Download Linear system

1
회로 이론 (2014)
EMLAB
Review of circuit theory I
2
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
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Linear system
i1 (t)
Linear system
L
1 (t)
i2 (t)
Linear system
L
 2 (t)
Ai1  Bi2 (t)
Linear system
L
A1 (t)  B 2 (t)
A system satisfying the above statements is called as a linear system.
Resistors, Capacitors, Inductors are all linear systems. An independent
source is not a linear system.
All the circuits in the circuit theory class are linear systems!
EMLAB
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Examples of Linear system
R1
output
Resistor
 R (t ) 1k
R1
i1 (t )
 R (t )  i1 (t )  R1
input
 C (t ) -
Capacitor
C1
C1
i2 (t )
1n
input
1 t
 C (t )   i2 (t ) dt
C0
EMLAB
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Inductor
 L (t) i2 (t )
 L (t )  L
di2
dt
EMLAB
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Kirchhoff’s Voltage law
 E  dr  0
 n ( t )  0
n
C
Sum of voltage drops along a closed loop should be equal to zero!
1 (t )   2 (t ) -  s (t )  0
1 (t) R1
 2 (t) C1

 s (t)
-
EMLAB
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Kirchhoff’s Current law
 J  da  0
 I n (t )  0
n
S
Sum of outgoing(incoming) currents from any node should be equal to zero!
1 (t )
i2 (t )
i1 (t )
 2 (t )
 0 (t)
R2
R1
R3
 0 - 1
R1
, i2 (t ) 
0 - 2
R2
, i3 (t ) 
0 -3
R3
i3 (t )

Current definition
i
n
i1 (t ) 
 3 (t )
a
 I n (t )  i1 (t )  i2 (t )  i3 (t )  0
R
b
a  b
R
 0 - 1  0 -  2
R1

R2

0 -3
R3
0
• 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.
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 - 0 1 - 0
6k

12k
 6m  0
 6m 
2 - 0 2 - 0
4k

4k
0
2
1
EMLAB
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Loop 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.
i1
i2
i3
4k  (i3  i2 )  4k  i3  0
6k  i1  12k  (i1  i2 )  0
i2  6m
EMLAB
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Superposition
• 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.
 L   L1   L 2

Circuit
L


=
Circuit
 L1

+
 L2
Circuit

Circuit with current
source set to zero(OPEN)

Circuit with voltage source
set to zero (SHORT CIRCUITED)
EMLAB
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Example
 02   01   02  4  24  20V
  01 
 01 
6k
 12V  4V
6k  6k  6k
  02 
 02  
1
6k
1
1

6k 12k
 6m  6k  
2
 36  24V
2 1
EMLAB
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Thevenin and Norton equivalent
Resistance obtained with voltage source
shorted and current source open
V
Open circuited voltage
measured by voltmeter
A
Short circuited current
measured by ammeter
EMLAB
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Example
RTh  2k || 2k  1k
VTh 
RTh  2k || 2k  1k
2k
 12V  6V
2k  2k
IN 
12V
 6mA
2k
EMLAB
Resistor – Input output relationship
14
i(t)
I_Probe1.i, A
8
i(t)
6
4
2
0
0
2
4
6
8
10
6
8
10
time, sec
v(t)
v(t)
20
Vout, V
 (t )  R  i (t )
15
10
5
0
0
2
4
time, sec
EMLAB
Capacitor – Input output relationship
15
i(t)
8
I_Probe1.i, A
6
4
2
0
i(t)
0
2
4
6
8
10
time, sec
t
v(t)
v(t)
25
1 t
 (t )  0 i ( )d
C
Vout, V
20
15
10
5
0
0
2
4
6
8
10
time, sec
EMLAB
Inductor – Input output relationship
16
i(t)
I_Probe1.i, A
8
6
4
2
0
0
i(t)
2
4
6
8
10
6
8
10
time, sec
v(t)
v(t)
4
di
 (t )  L
dt
Vout, V
3
2
1
0
-1
0
2
4
time, sec
EMLAB
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First order transient circuit
The voltage drop across a capacitor cannot
change instantaneously.
vC  iR  0
dvC
 0, vC (t  0)  VS
dt
vC  Ae st , Ae st (1  RCs )  0
 vC  RC
1

