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Chapter 4 and 5
Time Varying Circuits
Capacitors
Inductors
Time Constant TC
Waveforms
RC Circuits – Step Input
RL Circuits – Step Input
Grossman/ Melkonian
0
CAPACITORS:
Section 4.1
C =

Dielectric Insulator
 = permittivity
A
A
d
d
Metal plates of
area ‘A’
Where:
‘C’ is the capacitance in Farads. The farad is a large unit.
Typical values of capacitance are in F or pF.
‘’ is the permittivity of the dielectric medium
(8.854x10-12 F/m for air).
‘A’ is the cross-sectional area of the two parallel conducting
plates.
‘d’ is the distance between the two plates.
Grossman/ Melkonian
1
CAPACITORS:
+
Dielectric Insulator
 = permittivity
+
Vs
-
Metal plates
of area ‘A’
iC(t)
iC(t)
+
-
+
vC
vC
Standard Notations
(passive sign convention)
-
 When a voltage is applied to the plates, an electric field is produced
that causes an electric charge q(t) to be produced on each plate.
Grossman/ Melkonian
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CAPACITORS:
q (t) = CvC(t)
Charge of a capacitor
since: i(t) = dq(t)/dt
and:
Differentiating: dq(t) = C•dvC(t)
q(t) = CvC(t)
iC(t) = dq(t)/dt = C(dvC(t)/dt )
iC(t) = C[dvC(t)/dt]
i-v relationship for a
capacitor
Grossman/ Melkonian
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CAPACITORS:
 Integrating the i-v relationship of the capacitor:
t
t
 dvC(t) = 1/C iC(t)dt = vC(t)
 Letting VC(0) = V0 represent the initial voltage across the
capacitor at some time t = t0:
t
vC(t) = V0 + 1/C iC(t)dt
t0
t0
iC (t) = C[dvC(t)/dt]
Voltage across a
capacitor
Current through a capacitor
Grossman/ Melkonian
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CAPACITORS:
 Capacitor Power:
pC(t) = iC(t)•vC(t) = C[dvC(t)/dt]vC(t)
pC(t) = d/dt[½CvC2(t)]
Power associated with a capacitor
 Capacitor Energy:
Energy is the integral of power, therefore,
wC(t) = ½CvC2(t)
Energy associated with a capacitor
Grossman/ Melkonian
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CAPACITORS:
Example 1:
Given iC (t) = Io[e-t/Tc] for t  0 and vC(0) = 0V, find the
capacitor’s power and energy.
Power:
pc(t) = iC(t)•vC(t)
t
t
vC(t) = V0 + 1/C  iC(x)dx = 0V + 1/C  Io[e-x/Tc]dx
0
0
vc (t) = [(IoTc/C)(1-e-t/Tc)]V
pC(t) = [Ioe-t/Tc][(IoTC/C)(1-e-t/Tc)]W
iC(t)
vC(t)
pc(t) = Io2Tc/C (e-t/Tc - e-2t/Tc)
Grossman/ Melkonian
Power can be positive
or negative
6
CAPACITORS:
Example 1 cont.:
Energy:
wC(t) = ½CvC2(t)
wC(t) = [(IoTC)2/2C](1-e-t/Tc)2
Energy is always
positive
Waveforms:
iC(t)
Io
iC(t) = Io[e-t/Tc]
t
0
Grossman/ Melkonian
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CAPACITORS:
Example 1 cont.:
vC(t)
IoTc/C
vc(t) = [(IoTc/C)(1-e-t/Tc)]V
t
0
pC(t)
Io2Tc/4C
pc(t) = Io2Tc/C (e-t/Tc - e-2t/Tc)
0
Tcln2
t
Grossman/ Melkonian
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CAPACITORS:
Example 1 cont.:
wc (t)
Io2Tc2/2C
t
0
wC(t) = [(IoTC)2/2C](1-e-t/Tc)2
Grossman/ Melkonian
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CAPACITORS:
Series and Parallel Capacitors:
Series: Capacitors connected in series combine like resistors
connected in parallel.
1/CEQ = 1/C1 + 1/C2 + 1/C3 + 1/C4 +….+ 1/CN
C1
C2
C3
+
i(t)
vc(t)
C4
C7
C6
C5
Grossman/ Melkonian
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CAPACITORS:
Series and Parallel Capacitors:
Parallel: Capacitors connected in parallel add like resistors
connected in series.
CEQ = C1 + C2 + C3 + C4 + … + CN
iC1(t)
vc(t)
iC2(t)
C1
iC3(t)
C2
C3
Grossman/ Melkonian
C4
C5 . . .
CN
11
CAPACITORS:
Properties of the Capacitor:
 The voltage across a capacitor cannot change instantaneously.
 Current through a capacitor can change instantaneously.
 The capacitor stores energy in its electric field.
 The current through a capacitor is zero when the voltage across
the capacitor is constant. For example, when the capacitor is
fully charged. Therefore, a capacitor acts like an open circuit
when a DC voltage is applied.
 Capacitors in parallel add.
 Capacitors in series combine like resistors connected in
parallel.
Grossman/ Melkonian
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INDUCTORS:
vL(t)
+
-
iL(t)
Magnetic Field
Lines
 Inductors are typically made by winding a coil of wire around
an insulator or ferromagnetic material.
 Current flowing through an inductor creates a magnetic
field.
vL(t) = L[diL(t)/dt]
i-v relationship for an
inductor
 Where L is the inductance of the coil measured in henrys (H).
1 H = 1 V-s/A
Grossman/ Melkonian
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INDUCTORS:
 The basic results for the inductor can be derived in the same
way as we have done for the capacitor:
t
iL(t) = iL(0) + 1/L  vL(t) dt
Current through an inductor
pL(t) = iL(t)•vL(t) = d/dt[WL(t)]
WL(t) = 1/2LiL2(t)
Power associated with an inductor
Energy associated with an inductor
Grossman/ Melkonian
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INDUCTORS:
Properties of the Inductor:
 Inductor current is continuous. Cannot change instantaneously.
(Would require infinite power).
 The voltage across an inductor can change instantaneously.
 When the current through an inductor is constant, the voltage is
zero. Therefore, the inductor acts like a short circuit when a
DC source is applied.
 Inductors connected in series add like resistors connected in series.
 Inductors connected in parallel combine like resistors
connected in parallel
Grossman/ Melkonian
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TIME-DEPENDENT SIGNALS:
Section 4.2
 Sources that produce currents or voltages that vary with time
are called time-dependent signal sources.
 The sinusoidal waveform is one of the most important timedependent signals.
V(t) +-
i(t)
V(t), i(t)
Generalized time-dependent signals
Grossman/ Melkonian
+
~
-
Sinusoidal source
16
TIME-DEPENDENT SIGNALS:
 An important class of time-dependent signals is a periodic signal.
 A periodic signal satisfies the following equation:
x(t) = x(t + nT)
n = 1, 2, 3, …
where T is the period of the signal
 A generalized sinusiod is defined as follows:
x(t) = Acos(t + )
where
f = natural frequency = 1/T cycles/s or Hz
 = radian frequency = 2f rad/s
 = 2t/T = 360t/T rad
Grossman/ Melkonian
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TIME-DEPENDENT SIGNALS:
v(t)
v(t)
t
t
Damped sinusoid
Square wave
v(t)
v(t)
  Pulse width

