ECS 203 (ME2) - Part 2B
Dr.Prapun Suksompong
CHAPTER 6
Energy Storage Elements: Capacitors and Inductors
To this point in our study of electronic circuits, time has not been
important. The analysis and designs we have performed so far have been
static, and all circuit responses at a given time have depended only on
the circuit inputs at that time. In this chapter, we shall introduce two
important passive circuit elements: the capacitor and the inductor.
Capacitors and inductors, which are the electric and magnetic duals of
each other, differ from resistors in several significant ways. Unlike resistors,
which dissipate energy, capacitors and inductors do not dissipate but store
energy, which can be retrieved at a later time. They are called storage elements. Furthermore, their branch variables do not depend algebraically
upon each other. Rather, their relations involve temporal derivatives and
integrals. Thus, the analysis of circuits containing capacitors and inductors
involve differential equations in time.
6.1. Capacitors
A capacitor is a passive element designed to store energy in its electric
field. When a voltage source v is connected to the capacitor, the amount
of charge stored, represented by q, is directly proportional to v, i.e.,
q(t) = Cv(t)
where C, the constant of proportionality, is known as the capacitance
of the capacitor. The unit of capacitance is the farad (F) in honor of
Michael Faraday.
• 1 farad = 1 coulomb/volt.
6.1.1. Circuit symbol for capacitor of C farads:
71
72
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
6.1.2. Since i =
itor is
dq
dt ,
then the current-voltage relationship of the capac-
dv
.
dt
Note that in (6.2), the capacitance value C is constant (time-invariant) and
that the current i and voltage v are both functions of time (time-varying).
So, in fact, the full form of (6.2) is
d
i(t) = C v(t).
dt
(6.2)
i=C
Hence, the voltage-current relation is
Z
1 t
v(t) =
i(τ )dτ + v(to )
C to
where v(to ) is the voltage across the capacitor at time to . Note that capacitor voltage depends on the past history of the capacitor current. Hence,
the capacitor has memory.
6.1.3. The instantaneous power delivered to the capacitor is
d
p(t) = i(t) × v(t) = C v(t) v(t).
dt
6.1. CAPACITORS
73
The energy stored in the capacitor is
Z t
1
w(t) =
p(τ )dτ = Cv 2 (t).
2
−∞
In the above calculation, we assume v(−∞) = 0, because the capacitor
was uncharged at t = −∞.
6.1.4. Capacitors are commercially available in different values and
types. Typically, capacitors have values in the picofarad (pF) to microfarad
(µF) range.
6.1.5. Remarks:
(a) The word capacitor is derived from this element’s capacity to store
energy in an electric field.
(b) A capacitor is an open circuit to dc.
When the voltage across a capacitor is not changing with time
(i.e., dc voltage), its derivative wrt. time is dv
dt = 0 and hence the
dv
current through the capacitor is i(t) = C dt = C × 0 = 0.
(c) The voltage on a capacitor cannot change abruptly (since i = C dv
dt )
A discontinuous change in voltage requires an infinite current,
which is physically impossible.
(d) The ideal capacitor does not dissipate energy. It takes power from
the circuit when storing energy in its field and returns previously
stored energy when delivering power to the circuit.
Example 6.1.6. If a 10 µF is connected to a voltage source with
v(t) = 50 sin 2000t
determine the current through the capacitor.
74
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
Example 6.1.7. Determine the voltage across a 2-µF capacitor if the
current through it is
i(t) = 6e−3000t mA
Assume that the initial capacitor voltage (at time t = 0) is zero.
Example 6.1.8. Obtain the energy stored in each capacitor in the figure
below under dc conditions.
6.2. SERIES AND PARALLEL CAPACITORS
75
6.2. Series and Parallel Capacitors
We know from resistive circuits that series-parallel combination is a
powerful tool for simplifying circuits. This technique can be extended to
series-parallel connections of capacitors, which are sometimes encountered.
We desire to replace these capacitors by a single equivalent capacitor Ceq .
6.2.1. The equivalent capacitance of N parallel-connected capacitors
is the sum of the individual capacitance.
Ceq = C1 + C2 + · · · + CN
The equivalent capacitance of N series-connected capacitors is the the
reciprocal of the sum of the reciprocals of the individual capacitances.
1
1
1
1
=
+
+ ··· +
Ceq
C1 C2
CN
Example 6.2.2. Find the Ceq .
76
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
6.3. Inductors
An inductor is a passive element designed to store energy in its magnetic
field. Inductors find numerous applications in electronic and power systems. They are used in power supplies, transformers, radios, TVs, radars,
and electric motors.
6.3.1. Circuit symbol of inductor:
6.3.2. If a current is allowed to pass through an inductor, the voltage
across the inductor is directly proportional to the time rate of change of
the current, i.e.,
d
v(t) = L i(t),
dt
where L is the constant of proportionality called the inductance of the
inductor. The unit of inductance is henry (H), named in honor of Joseph
Henry.
• 1 henry equals 1 volt-second per ampere.
6.3. INDUCTORS
77
6.3.3. By integration, the current-voltage relation is
Z
1 t
i(t) =
v(τ ) dτ + i(to ),
L to
where i(to ) is the current at time to .
6.3.4. The instantaneous power delivered to the inductor is
d
p(t) = v(t) × i(t) = L i(t) i(t)
dt
The energy stored is
Z t
1
w(t) =
p(τ ) dτ = Li2 (t).
2
−∞
6.3.5. Like capacitors, commercially available inductors come in different values and types. Typical practical inductors have inductance values
ranging from a few microhenrys (µH), as in communication systems, to
tens of henrys (H) as in power systems.
6.3.6. Remarks:
(a) An inductor acts like a short circuit to dc.
The voltage across an inductor is zero when the current is constant.
(b) The current through an inductor cannot change instantaneously.
This opposition to the change in current is an important property
of the inductor. A discontinuous change in the current through an
inductor requires an infinite voltage, which is not physically possible.
(c) The ideal inductor does not dissipate energy. The energy stored in
it can be retrieved at a later time. The inductor takes power from
the circuit when storing energy and delivers power to the circuit
when returning previously stored energy.
78
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
Example 6.3.7. If the current through a 1-mH inductor is i(t) =
20 cos 100t mA, find the terminal voltage and the energy stored.
Example 6.3.8. Find the current through a 5-H inductor if the voltage
across it is
(
30t2 , t > 0
v(t) =
.
0,
t<0
In addition, find the energy stored within 0 < t < 5 s.
6.3. INDUCTORS
79
Example 6.3.9. The terminal voltage of a 2-mH inductor is v = 10(1−t)
V. Find the current flowing through it at t = 4 s and the energy stored in
it within 0 < t < 4 s. Assume i(0) = 2 A.
Example 6.3.10. Determine vC , iL and the energy stored in the capacitor and inductor in the following circuit under dc conditions.
80
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
Example 6.3.11. Determine vC , iL and the energy stored in the capacitor and inductor in the following circuit under dc conditions.
6.4. SERIES AND PARALLEL INDUCTORS
81
6.4. Series and Parallel Inductors
The equivalent inductance of N series-connected inductors is the sum
of the individual inductances, i.e.,
Leq = L1 + L2 + · · · + LN
The equivalent inductance of N parallel inductors is the reciprocal of
the sum of the reciprocals of the individual inductances, i.e.,
1
1
1
1
=
+
+ ··· +
Leq
L1 L2
LN
Note that
(a) inductors in series are combined in exactly the same way as resistors
in series and
(b) inductors in parallel are combined in the same way as resistors in
parallel.
tors in parallel.
