EENG 2610: Circuit Analysis Class 14: Sinusoidal Forcing Functions, Phasors, Impedance and Admittance Oluwayomi Adamo Department of Electrical Engineering College of Engineering, University of North Texas
Jan 21, 2016
EENG 2610: Circuit AnalysisClass 14: Sinusoidal Forcing Functions, Phasors, Impedance and Admittance
Oluwayomi AdamoDepartment of Electrical EngineeringCollege of Engineering, University of North Texas
AC Steady-State Analysis
Sinusoidal forcing function (f(t) is forcing function)
The natural response xc(t) is a characteristics of the circuit network and it is independent of the forcing function.
The forced response xp(t) depends on the type of forcing function. Why study sinusoidal forcing function?
This is the dominant waveform in electric power industry. Any periodic signal can be represented by a sum of sinusoids
(you will learn it in Fourier analysis) We will only concentrate on the steady-state forced response of
networks with sinusoidal forcing functions. We will ignore the initial conditions and the transient or natural
response
),()()(
tftaxdt
tdx ),()(
)()(212
2
tftxadt
tdxa
dt
txd )()()( txtxtx cp
Sinusoids Definition
XM is the amplitude ω is radian or angular frequency (unit: radian/second) θ is phase angle (unit: radian), T=2π/ω is period (unit: second) f=1/T is frequency (unit: Hertz),
)sin()( tXtx M
180radian
fT
22
In-phase and out-of-phase Any point on the waveform XMsin(ωt+θ) occurs θ radians
earlier in time than the corresponding point on the waveform XMsin(ωt). We say XMsin(ωt+θ) leads XMsin(ωt) by θ radians, or
XMsin(ωt) lags XMsin(ωt+θ) by θ radians.
In the more general situation, if
Then, x1(t) leads x2(t) by (θ1 - θ2) radians, or,
x2(t) lags x1(t) by (θ1 - θ2) radians.
If θ1 = θ2 ,the waveforms are identical and the functions are said in phase; otherwise, it is said out of phase.
)sin()(
)sin()(
22
11
tXtx
tXtx
M
M
Important Trigonometric Identities
)2
cos()sin(
)2
sin()cos(
tt
tt
)sin()sin(
)cos()cos(
tt
tt
sinsincoscos)cos(
sinsincoscos)cos(
sincoscossin)sin(
sincoscossin)sin(
2
2
22
sin21
1cos2
sincos)2cos(
cossin2)2sin(
)sin()sin(
)cos()cos(
180radian
jj
j
ee
ry
rxx
y
yxr
rejyx
1
sin
cos
tan 1
22
Rectangular formPolar form
Sinusoidal and Complex Forcing Functions Forcing function and circuit response
If we apply a constant forcing function (i.e., source function) to a network, the steady state circuit response is also a constant.
If we apply a sinusoidal forcing function to a linear network, the steady state circuit response will also be sinusoidal.
Sinusoidal source function If the input source is v(t)=Asin(ωt+θ), then the output will be
in the same sinusoidal form. For example, i(t)=Bsin(ωt+φ). That means: if the input source is sinusoidal function, we
know the form of the output response, and therefore the solution involves simply determining the values of the two parameters B and φ.
Learning Example
)()()( tvtRitdt
diL :KVL
tAtAtdt
di
tAtAti
tAti
cossin)(
sincos)(
)cos()(
21
21
or , statesteady In
tV
tRAALtRAAL
M
cos
cos)(sin)( 1221
MVRAAL
RAAL
12
21 0
algebraic problem
222221)(
,)( LR
LVA
LR
RVA MM
Determining the steady state solution canbe accomplished with only algebraic tools!
FURTHER ANALYSIS OF THE SOLUTION
)cos()(
cos)(
sincos)( 21
tAti
tVtv
tAtAti
M
write can one purposes comparisonFor
is voltageapplied The
is solution The
sin,cos 21 AAAA
222221)(
,)( LR
LVA
LR
RVA MM
1
222
21 tan,
A
AAAA
R
L
LR
VA M
1
22 tan,)(
)tancos()(
)( 122 R
Lt
LR
Vti M
voltagethe lags WAYScurrent AL the For 0L
90by voltagethe lagscurrent the inductor) (pure If 0R
Phasors
If the forcing function for a linear network is of the form v(t)=VMejωt, Then every steady-state voltage or current in the network will have the
same form and the same frequency ω; for example, a current will be of the form i(t)=IMej(ωt+φ).
In our circuit analysis, we can drop the factor ejωt, since it is common to every term in the describing equations.
}Re{
}Re{
}Re{
)cos()()(
tjM
tjjM
tjM
M
eV
eeV
eV
tVtv
M
M
I
V
I
VPhasors are defined as:
The magnitude of phasors are positive !
Sinusoidal signal:
)(
M
M
V
VV
)sin(cos
)sin(cos
jIeI
jVeV
Mj
M
Mj
M
Phasor Analysis (or Frequency Domain Analysis) The circuit analysis after dropping ejωt term is called phasor analysis
or frequency domain analysis. By phasor analysis, we have transformed a set of differential
equations with sinusoidal forcing functions in the time domain into a set of algebraic equations containing complex numbers in the frequency domain.
