emlab 1 회로 이론 (2014). emlab 2 review of circuit theory i 1.linear system 2.kirchhoff’s law...
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EMLAB
1
회로 이론 (2014)
EMLAB
2
Review of circuit theory I
1. Linear system
2. Kirchhoff’s law
3. Nodal & loop analysis
4. Superposition
5. Thevenin’s and Norton’s theorem
6. Resistor, Inductor, Capacitor
7. Operational amplifier
8. First and second order transient circuit
EMLAB
3
Linear system
(t)1(t)1iLinear systemL
(t)(t) 21 BA (t)21 BiAi Linear systemL
(t)2(t)2iLinear systemL
A system satisfying the above statements is called as a linear system. Resis-tors, Capacitors, Inductors are all linear systems. An independent source is not a linear system.
All the circuits in the circuit theory class are linear sys-tems!
EMLAB
4
R1
1k
Resistor
Examples of Linear system
-)(tR
)(1 ti
-)(tC
)(2 ti
11 )()( RtitR
t
02 )(
1)( dtti
CtC
C1
1n
Capacitor
input
output
R1
C1
input
EMLAB
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Inductor
-(t)L
)(2 ti
dt
diLtL
2)(
EMLAB
6
Kirchhoff’s Voltage law
-(t)1 -(t)2
-
(t)s
0)( n
n t 0C
drE
Sum of voltage drops along a closed loop should be equal to zero!
0)(-)()( 21 ttt s
R1 C1
EMLAB
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Kirchhoff’s Current law
0)( n
n tI 0S
daJ
(t)0
)(2 ti)(1 ti
)(3 ti3
303
2
202
1
101
321
-)(,
-)(,
-)(
0)()()()(
Rti
Rti
Rti
titititIn
n
Ri ba
ab
RCurrent definition
Sum of outgoing(incoming) currents from any node should be equal to zero!
• To define a current, a direction can be chosen arbitrarily.
• The value of a current can be obtained from a voltage drop along the direction of current divided by a resistance met.
R2
)(1 t )(2 t
)(3 t
R1
R3
0---
3
30
2
20
1
10 RRR
EMLAB
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Nodal analysis
• Unknowns : node voltages
• Kirchhoff’s current law is utilized to form matrix equations.
• For each node, the sum of out-going currents become zero.
1
2
0612
0-
6
0- 11 mkk
0
4
0-
4
0-6 22
kkm
EMLAB
9Loop analysis
• Unknowns : loop currents
• Matrix equations are formed by Kirchhoff’s voltage law.
• For each loop, the sum of voltage drops are equal to zero.
1i 2i 3i
0)(126 211 iikik
mi 62
04)(4 323 ikiik
EMLAB
10Superposition
• Superposition is utilized to simplify the original linear circuits.
• If a voltage source is eliminated, it is replaced by a short circuit connected to the original terminals.
• If a current source is eliminated, it is replaced by an open circuit.
Circuit Circuit Circuit
L
1L
2L+=
21 LLL
Circuit with currentsource set to zero(OPEN) Circuit with voltage source
set to zero (SHORT CIRCUITED)
EMLAB
11Example
VVkkk
k412
666
601
01
02
Vkm
kk
k 243612
266
121
61
61
02
V20244020102
EMLAB
12Thevenin and Norton equivalent
Open circuited voltagemeasured by voltmeter
Short circuited current measured by ammeter
Resistance obtained with voltage source shorted and current source open
AV
EMLAB
13Example
VVkk
kVTh 612
22
2
kkkRTh 12||2
kkkRTh 12||2
mAk
VIN 6
2
12
EMLAB
14Resistor – Input output relationship
2 4 6 80 10
5
10
15
0
20
time, sec
Vou
t, V
2 4 6 80 10
2
4
6
0
8
time, sec
I_P
robe
1.i,
A
VoutRR1R=2 Ohm
ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )
I_ProbeI_Probe1
i(t)
i(t)
v(t)v(t)
)()( tiRt
EMLAB
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t
2 4 6 80 10
2
4
6
0
8
time, sec
I_P
robe
1.i,
A
2 4 6 80 10
5
10
15
20
0
25
time, sec
Vou
t, V
VoutCC1C=1F
ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )
I_ProbeI_Probe1
i(t)
v(t)
t diC
t 0 )(1
)(
i(t)
v(t)
Capacitor – Input output relationship
EMLAB
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dt
diLt )(
2 4 6 80 10
2
4
6
0
8
time, sec
I_P
robe
1.i,
A
2 4 6 80 10
0
1
2
3
-1
4
time, sec
Vou
t, V
VoutLL1
R=L=1.H
ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )
I_ProbeI_Probe1
i(t)
v(t)
i(t)
v(t)
Inductor – Input output relationship
EMLAB
17First order transient circuit
The voltage drop across a capacitor cannot change instantaneously.
