chapter 6: ac circuitscc.ee.ntu.edu.tw/~ultrasound/ · ac circuits la stable, linear circuit...
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Chapter 6: AC Circuits
Chapter 6: Outline
Phasors and the AC Steady State
AC Circuits
l A stable, linear circuit operating in the steady state with sinusoidal excitation (i.e., sinusoidal steady state).
l Complete response = forced response + natural response.
l In the steady state, natural response à 0.
)cos( 11 θω +tA )cos( 22 θω +tA
Same frequency, different amplitudes and phases
Sinusoids
l Three parameters are needed to determine a sinusoid.l x(t)=Xmcos(ωt+φ)=Re[Xmej(ωt+φ)].l Xm: amplitude, ω=2πf=2π/Τ: angular frequency, φ:
phase angle (radian).
Phasors
l The three parameters can be represented by a rotating phasor in a two-dimensional plane.
l At a given time (e.g., t=0 ), the nonrotating phasor is represented by .
l The frequency information is not included.φ∠= mXX
AC Forced Response
l The forced response of any branch variable (current or voltage) is at the same frequency as the excitation frequency ω for a linear circuit.
l In other words, any branch variable has the general form y(t)=Ymcos(ωt+φy).
l Circuit analysis becomes manipulation of complex numbers.
Complex Numbers in the Complex Plane
ai
ar
Aa
Aa
φ
φ
sin
cos
=
=
1−≡j
Rotating phasor
( )
0 tan180
0 tan
10
1
2/122
<
−±=
>
=
+=
−
−
rr
ia
rr
ia
ir
aaa
aaa
aaA
φ
φ
Complex Addition and Subtraction
[ ] [ ] [ ][ ] [ ] [ ]BABA
BABAImImImReReRe
±=±±=±
Complex Conjugate
222*
*
AaaAA
ajaAA
ajaAA
ir
ira
ira
=+=⋅
⋅−=−∠≡
⋅+=∠=
φ
φ
Complex Multiplication
[ ] [ ]]Im[]Im[],Re[]Re[ real, is if
ImRe)()(
BkkBBkkBk
ABABbabajbabaBA riiriirr
==+=−⋅+−=⋅
Complex Division (Rationalization)
2222*
*
ir
riir
ir
iirr
aaabab
jaa
ababAABA
AB
+−
+++
==
Complex Number in Exponential Form
l Euler’s formula: l Complex number in exponential form:
ααα sincos jje ±=±
ajeAA
φ=
0901∠=
−∠=
+∠=⋅
j
BA
BA
BABA
ba
ba
φφ
φφ
Phasor Representation
l A sinusoid can be represented by a phasor:
==+=+ tjeXtjejXetjXetX ωωφφωφω ReReRe)cos(
l The sum of two sinusoids at the same frequency can be represented by another phasor. The new phasor is simply the sum of the two original phasors.
+=+=+ tjeXXtjeXtjeXtjeXtjeX ωωωωω )21(Re21Re2Re1Re
Phasor Representation
l The steady-state response of any branch variable in a stable circuit with a sinusoidal excitation will be another sinusoid at the same frequency (forced response in Chapter 5)
l Kirchhoff’s laws hold in phasor form (only additions are involved).
Phasor with Differentiation
[ ]XjX
Xejdt
dXedt
ationdifferenti
tjtj
tjXed
ω
ω φωφω
φω
→
=
=
++
+
ReReRe
Example 6.3: Parallel Network with an AC Voltage Source
VjCIdt
tdvCti
VI
vi
Vttv
Cc
C
RR
ω⋅=⇒=
=⇒=
∠=⇒+=
)()(
55
2030)204000cos(30)( 00
))(6.464000cos(71.6)(
)(6.4671.60 Atti
AIII CR
+=
∠=+=
Example 6.4: Parallel Network with an AC Current Source
VjCIV
I
Itti
CR ω⋅==
=⇒=
,5
34000cos3)(
))(6.264000cos(4.13)(
1.02.03
)1.02.0(3
0 Vttv
jV
VjIII CR
−=
⋅+=
⋅+=+==
Impedance and Admittance
Phasor Representation
l Under ac steady-state, both the voltage and the current of a branch are sinusoids at the same frequency.
