Chapter 7
Time domain analysis
Time domain analysis is associated with differential equations. The differential equations
resulting from analyzing RC and RL circuits are of the first order and hence the circuits are
collectively known as First order circuits. A first order circuit is characterized by a first
order differential equation. Typical applications of RL and RC circuits includes delay and
relay circuits, a photoflash unit and automobile ignition circuit.
The differential equations resulting from analyzing of RLC circuits are of the second order.
There are two ways to excite first order circuits
1. The first way is by initial conditions of the storage elements in the circuits. These are
referred to as source- free circuits and we assume that energy is initially stored in the
capacitive or inductive element.The energy causes current to flow in the circuit ad is
gradually dissipated in the resistors.Although source free circuits are by definition free
of independent sources, they may have dependebt sources.
2. The second way of exciting first order circuits is by independent sources.The indepen-
67
dent sources to be considered in this chapter are dc sources.
7.1
Source-free RC circuit
A source -free circuit occurs when its dc source is suddenly disconnected. The energy already
stored in the capacitor is released to the resistors. Consider a series combination of a resistor
and an initially charged capacitor as shown in Figure 7.1 Our objective is to determine the
Figure 7.1: A source-free circuit
circuit response which for academic reasons is assumed to be the voltage v(t) across the
capacitor.
Since the capacitor is initially charged, we assume the that at time t=0, the initial voltage is
v(0) = V0
with the corresponding value of energy stored as
1
ω(0) = CV02
2
Applying KCL at the top node of the circuit gives
ic + iR = 0
C
dv
v
dv
v
+ =
+
=0
dt R
dt RC
68
(7.1)
Equation 7.1 is a first differential equation since only the first derivative of v is involved.To
solve it we arrange the terms as
dv
1
=−
dt
v
RC
integrating both sides, we get
lnv = −
t
+ lnA
RC
where ln A is the integration constant.Thus
ln
v
t
=−
A
RC
taking powers of exponential e we get
−t
v(t) = Ae RC
But from the initial conditions, v(0)=A=V0 , hence
−t
v(t) = V0 e RC
This shows thatthe voltage response of the RC circuit is an exponential decay of the initial
voltage. Since the response is due to the initial energy stored and the physical characteristics
of the circuit and not due to some external voltage or current source.This is referred to as
the natural response of the circuit
Natural response of a circuit refers to the behavior (in terms of voltages and currents)of the
circuit itself, with no external sources of excitation.
The natural response is illustrated graphically in figure 7.2. At t=0,v=V0 which is the correct
initial condition.As t increases, the voltage decreases towards zero.The rapidity with which
the voltage decreases is expressed in terms of the time constantτ .
69
Figure 7.2: The voltage response of the RC circuit
τ is the time require for the response to decay by 36.8 % of the its initial value or by a factor
e−1 .Hence, at time t=τ
−t
V0 e RC = V0 e−1 = 0.368V0
or τ =RC
−t
v(t) = V0 e τ
The value of v(t)
is shown in the Table It is thus evident that v(t) is less than 1% of V0
V0
Figure 7.3:
after 5 time constants.The assumption is that the capacitor is fully discharged after 5 time
constants.
The smaller the time constant, the more rapidly a voltage decreases, i.e the faster the response.
Examples
70
7.2
Source free RL circuit
Figure 7.4:
The series connection of a resistor and an inductor is shown in figure 7.4.The circuit
response will be determined by assuming the current i(t) is through the inductor. The
inductor current is selected as a response so as to take advantage of the fact that the inductor
current does not change instantaneously.At t=0, we assume the inductor has an initial current
I0 so that i(0)=I0 . With the corresponding energy stored in the inductor as
1
ω0 = LI02
2
Applying KVL around the loop we have
VL + VR = 0
=L
di
di R
+ Ri =
+ i=0
dt
dt L
Rearranging terms and integrating gives,
Z
i(t)
I0
i(t)
lni|I0 = −
di
=−
i
Z
t
0
R
dt.