For two inductors in parallel (N = 2), Eq. (6.30) becomes
1
1
1
L 1 L2
=
+
or
Leq =
Leq
L1
L2
L1 + L 2
(6.31)
It is appropriate
at this point
to summarize the most important character82
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND
istics of the three basic circuit elements we have studied. The summary
is given in Table 6.1.
TABLE 6.1
INDUCTORS
Important characteristics of the basic elements.†
Relation
Resistor (R)
v-i:
v = iR
i-v:
i = v/R
Capacitor (C)
1 t
v=
i dt + v(t0 )
C t0
v2
R
p = i2R =
Series:
Req = R1 + R2
Parallel:
Req =
At dc:
Circuit variable
that cannot
change abruptly:
R1 R2
R1 + R 2
w=
v=L
dv
dt
i=
1 2
Cv
2
w=
i=C
p or w:
Inductor (L)
Ceq =
C1 C2
C1 + C2
1
L
di
dt
t
i dt + i(t0 )
t0
1 2
Li
2
Leq = L1 + L2
L1 L2
L1 + L 2
Ceq = C1 + C2
Leq =
Same
Open circuit
Short circuit
Not applicable
v
i
† Passive sign convention is assumed.
E X A M P L Example
E 6 . 1 1 6.4.1. Find the equivalent inductance Leq of the circuit shown
below.
Find the equivalent inductance of the circuit shown in Fig. 6.31.
6.5. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS
83
6.5. Applications: Integrators and Differentiators
Capacitors and inductors possess the following three special properties
that make them very useful in electric circuits:
(a) The capacity to store energy makes them useful as temporary voltage or current sources. Thus, they can be used for generating a
large amount of current or voltage for a short period of time.
(b) Capacitors oppose any abrupt change in voltage, while inductors
oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating
dc voltage into relatively smooth dc voltage.
(c) Capacitors and inductors are frequency sensitive. This property
makes them useful for frequency discrimination.
The first two properties are put to use in dc circuits, while the third
one is taken advantage of in ac circuits.
In this final part of the chapter, we will consider two applications involving capacitors and op amps: integrator and differentiator.
84
6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS
6.5.1. Integrator. An integrator is an op amp circuit whose output is
proportional to the integral of the input signal. We obtain an integrator by
replacing the feedback resistor Rf in the inverting amplifier by a capacitor.
This gives
d
1
vo (t) = −
vi (t),
dt
RC
which implies
Z t
1
vo (t) = −
vi (τ )dτ + vo (0).
RC 0
To ensure that vo (0) = 0, it is always necessary to discharge the integrators
capacitor prior to the application of a signal.
In practice, the op amp integrator requires a feedback resistor to reduce
dc gain and prevent saturation. Care must be taken that the op amp
operates within the linear range so that it does not saturate.
6.5. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS
85
6.5.2. Differentiator. A differentiator is an op amp circuit whose output is proportional to the differentiation of the input signal. We obtain a
differentiator by replacing the input resistor in the inverting amplifier by
a capacitor. This gives
d
vi (t).
dt
Differentiator circuits are electronically unstable because any electrical
noise within the circuit is exaggerated by the differentiator. For this reason, the differentiator circuit above is not as useful and popular as the
integrator. It is seldom used in practice.
vo (t) = −RC
ECS 203 (ME2) - Part 3A
Dr.Prapun Suksompong
CHAPTER 9
Sinusoids and Phasors
We now begins the analysis of circuits in which the voltage or current
sources are time-varying. In this chapter, we are particularly interested in
sinusoidally time-varying excitation, or simply, excitation by a sinusoid.
9.0.1. Some terminology:
(a) A sinusoid is a signal that has the form of the sine or cosine function.
• Turn out that you can express them all under the same notation
using only cosine (or only sine) function. We ill use cosine.
(b) A sinusoidal current is referred to as alternating current (AC).
(c) Circuits driven by sinusoidal current or voltage sources are called
AC circuits.
9.1. Sinusoids
9.1.1. Consider the sinusoidal signal (in cosine form)
x(t) = Xm cos(ωt + φ) = Xm cos(2πf t + φ),
where
Xm : the amplitude of the sinusoid,
ω: the angular frequency in radians/s (or rad/s),
φ: the phase.
• First, we consider the case when φ = 0:
109
110
9. SINUSOIDS AND PHASORS
• When φ 6= 0, we shift the graph of Xm cos(ωt) to the left “by φ”.
9.1.2. The period (the time of one complete cycle) of the sinusoid is
2π
T =
.
ω
The unit of the period is in second if the angular frequency unit is in radian
per second.
The frequency f (the number of cycles per second or hertz (Hz)) is
the reciprocal of this quantity, i.e.,
1
f= .
T
9.1.3. Standard form for sinusoid: In this class, when you are asked
to find the sinusoid representation of a signal, make sure that your answer
is in the form
x(t) = Xm cos(ωt + φ) = Xm cos(2πf t + φ),
where Xm is nonnegative and φ is between −180◦ and +180◦ .
• When the signal is given in the sine form, it can be converted into
its cosine form via the identity
In particular,
sin(x) = cos(x − 90◦ ).
Xm sin(ωt + φ) = Xm cos(ωt + φ − 90◦ ).
9.2. PHASORS
111
• Xm is always non-negative. We can avoid having the negative sign
by the following conversion:
− cos(x) = cos(x ± 180◦ ).
In particular,
−A cos(ωt + φ) = A cos(2πf t + φ ± 180◦ ).
Note that usually you do not have the choice between +180◦ or
−180◦ . The one that you need to use is the one that makes φ ± 180◦
falls somewhere between −180◦ and +180◦ .
9.2. Phasors
Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions. The tradoff is that
phasors are complex-valued.
9.2.1. The idea of phasor representation is based on Euler’s identity:
ejφ = cos φ + j sin φ,
From the identity, we may regard cos φ and sin φ as the real and imaginary
parts of ejφ :
cos φ = Re ejφ , sin φ = Im ejφ ,
where Re and Im stand for the real part of and the imaginary part of ejφ .
9.2.2. A phasor is a complex number that represents the amplitude and
phase of a sinusoid. Given a sinusoid v(t) = Vm cos(ωt + φ), then
n
o
j(ωt+φ)
v(t) = Vm cos(ωt+φ) = Re Vm e
= Re Vm ejφ · ejωt = Re Vejωt ,
where
V = Vm ejφ = Vm ∠φ.
V is called the phasor representation of the sinusoid v(t). In other
words, a phasor is a complex number that represents amplitude and phase
of a sinusoid.
9.2.3. Remarks:
• Whenever a sinusoid is expressed as a phasor, the term ejωt is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the (angular) frequency ω of the phasor.
112
9. SINUSOIDS AND PHASORS
• To obtain the sinusoid corresponding to a given phasor V, multiply
the phasor by the time factor ejωt and take the real part.
Equivalently, given a phasor, we obtain the time-domain representation as the cosine function with the same magnitude as the
phasor and the argument as ωt plus the phase of the phasor.
• Any complex number z (including any phasor) can be equivalently
represented in three forms.
(a) Rectangular form: z = x + jy.
(b) Polar form: z = r∠φ.
(c) Exponential form: z = rejφ
where the relations between them are
p
y
r = x2 + y 2 , φ = tan−1 ± 180◦ .
x
x = r cos φ, y = r sin φ.
Note that for φ, the choice of using +180◦ or −180◦ in the formula
is determined by the actual quadrant in which the complex number
lies.
• As a complex quantity, a phasor may be expressed in rectangular
form, polar form, or exponential form. In this class, we focus on
polar form.
