CHAPTER 8
COMPLETE RESPONSE OF RC & RL CIRCUITS
The order of the circuit is defined by the number of storage elements in the circuit. In
this chapter, we will find the complete response of a first-order circuit; a first-order
circuit means there is only one storage element in the circuit such as in an RC or RL
circuit. A second-order circuit means there are two storage elements in the circuit; for
example RLC, LC, L 2 or C2. This is the subject of Chapter 9.
There are two types of first-order circuits we will study in this chapter. These are
defined in terms of the sources (voltage or current) within the circuit:
Constant sources
  first-order response to a constant input source

Time-dependent sources  first-order response to a time varied input source
RESPONSE OF A FIRST-ORDER CIRCUIT - CONSTANT INPUT
There are two types of first-order circuit (RC and RL) and we will always want to
simplify them Thévenin and Norton equivalents:
RC circuit  convert into a Thévenin Equivalent circuit
RL circuit  convert into a Norton Equivalent circuit
RC CIRCUIT
A typical RC circuit is not in Thévenin form, however, a conversion leads to a nice
series RC circuit of the form
Our focus is analyzing the behavior of the voltage across the capacitor in a series RC
circuit. Due to the switch, there are three distinct time periods.
 t < 0 (Circuit #1): Before the switch is closed, the battery has charged the
capacitor to its initial steady state value vC(0) vinitial.
 0 < t < ∞ (Circuit #2): The switch is thrown, the voltage across the capacitor
starts either charging or discharging depending on the type of circuit. This
change is time dependent and is not a constant. This voltage is defined as
vC(0t∞) ≡ vC(t).
 t  ∞ (Circuit #3): When sufficient time has elapsed, the capacitor has
reached a new final steady state value defined by vC(∞) vfinal. Note that this
voltage is understood well since the capacitor is an open in its final steady state value, vC(∞) vfinal VTh ( VOC).
 Solving the circuit problem (Circuit #4)
There are two steady state values (vC(0) and vC(∞)) that can be analyzed using
Chapter 7 techniques, however, it is between these two steady-states, vC(0t∞),
that is the focus of this chapter.
That is, to understand the capacitor’s voltage, there are two effects that are
impressed on it:
 how it responses to the voltage source and the changes in voltage due to
switches being thrown.
8-1

how the capacitors naturally responding to its environment (how the resistors
are organized) by either charging or discharging.
When the capacitor's voltage is completely known it is called the complete response
of the capacitor's voltage and written as
v C (t  0)  v C (0 ) Initial steady-state value

v C (t)  Complete Response  v C (t)
Transit-state
v (t  )  V
Final steady-state value
oc
 C
Let's start analyzing an RC circuit. Every time a switch is thrown, a new circuit is
created and the capacitor responses to the sudden changes by either charging or
discharging. The natural response of the capacitor’s voltage is broken into two parts:
 BEFORE the switch is thrown, find the initial steady state voltage vC(0)
 AFTER the switch is thrown, convert the RC circuit into a Thévenin circuit, so that
now the circuit is a simple series RC circuit with known VOC and RTh values.
 Solve the first order (differential) equation for the capacitor’s voltage
vC(0t∞) ≡ vC(t).
 Undo the Thévenin circuit and solve the circuit problem and interpret.
BEFORE the switch is thrown, we solve the circuit for v C(0), so that VDR gives
v C (0 ) 
R3
VS
R1  R2  R3
AFTER the switch is thrown, the RC circuit is converted into a Thévenin equivalent
using standard techniques; the result is a series RC circuit.
we get
R3
vS;
RTh  R2 // R3
R 2  R3
To find the voltage vC(t), apply KVL to the Thévenin equivalent and recognize that the
current through the resistor is equivalent through the current through the capacitor
according to the capacitor equation:
Voc  RThiC  v C (t) i Cdv /dt
Voc 
C
C
Rewriting it, we find
Voc  RThC
dv C
 v C (t) 

dt
dv C
v
V
 C  oc
dt RThC RThC
This first-order response equation is a form we will encounter again with the RL circuit,
so it is useful to write this in a general form so that we can use it again. Defining the
following parameters, we rewrite it as
x  vC

dv C
v
V
dx x

  RThC
 C  oc

 k

dt
RThC RThC
dt 

k  Voc RThC
8-2
where  is the time constant of the circuit and k is some constant (sometimes called the
effective source term) to be defined. To integrate, I must isolate the x and t terms:
dx
(x  k)
dx
dt