t
t
Pulse train
Sawtooth wave
Grossman/ Melkonian
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TIME-DEPENDENT SIGNALS:
Sinusoidal Waveform
Radial frequency
Amplitude
Phase shift
x(t) = Acos(t + )
Time
A
T
t
Reference cosine
t
Arbitrary sinusoid
A
T
t
Grossman/ Melkonian
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TIME-DEPENDENT SIGNALS:
Root Mean Square
 The average value of a sinusoidal signal is zero, independent of
its amplitude and frequency. Therefore, another method must be
used to quantify the strength of a time-varying signal.
 The operation of computing the Root-Mean-Square of a waveform
is a method for quantifying the strength of a time-varying signal.
T
xrms =

1/T  x2 (t) dt
Root-mean-square
0
Grossman/ Melkonian
20
TIME-DEPENDENT SIGNALS:
Example 2:
Calculate the rms value of the sinusoidal voltage v(t) = Vsin(t).
T
vrms =

1/T  v2 (t) dt
0
T = 1/f = 2/
2/
vrms =  /2  V2sin2(t) dt
0
2/
vrms =

V2/2  1/2 - 1/2cos(2t) dt
0
Grossman/ Melkonian
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TIME-DEPENDENT SIGNALS:
Example 2 cont.:
vrms =  V2/2[t/2 - 1/4sin(2t)]
2/
0
vrms =  V2/2[(2/2 - /4sin(4/) – (0 – 0)]
0
vrms =  V2/2(/ - /4sin(4) – 0)
vrms =
V
2
vrms = 0.707V
Grossman/ Melkonian
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RC CIRCUIT, STEP RESPONSE:
Sections 4.3 (pgs 161 & 162), 5.1, 5.2, 5.3
 Assume we have the following series circuit containing a
voltage source, a resistor, and a capacitor. How do we
write an expression for the voltage across the capacitor?
Switch
t=0
VT
R
iC(t)
+
+
vC(t)
-
C
-
Grossman/ Melkonian
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RC CIRCUIT, STEP RESPONSE:
 Writing KVL for the circuit:
– VS + Ric(t) + vC(t) = 0
ic(t) = C(dvc(t)/dt )
RC(dvC(t)/dt) + vC(t) = VS
RC(dvC(t)/dt) + vC(t) = VS
Switch
t=0
VS
+
-
for t  0
First order linear differential
equation.
R
iC(t)
+
vC(t)
C
-
Grossman/ Melkonian
24
RC CIRCUIT, STEP RESPONSE:
 Since the circuit is linear, we can separate the solution
vC(t) into two components:
vC(t) = vN(t) + vF(t)
 Natural Response:
1. RC(dvN(t)/dt) + vN(t) = 0
t0
Natural Response: VS = 0
 The classical approach to solving this type of an equation is to try
a solution of the following form:
vN(t) = Kest
where K and s are constants
Grossman/ Melkonian
25
RC CIRCUIT, STEP RESPONSE:
 Substituting vN(t) = Kest into the above equation:
RCsKest + Kest = 0
Kest(RCs + 1) = 0
K = 0 is trivial solution
 RCs + 1 = 0
Characteristic Equation
 Solving the Characteristic Equation:
s = -1/RC
 Therefore, the Natural Response has the form:
vN(t) = Ke-t/RC
t>0
Grossman/ Melkonian
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RC CIRCUIT, STEP RESPONSE:
 Forced Response:
2. RC(dvF(t)/dt) + vF(t) = VS
t0
Forced Response
 Equation 2 requires the linear combination of vF(t) and its derivative
equal a constant VS. Setting vF(t) = VS satisfies this condition.
Combining the Natural and Forced Responses:
vC(t) = vN(t) + vF(t) = Ke-t/RC + VS
t>0
(General
Solution)
 Evaluating K using initial conditions (at t=0, the voltage across
the capacitor):
v(0) = V0 = Ke0 + VS = K + VS
This requires K = (V0 – VS)
Grossman/ Melkonian
27
RC CIRCUIT, STEP RESPONSE:
 Substituting into the general solution:
vC(t) = (V0 – VS)e-t/RC + VS
Decaying exponential
t>0
Constant
Important Values:
RC: Defined as the time constant, sometimes written as TC or .
The time constant depends only on fixed circuit parameters.
V0: The initial condition or initial voltage across the capacitor.
Sometimes written as Vi or vC(0)
VS: The Thevenin voltage seen by the capacitor as t. Sometimes
written as Vf or vC().
Grossman/ Melkonian
28
RC CIRCUIT, STEP RESPONSE:
Important Values cont.:
R: The Thevenin resistance or equivalent resistance seen by the
capacitor. Sometimes written as RT.
t(0-): The instant of time just prior to t=0 or the switch changing
its position.
t(0): The time t=0 or the instant the switch changes its position.
t(0+): The instant of time just after t=0 or just after the switch
changes its position.
Grossman/ Melkonian
29
RC CIRCUIT, STEP RESPONSE:
Important Concepts:
 The voltage across a capacitor cannot change instantaneously.
Therefore, vC(0-) = vC(0) = vC(0+).
 The current through a capacitor can change instantaneously.