t
1
s  
, vC  Ae RC
RC
The current through an inductor
cannot change instantaneously.
di
 VS (t  0), i (t  0)  0
dt
R
 t
V
di
iR  L  0, i (t )  Ae L  S
dt
R
R
 t
VS
 i (t )  [1  e L ]
R
iR  L
EMLAB
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Second order transient circuit
R
L

s
C
 out


di
1 t
L  iR   i (t )dt   C (0)   s
dt
C0
0
d 2i R di
i
 2

0
dt
L dt LC
( t  0)
( t  0)
 out
1t
  i (t )dt
C0
Normalized form
d 2i
di
2

2



0
0i  0
dt 2
dt
 2
1
R 
 0 

,  
LC
2

L
0 

 2
1
R 
 0 

,  
LC
20 L 

i (t )  e st  s 2  20 s  02  0
s1, 2  0  0  2  1
EMLAB
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Solution
Over-damped : ζ > 1
Under-damped : ζ <1
i (t )  K1e s t  K 2 e s t
i (t )  e  t  A1 cos d t  A2 sin d t 
i ( 0  )  K1  K 2  0
i (0 )  A1  0
1
L
2
di
(0 )  s1 K1  s2 K 2   s
dt
0
L
    1   2 
0
 d

di
(0 )  0 A1  d A2   s
dt
Critically damped : ζ = 1
i (t )  B1  B2 t e  t
0
i (0 )  B1
L
di
(0 )  0 B1  B2   s
dt
EMLAB
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Critically damped: ζ=1인 경우
d 2i
di
2
(
t
)

2

(
t
)


0
0 i (t )  0
2
dt
dt
1
 0t [e0t i (t )]  0  [e0t i (t )]  0
e
 e0t i (t )  B1t  B2
 i (t )  ( B1t  B2 )e 0t
EMLAB
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Transient response
Under-damped
  0.25
1.6
  1 Critically damped
1.4
1.2
Vout, V
1.0
0.8
0.6
  1.25
Over-damped
0.4
0.2
0.0
0
5
10
15
20
25
30
time, sec
EMLAB
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Ringing in digital logics
 in
s
R
s
L

C
 out

EMLAB
Contents : Circuit theory 2
23
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
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Sinusoidal input signal
t0
s
R
L

C
 out

0
( t  0)

d 2x
dx
2
(t )  20
(t )  0 x (t )  
dt 2
dt
Vs cos s t (t  0)
입력 전압의 형태
EMLAB
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Sinusoids
x (t )  X M sin  t
Dimensionless plot
As function of time
xlead (t )  X M sin  (t  t0 )
X M  amplitude or maximum value
  angular frequency (rads/sec)
 t  argument (radians)
T
f 
2

 Period  x(t )  x(t  T ), t
1 

 frequency in Hertz (cycle/sec )
T 2
  2 f
“Lead by t0”
 t0
t0
xlag (t )  X M sin  (t  t0 )
“Lag by t0”
EMLAB
27
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.)
60Hz, 220 Vrms
EMLAB
28
Solution of Differential Eq.
di
KVL : L (t )  Ri (t )  VS (t ), i (0  )  0
dt
To solve a differential equation, initial conditions
must be specified.
L
d [ih (t )  i p (t )]
dt
L
V0 cos t (t  0)
VS (t )  
(t  0)
 0
L
 R [ih (t )  i p (t )]  VS (t )
dih
(t )  Ri h (t )  0
dt
di p
dt
(t )  Ri p (t )  VS (t )
i(t )  ih (t )  i p (t )
ih (t ) : Homogeneou s solution
i p (t ) : Particular solution
EMLAB
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Solution method #1
di
L h (t )  Ri h (t )  0 
dt
L
di p
dt
ih (t )  Ae  Ae
st

R
t
L
(t )  Ri p (t )  VS (t )  i p (t )  B cos(t   )
For a particular solution, choose a trial function that might
produce VS(t) on entering the differential Eq.
 LB sin( t   )  RB cos(t   )  V0 cos  t
 ( R cos   L sin  ) B cos t  ( R sin   L cos  ) B sin t  V0 cos  t
B 
V0
L
, tan   
R cos   L sin 
R
 i p (t ) 
 i(t )  Ae
 L 
cos(t   ),    tan 

2
2
R


R  (L)
V0

R
t
L

1
 L 
cos(t   ),    tan 1 

2
2
R


R  (L)
V0
EMLAB
30
Simpler method for sinusoidal source case
L
di p
dt
j jt
jt
(t )  Ri p (t )  VS (t )  i p (t )  B cos(t   )  Re{Be e }  Re{Ie }
d
Re{Ie jt }  R Re{Ie jt }  V0 Re{e jt }
dt
 d