 
 
dt

(x  k)

 t

 exp  n(x  k)  exp    integration constant 
 

 t /  constant
t / 
 (x  k)  e e
 Ae

x(t)  k  Ae  t / 
Apply Boundary Conditions to determine A and k
To determine the complete response x(t), constants A and k must be evaluated with the
Boundary Conditions (BCs) established by the circuit. These BCs are the initial and
final steady state values (x(0), x(∞))of the circuit. These BCs are a very important part
to the complete response since it "fits" the solution to the particular circuit you are
solving; furthermore, it keeps the solution realistic and finite.
The value of x(t  0) x(0) implies the initial steady-state value that the circuit forced
on the storage element. That is, the circuit has had an infinite amount of time to reach its
initial steady-state value before the switch is thrown. Therefore, the complete response
must obey this initial state:
x(0 )  k  A  1
 A  x(0 )  k
After the switch is thrown, the storage element has had time to evolve and reach its new
final steady-state value x(t ∞)  x( ∞). Applying this requirement to the complete
response,
 x()  k

x(t  )  x()  k  Ae t/  



 A  x(0 )  x()
Gathering all the terms together, the first-order response takes on the form
x(t)  k  Ae t / 
A  x(0) x(  )
k x(  )

 x(t)  x( )  [x(0 )  x( )]e t / 
Now translating this into the RC complete response,
dv C
v
V
 C  oc


v C (t)

Voc
 [v C (0 )  Voc ] e  t/RThC
dt
RThC RThC x(t ) v (t )
Complete Response
Forced Response
Natural Response
C
Physical interpretation of the complete response to the RC circuit
1. Mathematically, the BCs are obviously met:
v C (t  0 )  Voc  [v C (0 )  Voc ]  e0  v C (0 )





v C (t  )  Voc  [v C (0 )  Voc ]  e  Voc
2. Charging of the capacitor. A better way to write the complete response is
v C (t)  Voc  [v C (0 )  Voc ] e t/   v C (0 )e t/   Voc 1  e  t/   Vinitiale  t/   Vfinal 1  e  t/ 



"natural"

" forced"
These two terms mean that

Vinitiale t/  
 discharges from initial steady-state value


 t/ 

 charges to the final steady-state value

 Vfinal 1  e


8-3
According to the table below,


the capacitor has discharged from its initial voltage of vC(0) to essentially zero
(0.7% of its original value) after 5 time constants (5).
the capacitor has charged to its final steady state value of VOC after 5.
v C (5)  Vinitiale5 +Vfinal 1  e5  0.007  Vinitial  0.993  Vfinal  Vfinal  Voc


Discharging of the capacitor
3. The charging/discharging depends ONLY on the circuit elements RTh and C.
Summary
The four step process for solving for the capacitor voltage is
1. Find the initial steady state voltage of the capacitor vC(0) before the switch is thrown
2. After the switch is thrown, find the Thévenin parameters (VOC & RTh) and the time
constant .
3. Put the voltage into solution form and interpret the solution:
vC(t)  Voc  [v C(0)  Voc ] et /RThC
4. Solve the circuit problem for the appropriate values.
Let’s apply this to the following example
Example 8.1
The circuit is in steady state before the switch closes at t = 0 s.
a. Describe the behavior of the capacitor for all time?
b. Determine the capacitor voltage vC(t) for t > 0 s.
8-4
Solution
Remember that there are at least four parts to solving this problem.
Step 1: Find vC(0) = vC(0+)
To find the steady state voltage across the capacitor before the switch
is thrown, we treat the capacitor as an open circuit and use whatever
method is most convenient to get vC(0). Since is circuit is completely
open, the voltage across the capacitor is the same as the battery: vC(0)
= 3V since v3 = v6 = 0.
Step 2: Find the Thévenin parameters (VOC & RTh) and 
The switch is thrown and the RC circuit must be converted into a Thévenin equivalent.
There are two circuits needed to find VOC & ISC:

Find VOC
Note that v6 = VOC. Using the VDR we get v 6 

6
 3V  2V  Voc
63
Find ISC so that RTH  VOC/ISC
Using the VDR to get the voltage across v6 to get the current through the 6
resistor, we find that v 6 
v
3
 3V  1.5V  i6   ISC  6  41 A
33
6
RTh =
Find the time constant 
Voc 2
  8  RTh
1
ISC
4
  RThC  8   0.05   0.4 s  
Step 3: Put the voltage vC (t ) into solution form and interpret the solution
vC (t)  Voc  [v C (0 )  Voc ] e t/   2  3  2 e t/0.4  2  e2.5t V
Putting this together, the capacitor voltage behaves as
3V
t0