 A capacitor acts like a short circuit the instant a voltage is
applied across its terminals.
 When a DC source is applied to a capacitor, it becomes an open
circuit as t.
vC(t) = (Vi – Vf)e-t/TC + Vf
Grossman/ Melkonian
t>0
30
RL CIRCUIT, STEP RESPONSE:
 Assume we have the following series circuit containing a voltage
source, a resistor, and an inductor. We can write an expression for
the current through the inductor using a similar procedure as that
applied previously for finding the voltage across a capacitor.
Switch
t=0
VT
+
-
R
iL(t)
+
vL(t)
L
-
Grossman/ Melkonian
31
RL CIRCUIT, STEP RESPONSE:
 Applying a similar procedure as that applied to finding the voltage
across a capacitor, we obtain the equation for the current through an
inductor when the input is a step function.
iL(t) = (I0 – IS)e-Rt/L + IS
Decaying exponential
t>0
Constant
Important Values:
L/R: Defined as the time constant, sometimes written as TC or .
The time constant depends only on fixed circuit parameters.
I0: The initial condition or initial current through the inductor.
Sometimes written as Ii or iL(0)
IS: The Norton current seen by the inductor as t. Sometimes
written as If or iL().
Grossman/ Melkonian
32
RL CIRCUIT, STEP RESPONSE:
Important Values cont.:
R: The Norton resistance or equivalent resistance seen by the
inductor. Sometimes written as RN.
t(0-): The instant of time just prior to t=0 or the switch changing
its position.
t(0): The time t=0 or the instant the switch changes its position.
t(0+): The instant of time just after t=0 or just after the switch
changes its position.
Grossman/ Melkonian
33
RL CIRCUIT, STEP RESPONSE:
Important Concepts:
 The current through an inductor cannot change instantaneously.
Therefore, iL(0-) = iL(0) = iL(0+).
 The voltage across an inductor can change instantaneously.
 A inductor acts like an open circuit the instant a voltage or
current is applied to its terminals.
 When a DC source is applied to a inductor, it becomes a short
circuit as t.
iL(t) = (Ii – If)e-t/TC + If
Grossman/ Melkonian
t>0
34
RC CIRCUIT, STEP RESPONSE:
Example 3:
 Calculate the RC time constant (TC), Vi, Vf, the voltage across
the capacitor, and the current through the capacitor for t > 0.
The switch has been open for a long time before closing.
Switch
1.5K
t=0
16V
+
-
iC(t)
1.5K
+
vC(t)
1.5F
-
Grossman/ Melkonian
35
RC CIRCUIT, STEP RESPONSE:
Example 3 cont.:
 RC time constant (TC):
TC = RTH•CEQ = 1.5K 1.5K•1.5F
TC = 1.125ms
Switch
1.5K
t=0
16V
+
-
iC(t)
1.5K
+
vC(t)
1.5F
Grossman/ Melkonian
36
RC CIRCUIT, STEP RESPONSE:
Example 3 cont.:
Vi = 0V
The capacitor has had time to discharge.
1.5K
Vf = 16V
1.5K + 1.5K
Voltage divider
Vf = 8V
Switch
t=0
16V
+
-
1.5K
iC(t)
1.5K
+
vC(t)
1.5F
Grossman/ Melkonian
37
RC CIRCUIT, STEP RESPONSE:
Example 3 cont.:
vC(t) = (Vi – Vf)e-t/TC + Vf
 vC(t) t > 0
vC(t) = (0V - 8V)e-t/TC
+ 8V
t>0
vC(t) = 8 - 8e-t/1.125ms V
Switch
+
-
t>0
1.5K
t=0
16V
t>0
iC(t)
1.5K
+
vC(t)
1.5F
Grossman/ Melkonian
38
RC CIRCUIT, STEP RESPONSE:
Example 3 cont.:
 iC(t) t > 0
iC(t) = C[dvC(t)/dt]
iC(t) = 1.5F d/dt[8 - 8e-t/1.125ms]A
iC(t) = 10.67e-t/1.125ms mA
t>0
t>0
Switch
t=0
16V
+
-
1.5K
iC(t)
1.5K
+
vC(t)
1.5F
Grossman/ Melkonian
39
RL CIRCUIT, STEP RESPONSE:
Example 4:
 Calculate the L/R time constant (TC), Ii, If, the current
through the inductor, and the voltage across the inductor
for t > 0. The switch has been closed for a long time.
Switch
5K
1K
t=0
14V
+
-
iL(t)
3k
+
5mH
vL(t)
Grossman/ Melkonian
40
RL CIRCUIT, STEP RESPONSE:
Example 4 cont.:
 L/R time constant (TC):
TC = LEQ/RN =
5mH
3k + 1K
TC = 1.25s
1K
Switch
t=0
14V
+
-
5K
iL(t)
3k
+
5mH
vL(t)
Grossman/ Melkonian
41
RL CIRCUIT, STEP RESPONSE:
Example 4 cont.:
Ii = 2.8mA
1/1K
1/5K + 1/3k + 1/1K
If = 0A
Ii = 1.83mA
Switch
5K
1K
t=0
14V
+
-
Source transformation
and current divider
iL(t)
3k
+
5mH
vL(t)
Grossman/ Melkonian
42
RL CIRCUIT, STEP RESPONSE:
Example 4 cont.:
 iL(t):
iL(t) = (Ii – If)e-t/TC + If
t>0
iL(t) = (1.83 - 0)e-t/1.25s +
0 mA
iL(t) = 1.83e-t/1.25s
Switch
mA
5K
+
-
t>0
1K
t=0
14V
t>0
iL(t)
3k
+
5mH
vL(t)
Grossman/ Melkonian
43
RL CIRCUIT, STEP RESPONSE:
Example 4 cont.:
 vL(t):
vL(t) = L[diL(t)/dt]
vL(t) = 5mH•d/dt[1.83x10-3•e-t/1.25s]
vL(t) = - 7.32e-t/1.25s V
Switch
5K
+
-
t>0
1K
t=0
14V
V
iL(t)
3k
+
5mH
vL(t)
Grossman/ Melkonian
