Complete Response I – Constant Input

The complete response of a non-homogeneous linear system due to both the external input and the initial condition can be found as the sum of the homogeneous and particular solutions of the non-homogeneous DE with a non-zero right-hand side for the external input:

$$= +$$ (107)

However, one cannot simply add Eqs.（79）and (91).

For the RC circuit, we have

$$\tau\frac{d}{dt} v_c(t)+v_c(t)=v_s(t)=V_s$$ (108)

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

Given an RC circuit shown above where the switch is closed at $t=0$, we want to find the voltage $v_C(t)$ across $C$ and voltage $v_R(t)$ across $R$ as a function of time for $t>0$.

First we consider a constant (DC) input $v(t)=V_s$ applied to the circuit at $t=0$, i.e., a step input

$$v(t)=V_s u(t) =\left\{ \begin{array}{ll}V_s & t\ge 0\\ 0 & t<0\end{array}\right.$$ (109)

The solution of this inhomogeneous DE is composed of two parts,

• Homogeneous solution (natural response) : $v_h(t)=A e^{-t/\tau}$ due to the initial condition $v_C(t)\big\vert _{t=0}=v_C(0)=V_0$.
• Particular solution (forced response) $v_p(t)=v_C(t)\vert _{t\rightarrow \infty}=V_s$ due to the DC input $V_s$. Or by phasor method, we can get

  $$V_c=V_s\frac{Z_c}{R+Z_c}=V_s\frac{1/j\omega C}{R+1/j\omega C} \stackrel{\omega\rightarrow 0}{\Longrightarrow} V_s$$ (110)

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

$$v_C(t)=v_h(t)+v_p(t)=V_s+A e^{-t/\tau}$$ (111)

The constant $A$ can then be determined from the initial condition $v_C(0)=V_0$:

$$v_C(0)=V_0=v_C(t)\big\vert _{t=0}=V_s+Ae^{0}=V_s+A, \;\;\;\;\; \;\;\;\;A=V_0-V_s$$ (112)

and the complete solution is

$$v_C(t)=V_s+A e^{-t/\tau}=V_s+(V_0-V_s) e^{-t/\tau}$$ (113)

The same result can also be obtained using the Laplace transform method. We note that

• When $t=0$, $v_C(t)=V_0$
• When $t\rightarrow\infty$, $v_C(t)\rightarrow V_s$
• When $0< t< \infty$, $v_C(t)$ decays exponentially from $V_0$ to $V_s$.

We can further find the current through $C$:

$$i(t)=C\frac{d}{dt}\,v_C(t)=-\frac{C}{\tau}(V_0-V_s)e^{-t/\tau} =\frac{V_s-V_0}{R}e^{-t/\tau}$$ (114)

and the voltage across $R$:

$$v_R(t)=R\; i(t)=(V_s-V_0) e^{-t/\tau}$$ (115)

We can verify that

$$v_C(t)+v_R(t)=V_s+(V_0-V_s) e^{-t/\tau}+(V_s-V_0) e^{-t/\tau} =V_s$$ (116)

The plots below show $v_C(t)$ (red) and $v_R(t)$ (green) under different initial conditions (purple) and inputs (blue).

• $V_0=2$, $V_s=1$, i.e., $V_0>V_s$

  

• $V_0=1$, $V_s=2$, i.e., $V_0<V_s$

  

Note that in the first case, after the switch is closed at $t=0$,

• $v_R(t)$ takes a negative value if $v_C(0)=V_0>V_s$ for $t>0$, although the voltage source $V_s>0$ is positive (both measured with respective to the bottom wire treated as the ground). 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 $0V$.
• Voltages on both sides of $C$ go through a discontinuous transition to drop (case 1) or jump (case 2) by $V_S-V_0$, however, the voltage $v_C(t)$ remains the same, as the voltage across $C$ does not change instantaneously. $v_R(t)$ drops from $v_R(t)=0$ for $t<0$ to $v_R(t)=V_s-v_c(0)=V_s-V_0$ at $t=0^+$, while $v_C(t)+v_R(t)$ (left side of $C$) drops from $V_0$ to $V_S$.