v C (t)  
2.5t
V t0
2  e
Interpretation: the capacitor’s voltage starts off at 3V and discharges to 2V. After 5 or 5
time constants, the capacitor has arrived at its final voltage for the circuit. The open
circuit voltage acts like the terminal velocity of an object falling in air: it does not change
it value again.
Example 8.2
Find i2(t) = i0(t) for t > 0 for the circuit shown.
8-5
Solution
Step 1: Find vC(0)
At the initial steady state, the capacitor acts like an open short and the resulting circuit is
To solve for the initial voltage of the capacitor, one can either use Mesh, Node or Simple
Circuits - I choose to use Simple Circuits because I feel like I know more about what is
happening in the circuit. Setting the ground to be zero at the bottom, one immediately
sees that the current source is in series with the left 2k-resistor, so we then know the
node voltage for the positive terminal of vC. That is, relative to the chosen ground,
vC ()  4V  v 2k  4  2k  4m  12V
To find the negative terminal of vC, I continue to follow the current through the ground
and see that at the bottom node, the current is split evenly between the two branches
since both branches have 4k resistance. So 2mA goes up the 2k, giving
vC ()  2k  2m  4V
Therefore, vC(0) is then
v C (0 )  v C ( )  v C ( )  12  ( 4)  16V  v C (0  )
Step 2: Find VOC, RTh and 
The switch has now been thrown and combining the resistors on the left side, the circuit
we solve is
Converting this RC circuit into a Thévenin equivalent, the two resulting circuits are
below.
Find VOC
The open circuit voltage is the same as the 2k: v2k = VOC.
Setting the ground at the bottom note that the positive node of VOC
is at 0V. If we follow the 4mA current source, when it reaches the
8-6
bottom node, it splits the current evenly between the two 4-resistor branches, so 2mA
goes up the first 2k-resistor. Therefore, from Ohm's law we know that the voltage drop
across the 2k resistor is -4V.
v 2k  2k  2mA  4V  Voc
Find RTh
Opening the current source,
RTh  2  4//2 
3 k
2
 RTh
Find L
The time constant is
1
1
2
1
20
1




Hz 
 RThC 3k 100
3

Step 3: Put the vC(t) into solution form and interpret:
vC (t)  Voc  [v C (0 )  Voc ] e t/   4  16  4  e20t/3  4  12e20t/3 V
Putting this together, the capacitor voltage behaves as
16 V t  0

v C (t)  
20t/3
V t0
4  12e
Step 4: Solving the circuit problem for i2k(t), from the circuit we see that the capacitor
and the 2k resistor in question are in parallel but with opposite polarity (vC = v2k).
Therefore, the current is
v (t)
i2k (t)   C  2  6e20t/3mA
2k
RL CIRCUIT
Having now studied the RC circuit in detail, it is almost a straight shot to understand the
RL circuit. Instead of using a Thévenin equivalent, a Norton equivalent is used to find the
current through an inductor. Consider the RL circuit shown below.
The typical process of analyzing the inductor’s current for all times is broken down
into 3 time periods:
 t < 0: BEFORE the switch is thrown, its initial steady state current is established
iL(0).
 0  t   : AFTER the switch is thrown, the inductor current in the inductor either
increases (stores) or decrease (decays) depending on the type of circuit. This
change is time dependent and is not a constant. This current is defined as
iL(0t∞) ≡ iL(t).
 t  ∞: When sufficient time has elapsed, the inductor has reached a new final
steady state value defined by iL(∞) ifinal IN ( ISC).
So similar to the RC circuit, the inductor goes from its initial to its final steady -state
values. The main difference now is that instead of analyzing a Thévenin circuit, we
focus a Norton circuit or a parallel RL circuit with a current source I SC. The complete
response is then
8-7
iL (t  0)  iL (0 )

iL (t)  Complete Response  iL (0  t  )
i (t  )  I
SC
L
Initial steady-state value
Final steady-state value
Let's start analyzing an RL circuit.
Following the problem solving strategies of the RC circuit, BEFORE the switch is
thrown, we determine the initial steady-state of the inductor’s current iL(0):
AFTER the switch is thrown, convert the RL circuit into a Norton RL
circuit using standard operating procedures:
v
ISC  S ;
RTh  R2 // R3
R2
To solve for the current of the inductor as a function of time, iL (t ) , apply KCL at node a
in the Norton equivalent circuit:
ISC  iRTh  iL (t)
Since parallel elements have equal voltages, and with the use of Ohm’s law for RTh, we
write
di
v (t)
L diL
vRTh  vL (t)  L L
 iRTh (t)  L 
dt
RTh
RTh dt
Inserting this in the KCL equation above, gives the first-order response equation:
I
diL
i
L diL
ISC  iRTh  iL (t) 
 iL (t) 
 L  SC
RTh dt
dt L / RTh L / RTh
Since this has the same form as that of the Thévenin RC circuit differential, it clearly
must have the same solution form:
dx x

 k 
 x(t)  x()  [x(0 )  x()]e t / L
dt L
That is, for the inductor current response to the Norton RL circuit,
ISC
diL
iL




iL (t)