9.2.4. Summary : By suppressing the time factor, we transform the
sinusoid from the time domain to the phasor domain. This transformation
is summarized as follows:
v(t) = Vm cos(ωt + φ) ⇔ V = Vm ∠φ.
Time domain representation ⇔ Phasor domain representation
9.2. PHASORS
113
9.2.5. Standard form for phasor: In this class, when you are asked
to find the phasor representation of a signal, make sure that your answer
is a complex number in polar form, i.e. r∠φ where r is nonnegative and φ
is between −180◦ and +180◦ .
Example 9.2.6. Transform these sinusoids to phasors:
(a) i = 6 cos(50t − 40◦ ) A
(b) v = −4 sin(30t + 50◦ ) V
Example 9.2.7. Find the sinusoids represented by these phasors:
(a) I = −3 + j4 A
◦
(b) V = j8e−j20 V
9.2.8. The differences between v(t) and V should be emphasized:
(a) v(t) is the instantaneous or time-domain representation, while V is
the frequency or phasor-domain representation.
(b) v(t) is time dependent, while V is not.
(c) v(t) is always real with no complex term, while V is generally complex.
9.2.9. Adding sinusoids of the same frequency is equivalent to adding
their corresponding phasors. To see this,
A1 cos (ωt + φ1 ) + A2 cos (ωt + φ2 ) = Re A1 ejωt + Re A2 ejωt
= Re (A1 + A2 ) ejωt .
114
9. SINUSOIDS AND PHASORS
9.2.10. Properties involving differentiation and integration:
(a) Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by jω. In other words,
dv(t)
⇔ jωV.
dt
To see this, suppose v(t) = Vm cos(ωt + φ). Then,
dv
(t) = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ − 90◦ + 180◦ )
dt
◦
= Re ωVm ejφ ej90 · ejωt = Re jωVejωt
Alternatively, express v(t) as
n
o
j(ωt+φ)
v(t) = Re Vm e
.
Then,
n
o
d
j(ωt+φ)
v(t) = Re Vm jωe
.
dt
(b) Integrating a sinusoid is equivalent to dividing its corresponding
phasor by jω. In other words,
Z
V
v(t)dt ⇔
.
jω
Example 9.2.11. Find the voltage v(t) in a circuit described by the
intergrodifferential equation
Z
dv
2 + 5v + 10 vdt = 20 cos(5t − 30◦ )
dt
using the phasor approach.
9.3. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS
115
9.3. Phasor relationships for circuit elements
9.3.1. Resistor R: If the current through a resistor R is
i(t) = Im cos(ωt + φ) ⇔ I = Im ∠φ,
the voltage across it is given by
v(t) = i(t)R = RIm cos(ωt + φ).
The phasor of the voltage is
V = RIm ∠φ.
Hence,
V = IR.
We note that voltage and current are in phase and that the voltage-current
relation for the resistor in the phasor domain continues to be Ohms law,
as in the time domain.
116
9. SINUSOIDS AND PHASORS
9.3.2. Capacitor C: If the voltage across a capacitor C is
v(t) = Vm cos(ωt + φ) ⇔ V = Vm ∠φ,
the current through it is given by
dv(t)
i(t) = C
⇔ I = jωCV = ωCVm ∠(φ + 90◦ ).
dt
The voltage and current are 90◦ out of phase. Specifically, the current
leads the voltage by 90◦ .
• Mnemonic: CIVIL
In a Capacitive (C) circuit, I leads V. In an inductive (L) circuit,
V leads V.
9.3. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS
117
9.3.3. Inductor L: If the current through an inductor L is
i(t) = Im cos(ωt + φ) ⇔ I = Im ∠φ,
the voltage across it is given by
di(t)
v(t) = L
⇔ V = jωLI = ωLIm ∠(φ + 90◦ ).
dt
9
Sinusoidsthe
andcurrent
Phasors
The voltage and current are 90◦ out ofCHAPTER
phase. Specifically,
lags the voltage by 90◦ .
relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2
summarizes the time-domain and phasor-domain representations of the
circuit elements.
Im
I
TABLE 9.2
Summary of voltage-current
relationships.
Element
Time domain
Frequency domain
R
v = Ri
L
v=L
di
dt
V = j ωLI
C
i=C
dv
dt
V=
E X A M P L E 9 . 8
V = RI
I
j ωC
Figure 9.14
118
9. SINUSOIDS AND PHASORS
9.4. Impedance and Admittance
Thus, we obtained the voltage current relations for the three passive
elements as
V = IR, V = jωLI, I = jωCV.
These equations may be written in terms of the ratio of the phasor voltage
to the phasor of current as
V
V
V
1
= R,
= jωL,
=
.
I
I
I
jωC
From these equations, we obtain Ohm’s law in phasor form for any type of
element as
V
or V = IZ.
Z=
I
Definition 9.4.1. The impedance Z of a circuit is the ratio of the
phasor voltage V to the phasor current I, measured in ohms (Ω).
As a complex quantity, the impedance may be expressed in rectangular
form as
Z = R + jX = |Z|∠θ,
with
p
X
|Z| = R2 + X 2 , θ = tan−1 , R = |Z| cos θ, X = |Z| sin θ.
R
R = Re {Z} is called the resistance and X = Im {Z} is called the reactance.
The reactance X may be positive or negative. We say that the impedance
is inductive when X is positive or capacitive when X is negative.
Definition 9.4.2. The admittance (Y) is the reciprocal of impedance,
measured in Siemens (S). The admittance of an element(or a circuit) is the
ratio of the phasor current through it to phasor voltage across it, or
1
I
Y= = .
Z V
9.4. IMPEDANCE AND ADMITTANCE
119
9.4.3. Kirchhoff ’s laws (KCL and KVL) hold in the phasor
form.
To see this, suppose v1 , v2 , . . . , vn are the voltages around a closed loop,
then
v1 + v2 + · · · + vn = 0.
If each voltage vi is a sinusoid, i.e.
vi = Vmi cos(ωt + φi ) = Re Vi ejωt
with phasor Vi = Vmi ∠φi = Vmi ejφi , then
Re (V1 + V2 + · · · + Vn ) ejωt = 0,
which must be true for all time t. To satisfy this, we need
V1 + V2 + · · · + Vn = 0.
Hence, KVL holds for phasors.
Similarly, we can show that KCL holds in the frequency domain, i.e., if
the currents i1 , i2 , . . . , in be the currents entering or leaving a closed surface
at time t, then
i1 + i2 + · · · + in = 0.
If the currents are sinusoids and I1 , I2 , . . . , In are their phasor forms, then
I1 + I2 + · · · + In = 0.
Major Implication: Since Ohm’s Law and Kirchoff’s Laws hold in
phasor domain, all resistance combination, analysis methods (nodal
and mesh analysis) and circuit theorems (linearity, superposition, source
transformation, and Thevenin’s and Norton’s equivalent circuits) that we
have previously studied for dc circuits apply to ac circuits!!!
Just think of impedance as a complex-valued resistance!!
In addition, our ac circuits can now effortlessly include capacitors and
inductors which can be considered as impedances whose values depend on
the frequency ω of the ac sources!!
120
9. SINUSOIDS AND PHASORS
9.5. Impedance Combinations
Consider N series-connected impedances as shown below.
The same current I flows through the impedances. Applying KVL
around the loop gives
V = V1 + V2 + · · · + VN = I(Z1 + Z2 + · · · + ZN )
The equivalent impedance at the input terminals is
V
Zeq =
= Z1 + Z2 + · · · + ZN .
I
In particular, if N = 2, the current through the impedance is
I=
V
.