The phasors are then simply transformed back to the time domain to yield the solution of the original set of differential equations.
Phasor representation:Time Domain Frequency Domain
)90()sin(
)cos(
AtA
AtA
Phasor Relationships for Circuit Elements We will establish the phasor relationships between voltage and
current for the three passive elements R, L, C.
IV R
eRIeV
eRIeV
tRitv
iv
iv
jM
jM
tjM
tjM
)()(
)()(
iMj
M
vMj
M
IeI
VeVi
v
I
V
Phasor diagram
IV Lj
eLIjeV
eIdt
dLeV
dt
tdiLtv
iv
iv
jM
jM
tjM
tjM
}{
)()(
)()(
iMj
M
vMj
M
IeI
VeVi
v
I
V
9011
190je
j
Voltage leads current by 90°
VI Cj
eCVjeI
eVdt
dCeI
dt
tdvCti
vi
vi
jM
jM
tjM
tjM
}{
)()(
)()(
iMj
M
vMj
M
IeI
VeVi
v
I
V
9011
190je
j
Current leads voltage by 90°
Definition of Impedance (unit: ohms): Impedance is defined as the ratio of the phasor voltage V to the
phasor current I at the two terminals of the element related to one another by the passive sign convention:
It’s important to note that: Resistance R and reactance X are real function of the frequency
of the forcing function ω, thus Z(ω) is frequency dependent. Impedance Z is a complex number; however, it is not a phasor,
since phasors denote sinusoidal functions.
)( ivM
M
iM
vM
z
I
V
I
V
Z
I
V
Z
reactance :
,resistance :
)()()(
X
R
jXR Z
)/(tan 1
22
RX
XRZ
z
z
z
ZX
ZR
sin
cos
Cj
Lj
Cj
Lj
C
L
RRR
11
ImpedanceEquationPhasor Element
Z
Z
IV
IVZIV
Passive element impedance:
Equivalent impedance if impedances are connected in series:
ns ZZZZ ...21
Equivalent impedance if impedances are connected in parallel:
np ZZZZ
1...
111
21
Two terminal input admittance:
esusceptanc : e,conductanc :
)()()(
siemens) :(unit 1
BG
jBG
YV
I
ZY
jXRjBG
11
ZY
2222 ,
XR
XB
XR
RG
In general, R and G are not reciprocals of one another. The purely resistive case is an exception.
KVL and KCL are both valid in frequency domain
Example 8.9: Determine the equivalent impedance of the network. Then compute i(t) for f = 60 Hz and f = 400 Hz.
V )30cos(50)( ttv
SPECIAL APPLICATION:IMPEDANCES CAN BE COMBINED USING THE SAME RULES
DEVELOPEDFOR RESISTORS
I 1V
1Z
2V
2ZI
21 ZZZs 1Z 2Z
V
I I
V
21
21
ZZ
ZZZ p
kks ZZ
kk
p ZZ
11
LEARNING EXAMPLE
current and impedance equivalent Compute
)30cos(50)(,60 ttvHzf
63
1050120
1,1020120
25,3050,120
jZjZ
ZV
CL
R
05.53,54.7 jZjZ CL
51.4525 jZZZZ CLRs
)(51.4525
3050A
jZ
VI
s
)(22.6193.51
3050A
))(22.91120cos(96.0)()(22.9196.0 AttiAI
RZR
LjZL
CjZC
1
LEARNING EXAMPLESERIES-PARALLEL REDUCTIONS
243 jZ
4/1
25.05.025.0
44
4
jYZ
jjjY
2
8
24
)2(44 jjj
jjZ
422622 jjjZ
21
)2(11 j
jZ
5.01
11 j
Z
2434 jZ
342
342234 ZZ
ZZZ
13 j
)(1.03.0
1.02.0
)(2.01.0
234
34
2
SjY
jY
SjY
)(4.08.0
)5.0(1
5.01
1
21
jZ
jZ
973.8847.36.08.32341 jZZZeq
222)4()2(
42
42
1
j
jY
20
24
24
134
j
jY
1.0
1.03.0
1.03.0
11
234234
j
jYZ
AC Steady-State Analysis
For relatively simple circuits (e.g., those with single source), use: Ohm’s law for AC analysis, i.e., V=IZ The rules for combining impedance Z (or admittance Y) KCL and KVL Current division and voltage division
For more complicated circuits with multiple sources, use: Nodal analysis Loop or mesh analysis Superposition Source exchange Thevenin’s and Norton’s theorem Software tools: MATLAB, PSPICE, …
LEARNING EXAMPLE COMPUTE ALL THE VOLTAGES AND CURRENTS
)48||6(4 jjZeq
28
4824832
28
48244
j
jj
j
jZeq
)(964.30604.9036.14246.8
45196.79
28
5656
j
jZeq
)(036.29498.2964.30604.9
60241 A
Z
VI
eq
S
)(036.29498.2036.14246.8
906
28
613 AI
j
jI
)(036.29498.2036.14246.8
565.26944.8
28
4812 AI
j
jI
3221 904906 IVIV
10582.158.1171.206.295.2 321 III
)(1528.7
)(42.7826.16
2
1
VV
VV
21
32
1
Vfor V law sOhm'
IIfor divider current Use
I Compute
,
,
Steady-State Power Analysis
Here we study powers in AC circuits: Instantaneous power Average power Maximum power transfer, Power factor, Complex power.