ISC
 [iL (0  )  ISC ] e  t/ L where   L / RTh
dt L / RTh  L / RTh 
Complete Response
Forced Response
Natural Response
Because of the same nature of the Thévenin RC circuit, the Norton RL circuit has all the
same physical traits and behaviors.
Example 8.3
The circuit shown is at steady state before the switch closes.
a. Explain the behavior of the inductor
current before any calculations?
Does it store or decay the current?
b. Find v(t) for t > 0.
8-8
Solution
Are interest is to determine the voltage across the 600 resistor, not the current of the
inductor. However, the inductor controls the current flows in the circuit. Therefore, we
must solve for the inductor current and then solve the circuit problem for v(t).
Step 1: Find iL(0)
At the initial steady state, the inductor shorts out the resistors so
that the current of the inductor is the same as the current source:
iL (0 )  0.5A
Step 2: Find ISC, RTh and 
The switch has now been thrown and the RL circuit must be
converted into a Norton equivalent. There are two resulting
circuit: the short circuit and the deactivated source one.
Find ISC
Replacing the inductor with a short circuit, the current through
the i400 = iSC. Using VDR and Ohm's law to get i400, we first
solve for v400:
600 400
v 400 
 100V   37.5V
600 400  400

 i400 
v 400
400
 93.8mA  iSC
Find RTh
By shorting out the voltage source
RTh  400  400 / /600  400  240  640  RTh
Find L
The inverse time constant is
1 RTh 640
1


 6400 Hz 
L
L
0.1
L
Step 3: Put the current iL(t) into solution form and interpret the solution
iL (t)  ISC  [iL (0  )  ISC ] e  t/   93.8mA   500  93.8  e 6400t
 93.75   406.2  e6400t mA
Putting this together, the capacitor voltage behaves as
500 mA
t0


iL (t)  
6400t
mA t  0

93.75   406.25  e
Interpretation: the inductor’s current starts off at 500mA and decays
to its final steady-state value of 93.75mA in 5 (0.78 ms). The short
circuit current value acts like the terminal velocity of an object falling
in air: it does not change its value again. Graphically, its looks like
the plot shows.
Step 4: Solve the circuit problem for the voltage across the 600 resistor
Up to this point, we have focused only on the current of the inductor; however, the
problem is about solving the voltage drop across the 600 resistor. If we look at the
Norton circuit, it tells us nothing about the 600 resistor, therefore, we need to go back
to the original circuit. Applying KVL to the right-handed loop in the circuit, we get
di
v 600  v 400  vL (t)  R400 iL (t)  L L
dt
Subing in the inductor current gives
8-9
v 600   400  93.8   406.2  e6400t  mA  (0.1)  406.2    6400  e 6400t  mA


v 600   37.5  97.5e6400t V
As this problem indicates, solving these types of problems will become time consuming
simply due to the fact that there are many steps in-between. It is very important to note
that these methods are only true for constant input sources NOT time dependent
sources. Look around and roughly one-third of you will make this mistake on our next
exam. Don’t be one of those!
SEQUENTIAL SWITCHING
When switches are thrown at various times (Sequential switching) several time
intervals are created with various time constants. As a result, one applies the general
solution form repeatedly over several sequent time intervals.
 Several time intervals will be created whenever a switch is thrown and therefore,
several time constants will describe the behavior of the circuit. We say that with
each new time constant, there is a new natural response to the circuit. We write
down that
t  0  Circuit-0; t  t1  Circuit-1; t  t 2  Circuit-2;
0

1
2
Our goal is to find the natural response of the circuit (or how the capacitor or
inductor behaves) as these switches are thrown at different times.
Consider the following physical example. Suppose that we have a circuit that has an
uncharged capacitor at t = 0 and that there are three switches in the circuit.
When a switch-1 is thrown, the capacitor begins to charge up according to the time
constant 1. At time t1, a second switch is thrown (switch-2) and the capacitor starts to
decay according to the new resistor-capacitor circuit, in other words, according to the
new time constant 2. At a later time, a third switch is thrown (switch-3) and the rate of
discharge occurs at a quicker pace (that is, according to the new time constant 3.)
Mathematically we expect:
Circuit 1: Switch is closed at t = 0 and the capacitor completely charges up. The natural
response of the circuit is
v C1(t1  t  0)  Voc1 1  e t / 1