Z1 + Z2
Because V1 = Z1 I and V2 = Z2 I,
Z1
Z2
V1 =
V, V2 =
V
Z1 + Z2
Z1 + Z2
which is the voltage-division relationship.
9.5. IMPEDANCE COMBINATIONS
121
Now, consider N parallel-connected impedances as shown below.
The voltage across each impedance is the same. Applying KCL at the
top node gives
1
1
1
+
+ ··· +
.
I = I1 + I2 + · · · + IN = V
Z1 Z2
ZN
The equivalent impedance Zeq can be found from
1
I
1
1
1
=
=
+
+ ··· +
.
Zeq
V Z1 Z2
ZN
When N = 2,
Z1 Z2
.
Zeq =
Z1 + Z2
Because
V = IZeq = I1 Z1 = I2 Z2 ,
we have
Z2
Z1
I1 =
I, I2 =
I
Z1 + Z2
Z1 + Z2
which is the current-division principle.
122
9. SINUSOIDS AND PHASORS
Example 9.5.1. Find the input impedance of the circuit below. Assume
that the circuit operates at ω = 50 rad/s.
Example 9.5.2. Determine vo (t) in the circuit below.
ECS 203 (ME2) - Part 3B
Dr.Prapun Suksompong
CHAPTER 10
Sinusoidal Steady State Analysis
10.1. General Approach
In the previous chapter, we have learned that the steady-state response
of a circuit to sinusoidal inputs can be obtained by using phasors. In this
chapter, we present many examples in which nodal analysis, mesh analysis,
Thevenin’s theorem, superposition, and source transformations are applied
in analyzing ac circuits.
10.1.1. Steps to analyze ac circuits, using phasor domain:
Step 1. Transform the circuit to the phasor or frequency domain.
• Not necessary if the problem is specified in the frequency domain.
Step 2. Solve the problem using circuit techniques (e.g., nodal analysis,
mesh analysis, Thevenin’s theorem, superposition, or source transformations )
• The analysis is performed in the same manner as dc circuit
analysis except that complex numbers are involved.
Step 3. Transform the resulting phasor back to the time domain.
10.1.2. ac circuits are linear (they are just composed of sources and
impedances)
10.1.3. The superposition theorem applies to ac circuits the same
way it applies to dc circuits. This is the case when all the sources in the
circuit operate at the same frequency. If they are operating at different
frequency, see Section 10.2.
123
124
10. SINUSOIDAL STEADY STATE ANALYSIS
10.1.4. Source transformation:
Vs = Zs Is ,
Is =
Vs
.
Zs
10.1.5. Thevenin and Norton Equivalent circuits:
VTh = ZN IN ,
ZTh = ZN
10.1. GENERAL APPROACH
125
Example 10.1.6. Compute V1 and V2 in the circuit below using nodal
analysis.
Example 10.1.7. Determine current Io in the circuit below using mesh
analysis.
The voltage-to-current converter, as shown in Figure 8-2, produces an output current
that depends on the input voltage and the resistor R. In particular, the output current
Iout = Vi/R
independent of the loading resistance RL.
126
10. SINUSOIDAL STEADY STATE ANALYSIS
The current-to-voltage converter, as shown in Figure 8-3, produces an output voltage
that depends on the input current and the resistor R. In particular, the output voltage
Example 10.1.8. Find the Thevenin
equivalent at terminals a-b of the
Vo = -IinR
circuit
below.
independent
of the size of the loading resistance RL.
R
V+
Vi
+
V+
Iin
Iout
V-
+
V-
RL
+
Vo
-
RL
R
Figure 8-2: Voltage-to-current converter.
Figure 8-3: Current-to-voltage converter.
4. An
integrating 10.1.9.
amplifier Op
is shown
in Figure
Example
Amp
AC 8-4a.
Circuits:
Find the (closed-loop) gain
of the circuit below.
+ vC C
R
iin
vi
iC
V+
X
+
V-
Figure 8-4a: Integrating amplifier
2
+
vo
-
10.2. CIRCUIT WITH MULTIPLE SOURCES OPERATING AT DIFFERENT FREQUENCIES
127
10.2. Circuit With Multiple Sources Operating At Different
Frequencies
A special care is needed if the circuit has multiple sources operating
at different frequencies. In which case, one must add the responses due
to the individual frequencies in the time domain. In other words, the
superposition still works but
(a) We must have a different frequency-domain circuit for each frequency.
(b) The total response must be obtained by adding the individual response in the time domain.
10.2.1. Since the impedance depend on frequency, it is incorrect to try
to add the responses in the phasor or frequency domain. To see this note
that the exponential factor ejωt is implicit in sinusoidal analysis, and that
factor would change for every angular frequency ω. In particular, although
(
!
)
X
X X
Vmi cos(ωt + φi ) =
Re Vi ejωt = Re
Vi ejωt ,
i
i
i
when we allow ω to be different for each sinusoid, generally
(
!
)
X
X X
Vmi cos(ωi t + φi ) =
Re Vi ejωi t 6= Re
Vi ejωi t .
i
i
i
Therefore, it does not make sense to add responses at different frequencies
in the phasor domain.
10.2.2. The Thevenin or Norton equivalent circuit (if needed) must be
determined at each frequency and we have one equivalent circuit for each
frequency.
128
10. SINUSOIDAL STEADY STATE ANALYSIS
Example 10.2.3. Find vo in the circuit below using the superposition
theorem.
ECS 203 (ME2) - Part 3C
Dr.Prapun Suksompong
CHAPTER 11
AC Power Analysis
Our effort in ac circuit analysis so far has been focused mainly on calculating voltage and current. The major concern in this chapter is power
analysis.
11.0.4. Power is the most important quantity in electric utilities, electronic and communication systems because such systems involve transmission of power (or energy) from one point to another.
Every industrial and household electrical device (every fan, motor, lamp,
pressing iron, TV, personal computer) has a power rating that indicates
how much power the equipment requires; exceeding the power rating can
do permanent damage to an appliance.
11.0.5. The most common form of electric power is 50-Hz (Thailand)
or 60-Hz (United States) ac power. The choice of ac over dc allowed highvoltage power transmission from the power generating plant to the consumer. (DC attenuation is high.)
11.1. Instantaneous Power
Definition 11.1.1. The instantaneous power p(t) absorbed by an
element is the product of the instantaneous voltage v(t) across the element
and the instantaneous current i(t) through it.
Assuming the passive sign convention as shown on Figure 1,
p(t) = v(t)i(t).
The instantaneous power is the power at any instant of time. It is the
rate at which an element absorbs energy.
129
130
11. AC POWER ANALYSIS
Consider the general case of instantaneous power absorbed by an arbitrary combination of circuit elements under sinusoidal excitation.
Figure 1. Sinusoidal source and passive linear circuit
Let the voltage and current at the terminals of the circuit be
v(t) = Vm cos(ωt + θv )
and
i(t) = Im cos(ωt + θi )
where Vm and Im are the amplitudes, and θv and θi are the phase of the
voltage and current, respectively. The instantaneous power absorbed by
the circuit is
p(t) = v(t)i(t) = Vm Im cos(ωt + θv ) cos(ωt + θi )
ej(ωt+θv ) + e−j(ωt+θv ) ej(ωt+θi ) + e−j(ωt+θi )
= Vm Im
2
2
1 j(2ωt+θv +θi )
j(θi −θv )
j(θv −θi )
−j(2ωt+θv +θi )
= Vm Im e
+e
+e
+e
4
1 ej(θv −θi ) + ej(θi −θv ) ej(2ωt+θv +θi ) + e−j(2ωt+θv +θi )
+
= Vm Im
2
2
2
1
= Vm Im (cos (θv − θi ) + cos (2ωt + θv + θi )) .