Device power ratings: Typically, electrical and electronic devices have peak
power or maximum instantaneous power ratings that cannot be exceeded without damaging the devices.
Instantaneous Power
Steady-state voltage and current:
)cos()(
)cos()(
iM
vM
tIti
tVtv
With passive sign convention
Instantaneous power:
)]2cos()[cos(2
)cos()cos(
)()()(
ivivMM
iMvM
tIV
tItV
titvtp
independent of time a function of time
)cos()(
)cos()(
iM
vM
tIti
tVtv
Statesteady In
)()()( titvtp Impedance to
SuppliedPower
ousInstantane
)cos()cos()( ivMM ttIVtp
)cos()cos(2
1coscos 212121
)2cos()cos(2
)( ivivMM tIV
tp
LEARNING EXAMPLE
)(),(
302
),60cos(4)(
tpti
Z
ttv
:Find
:Assume
)(302302
604A
Z
VI
))(30cos(2)( Atti
30,2
60,4
iM
vM
I
V
)902cos(430cos4)( ttp
constant Twice thefrequency
)cos(2
1
)]2cos()[cos(2
1
)cos()cos(1
)(1
0
0
0
0
0
0
ivMM
Tt
t ivivMM
Tt
t iMvM
Tt
t
IV
dttIV
T
dttItVT
dttpT
P
Average PowerSince p(t) is a periodic function of time, the average power:
In the equation, t0 is arbitrary, T=2π/ω is the period of the voltage or current.
For purely resistive circuit (i.e., Z = R+j0 ):R
VRIIVIVP M
MMMivMM
22
2
1
2
1
2
1)cos(
2
1
For purely reactive circuit (i.e., Z = 0 + jX ): 0)90cos(2
1 MM IVP
That’s why reactive elements are called lossless elements
For passive sign convention.
LEARNING EXAMPLE Determine the average power absorbed by each resistor, the total average power absorbed and the average power
supplied by the source
)(4534
45121 AI
WP 183122
14
If voltage and current are in phase
MMiv IVP2
1 2
12
1MRI
R
VM2
2
1
)(57.7136.537.265
4512
12
45122 A
jI
)(36.522
1 22 WP W7.28
Inductors and capacitors do not absorbpower in the average
WPtotal 7.2818
absorbedsupplied PP WP 7.46supplied
Verification
57.7136.545321 III
)(10.6215.8 AI
)cos(2 iv
MM IVP
)10.6245cos(15.8122
1suppliedP
Maximum Average Power Transfer Average power at the load:
LLivLL RIIVPLL
2
2
1)cos(
2
1
LTh
LocL
LTh
ocL
ZZ
ZVV
ZZ
VI
LLL
ThThTh
jXR
jXR
Z
Z
22
2
)()(2
1
LThLTh
LocL XXRR
RVP
For maximum average power transfer:
ThThThL jXR *ZZ
Th
ocThLL
ThocL
R
VRIP
R
82
1
)2/(2
2max,
VI
If the load is purely resistive (i.e., XL = 0): LLLL RjXR Z
0L
L
dR
dP For maximum average power transfer:
22ThThL XRR LLL RIP 2
max, 2
1
Effective or RMS Values The effective value of a periodic current (or voltage) is defined as a constant or
DC value, which would deliver the same average power to a resistor R. The 120 V AC electrical outlets in our
home is the rms value of the voltage: It is common practice to specify the voltage rating of AC electrical devices (such
as light bulb) in terms of the rms voltage.
RIP 2eff
The average power delivered to a resistor by DC effective current:
The average power delivered to a resistor by a periodic current:
Tt
tRdtti
TP
0
0
)(1 2
Effective (or rms) value of a periodic current:
Tt
tdtti
TII
0
0
)(1 2
rmseff
On using the rms values for the sinusoidal voltage and current, the average power:
)cos(rmsrms ivIVP
The power absorbed by a resistor:
R
VRIP
2rms2
rms
)377cos(170)( ttv
Hz60377602
V1201702120 rms
f
V
,2
rmsMV
V 2
rmsMI
I
LEARNING EXAMPLE Compute the rms value of the voltage waveform
32)2(4
210
104
)(
tt
t
tt
tv
3T
3
2
21
0
2
0
2 ))2(4()4()( dttdttdttvT
3
32
3
162)(
1
0
33
0
2
tdttv
)(89.13
32
3
1VVrms
Tt
trms dttx
TX
0
0
)(1 2