Circuit 2: When the switched is thrown, the natural response of the capacitor's voltage
changes and behaves according to a new time constant 2. That is, it decays
according to the new equation
v C2 (t 2  t  t1 )  Voc2e(t t1 )/ 2  constant
Circuit 3: Another switch is thrown such that the natural response of the capacitor
decays faster according to the new time constant 3:
v C3 (t  t 2 )  V3e(t t1 t2 )/ 3  constant
8-10
Note the change in slope of the curve.
Rough Outline to Sequential Switching:
1. The initial state of the circuit is typically that of a steady state situation. This is what I
called circuit-0.
2. When the switch is thrown, the first circuit to analyze is created. Label this circuit-1
with time constant 1 .The natural response of circuit-1 is
 t / 1

v C1(t)  Voc1  [v C1(0)  Voc1] e

 t / 1

iL1(t)  ISC1  [iL1(0)  ISC1] e
3. If a third switch exist or the switch is thrown back at some later time, Label this
circuit-2 with time constant 2 .The natural response of circuit-2 is
v C2 (t)  Voc2  [v C2 (0)  Voc2 ] e (t  t1 ) / 2


 (t  t1 ) / 2

iL2 (t)  ISC2  [iL2 (0)  ISC2 ] e
4. Draw plots of the voltages and currents with respect to time.
Example 8.4
a. Describe the behavior of the inductor current.
b. Find i12(t) for t > 0 for the circuit shown. The circuit is in steady state at t = 0.
Solution
There are three circuits that we must analyze to understand the behavior of the current
through the 12 resistor: initial
Circuit #1
Find iL1(0 < t < 0.051s)
Step 1A: Find iL1(0−)
Finding an equivalent resistance to get the source current
Req  18  2 / /6  19.5 A 
 i(0 ) 
52 V
19.5 A
 2.67 A
Using the CDR to find the inductor current, we get
 6 

i2 (0 )  
  2.67 A  2.00 A  iL (0 )
 6+2 
Step 2A: Find the Norton parameters (ISC1 , RTh1, L1) for 0 < t < 0.051s.
Immediately we can write ISC1 = 0 since there is not source in the circuit.
The Thévenin resistance is
RTh1  2  12 / /6  6 .
so that the inverse time constant is
1 RTh1 6
1


 6 Hz 
1
L
1H
1
Step 3A: Putting this into solution form:
iL1(0  t  0.051s)  ISC1  iL (0 )  ISC1 e t/ 1  2e6t A


8-11
Step 4A: Solve the circuit problem for the current through the 12 resistor
The inductor acts like a current source and provides the current for the resistors.
Using the CDR, we get
 6 
6t
i12 (0  t  0.051s)  
  2e
 6  12 
For the first time interval we get
0t

2A
2.67A 0  t
iL (0  t  0.051s)   6t
; i12 (0  t  0.051s)   2 6t
0  t  0.051s
0  t  0.051s
3 e
2e

Circuit #2
Find iL1(0 < t < 0.051s)
A second switch is thrown before the inductor has enough time to
complete decay away, and the new solution is
iL 2 (t  0.051s)  ISC-2  iL 1(0.051s)  ISC-2  e(t 0.051) / 2
Step 1B: Find the initial steady state value iL1(0.051s)
The inductor current at t = 0.051 s becomes the “initial steady state”
value for the second circuit:
iL 1(t  0.051s)  2e6t
 1.473A
t  0.051s
Step 2B: Find the Norton parameters (ISC2 , RTh2, L2) for t > 0.051s
Since there is no sources, and there are only two series resistors, we can write
ISC2  0 and RTh2  2  12  14
The inverse time constant is
1 RTh2 14
1


 14 Hz 
2
L
1H
2
Step 3B: Putting this into solution form:
iL 2 (t  0.051s)  ISC-2  iL 1(0.051s)  ISC-2  e(t 0.051) / 2  0  1.473  0  e14(t 0.051)
Step 4B Solve the circuit problem for the current through the 12 resistor
Since the inductor and the 12 resistor are in series, they have the same current:
iL 2 (t  0.051s)  i12 (t  0.051s)  1.473 e14(t 0.051) A
Summarizing the two time intervals, we get
0t
2A
 6t
iL (t)  2e
0  t  0.051s
;
1.473 e14(t 0.051) A 0  t  0.051s