44
RC CIRCUIT, STEP RESPONSE:
Example 5 :
 The switch has been in position ‘A’ for a long time. At t=0, the
switch moves to position ‘B’. Calculate Vi, Vf, TC, and vC(t),
iC(t), and iR(t) for t > 0.
800
t=0
A
B
10V
+
-
+
-
iC(t)
iR(t)
6V
400
+
vC(t)
2F
-
Grossman/ Melkonian
45
RC CIRCUIT, STEP RESPONSE:
Example 5 cont.:
Vi = 10V
Vf = 6V
400
800 + 400
Vi = 3.34V
Voltage divider
Vf = 2V
Voltage divider
400
800 + 400
800
t=0
A
B
10V
+
-
+
-
iC(t)
iR(t)
6V
400
+
vC(t)
2F
Grossman/ Melkonian
46
RC CIRCUIT, STEP RESPONSE:
Example 5 cont.:
 TC:
TC = RTH•CEQ = 800 400•2F
TC = 533.4 s
800
t=0
A
B
10V
+
-
+
-
iC(t)
iR(t)
6V
400
+
vC(t)
2F
Grossman/ Melkonian
47
RC CIRCUIT, STEP RESPONSE:
Example 5 cont.:
 vC(t) t > 0
vC(t) = (Vi – Vf)e-t/TC + Vf
t>0
vC(t) = (3.34V - 2V)e-1875t + 2V
vC(t) = 2 + 1.34e-1875t V
t>0
t>0
800
t=0
A
B
10V
+
-
+
-
iC(t)
iR(t)
6V
400
+
vC(t)
2F
Grossman/ Melkonian
48
RC CIRCUIT, STEP RESPONSE:
Example 5 cont.:
 iC(t) t > 0
iC(t) = C[dvC(t)/dt]
iC(t) = 2F d/dt[2 + 1.34e-1875t]A
iC(t) = -5.01e-1875t mA
A
B
+
-
t>0
800
t=0
10V
t>0
+
-
iC(t)
iR(t)
6V
400
+
vC(t)
2F
Grossman/ Melkonian
49
RC CIRCUIT, STEP RESPONSE:
Example 5 cont.:
 iR(t) t > 0
ix(t) = vC(t)/400 =
2 + 1.34e-1875t
400
= 5 + 3.35e-1875t mA t > 0
iR(t) = ix(t) + iC(t) = 5 + 3.35e-1875t - 5.01e-1875t mA
iR(t) = 5 – 1.66e-1875t mA
t>0
800
t=0
A
B
10V
+
-
t>0
+
-
iC(t)
iR(t)
6V
400
ix(t)
Grossman/ Melkonian
+
vC(t)
2F
50
RC CIRCUIT, SINUSOIDAL INPUT:
Section 4.3
 Forced Response of Circuits Excited by Sinusoidal Sources:
 Consider the following circuit. If vS(t) = Vcos(t) we obtain the
following differential equation:
RC(dvC(t)/dt) + vC(t) = Vcos(t)
t>0
 This equation is similar to the first order differential equation
found previously for a step input.
iC(t)
vS(t)
+~
-
R
+
vC(t)
C
-
Grossman/ Melkonian
51
RC CIRCUIT, SINUSOIDAL INPUT:
 As with the step response, we find the natural and forced
response:
Natural Response
vN(t) = Ke-t/RC
t>0
R
iC(t)
vS(t)
+~
-
+
vC(t)
C
-
Grossman/ Melkonian
52
RC CIRCUIT, SINUSOIDAL INPUT:
Forced Response
 The forced response depends on both the circuit and the nature of
the forcing function. It is the particular solution to the following
equation:
RC(dvF(t)/dt) + vF(t) = Vcos(t)
t>0
 This equation requires vF(t) plus RC times its first derivative add to
produce a cosine function. Try a solution of a general sinusoid.
vF(t) = Acos(t) + Bsin(t)
 Substituting this into the differential equation we obtain:
RCd/dt[Acos(t) + Bsin(t)] + Acos(t) + Bsin(t) = Vcos(t)
S
Grossman/ Melkonian
t>0
53
RC CIRCUIT, SINUSOIDAL INPUT:
 Performing the differentiation:
RC[-Asin(t) + Bcos(t)] + Acos(t) + Bsin(t) = Vcos(t)
t>0
 Rearranging the equation:
[RCB + A – V]cost + [-RCA + B]sint = 0
 The left side of the equation is zero for all t > 0 when the
coefficients of the cosine and sine terms are zero.
A + (RC)B = V and -(RC)A + B = 0
 Solving the two equations yields:
A =
V
1 + (RC)2
and
B =
Grossman/ Melkonian
RCV
1 + (RC)2
54
RC CIRCUIT, SINUSOIDAL INPUT:
 Combining the forced and natural responses:
vc(t) =
Ke-t/RC
V
+
1+
(RC)2
(cost + RCsint)
t>0
 The initial condition requires:
v(0) = V0 = K +
V
K = V0 -
1 + (RC)2
V
1 + (RC)2
 Substituting this value for K into the equation for the voltage across
the capacitor:
vc(t) = V0 -
V
1 + (RC)2
e-t/RC
+
V
1+
(RC)2
Natural Response
(cost + RCsint) t > 0
Forced Response
Grossman/ Melkonian
55
RC CIRCUIT, SINUSOIDAL INPUT:
Example 6:
Calculate the voltage across the capacitor for t > 0. V0 = 0V.
vc(t) = V0 -
V
1 + (RC)2
e-t/RC
+
Switch
V
1+
(RC)2
500
t=0
vS(t) = 5cos(2000t)V
(cost + RCsint) t > 0
+
iC(t)
+
vC(t)
~
-
2F
-
Grossman/ Melkonian
56
RC CIRCUIT, SINUSOIDAL INPUT:
Example 6 cont.:
V0 = 0V, V = 5V,  = 2000, R = 500, C = 2F, TC = 1ms
vc(t) = 0 -
5
1 + (2)2
e-1000t
+
5
1+
(cos2000t + 2sin2000t) t > 0
(2)2
vc(t) = -e-1000t + (cos2000t + 2sin2000t)V
t>0
Switch
t=0
vS(t) = 5cos(2000t)V
500
+
+
vC(t)
~
-
2F
iC(t)
Grossman/ Melkonian
57
PHASORS:
Section 4.4
 A phasor is the representation of a sinusoidal signal in the
frequency domain. This phasor representation eliminates the
need for solving differential equations.
 Euler’s Identity is the basis of phasor notation.
ej = cos() + jsin()
 Euler’s Identity is a trigonometric relationship in the complex plain.
Im
j
sin
1