=+= tjeVvtmVtv ωφω Re)cos()(
=+= tjeIitmIti ωφω Re)cos()(
Resistors
l Current and voltage are collinear (in phase).
=×== tjeIRtjeIRtjeVv ωωω ReReRe
imRIvmVIRV φφ ∠=∠==
ivmRImV φφ == ,
Inductors
l Current lags voltage by 90 degrees.
==×=== tjeVtjeILj
dt
tjdeILtjeIdtdL
dtdiLv ωωω
ωω ReReReRe
ILjV ω=
o90 , +== ivmLImV φφω
Capacitors
l Current leads voltage by 90 degrees.
==×=== tjeItjeVCj
dt
tjdeVCtjeVdtdC
dtdvCi ωωω
ωω ReReReRe
ICjI
CjV
ωω−== 1
o90 , −== ivCmI
mV φφω
Phasor Relations (Resistor)
Phasor Relations (Inductor)
Phasor Relation (Capacitor)
Impedance
l In general, we can define a quantity Z and Ohm’s law for ac circuits as IZV =
o90/1/1
o90
−∠==
∠==
=
CCjcZ
LLjLZ
RRZ
ωω
ωω
Time Domain vs. Frequency Domain
Admittance
l Similarly, another quantity admittance Y can be defined.
ZY /1≡
VYI =
Equivalent Impedance and Admittance
)21( NVVVV +++= L
•Series equivalent impedance:
•Parallel equivalent impedance:
NZZZserZ +++= L21
NYYYparY +++= L21
)21( NIIII +++= L
Load Network
IVeqZ /≡
Impedance and Admittance
l Impedance and admittance are complex functions of frequency.
[ ] [ ] )()(ImRe)( ωωω jXRZjZjZZ +=+==
Resistance (Ω) Reactance (Ω)
l Inductors and capacitors are reactive elements, inductive reactance is positive and capacitive reactance is negative.
Impedance and Admittance
Conductance(Siemens)
Susceptance(Siemens)
l Inductors and capacitors are reactive elements, inductive reactance is positive and capacitive reactance is negative.
[ ] [ ] )()(ImRe)( ωωω jBGYjYjYY +=+==
Impedance Triangle
Example 6.6: Impedance Analysis of a Parallel RC Circuit.
0
0
0
6.4671.6
6.26224.015.02.010
151
6.2647.410||5
∠===
∠=+=−
+=
−∠Ω=−=
AYVZV
I
Sjj
Y
jZ
Example 6.7: Frequency Dependence of a Parallel RC Network.
2
2
2
2
2
)(1]Im[)(
)(1]Re[)(
)(1)(
CRCR
ZX
CRR
ZR
CRCRjR
ZZZZ
jZCR
CR
ωω
ω
ωω
ωω
ω
+−==
+==
+−
=+
=
Example 6.8: AC Ladder Calculationsl AC ladder networks can be analyzed by series-parallel
reduction (by replacing resistance with impedance).
Cjω1
Ljω
)(50000cos10)( mAtti =
Example 6.8: (Cont.)
0
0
0
9.364010)24(
24
7.794.8910)24(
101.53806448
−∠=+−
−=
∠=+−
=
∠=+==
VVjj
jV
VVjj
jV
VjIZV
C
L
AC Circuit Analysis
AC Circuit Analysis
l Sources at the same frequency:– Phasor transform method: the time domain sinusoids
are transformed to the frequency domain and represented by phasors.
Time domain à Frequency domain
AC Circuit Analysis
l Sources at the same frequency:– With the transformation, all resistive circuit analysis
techniques are applicable. Resistance is replaced by impedance and conductance is replaced by admittance.