2.67A

i12 (t)   32 e6t A

14(t  0.051)
A
1.473 e
0t
0  t  0.051s
0  t  0.051s
8-12
By looking at the time constants for the two time intervals, note that 1 = 1/6 s while 2 =
1/14 s. The inductor decayed at a much faster rate after the second switch was thrown.
UNIT STEP SOURCE and PULSE SOURCE
We now consider circuits with special sources that can ideally parameterize a switch.
Consider the circuit
Before the switches are thrown, the circuit looked as
and the power supply was not connected. This resulted in the open circuit voltage being
zero v(t) = 0. At t = t0, the switches are thrown and the open circuit voltage is v(t) = V0.
0 t  t 0
We can summarize this circuit with v(t)  
. Physically, this appears like
 V0 t  t 0
turning on a light switch in your kitchen. That is, the voltage increases to V0
instantaneously from zero volts. Mathematically, this instantaneous jump is represented
by a UNIT STEP function defined as follows:
0 t  t 0 (off)
UNIT STEPfunction  u(t  t 0 )  
1 t  t 0 (on)
Another property of the unit step function is that if the argument is less than zero, we can
write
1 t 0  0 (on)
u(0  t 0 )  
1 t 0  0 (off)
The previous circuit can now be written into the equivalent circuit with the voltage source
as
0 t  t 0
V0u(t  t 0 )  
 V0 t  t 0
8-13
Pulse Voltage Source
If you think about a light switch and the process of turning it on and off, there is a sudden
increase and decrease in voltage. The unit step function can be used to define a pulse
wave form is. Mathematically and graphically the pulse waveform is defined as two unit
step function “turning on” at different times.
0 t  t 0

v(t)   V0 t 0  t  t1
0 t  t
1

The way we get the pulse wave form is by using a series of unit step functions. First, we
apply V0u(t  t0 ) ; at a later time we apply V0u(t  t1 ) . That is, the superposition of these
two sources gives
RESPONSE of FIRST-ORDER CIRCUITS to NONCONSTANT SOURCES
Up to now, we have assumed only constant sources. That is, the first order equation is
dx x
  k  constant
dt 
However, now we move on to looking at the response when the source is time
dependent. Then the first order equation looks like
K1  K 2e at

dx x
where
  f(t)  constant 
 f(t)  effective source  K1  K 2 sin t (cost)
dt 

n
K1  K 2 t
Interpretation of time dependent Responses
When a constant source is driving the circuit, we have seen that the behavior of the
storage element is forced to a constant final steady-state value, given by
x(t)
 x(0 )e t/   x()(1  e  t/  )
complete response
natural response
force response
On the other hand, when a time dependent source is driving the circuit, the storage
element parameter (vC and iL) is source dependent and must have the same form
as the time dependent source. That is, the complete response still has the same basic
structure,
8-14
x(t)
complete response
 x(0 )e t/  
x f (t)
natural response
force response
 xn  x f
however, the forced response of the storage element parameter must take on the same
form as the time dependent source:
K1  K 2 t n

B1  B2 t n



f(t)  K1  K 2eat
 x f  B1  B2e at
 
K sin t (cos t)
B sin t  B cos t)
2


 1
As a quick overview, there is a series of steps that must be taken to solve for the
response to time dependent source (say this, do not write this down):
 derive using SCT the first order equation
 generate a general natural and force responses (homogeneous & particular
solutions to the first order equation)
 Solve for the forcing response constants; solve for the BC and apply it to get the
natural response constant.
 Solve the circuit problem
Problem Solving Strategies
Storage Element Condition ≡ SEC  iC = C dvC/dt and vL = L diL/dt
Step 1: Apply Simple Circuit ideas (usually KVL or KCL) and SEC to derive the first
order equation:
dx x
  f(t)
dt 
SCT writes an equation that connects x and dx/dt (i.e. vC & iC or vL & iL) while
SEC writes down a second equation that connects x and dx/dt.
Step 2: The general response equation is of the form
x(t)  xnatural (t)  x force (t)  Ae-t/  + B1x f1 + B2 x f2  B3
The forcing response has the same form as the effective source term f(t)
in the first order equation:
-at


K1  K 2e

f(t)  


K1  K 2 sin(t) (or cos(t)) 

B1  B2e-at

 x f (t)  

B1 sin(t)  B2 cos(t)  B3
Natural Response xn(t)
From the first order equation pick-off the time constant  and write down the natural
response (the constant A is determined via BCs):
xn  Ae t/ 
Details: one derives the natural response when the effective source is zero. Integrating
the first order equation yields
dxn xn

 f(t)
 0 

dt

f (t)0

dxn
1
   dt  xn  Ae t/ 
xn

Forcing Response xf(t) (particular or inhomogeneous solution)
The storage element parameter (vC or iL) is source dependent and must have the same
form as the effective source. Physically it means how the source is forcing the storage
element to store energy (the E-field for the capacitor and the B-field for the inductor).
Now we take on the full first order equation and solve it by the easiest possible way: we
assume that the forcing response have the same form as the source:
8-15
dx f x f