ej = cos + jsin
-1
1
Re
cos
-j
Grossman/ Melkonian
58
PHASORS:
 The relationship between rectangular and polar forms is as
follows:
Aej = Acos + jAsin = A
Rectangular form
Polar form
 Mathematical Representations:
Acos(t + )
Time-domain
V(j) = Aej
Frequency-domain
A
Polar form
 Multiplication and Division is performed in polar form.
Addition and Subtraction is performed in rectangular form.
Grossman/ Melkonian
59
PHASORS:
Example 7:
Construct the phasors for the following signals:
v1(t) = 10cos(1000t - 45)
v2(t) = 5cos(1000t + 30)
Solution:
v1(t) = 10e-j45 = 10cos(- 45) + j10sin(-45)
Polar form
10-45 V
Rectangular form
7.07 - j7.07 V
Grossman/ Melkonian
60
PHASORS:
Example 7 cont.:
v2(t) = 5ej30 = 5cos(30) + j5sin(30)
Polar form
530 V
Rectangular form
4.33 + j2.5 V
 Use the additive property of phasors to find the sum of v1(t) and v2(t).
v(t) = v1(t) + v2(t) = (7.07 - j7.07) + (4.33 + j2.5) V
v(t) = 11.4 - j4.57 V
Rectangular form
= 12.28-21.8 V
Polar form
= 12.28cos(1000t - 21.8) V
Time-domain
Grossman/ Melkonian
61
PHASORS:
Example 8:
Use the derivative property of phasors to find the time derivative
of v(t) = 15cos(200t - 30) V.
Solution:
j t  