ProportionalityThévenin-NortonNode AnalysisMesh Analysis…
AC Circuit Analysis
l Sources at the same frequency:– After analysis, the resultant phasors are converted back
to the time-domain sinusoids.
Frequency domain à Time domain
AC Circuit Analysis
l Sources at different frequencies:– Due to the linearity, the proportionality method is still
applicable.– The phasor analysis is performed at each individual
frequency
Example 6.9: AC Network with a Controlled Source
)(1000cos20 Vtv =
2/2636
3132)4/(
812)01(01Let
1
21
11
1
02
jIVZjVVIV
jIIIjjVI
jVjI
x
−==+=+−=
+−=+=+−=−=
+=+=∠=
Ati
AIIII
AZVI
)3.1051000cos(4.11
3.1054.11)/(
9010/
01
011
0
+=
∠==
∠==
Use proportionality
Example 6.10: Phase-Sift Oscillator
l Oscillator: Generator a sinusoidal output without an independent input source with initial stored energy.
l Design goal: At one particular frequency, vout=vin.
vx=vin/K
Example 6.10: (Cont.)
RC
LjCR
LVV
I
x
in /2
1
,1let andality proportion Use
2
2
−+
+=
=
ωω
+===
=
=
21,when
/2
/2 :requiresn Oscillatio
CRL
KVV
LC
CL
x
inOSC
OSC
ωω
ω
ωω
Superposition
l An ac source network is any two-terminal network that consists entirely of linear elements and sources. If there are more than one independent source, all of them must be at the same frequency so that the phasor method can be applied.
Frequency Domain Thévenin Parameters
l Frequency-domain Thevenin parameters: – the open-circuit voltage phasor:– the short-circuit current phasor:– Thévenin impedance:
ocV
ocV
scI
scIocVtZ /=
Example 6.11: Application of an AC Norton Network.
Thévenin parameters:
0
0
0
8.81495.0
13.89.91040280
280
7.73202028040
−∠==
−∠=+
=
∠Ω=−=
AZ
VI
Vj
V
jjZ
t
OCSC
OC
t
V Maximize
Example 6.11: (Cont.)
08.814.35
048.00
)048.0()014.0(
minimum is if maximum is
/ and 11
−∠=
+=
−++=
=+=
VV
SjY
BjGY
YV
YIVZZ
Y
eq
eq
eqSCt
eq
Yeq
AC Mesh Analysis
l By using phasors, impedance and admittance, node analysis and mesh analysis are still applicable assuming all independent sources are at the same frequency.
l AC mesh analysis:
[ ][ ] [ ]sVIZ = or [ ][ ] [ ]sVIZZ~~ =−
with controlled sources
Example 6.12: Systematic AC Mesh Analysis
Find i1
Two sources at the same frequency
Example 6.12: (Cont.)
)3.10410cos(9.3)(
3.1049.3
3615)10(2615681048
:equationmesh Single
01
01
1
+=
∠=
+=−−+=Ω−=−+=
=
tti
AI
VjjjVjjjZ
VIZ
S
S
AC Node Analysis
[ ][ ] [ ]sIVY =
[ ][ ] [ ]sIVYY~~
=−
or
with controlled sources
Example 6.13: Systematic AC Node Analysis
Find V1 and Z1
1 2
Constraint equation: I=(V-V1)/j4
Example 6.13: (Cont.)
[ ]
[ ]
Ω+=∠Ω==
−∠=
−∠=
−=
++−−−
−
+
−+
−+
=
+−−−
=
39.192.88.803.9
1.3115.1
3.224.10
20180
3310226
05.005.0
69
24/2
332256
201
011
0
01
2
1
2
1
jI
VZ
AI
VV
jj
VV
jjj
VV
jj
jj
IjVI
I
jj
Y
s
Chapter 6: Problem Set
l 7, 17, 24, 32, 36, 41, 44, 47, 51, 53, 57, 59