 f(t) 
 x f  f(t)
dt

where
K1  K 2 t n

B1  B2 t n



f(t)  K1  K 2e at
 x f  B1  B2e  at
 
K  K sin t (cos t)
B sin t  B cos t  B
2
2
3
 1

 1
Picking the correct force response, the most the general complete response is
x(t)  Ae t/   B1x f1  B2 x f 2  B3
Step 3: Solving for Constants B1, B2, and A
 Substitute the general response into the first order equation (Step 1) to find
the constants B1 and B2.
 Solve for the BC [vC(0) or iL(0)] and apply it to the general response
equation solve for the constant A.
Step 4: Solve the circuit problem for the appropriate variable.
Let's consider a straightforward example and go through the above steps to get an idea
as to what to do.
Example
Find vC(t) for t > 0 when vC(0) = 0.
Solution
Our goal is to understand how the capacitor’s voltage vC(t) will
response to the sine voltage source. Applying the four step
process to this circuit is as follows.
Step 1: Derive the first-order equation to the circuit
Since this is a series circuit, apply KVL to get the vC equation:
 Apply SCT to write an equation that has both vC and iC:
v S  v c  vR v Ri  v C  RiC
R
C

Apply SEC to the above equation (this connects vC and iC):
dv
v S  v c  RiC i C dvC  v C  RC C
C
dt
dt
Cleaning this up a bit leads to
dv C v C
v

 S
dt
RC RC RC1010m1/10

v S 10sin20t
dv C
 10v C  100 sin 20t
dt
effective source
Step 2: Find the general response vC(t) = vn + vf
Natural Response
Pick-off the time constant 1/ = 10 and write the natural response:
vn (t)  Ae10t
Forcing Response
The first-order response equation with the effective source is
dv C
 10v C  100 sin20t
dt
Since the forcing response has to be the same form as the effective source, then the
most general solution is
v effective (t) =100 sin20t  v f =B1 sin20t  B2 cos20t
source
8-16
The complete response is
v C (t)  v f  v n vn  Ae10 t
v f B1 sin 20t B2 cos 20t
 Ae10t  B1 sin20t  B2 cos 20t
Step 3: Find the Constants B1, B2, and A
Find the Constants B1 and B2
Substituting the assumed general complete response into the first-order response
equation derive in Step 1, we get
dv C
 10v C
dt
vf
 100 sin 20t
=B1 sin 20t  B2 cos 20t
v f  B1 sin 20t  B2 cos 20t  Ae 10t


  dv
f
 20 B1 cos 20t  B2 sin 20t   10Ae 10t

 dt
Substituting everything in gives
The derive term
dv C
 10v C  20 B1 cos20t  B2 sin20t   10Ae10t
dt
 10B1 sin20t  10B2 cos20t  10Ae10t  100 sin20t
Hint: the A-terms will always cancel out here in this step.
Collecting like sine and cosine terms:
20B2  10B1  sin20t  20B1  10B2   cos20t  100sin20t
The only way this equation can be true is if all of the sine/cosine terms on each side of
the equation are equal.
 20B2  10B1  sin20t  100 sin20t 
20B2  10B1  100
  
20B1  10B2   cos 20t  0
 20B1  10B2  0


One finds that we get a 2 x 2 matrix for constants B1 and B2:
 10 20   B1  100 

   
  B1  2, B2  4
 20 10   B2   0 
Substituting in the constants into the general complete response gives
v C (t)  v f  vn B1 2  Ae10t  2sin20t  4cos20t
B2 4
Find the Constant A
There is only one constant left to solve for – A. Solve for the BC vC(0−) (= 0).
Substituting this into the complete response gives
vC (0 )  0  Ae0  2sin0  4cos0  A  4 
 A4
The complete first-order response is


v C (t)  4e10t  2sin20t  4cos 20t V
Interpret the solution
Example 8.5
Find v24(t) for t > 0 for the circuit shown. Assume steady state at t = 0.
8-17
Solution
We need to solve for v24(t). Since this is in series with the 8mH-inductor, v24 = 24iL(t).
Therefore, our immediate goal is to solve for the inductor’s current iL(t) and how it
responds to the exponential voltage source. Applying the four step process to do this.
Step 1: Derive the first-order response to the circuit
Solving for the inductor’s current iL at t = 0+, the series circuit implies we use KVL:

Apply SCT to write an equation that has both vL and iL:
v1  v  vL v Ri  25e5t  24iL  vL (t)
L

Apply SEC to the above equation (this connects vL and iL):
v1  v  v L
vL L
diL
dt
 25e5t  24iL  0.008
diL
dt
Rewriting this last step gives us the first-order response:
diL
 3000iL  3125e5t
dt
Step 2: Find the natural and forcing response iL(t) = in + if.
Natural Response
Pick-off the 1/ term, and put into the natural response form:
1

 3000 
 in (t)  Ae3000t
Forcing Response
The first-order response equation with the effective source is
diL
iL