L
Rt
i(t)
Rt
Rt t
|0 =⇒ lni(t) − lnI0 = −
+ 0 = ln
=−
L
L
I0
L
Rt
i(t) = I0 e− L
71
Figure 7.5:
This is the natural response of the RL circuit with an exponential decay of the initial current
as in figure 7.5 It is evident that
τ=
L
R
is a time constant for the RL circuit
Examples
7.3
Step response of RC circuit
When the dc source of an RC circuit is suddenly applied, the voltage or current source can
be modeled as a step function and the response is known as a step response.
The step response of a circuit is its behaviour when the excitation is the step function,
which may be a voltage or a current source.
Sudden application of a dc voltage or current source results to a step response. The circuit in
figure7.6a has a step response due to sudden application of of a dc voltage or current source
and hence can be replaced by circuit in 7.6b. U(t) represents step response. We select the
72
Figure 7.6:
capacitance voltage as the response to be determined.We assume an initial voltage V0 on the
capacitor since the voltage of a capacitor can’t change instantaneously.
V (0− ) = V (0+ ) = V0
Which is the voltage across the capacitor just before switching and voltage immediately after
switching respectively.Applying KCL
C
dv v − Vs U (t)
+
=0
dt
R
dv
v
Vs
+
=
u(t)
dt RC
RC
where v is the voltage across the capacitor.For t>0,equation 7.2 becomes,
dv
v
Vs
+
=
dt RC
RC
dv
v − Vs
=−
dt
RC
73
(7.2)
or
dv
dt
=−
v − Vs
RC
Integrating both sides and introducing initial conditions
v(t)
ln(v − Vs )|V0 = −
t t
|
RC 0
ln(v(t) − Vs ) − ln(V0 − Vs ) = −
t
+0
RC
or
ln
−t
v − Vs
=eτ
V0 − Vs
−t
v − Vs = (V0 − Vs )e τ
−t
v(t) = Vs + (V0 − Vs )e τ
t> 0
Thus



v(t) = 

V0 ,
t<0 


−t
Vs + (V0 − Vs )e τ , t > 0
This is known as the complete response of the RC circuit to a sudden application of a dc
voltage source assuming the capacitor is initially charged i.e Vs > V0 as shown in figure 7.7
If the capacitor is initially uncharged, the capacitor initially has got V0 =0 as in figure 7.8



v(t) = 

0,
t<0 


−t
Vs (1 − e τ ), t > 0
In general, finding the step response of RC and RL circuit can be determined as follows. It
is evident that v(t) has two components. Thus, we may write.
v = vf + vn
74
Figure 7.7: response of RC circuit with initially charged capacitor
Figure 7.8: response of RC circuit with initially uncharged capacitor
where
vf = Vs
and
−t
vn = (V0 − Vs )e τ
vn is the natural response of the circuit.Since it also decays to almost zero after five time
constants.It is also referred to as transient response because it is a temporary response that
will die out with time. Now vn is known as the forced response because it is produced by the
circuit when an external ’force’ is applied ( a voltage source ).It represents what the circuit
is forced to do by the input excitation.It is also known as the steady state response because
it remains a long time after the circuit is excited.
75
In summary:
The natural repsonse or transient response is the circuit’s temporary response that will die
out with time.
The forced response or steady state response is the behavior of the circuit a long time after
an external excitation is applied.
The complete response of the circuit is the sum of the natural response and the forced
response. Hence
−t
v(t) = v(∞) + [v(0) − v(∞)]e τ
Where v(0) is the initial voltage at t=0+ and v(∞) is the final voltage or steady state value.
Example
7.4
Step response of R-L circuit
The R-L circuit in figure 7.9a may be replaced with the circuit in figure 7.9b. Our aim is to
find the inductor current i as the circuit response. Using the complete response developed,
we let the response be the sum of the natural current and the forced current.
i = in + if
The natural response is always a decaying exponential.