 Re  L Ie jt  RIe jt   Re{V0 e jt }
 dt

L
 d

 Re  L Ie jt  RIe jt  V0 e jt   0
 dt

d jt
Ie  RIe jt  V0 e jt  0
dt
 jLIe jt  RIe jt  V0 e jt
L
e jt  cos t  j sin t
Complex Polar
x  jy  re j
r  x 2  y 2 ,   tan 1
y
x
x  r cos  , y  r sin 
 V0 e jt 
V0
jt
I 
, i p (t )  Re{Ie }  Re 

jL  R
R

j

L


j jt


 V0e jt 
 V0e e

 i p (t )  Re 
  Re  2

2
R

j

L



 R  (L) 

V0
R  (L)
2
2
cos(t   )
EMLAB
31
Phasor
j


 V0e jt 
 V0e
jt 
 i(t )  Re 
e   Re I e jt
  Re  2
2


 R  jL 
 R  (L)


V0
Phasor : I 

R  jL
V0e j
R  (L)
2
2

V0
R  (L)
2
2


•
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
Phasor - resistor
R  i(t)  VS (t)
I_Probe1.i, A
Vout, V
2
1
0
Relationship between sinusoids
-1
 (t )
-2
0
2
4
6
8
10
12
time, sec
14
16
18
20
i (t )
t
EMLAB
33
Phasor - inductor
Relationship between sinusoids
v(t )  L
di (t )
dt
 (t )
d
Re{Ve }  L Re{ Ie j t }
dt
V  j L I
i (t )
j t

t
2
di (t )
v(t )  L
dt
Re{Ve j t }  L
0
d
Re{ Ie j t }
dt
V  j L I
 v(t )  Re{Ve jt }  Re{ j L Ie jt }
  L Re{ Ie j (t  / 2) }   LI cos(t   / 2)
EMLAB
34
Phasor - capacitor
Relationship between sinusoids
i (t )
 (t )
i (t )  C
dv (t )
dt
d
Re{Ie }  C Re{Ve j t }
dt
 I  j C V
j t
V 
0

t
2
I
j C
EMLAB
35
Examples
L  0.05H , I  4  30( A), f  60 Hz
C  150  F , I  3.6  145, f  60 Hz
Find the voltage across the inductor
Find the voltage across the inductor
  2 f  120
V  jLI
V  120  0.05 190  4  30
V  2460
v (t )  24 cos(120  60)
  2 f  120
I  jCV  V 
V
I
jC
3.6  145
120 150 106 190
200
V
  235

v (t ) 
200

cos(120 t  235)
EMLAB
36
Impedance and Admittance
I M  i

Z z
VM  v
AC circuit

Impedance :
Admittance :
Element
V VM  v VM


( v   i )  Z M  z  R  jX
I
I M  i I M
I
Y   YY Y  G  jB
V
Z
Impedance
Admittance
R
ZR  R
YR 
L
Z L  j L
YL 
C
ZC 
1
j C
1
R
1
j L
YC  j C
EMLAB
37
Series / parallel combination
Z1
Z2
Y1 
1
Z1
Y2 
1
Z2
Y3 
1
Z3
Z3
Yeq  Y1  Y2  Y3
Z eq  Z1  Z 2  Z 3
Yeq 
1
1 1 1
 
Y1 Y2 Y3
Z eq 
1
1
1
1


Z1 Z 2 Z 3
EMLAB
38
Example
Find the current i(t) in the network in Fig. E8.8.
Y1
Y2
Y1  j 377  50  10 6  j 0.0189
Y2 
1
 0.0319  j 0.024
3
20  j 377  40  10
Yeq  Y1  Y2  0.0319  j 0.0052  0.0323  9.25
I  YeqV  0.0323  9.25  120  30  3.88  39.25
i (t )  3.88 cos(377t  39.25)
EMLAB
39
Example 8.15
Super node
V1  V2  60
V1
V
V
 20  2  2  0
1  j1
1 1 j
V2  6 V2
V
  2 2
1  j1 1 1  j
 1
1
V2 
1
1
 1  j1

6
  2 
j
1  j1
V2  2.5  j1.5
I0 
V2
 2.5  j1.5 [ A]
1
EMLAB