2
Alternatively, we can apply the trigonometric identity
1
cos A cos B = {cos(A − B) + cos(A + B)}
2
to directly arrive at the same result which is
1
1
p(t) = Vm Im cos(θv − θi ) + Vm Im cos(2ωt + θv + θi ) .
|2
{z
} |2
{z
}
constant term
time-dependent term
11.1. INSTANTANEOUS POWER
131
This shows that the instantaneous power has two parts.
• First Part: a constant or time-independent term. Its value depends on the phase difference between the voltage and the current.
• Second Part: a sinusoidal function whose frequency is 2ω, which
is twice the angular frequency of the voltage or current.
11.1.2. Consider the sketch of p(t), we observe that
(a) p(t) is periodic and has a period of To = T2 , where T = 2π
ω is the
period of the voltage and the current
(b) p(t) may become positive for some part(s) of each cycle and negative
for the rest of the cycle.
• When p(t) is positive, power is absorbed by the circuit.
• When p(t) is negative, power is absorbed by the source.
– In this case, power is transferred from the circuit to the
source.
– This is possible because of the storage elements (capacitors and inductors) in the circuit.
132
11. AC POWER ANALYSIS
11.2. Average Power
The instantaneous power changes with time and is therefore difficult to
measure. The average power is more convenient to measure.
Definition 11.2.1. The average power is the average of the
instantaneous power over one period.
Thus, the average power is given by
Z
1 T
1
P =
p(t)dt = Vm Im cos(θv − θi ).
T 0
2
Since cos(θv − θi ) = cos(θi − θv ), what is important is the difference in
the phases of the voltage and the current. Note that p(t) is time varying
while P does not depend on time.
11.2.2. Using the phasor forms of v(t) and i(t), which are V = Vm ∠θv
and I = Im ∠θi , we obtain
1
1
P = Vm Im cos(θv − θi ) = Re{VI∗ }.
2
2
11.2.3. Two special cases:
Case 1: When θv = θi , the voltage and the current are in phase. This implies
a purely resistive circuit or resistive load R, and
1
P = Vm Im
2
This shows that a purely resistive circuit (e.g. resistive load
(R)) absorbs power all times.
Case 2: When θv − θi = ±90◦ , we have a purely reactive circuit, and
1
P = Vm Im cos(90◦ ) = 0
2
showing that a purely reactive circuit (e.g. a reactive load L
or C) absorbs no average power.
11.2. AVERAGE POWER
133
11.2.4. From Ohm’s law, we have two more formula.
(a) The average power absorbed by an impedance Z when a voltage V
is applied across it is
1
1
V∗
1
1
1
Re{Z}
∗
P = Re{VI } = Re{V ∗ } = |V|2 Re{ ∗ } = |V|2
2
2
Z
2
Z
2
|Z|2
(b) The average power absorbed by an impedance Z when a current I
flows through it is
1
1
1
P = Re{VI∗ } = Re{IZI∗ } = |I|2 Re{Z}
2
2
2
Example 11.2.5. Calculate the average power absorbed by an impedance
Z = 30 − j70 Ω when a voltage V = 120∠0◦ is applied across it.
Example 11.2.6. A current I = 10∠30◦ flows through an impedance
Z = 20∠ − 22◦ Ω. Find the average power delivered to the impedance.
Example 11.2.7. For the circuit shown below, find the average power
supplied by the source and the average power absorbed by the resistor.
134
11. AC POWER ANALYSIS
11.3. Maximum Average Power Transfer
11.3.1. Optimal Load Impedance: Before the midterm, we solved
the problem of maximizing the power delivered by a power-supplying resistive network to a load RL . Representing the circuit by its Thevenin
equivalent, we proved that the maximum power would be delivered to the
load if the load resistance is equal to the Thevenin resistance RL = RT h .
We now extend that result to ac circuits.
Consider an ac circuit which is connected to a load ZL and is represented
by its Thevenin equivalent.
In a rectangular form, the Thevenin impedance ZTh and load impedance
ZL are
ZTh = RT h + jXT h
ZL = RL + jXL
The current through the load is
I=
VTh
,
ZL + ZTh
and the average power delivered to the load is
1
1
RL
P = |I|2 RL = VTh 2 2
2
2
(RL + RT h ) + (XL + XT h )2
Our objective is to adjust the load parameter RL and XL so that P is
∂P
∂P
maximum. To do this we set ∂R
and ∂X
equal to zero.
L
L
11.3. MAXIMUM AVERAGE POWER TRANSFER
Setting
Setting
∂P
∂XL
∂P
∂RL
135
= 0 gives
= 0 gives
XL = −XT h .
q
RL = RT2 h + (XT h + XL )2
Hence, to get the maximum average power transfer, the load impedance
ZL must selected so that
i.e.,
XL = −XT h and RL = RT h ,
ZL = RL + jXL = RT h − jXT h = Z∗Th
That is, for the maximum average power transfer, the load
impedance ZL must be equal to the complex conjugate of the
Thevenin impedance ZTh .
When ZL = Z∗Th , the maximum average power is
Pmax
|VTh |2
.
=
8RT h
Example 11.3.2. Determine the load impedance ZL that maximizes
the average power drawn from the circuit below. What is the maximum
average power?
136
11. AC POWER ANALYSIS
11.3.3. Optimal Purely Resistive Load: In a situation in which the
load must be purely real; that is XL must be 0. Then,
1
1
RL
P = |I|2 RL = VT2 h 2
2
(RL + RT h )2 + (XT h )2
Setting
∂P
∂RL
= 0 gives
q
RL = RT2 h + XT2 h = |ZTh |.
Hence, for maximum average power transfer to a purely resistive load, the load impedance is equal to the magnitude of the
Thevenin impedance ZTh . In which case, the maximum average power
is
1
1
P = VTh 2 4
|ZTh | + RT h
Note that
|ZTh | + RT h ≥ RT h + RT h = 2RT h
Hence,
2
V
1
1 Th
1
VTh 2 ≤
.
4
|ZTh | + RT h
8 RT h
11.4. EFFECTIVE OR RMS VALUE
137
11.4. Effective or RMS Value
The idea of effective value arises from the need to measure the effectiveness of an ac voltage or current source in delivering power to a resistive
load.
Definition 11.4.1. The effective value Ieff of a periodic current
i(t) is the dc current that delivers the same average power to a resistor as
the periodic current.
11.4.2. Consider the following ac and dc circuits,
our objective is to find the current Ieff that will transfer the same power
to the resistor R as the sinusoid current i
The average power absorbed by the resistor in the ac circuit is
1
P =
T
Z
0
T
R
i (t)Rdt =
T
2
Z
T
i2 (t)dt.
0
Q: Where don’t we have the factor of 12 ?
The power absorbed by the resistor in the dc circuit is
2
P = Ieff
R.
138
11. AC POWER ANALYSIS
Equating the two expressions and solving for Ieff , we obtain
s
Z
1 T 2
Ieff =
i (t)dt
T 0
11.4.3. Similarly, the effective value of the voltage is found in the same
way as current; that is,
s
Z
1 T 2
Veff =
v (t)dt
T 0
11.4.4. This indicates that the effective value is the square root of the
mean (or average) of the square of the periodic signal. Thus, the effective
value is often know as the root mean square, or rms value for short.
We write
Ieff = Irms , Veff = Vrms
11.4.5. Note that the rms value of a constant is the constant itself.
11.4.6. AC Circuit: For a sinusoid x(t) = Xm cos(ωt + θx ), the effective value or rms value is
s
Z
Xm
1 T 2
Xrms =
Xm cos2 (ωt + θx )dt = √ .