In general, neither the voltage across a capacitor nor the current through an inductor can be changed instantaneously as it takes time for them to build up:

$$v_C(t)=\frac{1}{C}\int i(t) dt,\;\;\;\;\;\;\;i_C(t)=\frac{1}{L}\int v(t) dtt$$ (117)

Therefore the capacitor can behave like a temporary voltage source, and, similarly, an inductor can behave like a temporary current source.

Example 0 (homework): When an RC circuit with zero initial voltage $v_C(0)=0$ is charged by a DC voltage $V_s=1$. Find energy $W_R$ is consumed by $R$ and energy $W_C$ is stored in $C$.

A Shortcut Method:

Observing the complete solution $v_C(t)=V_s+(V_0-V_s) e^{-t/\tau}$ obtained above, we see that

• When $t=0$, $v_C(t)=v_C(0)=V_0$ is the initial condition
• When $t\rightarrow\infty$, $v_C(t)=v_C(\infty)=V_s$ is the steady state response.

We can therefore generalize the complete solution obtained above to all first-order systems, i.e., their responses to a step input, a constant input that is turned on at moment $t=0$, always take the same form:

$$f(t)=f(\infty)+[f(0)-f(\infty)] e^{-t/\tau}$$ (118)

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

• The steady state response $f(\infty)$: as discussed in previous section for steady state response.
• The initial value $f(0)$: 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 $f(0)$;
• The time constant of the system $\tau$: 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 $\tau=L/R$.

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 and the steady state values of the response decays exponentially. This term is the transient response of the system.

Example 1: $V_{in}=2\,V$, $R_1=R_2=2\,k\Omega$, $C=10^{-6}\,F=1\,\mu F$, $v_C(0)=2$. Find $v_C(t)$.

• Initial value $v_C(0)=2$.

• Find steady state value $v_C(\infty)$:

  $$v_C(\infty)=V_s\frac{R_2}{R_1+R_2}=\frac{V_s}{2} =1\,V$$ (119)

• Find equivalent resistance $R$:

  $$R=R_1 \vert\vert R_2=\frac{R_1 R_2}{R_1+R_2}=1\;k\Omega$$ (120)

• Find time constant

  $$\tau=RC=1000\times 10^{-6}= 10^{-3}\,sec$$ (121)

• Find the complete response

  $$v_C(t)=v_C(\infty)+[v_C(0)-v_C(\infty)]e^{-t/\tau} =1+(2-1)e^{-t/10^{-3}}=1+e^{-1000\,t}$$ (122)

  In particular,

  $$v_C(0)=\left[1+e^{-1000\,t}\right]_{t=0}=2,\;\;\;\;\; v_C(\infty)=\left[1+e^{-1000\,t}\right]_{t=\infty}=1$$ (123)

Example 2:

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

Consider node voltage method. Applying KCL to node $c$ we get

$$\frac{V_0-v_c(t)}{R_1}=C\dot{v}_c(t)+\frac{v_c(t)}{R_2}$$ (124)

i.e.,

$$\frac{R_1R_2}{R_1+R_2}\,C\;\dot{v}_c(t)+v_c(t) =\tau\,\dot{v}_c(t)+v_c(t)=V_0\frac{R_2}{R_1+R_2}$$ (125)

where

$$\tau=C\frac{R_1R_2}{R_1+R_2}=C(R_1\vert\vert R_2) =0.5\times 10^{-6}\times 10^3=5\times 10^{-4}$$ (126)

To find the initial value $v_C(0)$, we assume the circuit is in steady state before $t=0$, i.e., $v_C(0_-)=-5V$. Also, as the voltage across $C$ does not change instantaneously (unless $R=0$ therefore $\tau=RC=0$), we have $v_C(0_+)=v_C(0_-)=-5\,V$. At $t=0_+$, $V_d$ drops from $10\,V$ to $0\,V$, $V_c$ also drops from $5\,V$ to $-5\,V$.

