Download Lecture 01

ECE 342
1. Networks & Systems
Jose E. Schutt-Aine
Electrical & Computer Engineering
University of Illinois
[email protected]
ECE 342 – Jose Schutt-Aine
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What is Capacitance?
1
Voltage=0
No Charge
No Current
2
Voltage build up
Charge build up
Transient Current
ECE 342 – Jose Schutt-Aine
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Voltage = 6V
Charge=Q
No Current
2
What is Capacitance?
4
Voltage=6V
Charge=Q
No Current
5
Voltage decaying
Charge decaying
Transient Current
ECE 342 – Jose Schutt-Aine
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Voltage=0
No Charge
No Current
3
Capacitance
Relation: Q = Cv
Q: charge stored by capacitor
C: capacitance
v: voltage across capacitor
i: current into capacitor
dv dQ
i(t )  C 
dt dt
1 t
v(t )   i ( )d
C 0
ECE 342 – Jose Schutt-Aine
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What is Inductance?
- Current in wire produces magnetic field
- Flux is magnetic field integrated over area
Total Flux Linked
Inductance 
Current
ECE 342 – Jose Schutt-Aine
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Inductance
d
LN
di

L
I
B  Hdv
L
2
I
v
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Inductance
Relation:  = Li
: flux stored by inductor
L: inductance
i: current through inductor
v: voltage across inductor
di d 
v(t )  L 
dt dt
1 t
i (t )   v( )d
L 0
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Low-Pass Circuit
dvo
iR (t )  C
 iC (t )
dt
vi (t )  RiR (t )  vo (t )
dvo
vi (t )  RC
 vo (t )
dt
Need to solve for vo(t)
ECE 342 – Jose Schutt-Aine
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Low-Pass Circuit
dvo
iR (t )  C
 iC (t )
dt
vi (t )  RiR (t )  vo (t )
dvo
vi (t )  RC
 vo (t )
dt
Need to solve for vo(t)
ECE 342 – Jose Schutt-Aine
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Reactive Circuit
iC (t )  iR (t )  iL (t )
diL
vx  L
dt
Need to solve for vx(t)
d  vi  vx 
iC  C
dt
ECE 342 – Jose Schutt-Aine
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Laplace Transforms
The Laplace transform F(s) of a function f(t) is
defined as:

L  f (t )   f (t )e st dt  F ( s)
0
To a mathematician, this is very meaningful;
however circuit engineers seldom use that
integral
The Laplace transform provides a conversion
from the time domain into a new domain, the s
domain
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Laplace Transforms
The Laplace transform of the derivative of a
function f(t) is given by

df (t )  st
 df (t ) 
L
e dt  sF ( s)

 dt  0 dt
Differentiation in the time domain becomes a
multiplication in the s domain. That is why Laplace
transforms are useful to circuit engineers
s = jw
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Low-Pass Circuit
Time
domain
s
domain
vo (t )  Vo ( s )
dvo (t )
 sVo ( s )
dt
Time domain
s domain
dvo
vi (t )  RC
 vo (t )
dt
Vi (s)  sRCVo (s)  Vo (s)
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Low-Pass Circuit - Solution
Vi ( s )
Vo ( s )
1
Vo ( s ) 


1  sRC
Vi ( s ) 1  sRC
Pole exists at s  1/ RC
Assume that Vi(s)=Vd/s
Vd
A
B
Vo ( s ) 
 
s 1  sRC  s  s  1/ RC 
Inversion
vo (t )  Au(t )  Bet / RC
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Reactive Circuit
diL
vx  L
dt
d  vi  vx 
iC  C
dt
Vx (s)  sLI L
IC  sC (Vi  Vx )
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Reactive Circuit - Solution
Vx Vx
sC (Vi  Vx )  
R sL
1 1 