The current through an inductor cannot change instantaneously.
tRC
C
ststC
SCC
C
C
AevRC
s
RCsAeAev
Vtvdt
dvRCv
iRv
1
,1
0)1(,
)0(,0
0
]1[)(
)(,0
0)0(),0(
tL
RS
St
L
R
S
eR
Vti
R
VAeti
dt
diLiR
titVdt
diLiR
EMLAB
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Second order transient circuit
02 2002
2
idt
di
dt
id 0
)0(0
)0()0()(
1
2
2
0
LC
i
dt
di
L
R
dt
id
t
tdtti
CiR
dt
diL s
t
C
Normalized form
L
R
LC 0
20 2
,1
02)( 200
2 sseti st
12002,1 s
LR
C
outst
out dttiC 0
)(1
L
R
LC 0
20 2
,1
EMLAB
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tsts eKeKti 21
21)(
0)0( 21 KKi
sKsKsdt
diL 2211)0(
tAtAeti ddt sincos)( 21
0
0)0( 1 Ai
sd AAdt
diL 210)0(
tetBBti 0
21)(
1)0( Bi
sBBdt
diL 210)0(
Solution
Critically damped : ζ = 1
Under-damped : ζ <1Over-damped : ζ > 1
2
0 1 d
EMLAB
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t
t
ttt
eBtBti
BtBtie
tietiee
titdt
dit
dt
id
0
0
00
0
)()(
)(
0])([0])([1
0)()(2)(
21
21
2002
2
Critically damped: ζ=1 인 경우
EMLAB
5 10 15 20 250 30
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0.0
1.6
time, sec
Vou
t, V
21
25.1
25.0
1
Transient response
Critically damped
Under-damped
Over-damped
EMLAB
22
LR
C
outs
Ringing in digital logics
in s
EMLAB
23Contents : Circuit theory 2
1. AC steady-state analysis : 60Hz sinusoidal input signal
• Power factor
2. Magnetically coupled networks : transformer
3. Poly-phase circuits : power distribution
• Single phase two wire
• Three phase 4 wire power distribution
4. Arbitrary input signal
• Fourier series and Fourier transform
• Laplace transform
5. Two-port network : black box
EMLAB
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Chapter 8
AC steady-state analysis
EMLAB
25
)0(cos
)0(0)()(2)( 2
002
2
ttV
ttxt
dt
dxt
dt
xd
ss
Sinusoidal input signal
입력 전압의 형태
LR
C
outs
0t
EMLAB
26Sinusoids
tXtx M sin)(
(radians)argument
(rads/sec)frequency angular
valuemaximumor amplitude
t
XM
Dimensionless plot As function of time
tTtxtxT ),()( Period 2
)(cycle/sec Hertzin frequency 2
1
Tf
f 2
0t
)(sin)( 0ttXtx Mlag
“Lag by t0”
)(sin)( 0ttXtx Mlead
0t
“Lead by t0”
EMLAB
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AC (Alternating Current)• Easy to generate ( 교류 전압은 만들기 쉽다 .)
• Easy to change voltage levels. ( 전압을 변화하기도 쉽다 .)
• Less damage on human compared with DC ( 직류에 비해 덜 위험하다 .)
rmsV 220 Hz,60
EMLAB
28
0)0(),()()( :KVL itVtRitdt
diL S
)0(0
)0(cos)(V 0
S t
ttVt
0)()( tRit
dt
diL h
h
)()()( tVtRitdt
diL Sp
p
Solution of Differential Eq.