The homogeneous solution is $v_h(t)=A e^{-t/\tau}$ and the particular (steady state) solution is $v_p(t)=V_0/2$. The complete solution is

$$v_c(t)=v_h(t)+v_p(t)=Ae^{-t/\tau}+\frac{V_0}{2}$$ (127)

The coefficient $A$ can be found by equating $v_c(t)$ evaluated at $t=0$ and the initial condition $v_c(0)=-V_0/2$:

$$v_c(t)\bigg\vert _{t=0}=V_s+Ae^{-t/\tau}\bigg\vert _{t=0}=\frac{V_0}{2}+A =v_c(0)=-\frac{V_0}{2}$$ (128)

i.e., $A=-V_0$. Now the solution is

$$v_c(t)=Ae^{-t/\tau}+\frac{V_0}{2} =\frac{V_0}{2}-V_0 e^{-t/\tau}=5-10\,e^{-2000\,t}$$ (129)

Example 3:

Resolve the circuit above using the short-cut method:

• The initial condition is $v_C(0_+)=v_C(0_-)=-5V$, same as that found previously;
• Find steady state value $v_C(\infty)$:

  $$v_C(\infty)=V_{R_2}=V_0\frac{R_2}{R_1+R_2}=\frac{V_0}{2}=5V$$ (130)

• Find equivalent resistance $R$:

  $$R=R_1 \vert\vert R_2=\frac{R_1 R_2}{R_1+R_2}=1\;k\Omega$$ (131)

• Find time constant

  $$\tau=RC=1000\times 0.5\times 10^{-6}=5\times 10^{-4}$$ (132)

• Find the complete response

  $$v_C(t)=v_C(\infty)+[v_C(0)-v_C(\infty)]e^{-t/\tau} =5+(-5-5) e^{-t/0.0005}=5-10 e^{-2000t}\;(V)$$ (133)

  In particular,

  $$v_C(0)=\left[5-10 e^{-2000t}\right]_{t=0}=-5,\;\;\;\;\; v_C(\infty)=\left[5-10 e^{-2000t}\right]_{t=\infty}=5$$ (134)

• Find current $i_C(t)$ through $C$:
  

  $$i_C(t) = C\frac{dv_C(t)}{dt}=C\frac{d}{dt}(5-10 e^{-t/\tau}) =C\,\frac{10 e^{-t/\tau}}{\tau}=10 e^{-2000 t}\;(mA)$$

  $$= \frac{10 e^{-t/\tau}}{R}=10 e^{-2000 t}\;(mA)$$ (135)

  
  
• Find voltages $v_1(t)$ and $v_2(t)$ across $R_1$ and $R_2$, respectively:

  $$v_2(t)=v_C(t)=5-10 e^{-2000t}\;(V)$$ (136)

  $$v_1(t)=V_0-v_C(t)=10-(5-10 e^{-2000t})=5+10 e^{-2000t}$$ (137)

• Find currents $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$, respectively:

  $$i_1(t)=\frac{v_1(t)}{R_1}=2.5+5 e^{-2000t} \; (mA)$$ (138)

  $$i_2(t)=\frac{v_C(t)}{R_2}=2.5-5 e^{-2000t}\; (mA)$$ (139)

• Verify current $i_C(t)$:

  $$i_C(t)=i_1(t)-i_2(t)=10 e^{-2000t} \; (mA)$$ (140)

Example 4:

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 (with respect to node b as ground), and the voltage at node c drops from 5V to -5V (voltage across a capacitor cannot change instantaneously), 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 $v_2(t)$:

  $$v_1(t)=v_1(\infty)+[v_1(0_+)-v_1(\infty)]e^{-t/\tau} =5+(15-5)e^{-t/\tau}=5+10e^{-t/\tau}\;V$$ (141)

  $$v_2(t)=v_2(\infty)+[v_2(0_+)-v_2(\infty)]e^{-t/\tau} =5+(-5-5)e^{-t/\tau}=5-10e^{-t/\tau}\;V$$ (142)

  We see that $v_1(t)+v_2(t)=V_0=10V$ for all $t$.
• Find $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$:

  $$i_1(t)=\frac{v_1(t)}{R_1}=2.5+5e^{-t/\tau}\;mA$$ (143)

  $$i_2(t)=\frac{v_2(t)}{R_2}=2.5-5e^{-t/\tau}\;mA$$ (144)

• Find current $i_C(t)$ through capacitor $C$:

  $$i_C(t)=i_1(t)-i_2(t)=10 e^{-t/\tau}\; mA$$ (145)

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

We see that $V_{cd}(t)+V_{db}(t)=V_{cb}(t)$ and $V_{cb}(t)+V_{ac}(t)=V_{ab}(t)=10\,V$.