−t
i = Ae τ , τ =
L
R
A forced response, value of i (current) a long time after the switch in figure 7.9a is closed.
The natural response essentially decays after 5 time constants(5τ ). At that time the inductor
76
Figure 7.9: R-L circuit with a step input voltage
becomes short circuited and voltage across it is zero. The entire source voltage appears across
the resistor R. Thus the forced response is
if =
Vs
R
This gives
−t
i = Ae τ +
Vs
R
To determine the constant A in from the initial value of i, Let I0 be the initial current
through the inductor, which may come from a source other than Vs .Since the current though
the inductor can’t change instantaneously
i(0− ) = i(0− ) = I0
Thus at t=0
I0 = A +
Vs
R
A = I0 −
Vs
R
A is obtained as
77
thus
i(t) =
Vs
Vs −t
+ I0 − e τ
R
R
This is the complete response of R-L circuit as illustrated in figure7.10. Examples
Figure 7.10: Total response of the R-L circuit with initial inductor current
7.5
Applications
Various devices in which R-L and R-C circuits find applications include filtering in dc power
supplies, smoothing circuits in digital communications, differentiators, Integrators, delay
circuits, relay circuits, photo flash unit and automobile ignition.
7.5.1
Delay circuit
Figure 7.11: Delay circuit
78
An R-C circuit can be used to provide various time delays. It consists of an RC circuit
with the capacitor connected in prallel with a neon lamp. The voltage source an provide
enough voltage to fire the lamp. When the switch is closed, the capacitor voltage increases
gradually towards 110 V at a rate determined by the circuits time constant, (R1 +R2 )C.The
lamp will act as an open circuit and not emit light until the voltage across it exceeds a
particular voltage level, say 70 V. When the voltage level is reached,the lamp fires and the
capacitor discharges through it.Due to the low resistance of the lamp when on,the capacitor
voltage drops fast and the lamp turns off.The lamp acts again as an open circuit and the
capacitor recharges. By adjusting R2 we can introduce either short or long time delays into
the circuit.The warning blinkers commonly found on road construction sites are examples of
RC delay circuits.
7.5.2
Photo-flash unit
An electronic flash unit provides a common example of an RC circuit. This application
exploits the ability of the capacitor to oppose any abrupt change in voltage. The simplified
circuit of a photo-flash unit consists of a high-voltage dc supply, a current limiting large
resistor R1 and a capacitor C in parallel with the flash lamp of low resistance R2 . When the
Figure 7.12: Flash unit providing slow charge in position 1 and fast discharge in position 2
79
switch is in position 1, the capacitor charges slowly due to the large time constant(τ1 =R1 C.
The charging time is approximately five times the constant.
t = 5R1 C
When the switch is in position 2, the capacitors will is discharged.The low resistance R2 ,of
the photo lamp permits a high discharge current with peak
I2 =
Vs
R2
in a short duration.Discharging takes place in approximately five time the time constants.
Such a circuit finds application in electric spot welding and the radar transmitter tube.
7.5.3
Relay circuit
Figure 7.13: Relay circuit
A magnetically controlled switch is called a relay.It is an electromagnetic device used
to open and close a switch that controls another circuit.The coil circuit is an R-L circuit
where R and L are the resistance and inductance of the coil. When the switch is closed, the
80
coil circuit is energized. The coil current gradually increases and produces a magnetic field.
Eventually the magnetic field is sufficiently strong to pull the movable contact in the other
circuit and close switch S2. At this point, the relay is pulled in.The time interval between the
closure of the switches S1 and S2 is called the relay delay time. Relays are used for switching
high power circuits.
7.5.4
Automobile ignition circuit
The ability of inductors to oppose rapid change in current makes them useful for arc or spark
generation.An automobile ignition system takes advantage of this feature.The gasoline engine
of an automobile requires the fuel-air mixture in each cylinder be ignited at proper times.
