If $f(0_-)\ne f(0_+)$, then use $f(0_+)$ for $f(0)$;
• When there is only one resistor in the circuit, the time constant is $\tau=RC$ or $\tau=L/R$. When there are multiple resistors, the time constant can be found by:
  • Remove $C$ or $L$ so that the rest of the circuit ($t>0$) is a one port network.
  • Find the equivalent resistance $R$ of the network by turning off all energy sources (short-circuit for voltage source, open-circuit for current source).
  • Find time constant $\tau=RC$ or $\tau=L/R$.

Example 1:

Find $v_C(t)$ after the switch is closed at $t=0$. Assume $V_0=10V$, $R_1=R_2=2K\Omega$, $R_3=5K\Omega$, $C=0.5\mu F$, and the circuit was in steady-state at $t=0$.

• Find initial value $v_C(0)$. The circuit has been in steady state before $t=0$, and voltage across a capacitor cannot not change discontinuously, therefore we get $v_C(0_+)=v_C(0_-)$:
  
  $$v_C(0)=v_C(0_+)=-V_{R_1}=-V_0\frac{R_1}{R_1+R_2}=-V_0/2=-5V$$
  
  

• Find steady-state value $v_C(\infty)$:
  
  $$v_C(\infty)=V_{R_2}=V_0\frac{R_2}{R_1+R_2}=V_0/2=5V$$
  
  

• Find equivalent resistance $R$:
  
  $$R=R_1 \vert\vert R_2=\frac{R_1 R_2}{R_1+R_2}=1\;k\Omega$$
  
  

• Find time constant
  
  $$\tau=RC=1000\times 0.5\times 10^{-6}=5\times 10^{-4}$$
  
  

• Find the complete response
  
  $$v_C(t)=v_C(\infty)+[v_C(0)-v_C(\infty)]e^{-t/\tau} =5+(-5-5) e^{-t/0.0005}=5-10 e^{-2000t}\;(V)$$
  
  

• Find current $i_C(t)$ through $C$:
  
  $$i_C(t)=C\frac{dv_C(t)}{dt}=C\frac{d}{dt}(5-10 e^{-t/\tau}) =\frac{10 e^{-t/\tau}}{R}=10 e^{-2000 t}\;(mA)$$
  
  

• Find voltages $v_1(t)$ and $v_2(t)$ across $R_1$ and $R_2$, respectively:
  
  $$v_2(t)=v_C(t)=5-10 e^{-2000t}\;(V)$$
  
  
  
  
  $$v_1(t)=V_0-v_C(t)=10-(5-10 e^{-2000t})=5+10 e^{-2000t}$$
  
  

• Find currents $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$, respectively:
  
  $$i_1(t)=\frac{v_1(t)}{R_1}=2.5+5 e^{-2000t} \; (mA)$$
  
  
  
  
  $$i_2(t)=\frac{v_C(t)}{R_2}=2.5-5 e^{-2000t}\; (mA)$$
  
  

• Verify current $i_C(t)$:
  
  $$i_C(t)=i_1(t)-i_2(t)=10 e^{-2000t} \; (mA)$$
  
  

Example 1a:

In the same circuit above, find the voltages $v_1(t)$ and $v_2(t)$ across and currents $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$, respectively.

• Find $v_1(\infty)=v_2(\infty)=5\;V$
• Find $v_1(0_+)$ and $v_2(0_+)$. Before $t=0$, the circuit is in steady state, i.e., $v_1(0_-)=v_2(0_-)=5\;V$. However, after the switch closes at $t=0$, the voltage at node d drops from 10V to 0V, and the voltage at node c drops from 5V to -5V with respect to node b as ground (voltage across a capacitor cannot change suddenly), i.e., $v_1(0_+)=15V$ and $v_2(0_+)=-5V$;
• Find $\tau=RC=5\times 10^{-4}$ (same as before);
• Find $v_1(t)$ and $v_2(t)$:
  
  $$v_1(t)=v_1(\infty)+[v_1(0_+)-v_1(\infty)]e^{-t/\tau} =5+(15-5)e^{-t/\tau}=5+10e^{-t/\tau}\;V$$
  
  
  
  
  $$v_2(t)=v_2(\infty)+[v_2(0_+)-v_2(\infty)]e^{-t/\tau} =5+(-5-5)e^{-t/\tau}=5-10e^{-t/\tau}\;V$$
  
  
  We see that $v_1(t)+v_2(t)=V_0=10V$.
• Find $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$:
  
  $$i_1(t)=\frac{v_1(t)}{R_1}=2.5+5e^{-t/\tau}\;mA$$
  
  
  
  
  $$i_2(t)=\frac{v_2(t)}{R_2}=2.5-5e^{-t/\tau}\;mA$$
  
  

• Find current $i_C(t)$ through capacitor $C$:
  
  $$i_C(t)=i_1(t)-i_2(t)=10 e^{-t/\tau}\; mA$$
  
  

Note that when $t=0_+$, $v_2(t)=V_c=-5V$, lower than ground voltage $V_b$!