V j  Ae
d j t  
V '  j   A e
dt
j t   d
 j t     jAe j t  
 Ae
dt
 j V  j  
Grossman/ Melkonian
62
PHASORS:
• Solution (cont.)
The phasor form of the sinusoid is V(j) = 15-30 V.
The time derivative is found by multiplying V(j) by j.
dv(t)/dt = j15-30 = j20015-30
= 2009015-30 = 300060 V/s
dv(t)/dt = 3000cos(200t + 60) V/s
Grossman/ Melkonian
63
IMPEDANCE:
 Using phasor notation we will analyze the ideal resistor, inductor and
capacitor. When dealing with AC signals, we will define a new
parameter called impedance. Impedance can be viewed as complex
resistance. The concept of impedance shows that certain parameters
of inductors and capacitors are frequency-dependent.
The Resistor:
Ohm’s law dictates the relationship v = iR. If the source is sinusoidal,
vS(t) = Acos(t), then the current through the resistor is:
i(t) = vS(t)/R = (A/R)cost
Converting vS(t) and i(t) to phasor notation:
VZ(j) = A0
I(j) = (A/R)0
Grossman/ Melkonian
64
IMPEDANCE:
 Therefore, the impedance of a resistor is found to be the ratio of the
phasor voltage across it to the phasor current flowing through it.
ZR(j) =
VZ(j)
I(j)
= R
 The relationship between VZ and I viewed in the complex plane is
shown below:
Imaginary
I
Relationship between
VZ & I for the resistor.
V
Real
Grossman/ Melkonian
65
IMPEDANCE:
The Inductor:
vL(t) = L[diL(t)/dt]
vL(t) = vS(t)
iL(t) = 1/L  vL(t)dt
iL(t) = i(t)
For the circuit
below
We can write:
iL(t) = i(t) = 1/L vL(t)dt
if vS(t) = Acos(t)
then, iL(t) = 1/L  Acos(t)dt = (A/L)sin(t)
i(t)
vS(t)
+
~
+
L
-
vL(t)
Grossman/ Melkonian
66
IMPEDANCE:
vS(t) = vL(t) = Acos(t)
i(t) = iL(t) = (A/L)sin(t)
= (A/L)cos(t - /2)
Note that the inductor current is dependent
on the radian frequency, , of the source
and is shifted in phase by 90° with respect
to the voltage.
Converting to phasor notation:
Imaginary
I
VZ(j) = A0
I(j) = (A/L)-/2
-/2
Real
Therefore:
ZL(j) =
VZ(j)
I(j)
V
Relationship between VZ
& I viewed in the complex
plane (for the inductor).
= L/2 = jL
Grossman/ Melkonian
67
IMPEDANCE:
The Capacitor:
iC(t) = C[dvC(t)/dt]
iC(t) = i(t)
vC(t) = 1/C  iC(t)dt
vC(t) =
vS(t)
We can write:
iC(t) = C[dvC(t)/dt] = C[dAcos(t)/dt]
= -C(Asint)
= CAcos(t + /2)
vS(t)
~
-
{for vS(t) = Acos(t)}
Note that the capacitor current is dependent on the
radian frequency, , of the source and is shifted in
phase by 90° with respect to the voltage.
i(t)
+
For the circuit
below
C
+
vC(t)
Grossman/ Melkonian
68
IMPEDANCE:
Converting to phasor notation:
VZ(j) = A0
I(j) = CA/2
Therefore:
ZC(j) =
VZ(j)
I(j)
1
= 1/(C)-/2 = -j/(C) = (jC)
Imaginary
I
Relationship between VZ & I
viewed in the complex plane
(for the capacitor).
V
/2
Real
Grossman/ Melkonian
69
IMPEDANCE:
 Impedance of a circuit element is defined as the sum of the real
and imaginary parts.
Z(j) = R(j) + jX(j)
Impedance
Im
L
ZL
/2
-/2
R
ZR
Re
ZC
-
ZR = R
ZL = jL
ZC =
1
C
Grossman/ Melkonian
1
jC
70
IMPEDANCE:
 KCL, KVL, voltage division, current division, series connections,
and parallel connections apply to circuits in the frequency domain
or in phasor form.