T 0
2
In particular, for i(t) = Im cos(ωt + θi ) and v(t) = Vm cos(ωt + θv ), we have
Vm
Im
Irms = √ and Vrms = √ .
2
2
and the average power can be written in terms of the rms values as
1
P = Vm Im cos(θv − θi ) = Vrms Irms cos(θv − θi ) .
2
11.4.7. Similarly, the average power absorbed by a resistor R can be
written as
2
Vrms
2
P = Irms Vrms = Irms R =
.
R
11.5. APPARENT POWER AND POWER FACTOR
139
11.5. Apparent Power and Power Factor
Definition 11.5.1. The apparent power S (in VA) is the product of
the rms values of voltage and current.
S = Vrms Irms
Hence, the average power P = S cos(θv − θi ).
Definition 11.5.2. The power factor (pf) is the ratio of the average
power to the apparent power.
P
pf = = cos(θv − θi ).
S
Hence,
average power P = apparent power S × power factor pf.
The angle θv − θi is called the power factor angle which is equal to
the angle of the load impedance if V = Vm ∠θv is the voltage across the
load and I = Im ∠θi is the current through it. This is evident from the fact
that
V Vm ∠θv
Vm
Z=
=
=
∠(θv − θi ).
I
Im ∠θi
Im
Alternately, define Vrms = √V2 = Vrms ∠Θv and Irms = √I2 = Irms ∠Θi .
The impedance can then be written as
Vrms
V Vrms
=
=
∠(θv − θi ).
Z=
I
Irms
Irms
The power factor is the cosine of the phase difference between
the voltage and current. It is also the cosine of the angle of the
load impedance.
11.5.3. The value of the power factor pf ranges between 0 and 1.
• For a purely resistive load, the voltage and current are in phase, so
that θv − θi = 0 and pf =1. This implies that the average power is
equal to the apparent power .
• For a purely reactive load, θv − θi = ±90◦ . Hence, pf = 0. In this
case the average power is zero.
• In between these two extreme cases, pf is said to be leading or
lagging. Leading power factor means that current leads voltage
which implies a capacitive load. Lagging power factor means
that current lags voltage, implying an inductive load.
140
11. AC POWER ANALYSIS
11.6. Complex Power
The complex power (in VA) S is the product of the rms voltage phasor
and the complex conjugate of the rms current phasor. S is a complex
quantity whose real part is the real or average power P and imaginary
part is the reactive power Q.
Complex power: S
2
Vrms
1 ∗
∗
2
S = P + jQ = VI = Vrms Irms = Vrms Irms ∠(θv − θi ) = Irms Z = ∗ .
2
Z
Note that all previously studied quantities can be derived from the complex power. That is,
The apparent power S is the magnitude of the complex power S, i.e.,
p
S = |S| = Vrms Irms = P 2 + Q2 .
The real or average power P is
2
P = Re {S} = S cos(θv − θi ) = Irms
R.
The reactive power Q is
2
Q = Im {S} = S sin(θv − θi ) = Irms
X.
The power factor pf is
P
pf = = cos(θv − θi ) = cos(phase of S).
S
ECS 203 (ME2) - Part 4A
Dr.Prapun Suksompong
CHAPTER 7
First-Order Circuits
In Chapter 6, we have studied the relationships between current and
voltage of capacitors/inductors. These elements can be used to store energy
and release energy when needed. In this chapter, we will see how the
voltage or current behaves during the charging/discharging of these storage
elements.
First, we will examine two types of simple circuits with a storage element:
(a) A circuit with a resistor and one capacitor (called an RC circuit);
and
(b) A circuit with a resistor and an inductor (called an RL circuit).
These circuits may look simple but they find continual applications in
electronics, communications, and control systems.
7.0.1. Applying Kirchhoff’s laws to the RC and RL circuits produce
first order differential equations. Hence, the circuits are collectively
known as first-order circuits.
7.0.2. There are two ways to excite the circuits.
(a) By initial conditions of the storage elements in the circuit.
• Also known as source-free circuits
• Assume that energy is initially stored in the capacitive or inductive element.
• This is the discharging process.
(b) By using independent sources
• This is the charging process
• For this chapter, we will consider independent dc sources.
87
88
7. FIRST-ORDER CIRCUITS
Before we start our circuit analysis, it is helpful to consider one mathematical fact which we will use through out this chapter:
7.0.3. The solution of the first-order differential equation
d
x(t) = ax + b
dt
is given by
b
b
(7.3)
x(t) = ea(t−t0 ) x(t0 ) +
− .
a
a
In addition, if a < 0, the solution is given by
(7.4)
x(t) = ea(t−t0 ) (x(t0 ) − x(∞)) + x(∞).
where x(∞) = limt→∞ x(t).
The results above take a bit of effort to proof. However, if you have
MATLAB, you can get 7.3 using one line of code:
dsolve(’Dx = a*x+b’,’x(t0) = x0’, ’t’)
When a < 0, from (7.3), x(∞) = −b/a. We can then get (7.4) by substituting b/a in (7.3) with x(∞).
7.0.4. It turns out that you only have two kinds of plots to worry about
when dealing with (7.4):
7.1. SOURCE-FREE RC CIRCUITS
89
7.1. Source-Free RC Circuits
A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the
resistor.
Consider a series combination of a resistor an initially charged capacitor.
We assume that at time t = 0, the initial voltage is v(0) = Vo . (Hence, the
initial stored energy is w(0) = 21 CVo2 .)
Applying KCL, we get
i C + iR
dv
v
C +
dt R
dv
v
+
dt CR
dv
dt
= 0
= 0
= 0
1
= −
v
CR
Hence,
t
t
v(t) = Vo e− RC = Vo e− τ ,
where τ = RC.
This shows that the voltage response of 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, it is called the natural response of the circuit.
90
7. FIRST-ORDER CIRCUITS
7.1.1. Remarks:
(a) τ = RC is the time constant which is the time required for the
response to decay to 36.8 percent of its initial value. (e ≈ 2.718 and
hence 1/e ≈ 0.368.)
(b) The smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response.
7.1. SOURCE-FREE RC CIRCUITS
91
(c) The voltage v(t) is less than 1 percent of V0 after 5τ (five time
constants). Thus, it is customary to assume that the capacitor is
fully discharged (or charged) after five time constants. In other
words, it takes 5τ for the circuit to reach its final state or steady
state when no changes take place with time.
−t
Vo τ
(d) iR (t) = v(t)
R = Re .
2 −2t
(e) pR (t) = viR = VRo e τ
Rt
−2t
(f) wR (t) = 0 p(µ)dµ = 12 CVo2 (1 − e τ ) = wC (0) − wC (t)
• Notice that as t → ∞, wR → 21 CV02 , which is the same as
wC (0), the energy initially stored in the capacitor.
• The energy that was initially stored in the capacitor is eventually dissipated in the resistor.
7.1.2. In summary: The key in working with a source-free RC circuit
is to determine:
1. The initial voltage v(0) = V0 across the capacitor.
2. The time constant τ (= RC)
92
7. FIRST-ORDER CIRCUITS
7.1.3. There are basically three ways that we can extend what we have
found.
(a) The initial condition v(t0 ) can be given at non-zero t0 . This has
already been taken care of by (7.4).
(b) There can be more than one resistors. In which case, you first simplify the circuit by finding the equivalent resistance at the capacitor
terminals. See Example 7.1.4.
(c) The initial value of the voltage v(t0 ) might not be given but condition(s) of the circuit before time t0 is given instead. In which case,
you may find v(t−
0 ) by considering the long-term behavior of the circuit before t0 . In such situation, you should review two properties
of capacitors:
(i) For DC, capacitor becomes open circuit.