This is achieved by means of a spark plug which consists of a pair of electrodes separated by
an air gap. By creating a large voltage between the electrodes, a spark id formed across the
Figure 7.14: Circuit for an automobile ignition system
air gap, thereby igniting the fuel.The large voltage is achieved by means of an inductor (The
spark coil)L. Since voltage across the inductor is
v=L
81
di
dt
we can make
di
dt
large by creating a large change in current in a very short time.When the ignition switch is
closed,the current through the inductor increases gradually and reaches the final value of
i=
Vs
R
where Vs =12 V.Again time taken for the inductor to charge is five times the time constant
of the circuit.
t=5
L
R
When the switch suddenly opens, a large voltage is developed across the inductor causing
a spark or arc in the air gap.The spark continues until the energy stored in the inductor is
dissipated in the spark discharge.
7.6
Second order circuits
A second-order circuit is characterized by a second order differential equation.It consists of
of resistors and the equivalent of two energy storage elements.
Initial and final values
This section focuses on determining initial and final values of derivatives.i.e
dv
dt
and
di
dt
82
Unless stated otherwise, v denotes capacitor voltage and i is the inductor current.V and i
are defined strictly according to the passive sign convention. Capacitor voltage is always
continuous so that
v(0+ ) = v(0− )
and the inductor current is always continuous so that
i(0+ ) = i(0− )
In finding initial conditions, we focus on those variables that cannot change abruptly i.e
capacitor voltage and inductor current.This is illustrated with use of examples.
7.6.1
Source free series RLC circuit
Figure 7.15: Source -free RLC circuit
Consider RLC circuit shown in figure 7.15, the circuit is being excited by the energy
initially stored in the capacitor and the inductor.the energy is represented by the initial
capacitor voltage V0 and initial inductor current I0 Thus at t=0
1
v(0) =
C
Z
0
idt = V0
−∞
i(0) = I0
83
Applying KVL
di
1
Ri + L +
dt C
Z
t
idt = 0
−∞
Differentiating to eliminate the integral, we get,
d2 i R di
i
+
+
=0
dt
L dt LC
This is a second order differential equation which results from RLC circuits.Solving second
order differential equation requires we have two initial values of i and v. Using
Ri(0) + L
di(0)
+ V0 = 0
dt
or
i(0) = I0
1
di(0)
= − (RI(0) + V0 )
dt
L
With the two initial conditions, the second order differential equation can be solved.From
first order systems, the solution is of exponential form. So we let,
i = Aest
Where A and s are constants to be determined.Hence substituting and carrying out the
differentiation.
As2 est +
AR st
A st
se +
e =0
L
LC
or
Aest (s2 +
R
A
s+
)=0
L
LC
Since
i = Aest
84
is the assumed solution we are trying to find,only the equation in parentheses can be zero.
s2 +
R
A
s+
=0
L
LC
This quadratic equation is known as the characteristic equation of the differential.Since the
roots of equation dictate the character of i. The two roots are
R
+
S1 = −
2L
r
R
S2 = −
−
2L
r
(
(
1
R 2
) −
2L
LC
1
R 2
) −
2L
LC
A more compact way of expressing the roots is
q
S1 = −α + α2 − ω02
q
S2 = −α − α2 − ω02
Where
R
2L
1
ω0 = √
LC
α=
The roots s1 and s2 are known as natural frequencies because they are associated with natural
response of the circuit.ω0 is known as the resonant frequency or strictly as the undamped
natural frequency.The characteristic equation can be written as
s2 + 2αs + ω02
7.6.2
Source free Parallel RLC circuit
Parallel RLC circuit find many practical applications in communication networks and filter
designs. Assume initial inductor current I0 and initial capacitor voltage V0 .