For the circuit below, the elements are connected in series, therefore:
1. The same phasor current I flows through each impedance.
2. The voltage across the series connection can be written as:
V = V 1 + V2 + . . . + VN
= Z1I + Z2I + . . . + ZNI
I
Rest
of the
circuit
+
V
+ V1
ZN
Z2
Z1
-
+ V2
-
+ VN
-
Grossman/ Melkonian
71
IMPEDANCE:
3. The equivalent impedance is
ZEQ = Z1 + Z2 + . . . + ZN
 In general, the equivalent impedance is a complex quantity of the
form:
ZEQ = R + jX
Where R is the real part and X is the imaginary part. The real part
is called resistance and the imaginary part is called reactance.
I
Rest
of the
circuit
+
V
+ V1
ZN
Z2
Z1
-
+ V2
-
+ VN
-
Grossman/ Melkonian
72
IMPEDANCE:
 For passive circuits:
Resistance is always positive
Capacitive reactance is always negative
XC = -1/C ohms ()
Inductive reactance is always positive
XL = L ohms ()
 The phasor voltage across the kth element in a series connection is:
Z
V k = Zk I k = V k
ZEQ
I
Rest
of the
circuit
+
V
ZN
Z2
Z1
+ V1
Phasor version of
Voltage Divider Rule
-
+ V2
-
+ VN
-
Grossman/ Melkonian
73
IMPEDANCE:
 For the circuit below, the elements are connected in parallel,
therefore:
1. The same phasor voltage V is across each impedance.
2. The phasor current I can be written as:
I = I1 + I 2 + . . . + I N
= V/Z1 + V/Z2 + . . . + V/ZN
I
Rest
of the
circuit
+
V
I1
I2
Z1
IN
Z2
ZN
Grossman/ Melkonian
74
IMPEDANCE:
 The equivalent impedance of the parallel connection is:
1/ZEQ = 1/Z1 + 1/Z2 + . . . + 1/ZN
 The phasor current through the kth element in a parallel connection
is:
Ik = I
1/Zk
1/Z1 + 1/Z2 + . . . + 1/ZN
I
Rest
of the
circuit
+
V
I1
I2
Z1
IN
Z2
ZN
Grossman/ Melkonian
75
IMPEDANCE:
Example 9:
Find the impedance (Z) of the elements in the rectangular box. The
circuit is in steady-state. vs(t) = 50cos(4000t - 20)V and i1(t) =
0.5cos(4000t)A.
Solution:
Phasor representation of signals:
Vs = 50-20 V and I1 = 0.50 A
50
i1(t)
+
v1(t)
-
i3(t)
i2(t)
vs(t)
+
+
~
-
v2(t)
100
Grossman/ Melkonian
76
IMPEDANCE:
Example 9 cont.:
Z = V 2 / I2
V2 = Vs - V1 and V1 = I1•Rs = 0.50•50 = 250 V
V2 = 50-20 - 250 = (46.98 - j17.10) - (25 + j0)
V2 = 21.98 - j17.10 = 27.85-37.88
50
i1(t)
+
v1(t)
-
i3(t)
i2(t)
vs(t)
+
+
~
-
v2(t)
100
Grossman/ Melkonian
77
IMPEDANCE:
Example 9 cont.:
VL = V2 = 27.85-37.88
I3 = VL/1000 = [27.85-37.88]/[1000] = 278.5-37.88 mA
I2 = I1 - I3 = (0.5 + j0) - (0.2198 - j0.1710) = 0.2802 +j0.1710 A
I2 =0.32831.39
50
i1(t)
+
v1(t)
-
i3(t)
i2(t)
vs(t)
+
+
~
-
v2(t)
-
Grossman/ Melkonian
100
+
vL(t)
78
IMPEDANCE:
Example 9 cont.:
Z = V 2 / I2 =
27.85-37.88 V
0.32831.39 A
Z = 84.9-69.3 = 30.1 -j79.4 
50
i1(t)
+
v1(t)
-
i3(t)
i2(t)
vs(t)
+
+
~
-
v2(t)
-
Grossman/ Melkonian
100
+
vL(t)
-
79
Transient Analysis of Second Order
Circuit
iL(t)
ic(t)
is(t)
RT
+
+
C vL(t) L
vc(t)
vT(t)
-
KCL:
-
is t   iC t   iL t   0
Grossman/ Melkonian
80
Transient Analysis of Second Order
Circuit
• KVL: vC(t) = vL(t)
vT t   vC t 
dvC t 
C
 iL t   0
RT
dt
diL t 
vC t   L
dt
1
L diL t 
d 2iL t 
vT t  
 LC
 iL t 
2
RT
RT dt
dt
d 2iL t  L diL t 
1
LC