(ii) Voltage across the capacitor cannot change instantaneously
See Example 7.1.5.
Example 7.1.4. Let vC (0) = 15 V. Determine vC (t), vx (t), and ix (t) for
t > 0.
Remark: The answer is the same if we replace t > 0 in the above
question by t ≥ 0.
7.2. SOURCE-FREE RL CIRCUITS
93
In Example 7.1.4, you may ask how can we have the initial voltage of
15 V across the capacitor in the first place. The next example suggest a
way to get the initial voltage.
Example 7.1.5. The switch in the circuit below has been closed for a
long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the
initial energy stored in the capacitor.
94
7. FIRST-ORDER CIRCUITS
7.2. Source-Free RL Circuits
Consider the series connection of a resistor and an inductor. Assume
that the inductor has an initial current Io or i(0) = Io .(hence, the energy
stored in the inductor is w(0) = 21 LIo2 .)
Applying KVL, we get
vL + vR = 0
di
L + Ri = 0
dt
di R
+ i = 0
dt L
Hence,
i(t) = Io e
−Rt
L
−t
= Io e τ .
Note that:
(a) τ = RL is the time constant which is the time required for the response to decay to 36.8 percent of its initial value.
−t
(b) vR (t) = iR = Io Re τ .
−2t
(c) p(t) = vRRi = Io2 Re τ .
−2t
t
(d) wR (t) = 0 p(µ)dµ = 12 LIo2 (1 − e τ ) = wL (0) − wL (t) .
7.2. SOURCE-FREE RL CIRCUITS
95
7.2.1. In summary: The key in working with a source free RL circuit
is to determine:
(a) The initial current i(0) = Io .
(b) The time constant τ (= RL ).
• In general, R is the Thevenin equivalent resistance at the terminal of the inductor.(We take out the inductor L and find
R = RT H at its terminals.)
Example 7.2.2. The switch in the circuit below has been closed for a
long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.
96
7. FIRST-ORDER CIRCUITS
Example 7.2.3. Determine i, io and vo for all t in the circuit. Assume
that the switch was closed for a long time.
7.3. UNIT STEP FUNCTION
97
7.3. Unit step function
We use the step function to represent an abrupt change in voltage or
current, like the changes that occur in the circuits of control systems and
digital computers.
The unit step function u(t) is 0 for negative values of t and 1 for positive
values of t. In mathematical terms,
u(t) =
0, t < 0
1, t > 0
For example, here is a voltage source of V0 u(t)
Similarly, here is a current source of I0 u(t)
98
7. FIRST-ORDER CIRCUITS
We can also use the step functions to represent the gate function (pulse)
which may be regarded as a step function that switches on at one value of
t and switches off at another value of t. The gate function below switches
on at t = 2 s and switches off at t = 5 s.
It consists of the sum of two unit step functions:
v(t) = 10u(t − 2) − 10u(t − 5).
When the dc source of an RC circuit is suddenly applied (i.e., this
happens when the capacitor is being charged), the voltage or current source
can be modeled as a step function, and the response is known as a step
response.
7.4. STEP RESPONSE OF AN RC CIRCUIT
99
7.4. Step Response of an RC Circuit
7.4.1. Consider an RC circuit with voltage step input below:
The voltage across the capacitor is v. We assume an initial voltage
V0 on the capacitor. Since the voltage of a capacitor cannot change instantaneously v(0−) = v(0+) = V0 , where v(0−) is the voltage across the
capacitor just before switching and v(0+) is its voltage immediately after
switching.
For t > 0, applying KCL, we have
dv v − Vs
C +
= 0
dt
R
dv
v
Vs
+
=
dt RC
RC
dv
1
Vs
= −
v+
dt
RC
RC
The solution is
−t
v(t) = Vs + (Vo − Vs )e τ , t ≥ 0.
100
7. FIRST-ORDER CIRCUITS
7.4.2. In general, the step response of an RC circuit whose switch
changes position at time t = 0 can be written as
−t
v(t) = v(∞) + (v(0) − v(∞))e τ ,
t > 0.
where v(0) is the initial voltage and v(∞) is the final or steady-state value.
We obtain
(a) v(0) from the given circuit for t < 0 and
(b) v(∞) and τ from the circuit for t > 0.
7.4.3. If the switch changes position at time t = to instead of at t = 0,
the response becomes
v(t) = v(∞) + (v(to ) − v(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial value at t = to .
7.4.4. The key in finding the step response of an RC circuit is to determine three values:
1. The initial capacitor voltage v(0) or V (t0 ).
2. The final capacitor voltage v(∞).
3. The time constant τ .
7.4. STEP RESPONSE OF AN RC CIRCUIT
101
Example 7.4.5. The switch in the circuit below has been in position A
for a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0
and calculate its value at t = 1 s and 4 s.
Example 7.4.6. Find v(t) for t > 0 in the circuit below. Assume the
switch has been open for a long time and is closed at t = 0. Numerically
evaluate v(t) at t = 0.5.
102
7. FIRST-ORDER CIRCUITS
Example 7.4.7. 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. A simplified circuit is shown
below. It consists essentially of a high-voltage dc supply, a current-limiting
large resistor R1 , and a capacitor C in parallel with the flashlamp of low
resistance R2 .
• When the switch is in position 1, the capacitor charges slowly due
to the large time constant (τ = R1 C). The capacitor voltage rises
gradually from zero to Vs , while its current decreases gradually from
I1 = Vs /R1 to zero.
• With the switch in position 2, the capacitor voltage is discharged.
The low resistance R2 of the photolamp permits a high discharge
current with peak I2 = Vs /R2 in a short duration.
This simple RC circuit provides a short-duration, high current pulse.
Such a circuit also finds applications in electric spot welding and the radar
transmitter tube.
7.4. STEP RESPONSE OF AN RC CIRCUIT
103
Example 7.4.8. An RC circuit can be used to provide various time
delays. Consider the circuit below. It basically consists of an RC circuit
with the capacitor connected in parallel with a neon lamp. The voltage
source can provide enough voltage to fire the lamp.
• When the switch is closed, the capacitor voltage increases gradually
toward 110 V at a rate determined by the circuit’s 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 level, say 70 V.
• When the voltage level is reached, the lamp fires (goes on), 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 and make the lamp fire, recharge, and fire repeatedly
every time constant τ = (R1 + R2 )C, because it takes a time period
τ to get the capacitor voltage high enough to fire or low enough to
turn off.
• The warning blinkers commonly found on road construction sites
are one example of the usefulness of such an RC delay circuit.
104
7. FIRST-ORDER CIRCUITS
7.5. Step Response of an RL Circuit
Here, our goal is to find the inductor current i as the circuit response.
7.5.1. In general, the step response of an RL circuit whose switch
changes position at time t = 0 can be written as
−t
i(t) = i(∞) + (i(0) − i(∞))e τ ,
t > 0.
where i(0) is the initial current and i(∞) is the final or steady-state value.
If the switch changes position at time t = to instead of at t = 0, the
response becomes
i(t) = i(∞) + (i(to ) − i(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial inductor current value at time t = to .
Example 7.5.2. Find i(t) in the circuit below for t > 0. Assume that
the switch has been closed for a long time.
7.5. STEP RESPONSE OF AN RL CIRCUIT
105
Example 7.5.3. The switch in the circuit below has been closed for a
long time. It opens at t = 0. Find i(t) for t > 0.
Example 7.5.4. At t = 0, switch 1 in the circuit below is closed, and
switch 2 is closed 4 s later. Find i(t) for t > 0.