sCVi  Vx  sC   
R sL 

sCVi
Vi
Vx 

1 1  
1
1 

 2
 sC    1 

R sL   sRC s LC 

Exercise: Use inversion to solve for vx(t)
ECE 342 – Jose Schutt-Aine
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Table of Laplace Transforms
Function
u (t   )
e  t
cos( t )
Transform
1  s
e
s
1
s 
s
s2   2
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Capacitance – s Domain
Assume time-harmonic behavior of
voltage and current variables jw = s
i(t ), v(t ) ~ e
jwt
i (t )  I ( s )
I ( s)  sCV (s)
v(t )  V ( s )
The capacitor is a reactive element
d
1
s
Reactance : X C 
dt
sC
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Inductance – s Domain
Assume time-harmonic behavior of
voltage and current variables jw = s
i(t ), v(t ) : e
jwt
i (t )  I ( s )
V ( s)  sLI (s)
v(t )  V ( s )
The inductor is a reactive element
d
s
Reactance
:
X

sL
L
dt
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Zth
Thevenin Equivalent
Network
Vth
+
-
• Principle
– Any linear two-terminal network consisting of current or voltage
sources and impedances can be replaced by an equivalent circuit
containing a single voltage source in series with a single impedance.
• Application
– To find the Thevenin equivalent voltage at a pair of terminals, the load
is first removed leaving an open circuit. The open circuit voltage across
this terminal pair is the Thevenin equivalent voltage.
– The equivalent resistance is found by replacing each independent
voltage source with a short circuit (zeroing the voltage source),
replacing each independent current source with an open circuit
(zeroing the current source) and calculating the resistance between
the terminals of interest. Dependent sources are not replaced and can
have an effect on the value of the equivalent resistance.
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Network
Norton Equivalent
Ith
Gth
• Principle
– Any linear two-terminal network consisting of current or voltage
sources and impedances can be replaced by an equivalent circuit
containing a single current source in parallel with a single impedance.
• Application
– To find the Norton equivalent current at a pair of terminals, the load is
first removed and replaced with a short circuit. The short-circuit
current through that branch is the Norton equivalent current.
– The equivalent resistance is found by replacing each independent
voltage source with a short circuit (zeroing the voltage source),
replacing each independent current source with an open circuit
(zeroing the current source) and calculating the resistance between
the terminals of interest. Dependent sources are not replaced and can
have an effect on the value of the equivalent resistance.
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Thevenin Equivalent - Example
Circuit for calculating impedance
Rth 
Calculating Thevenin voltage
( 2 )( 3 )
 1.2 k
23
KCL law at node A’ gives 1+Iy = Ix
Also KVL gives 10=2Iy+3Ix
Combining these equations gives Ix = 2.4 mA
From which we calculate
Vx = Vth=3(2.4)=7.2 V
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Thevenin Example
Find voltage across R4
Define G1  1/ R1 ,G2  1/ R2 , and G3  1/ R3 ,
 G1  G2  v1  G2v2  is
G2v1   G2  G3  v2  0
From which
ECE 342 – Jose Schutt-Aine
v2 
is
 Vth
5
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Thevenin Equivalent
Circuit for calculating impedance
(current source is replaced with open circuit)
R2
R1
R3
Zth
( R1  R2 )R3 2
Rth  ( R1  R2 ) P R3 
 
R1  R2  R3 5
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Circuit with Thevenin Equivalent
ZTH
VTH
R4is / 5
vo ( t ) 
2 / 5  R4
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Thevenin Example
Need to find the current i1(t) in resistor R3
R1
10 
V
100 V
L1
L2
1H
1H
R2
10 
ECE 342 – Jose Schutt-Aine
i1(t)
R3
10 
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Thevenin Example
Thevenin voltage calculation
10 +s
100/s
+
-
10
10 +s
Thevenin Network
10( 100 / s )
1,000
Vth 

10  s  10 s  s  20 
ECE 342 – Jose Schutt-Aine
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Thevenin Example
Calculating the impedance
10 +s
10
10( s  10 )
Zth 
 s  20 
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Thevenin Example
Zth
I1
Vth
+
-
ZLoad = 10 +s
Vth ( s )
1,000
I 1( s ) 

Z th ( s )  Z Load 10  s  10 
 ( s  10 )
s  20
1,000
I1( s ) 
s s 2  40s  300


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Thevenin Example
Expanding using partial fraction
K3
K1
K2
I 1( s ) 


s ( s  10 ) ( s  30 )
With K1 = 3.33, K2 = -5, K3 = 1.67
The time-domain current i1(t) is:
i1( t )  3.33  5e
10t
 1.67e
30t
ECE 342 – Jose Schutt-Aine
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