 3125e5t
dt 1/ 3000
Assuming the solution has the same form as the effective source, we write
v effective (t)  3125e5t 
 if =B1e5t
source
The complete response is
iL (t)  if  in in  Ae3000 t  B1e5t  Ae3000t
if =B1e5 t
Step 3: Find the Constants B1, B2, and A
Find the Constants B1 and B2
Substituting the assumed solution into the first-order equation derived in Step 1, we have
8-18
diL
 3000iL
dt
if
 3125e5t
=Ae
3000 t
 B1e
5 t
iL  B1e5t  Ae3000t

  di
L
 5B1e5t  3000Ae3000t

 dt


diL
 3000iL  5B1e5t  3000Ae3000t  3000 B1e5t  Ae3000t  3125e5t
dt
Solving for B1 and the forcing response, we get
B1  3000  5  e5t  3125e 5t 
B1 
3125
2995
 1.04
The general form of the complete response is now
iL (t)  if  in B 1.04  1.04e5t  Ae3000t
1
Find the Constant A
Solve for the BC iL(0):
Using the CDR,
6
  5   1 mA
6  24
Substituting this into the complete response gives
iL (0 ) 
iL (0)  1.0  A  1.04 
 A  2.04
The complete first-order response is


iL (t)  10.4e5t  2.04e3000t mA
As expected, the current of the inductor goes to zero fairly quickly. The natural response
immediately dies off at e-3000t where the force response is much slower in comparison at
e-5t.
Step 4: Solve the circuit problem for the voltage across the 24 resistor.
Note that because all the elements are in series, the current through the 24 resistor is
the same as the inductor current. Using Ohm’s law we get


v 24 (t)  24iL (t)  48.96e5t  24.98e3000t mV
The voltage across the 24 resistor dies off from roughly 25 voltages down to zero in 5
= 1 s.
Example 8.6
Find vC(t) for t > 0.
8-19
Solution
Step 1: Derive the first-order response to the circuit
We need to solve for the cap's voltage vC(t) when the switch is thrown open. Since a
dependent source is in the circuit, a dependent condition will have to be used.

KVL: v12  8e5t  vC  0 
 12iX  8e5t  vC  0

Dependent Condition & Capacitor equation condition to convert ix to vC; use KCL:
1 dv C 
1 dv C
iX  2ix  iC i  1 dvC 
 3ix  36
 ix  108
C C
dt
dt
dt
This condition connects iC(t) and vC(t).
Rewriting this last step gives us the first-order response:
dv
12iX  8e5t  v C  0 1 dvC 
 C  9v C  72e5t
ix  108
dt
dt
Step 2: Find the natural and forcing response iL(t) = in + if.
Natural Response
Pick-off the time constant 1/ = 9 and write the natural response:
vn (t)  Ae9t
Forcing Response
Need to solve the first-order response equation with the effective source:
dv C
 9v C  72e5t
dt
Since the effective source is exponential in form, so is the forcing response:
v f  B2e5t
The most general form of the complete response is
v C (t)  B2e5t  Ae9t
Step 3: Find the Constants B1, B2, and A
Find the Constants B1 and B2
Substituting the assumed solution into the first-order response equation derive in Step 1,
we have
v C (t)  B2e5t  Ae 9t
dv C

5t
 9v C
 72e
  dv
C
dt
 5B2e5t  9Ae9t
v C (t ) B2e 5 t  Ae9 t

 dt
dv C
 9v C  5B2e5t 9Ae9t  9 B2e5t  Ae9t  72e5t
dt
Solving for B2 and the forcing response, we get

B2  9  5  e5t  72e5t 
 B2 

72
4
 18
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The general form of the complete response is now
v C (t)  18e5t  Ae9t
Find the Constant A
There is only one constant left to solve for – A. The initial condition of the capacitor is
given by the circuit at t < 0.
Because of the CCCS, I must find the controlling current ix. So the easiest way to solve
for vC(0) is to observe that the 12-resistor is parallel with the capacitor and so there
voltages are equivalent. That is,
v C (0 )  v12  12ix
Solving for the controlling current ix will be our focus. Using KCL at the top node, the
current through the 3 resistor is 3ix, now applying a KVL loop around the outside yields
KCL 
 i3   ix  2ix  3ix
KVL 
 12ix  9ix  38.5  0 
 ix 
38.5
 1.83A
21
Therefore, the initial voltage of the capacitor is
v C (0 )  12ix  12  1.83 V  22V  v C (0  )
Substituting this into the first-order response equation gives
v C (0 )  22  18  A 
 A  4V
The complete response of the capacitor in this circuit is then
v C (t)  18e5t  4e9t
As expected, the capacitor's voltage goes to zero fairly quickly. The natural response
immediately dies off at e-9t where the force response is much slower in comparison at e-5t.
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