### 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:

 Complete SolutionHomogeneous SolutionParticular Solution (106)

Specifically in the RC circuit, we have

 (107)

We want to solve the DE for as the circuit's complete response to the input as well as the initial condition , after a switch is closed at time moment .

Given an RC circuit shown above where the switch is closed at , we want to find the voltage across and voltage across as a function of time for .

First we consider a constant (DC) input applied to the circuit at , i.e., a step input

 (108)

The solution of this inhomogeneous DE is composed of two parts,
• Homogeneous solution (natural response) : due to the initial condition .
• Particular solution (forced response) due to the DC input . Or by phasor method, we can get

 (109)

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

 (110)

The constant can be determined from the initial condition :

 i.e., (111)

and the complete solution is

 (112)

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

In particular, if , the solution is

 (113)

The current through is

 (114)

The voltage across is

 (115)

We can further verify that

 (116)

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

• , , i.e.,

• , , i.e.,

Note that in the first case, after the switch is closed at ,

• takes a negative value if for , although the voltage source 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 drops from to , causing the voltage on its right side to also drop from 0 to , lower than the ground level of .
• Voltages on both sides of go through a discontinuous transition to drop (case 1) or jump (case 2) by , however, the voltage remains the same, as the voltage across does not change instantaneously. drops from for to at , while (left side of ) drops from to .

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:

 (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 is charged by a DC voltage . Find energy is consumed by and energy is stored in .

A Shortcut Method:

Observing the complete solution obtained above, we see that

• When , is the initial condition
• When , 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 , always take the same form:

 (118)

in terms of three essential components of the system's response:
• The steady state response : as discussed in previous section for steady state response.
• The initial value : Denote the value of immediately before and after the moment by and , respectively. If , then use for ;
• The time constant of the system : When there is only one resistor in the circuit, the time constant is or . When there are multiple resistors, the time constant can be found by:
• Remove or so that the rest of the circuit () is a one port network.
• Find the equivalent resistance of the network by turning off all energy sources (short-circuit for voltage source, open-circuit for current source).
• Find time constant or .

In particular, note that

• when , the initial condition;
• when , the steady state response;
• when , the difference between the initial and the steady state values of the response decays exponentially. This term is the transient response of the system.

Example 1: , , . Find .

• Find initial value .

• Find steady state value :

 (119)

• Find equivalent resistance :

 (120)

• Find time constant

 (121)

• Find the complete response

 (122)

In particular,

 (123)

Example 2:

In the circuit below, , , , , the circuit is in steady state when . Find after the switch is closed at .

Consider node voltage method. Applying KCL to node we get

 (124)

i.e.,

 (125)

where

 (126)

To find the initial value , we assume the circuit is in steady state before , i.e., . Also, as the voltage across does not change instantaneously (unless therefore ), we have . At , drops from to , also drops from to .

The homogeneous solution is and the particular (steady state) solution is . The complete solution is

 (127)

The coefficient can be found by equating evaluated at and the initial condition :

 (128)

i.e., . Now the solution is

 (129)

Example 3:

Resolve the circuit above using the short-cut method:

• The initial condition is , same as that found previously;
• Find steady state value :

 (130)

• Find equivalent resistance :

 (131)

• Find time constant

 (132)

• Find the complete response

 (133)

In particular,

 (134)

• Find current through :
 (135)

• Find voltages and across and , respectively:

 (136)

 (137)

• Find currents and through and , respectively:

 (138)

 (139)

• Verify current :

 (140)

Example 4:

In the same circuit above, find the voltages and across and currents and through and , respectively.

• Find
• Find and . Before , the circuit is in steady state, i.e., . However, after the switch closes at , 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., and ;
• Find (same as before);
• Find and :

 (141)

 (142)

We see that .
• Find and through and :

 (143)

 (144)

• Find current through capacitor :

 (145)

Note that when , , lower than ground voltage !