 iL t  
vT t 
2
dt
RT dt
RT
Grossman/ Melkonian
81
Transient Analysis of Second Order
Circuit
• Knowing iL(t):
diL t 
vC t   vL t   L
dt
dvC t 
d 2iL t 
iC t   C
 LC
dt
dt 2
Grossman/ Melkonian
82
Transient Analysis of Second Order
Circuit
• Another formulation:
is t   iC t   iL t   0
vT t   RT is t   vC t   0
vT t   RT iC t   iL t   vC t 
dvT t 
 di t  di t   dv t 
 RT  C  L   C
dt
dt 
dt
 dt
di t  1
di t 
vC t   vL t   L L  L  vC t 
L
dt
dt
d 2 vC t 
diC t 
dvC t 
C

iC t   C
dt 2
dt
dt
d 2 vC t  dvC t  RT
dvT t 



t
v


RT C
C
dt
L
dt
dt 2
d 2 vC t  L dvC t 
L dvT t 



t
v


LC
C
RT dt
RT dt
dt 2
Grossman/ Melkonian
83
Solution of Second Order Circuit
• Generalized second order DEQ:
d 2 xt 
dxt 
a2
 a1
 a0 xt   b0 f t 
2
dt
dt
1 d 2 xt  2 dxt 

 xt   K s f t 
2
2
n dt
n dt
• Natural Frequency:
a0
n 
a2
• Damping Ratio
a1
 
2
• DC gain
b0
Ks 
a0
Grossman/ Melkonian
1
a0 a 2
84
Natural Response of a SecondOrder System
• Solution is known to be of the form:
• xN(t) = est
1
n2
1

2
n
s 2e st 
s2 
2
n
2
n
se st  e st  0
s 1  0
x N t   1e s1t   2 e s2t
1
s1, 2   n 
2
2 
2
n
 4n2   n  n  2  1
Grossman/ Melkonian
85
Roots of Second Order System
1. Real and Distinct Roots:  > 1
s1, 2  n  n  2  1
•
Over-damped Response
2. Real and Repeated Roots:  = 1
s1, 2  n  n
•
Critically Damped Response
3. Complex Conjugate Roots:  < 1
s1, 2  n  jn 1   2
•
Under-damped Response
Grossman/ Melkonian
86
Response of the Second Order
System as function of 
Grossman/ Melkonian
87
Roots of Second Order System
• Real and Distinct Roots:  > 1
Over-damped Solution:
xN t   1e s1t   2 e s2t  1e
xN t   1e
1 

t
1
  2e

    2 1  t
n
n


  2e
    2 1  t
n
n


t
2
1
 n  n  2  1
& 2 
1
 n  n  2  1
Grossman/ Melkonian
88
Overdamped Solution
Natural response of overdamped second-order system for
α 1 = α 2 = 1; ζ = 1.5; ω n = 1
Grossman/ Melkonian
89
Roots of Second Order System
• Real and Repeated Roots:  = 1
Critically-damped Solution:
x N t   1e   2 e
x N t   1e
s1t
s2t
t
t



 1e
  2e ,

Grossman/ Melkonian
 n t
  2e

 n t
1
n
90
Critically Damped Solution
Natural response of a critically damped second-order system for
α 1 = α 2 = 1; ζ = 1; ω n = 1
Grossman/ Melkonian
91
Roots of Second Order System
• Complex Conjugate Roots:  < 1
Under-damped Solution:
   j 1  t
 

x N t   1e s t   2 e s t  1e
  2e
2
1
n
2
n
n
 j n 1 2  t

1   2  
x N t   e
 n t
 e jn

1 2 t

e
 j n 1 2 t



x N t   2e  n t cos n 1   2 t
d  n 1   2
Damped Natural Frequency
Grossman/ Melkonian
92
Underdamped Solution
Natural response of an underdamped second-order system for
α 1 = α 2 = 1; ζ = 0.2; ω n = 1
Grossman/ Melkonian
93
Forced Response
•
Solution to the equation:
1 d 2 xt  2 dxt 

 xt   K s f t 
2
2
n dt
n dt
1. Constant Input:
f(t) = F for t ≥ 0 ⇒ xF(t) = KsF
–
t≥0
DC steady-state solution
Grossman/ Melkonian
94
Complete Solution
• Overdamped case ( > 1):
xt   xN t   xF t   1e
    2 1  t
n
n


  2e
    2 1  t
n
n


 x 
• Critically Damped case ( = 1):
• Underdamped case ( < 1)
Grossman/ Melkonian
95