ECS 203 (ME2) - Part 4B
Dr.Prapun Suksompong
CHAPTER 8
Second-Order Circuits
In this chapter, we will consider circuits containing two storage elements.1 They are known as second-order circuits because their responses are described by differential equations that contain second derivatives.
Definition 8.0.5. A second-order circuit is characterized by a
second-order differential equation. It consists of resistors and the
equivalent of two energy storage elements.
8.0.6. Second-order differential equations that we will focus on will have
the following general form:
d
d2
x(t)
+
c
x(t) + c2 x(t) = f (t).
(8.5)
1
dt2
dt
In this chapter2 f (t) will be some constant C. Hence, (8.5) becomes
d
d2
x(t)
+
c
x(t) + c2 x(t) = C.
(8.6)
1
dt2
dt
The solution to (8.6) has two components: the transient response xt (t) and
the steady-state response xss (t); that is;
(8.7)
x(t) = xt (t) + xss (t).
(a) The transient response xt (t) is the component of the total response that dies out with time. To get this, we find the the solution
of (8.6) when C = 0. This part of the solution will have two constants which can be determined by the initial conditions.
(b) It turns out that for (8.6), the steady-state response xss (t) is
given by C/c2 . However, we will use that fact that as t → ∞, the
transient response xt (t) will die out and hence we can use for find
xss (t) from x(∞).
1A second-order circuit may have two storage elements of different type or the same type (provided
elements of the same type cannot be represented by an equivalent single element).
2In Chapter 9 and 10, we work with (8.5) with sinusoidal f (t). Moreover, we shall limit our interest
to the steady-state part of the solution.
103
104
8. SECOND-ORDER CIRCUITS
8.1. Finding Initial and Final Values
This section is explicitly devoted to the subtleties of getting v(0), i(0),
i(∞) and v(∞). Unless otherwise stated in this chapter, v
denotes capacitor voltage, while i is inductor current. We will write vC
and iL when we want to be explicit.
dv(0) di(0)
dt , dt ,
8.1.1. There are two key points to keep in mind in determining the
initial conditions.
(a) v and i are defined strictly according to the passive sign convention.
(b) There is no jump in the capacitor’s voltage and the inductor’s current:
(i) the capacitor voltage is always continuous so that
−
vC (t+
0 ) = vC (t0 ) = vC (t0 ),
and
(ii) the inductor current is always continuous so that
−
iL (t+
0 ) = iL (t0 ) = iL (t0 ),
where t = t−
0 denotes the time just before the (switching) event and
+
t = t0 is the time just after the (switching) event, assuming that
the (switching) event takes place at time t = t0 .
Example 8.1.2. The switch in the figure below has been closed for a
long time. It is open at t = 0. Find i(0+ ), v(0+ ), dtd i(0+ ), dtd v(0+), i(∞),
and v(∞).
8.2. GENERAL STEP RESPONSE OF SECOND-ORDER CIRCUITS
105
8.2. General Step Response of Second-order circuits
8.2.1. Given a second-order circuit, we determine its step response
x(t) (which may be capacitor-voltage or inductor-current) by taking the
following four steps.
Step 1: Determine the initial conditions x(0) and dx(0)
dt and the final value
x(∞).
Step 2: Find the natural response xn (t) by turning off independent sources
and applying KCL and KVL. Once a second-order differential equation is obtained, we determine its characteristic roots. As a general
rule, you will confront with a differential equation of the form
d
d2
x(t)
+
2α
x(t) + ω02 x(t) = 0
2
dt
dt
whose general form of the characteristic equation is
s2 + 2αs + ωo2 = 0.
The two roots of the characteristic equation are
q
q
2
2
s1 = −α + α − ω0 , s2 = −α − α2 − ω02 .
We can infer that there are three types of solutions:
(i) If α > ωo , we have the overdamped case,
xn (t) = A1 es1 t + A2 es2 t .
(ii) If α = ωo , we have the critically damped case,
xn (t) = (A2 + A1 t)e−αt .
(iii) If α < ωo , we have the underdamped case,
xn (t) = e−αt (B1 cos ωd t + B2 sin ωd t),
p
where ωd = ω02 − α2 .
Depending on whether the response is overdamped (distinct
roots), critically damped (repeated roots), or underdamped
(complex conjugated roots), we obtain xn (t) with two unknown
constants.
Step 3: We obtain the force response as
xf (t) = x(∞).
106
8. SECOND-ORDER CIRCUITS
Step 4: The total response is now found as the sum of the natural response
and the 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) and dx(0)
dt , determined in Step 1.
Example 8.2.2. Find the complete response v for t > 0 in the circuit.
Vs := 12
ω := 1
R1 := 4
L := 1
C :=
1
R2 := 2
2
t0 := 0
t1 := 2
Given
v1( t) =
d
v0( t)
dt
d
v1( t) +
dt
⎛ 1 + R1 ⎞ ⋅ v1( t) +
⎜
⎟
⎝ C⋅ R2 L ⎠
Initial Conditions:
⎡ 1 ⋅ ⎛ 1 + R1 ⎞⎤ ⋅ v0( t) = 1 ⋅ Vs
⎢
⎜
⎟⎥
R2 ⎠⎦
L⋅ C
⎣ L⋅ C ⎝
v1( t0) = −12
v0( t0) = 12
⎛ v0 ⎞ := Odesolve⎡⎛ v0 ⎞ , t , t1⎤
⎜ ⎟
⎢⎜ ⎟
⎥
⎝ v1 ⎠
⎣⎝ v1 ⎠
⎦
v ( t) := v0( t)
10
v( t)
− 2⋅ t
4+ 12⋅ e
− 3⋅ t
− 4⋅ e
5
0
0
0.5
1
t
1.5
2
Vs := 12
ω := 1
R1 := 4
L := 1
C :=
1
R2 := 2
2
vs ( t) := Vs⋅ cos( ω⋅ t)
t0 := 0
t1 := 30
Given
v1( t) =
d
v0( t)
dt
d
v1( t) +
dt
⎛ 1 + R1 ⎞ ⋅ v1( t) +
⎜
⎟
⎝ C⋅ R2 L ⎠
Initial Conditions:
⎡ 1 ⋅ ⎛ 1 + R1 ⎞⎤ ⋅ v0( t) = 1 ⋅ vs( t)
⎢
⎜
⎟⎥
R2 ⎠⎦
L⋅ C
⎣ L⋅ C ⎝
v1( t0) = −12
v0( t0) = 12
(assumed)
⎛ v0 ⎞ := Odesolve⎡⎛ v0 ⎞ , t , t1⎤
⎜ ⎟
⎢⎜ ⎟
⎥
⎝ v1 ⎠
⎣⎝ v1 ⎠
⎦
v ( t) := v0( t)
10
5
v( t)
− 2⋅ t
4+ 12⋅ e
− 3⋅ t
− 4⋅ e
0
−5
0
10
20
t
30
AC Analysis
R1 := 4
L := 1
C :=
1
2
R2 := 2
parallel( x , y ) :=
g :=
x⋅ y
x+y
ZC :=
1
parallel( R2 , ZC)
g→
parallel( R2 , ZC ) + R1 + ZL
2
g →
25
ZL := j ⋅ ω⋅ L
j ⋅ ω⋅ C
arg( g ) ⋅
180
π
1
5
−
⎛ 1 ⎞ ⋅i
⎜ ⎟
⎝5⎠
= −45
10
5
v( t)
g ⋅ Vs⋅ cos( ω ⋅ t+ arg( g) )
0
−5
0
10
20
t
30