Complete Solution 2: Sinusoidal input

If the input voltage is sinusoidal $v(t)=V_s cos(\omega t+\psi)$, the DE becomes

$$v_R(t)+v_C(t)=\tau\frac{d}{dt} v_c(t)+v_c(t)=V_s cos(\omega t+\psi)$$

The homogeneous solution is the same as before $v''_C(t)=A e^{-t/\tau}$. The particular solution $v'_C(t)$ as the steady-state response to $v(t)$ can be found by the phasor method. The input is

$$v(t)=V_s cos(\omega t)=Re[\dot{V}\sqrt{2}e^{j\omega t}]$$

where $\dot{V}=V_s\; e^{j\psi}/\sqrt{2}$. The steady-state voltage across $C$ is (voltage divider)

$$\dot{V}_C=\dot{V} \frac{Z_C}{R+Z_C} =\dot{V} \frac{1/j\omega... ...tau+1} =\frac{\dot{V}\;e^{-j\phi}}{\sqrt{(\omega \tau)^2+1}}$$

where $\phi=\tan^{-1} \omega \tau$. The steady state voltage $v'_C(t)$ in time domain is:

$$v'_C(t)=Re[\frac{V_s e^{j(\omega t+\psi)}\;e^{-j\phi}}{\sqrt... ...= \frac{V_s}{\sqrt{(\omega \tau)^2+1}} cos(\omega t+\psi-\phi)$$

The complete solution is then the sum of both homogeneous and particular solutions:

$$v_C(t)=v'_C(t)+v''_C(t)= \frac{V_s}{\sqrt{(\omega \tau)^2+1}}cos(\omega t+\psi-\phi)+A e^{-t/\tau}$$

From the initial condition $v_C(0)=V_0$, we have

$$v_C(0)=V_0=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}cos(\psi-\phi)+A$$

Solving for $A$ we get

$$A=V_0-\frac{V_s}{\sqrt{(\omega \tau)^2+1}}cos(\psi-\phi)$$

Substituting $A$ back to $v_C(t)$, we get

$$v_C(t)=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}cos(\omega t+\psi-... ...frac{V_s}{\sqrt{(\omega \tau)^2+1}}cos(\psi-\phi)] e^{-t/\tau}$$

This result can also be generalized for any non-constant input:

$$f(t)=f_{\infty}(t)+[f(0)-f_{\infty}(0)]e^{-t/\tau}$$

where $f_{\infty}(t)$ is the steady state response of the system.

If the initial voltage on $C$ is zero $V_0=0$, then

$$v_C(t)=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}[cos(\omega t+\psi-\phi) -cos(\psi-\phi) e^{-t/\tau}]$$

Note that the initial magnitude of the transient component varies depending on the angle $\psi-\phi$. When this angle is either $\pm 90^\circ$, $cos(\psi-\phi)=0$, i.e., the transient component disappears altogether. However, when $\psi-\phi$ is either $0^\circ$ or $180^\circ$, $cos(\psi-\phi)=\pm 1$, i.e., the magnitude of the transient component reaches maximum (positive or negative), and if $\tau$ value is large and therefore the transient component decays slowly, the magnitude of the initial voltage $v_C(t)$ could be close to twice as much as the peak of steady-state. The three cases for $\psi-\phi$ to be $0^\circ$, $90^\circ$, and $180^\circ$ are shown below:

Example 2:

An electromagnet, modeled by a resistor $R=20\Omega$ and $L=0.3H$, is powered by sinusoidal voltage of $120V$ and $60Hz$. Find the current through the circuit when the switch is closed at $t=0$ when the phase angle happens to be $\psi=10^\circ$, i.e., $v(t)=120\sqrt{2}\; \cos(6.28\times 60 t+10^\circ)$.

• Find initial value: $i(0)=0$.
• Find impedance of circuit:
  
  $$Z=R+j\omega L=20+j6.28\times 60 \times 0.3=20+j113=114.8\angle{80^\circ}$$
  
  

• Find steady-state value $i_\infty(t)$ by phasor method:
  
  $$\dot{V}=120 \angle{10^\circ},\;\;\;\;\dot{I}=\frac{\dot{V}}{... ...gle{10^\circ}}{114.8\angle{80^\circ}} =1.05\angle{-70^\circ}$$
  
  
  
  
  $$i_\infty(t)=1.05\sqrt{2}\;\cos(6.28\times 60 t -70^\circ), \... ...=1.05\sqrt{2}\;\cos(-70^\circ) =1.05\sqrt{2}\times 0.342=0.51$$
  
  

• Find time constant $\tau=L/R=0.3/20=0.015$.
• Find current
  
  $$i(t)=i_{\infty}(t)+[i(0)-i_{\infty}(0)]e^{-t/\tau} =1.05\sqrt{2}\;\cos(6.28\times 60 t -70^\circ)-0.51 e^{-t/0.015}$$