85
Figure 7.16: Source -free parallel RLC circuit
Z
1
i(0) = I0 =
L
0
v(t)dt
∞
v(0) = V0
Since the three elements are in parallel, they have same voltage across them.According to positive sign convention, the current is entering each element and leaving the top node.Applying
KCL
v
1
+
R L
Z
t
vdt + C
−∞
dv
=0
dt
Taking derivative and dividing by C gives,
d2 v
1 dv
1
+
+
v=0
2
dt
RC dt LC
We obtain the characteristic equation by replacing the first derivative by s and second derivative by s2 and obtained as
s2 +
1
1
s+
=0
RC
LC
The roots of the characteristic equation are
s1,2
1
=−
±
2RC
r
(
1 2
1
) −
2RC
LC
or
s1,2
q
= −α ± α2 − ω02
86
There are three possible solution to this depending on whether,
α > ω0
α = ω0
α < ω0
7.6.3
Step response of series RLC circuit
Figure 7.17: Step voltage applied to a series RLC circuit
The step response is obtained by sudden application of a dc source. Consider the series
RLC circuit in figure 7.17. Applying KVL around the loop for t greater than zero
L
di
+ Ri + v = Vs
dt
but
i=C
dv
dt
substituting for i and rearranging
d2 v R dv
v
Vs
+
+
=
dt2
L dt LC
LC
which is the same as the equation developed for a source free RLC circuit . However the
variable is different.The characteristic equation for the RLC series circuit is not affected by
87
the presence of the dc source.The solution has two components, the natural response and the
forced response.
v(t) = vn (t) + vf (t)
The natural response is the solution when we set Vs =0 The natural response for the overdamped, underdamped and critically damped cases are
Vn (t) = A1 es1 t + A2 es2 t − overdamped
Vn (t) = (A1 + A1 )e−αt − criticallydamped
Vn (t) = (A1 cosωd t + A2 sinωd t)e−αt − underdamped
The forced response is the steady state or final value v—(t). The final value is of the capacitor
voltage is same as the source voltage.The complete solution is
Vn (t) = Vs + A1 es1 t + A2 es2 t − overdamped
Vn (t) = Vs + (A1 + A2 )e−αt − criticallydamped
Vn (t) = Vs + (A1 cosωd t + A2 sinωd t)e−αt − underdamped
7.6.4
Step response of a parallel RLC circuit
Figure 7.18: Parallel RLC circuit with applied current.
88
Consider the parallel RLC circuit in figure 7.18. We want to find i due to a sudden
application of a dc current.Applying KCL at top of the node for t greater than zero.
v
dv
+i+C
= Is
R
dt
But
v=L
di
dt
This gives
1 di
i
Is
d2 i
+
+
=
2
dt
RC dt LC
LC
The complete solution consists of the natural response and forced response.that is
i(t) = in (t) + if (t)
The forced response is the steady state or final value of i. The final value of the current
through the inductor is the same as the source current Is .Thus
i(t) = Is + A1 es1 t + A2 es2 t − overdamped
i(t) = Is + (A1 + A2 )e−αt − criticallydamped
i(t) = Is + (A1 cosωd t + A2 sinωd t)e−αt − underdamped
7.6.5
General second order circuits
Given a second order circuit, we determine its step response x(t) which may be voltage or
current by taking the following four steps.
1. We first determine the initial conditions x(0)and
89
dx(0)
dt
and the final value is x(∞)
2. We find the natural response xn t by turning off independent sources and applying
KCL and KVL. Once a second order DE is obtained.We determine its characteristic
roots.Depending on whether the response is overdamped, critically damped or under
damped, we obtain the natural response with two unknown constants as we did in
previous sections.
3. We obtain the forced response as
xf (t) = x(∞)
where x(∞)is the final value.
4. The total response is found as a sum of the natural response and forced response.
x(t) = xn (t) + xf (t)
We finally determine the constants associated with the natural response by imposing
the initial conditions.
x(0),
dx(0)
dt
determined in step 1
This general procedure is applicable to any second -order circuit.
90