To solve a differential equation, initial conditions must be specified.
solution Particular:)(
solution sHomogeneou:)(
)()()(
ti
ti
tititi
p
h
ph
)()]()([)]()([
tVtitiRdt
titidL Sph
ph
EMLAB
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0)()( tRitdt
diL h
h
)()()( tVtRitdt
diL Sp
p
tL
Rst
h AeAeti
)(
)cos()( tBtip
tVtBLRtBLR
tVtRBtLB
cossin)cossin(cos)sincos(
cos)cos()sin(
0
0
R
L
LR
VB
tan,sincos
0
R
Lt
LR
Vtip
1
22
0 tan),cos()(
)(
R
Lt
LR
VAeti
tL
R
1
22
0 tan),cos()(
)(
Solution method #1
For a particular solution, choose a trial function that might produce VS(t) on entering the differential Eq.
EMLAB
30Simpler method for sinusoidal source case
)()()( tVtRitdt
diL Sp
p }Re{}Re{)cos()( tjtjjp eeBetBti I
}Re{Re
}Re{}Re{}Re{
0
0
tjtjtj
tjtjtj
eVeRedt
dL
eVeRedt
dL
II
II
0Re 0
tjtjtj eVeRe
dt
dL II
LjR
eVeti
RLj
V
eVeReLj
eVeRedt
dL
tjtj
p
tjtjtj
tjtjtj
00
0
0
Re}Re{)(,
0
II
II
II
)cos()()(
ReRe)(22
0
22
00
t
LR
V
LR
eeV
LjR
eVti
tjjtj
p
tjte tj sincos
sin,cos
tan, 122
ryrxx
yyxr
rejyx
PolarComplexj
EMLAB
31
tjtjjtj
eeLR
eV
LjR
eVti
IRe
)(ReRe)(
22
00
Phasor
22
0
22
00
)()(I:Phasor
LR
V
LR
eV
LjR
V j
• With sinusoidal source function, it is simpler to use a trial solution ~ Re{I ejwt}.
• The complex coefficient of the exponential function is called as a phasor.
EMLAB
32
VoutRR1R=2 Ohm
ItSineSRC1
Phase=0Damping=0Delay=0 nsecFreq=0.16 HzAmplitude=1 AIdc=0 mA
I_ProbeI_Probe1
Phasor - resistor
2 4 6 8 10 12 14 16 180 20
-1
0
1
-2
2
time, sec
Vou
t, V
I_P
robe
1.i,
A
(t)(t) SViR
)(t
t
)(ti
Relationship between sinusoids
EMLAB
33
ILjV
Iedt
dLVe
dt
tdiLtv
tjtj
}Re{}Re{
)()(
Relationship between sinusoids
)2/cos(}Re{
}Re{}Re{)(
}Re{}Re{
)()(
)2/(
tLIIeL
IeLjVetv
ILjV
Iedt
dLVe
dt
tdiLtv
tj
tjtj
tjtj
Phasor - inductor
)(t
t
)(ti
0
2
EMLAB
34
Relationship between sinusoids
Cj
Cj
edt
dCe
dt
tdvCti
tjtj
IV
VI
VI
}Re{}Re{
)()(
Phasor - capacitor
)(t
t
)(ti
0
2
EMLAB
35Examples
inductor theacross voltage theFind
60),(304,05.0 HzfAIHL
1202 fLIjV
30490105.0120V 6024V
)60120cos(24)( tv
inductor theacross voltage theFind
60,1456.3,150 HzfIFC
1202 f
Cj
IVCVjI
90110150120
1456.36
V
235200
V
)235120cos(200
)( ttv
EMLAB
36Impedance and Admittance
jXRZI
V
I
VzMiv
M
M
iM
vM
)(I
VZ
AC circuitvMV
iMI
zZ
Impedance :
jBGY YY V
IYAdmittance :
Element Impedance Admittance
R
L
C
LjL Z
CjC 1
Z
LjYL
1
CjYC
RR ZR
YR
1
EMLAB
37
1Z
2Z
3Z
321 ZZZZeq
321 YYYYeq
11
1
ZY
22
1
ZY
33
1
ZY
321
1111
ZZZ
Zeq
321
1111
YYY
Yeq
Series / parallel combination
EMLAB
38Example
Find the current i(t) in the network in Fig. E8.8.
25.90323.00052.00319.021 jYYYeq
0189.01050377 61 jjY
024.00319.0104037720
132 j
jY
1Y
2Y
25.3988.33012025.90323.0VYI eq
)25.39377cos(88.3)( tti
EMLAB
39Example 8.15
0621 VV
011
0211
221
j
VV
j
V
21111
6 222
j
VV
j
V
11
62
1
11
11
12 jjj
V
5.15.22 jV
][5.15.21
20 Aj
VI
Super node