eleg 205 fall 2017 lecture #14mirotzni/eleg205/lecture15.pdf · chapter 10: steady-state....
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ELEG 205Fall 2017
Lecture #15
Mark Mirotznik, Ph.D.Professor
The University of DelawareTel: (302)831-4221
Email: mirotzni@ece.udel.edu
Chapter 10: Steady-StateSinusoidal Analysis
)cos()( φω += tAtV
Review of Complex NumbersHow can we convert a complex number from rectangular to polar notation?
22ir AAA +=
)(tan 1
r
i
AA−=φ
Review of Complex NumbersHow can we convert a complex number from polar to rectangular notation?
)sin(
)cos(
φ
φ
AA
AA
i
r
=
=
Review of Complex NumbersAj
ir eAjAAA φ⋅=+=Bj
ir eBjBBB φ⋅=+=Summary
( )BAjeBA
C φφ −⋅=
( )BAjeBAC φφ +⋅=( ) ( )iirr BAjBAC +++=
( ) ( )iirr BAjBAC −+−=
Addition and Subtraction Multiplication and Division
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
485
832π
π
j
j
je
jeC−
+
+−=
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
485
832π
π
j
j
je
jeC−
+
+−=
42185
83)sin(2)cos(2ππ
ππjj
ee
jjC−
+
+−+=
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
42185
83)sin(2)cos(2ππ
ππjj
ee
jjC−
+
+−+=
485
83)0(2)1(2πj
e
jjC+
+−⋅+−⋅=
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
)4
sin(8)4
cos(85
85ππ j
jC++
+−=
485
83)0(2)1(2πj
e
jjC+
+−⋅+−⋅=
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
657.5657.1085
228
2285
85j
j
j
jC++−
=++
+−=
)4
sin(8)4
cos(85
85ππ j
jC++
+−=
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
657.5657.1085j
jC++−
=
−
−
−
⋅+
⋅+=
657.10657.5tan
22
58tan
22
1
1
657.5657.10
85j
j
e
eC
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
−
−
−
⋅+
⋅+=
657.10657.5tan
22
58tan
22
1
1
657.5657.10
85j
j
e
eC
488.0
0122.1
065.12434.9
j
j
eeC⋅⋅
=−
Review of Complex Numbers: ExampleConvert to single complex number in polar notation
5.17819.0 jeC −⋅=
488.0
0122.1
065.12434.9
j
j
eeC⋅⋅
=−
Phasors
)sin()cos( xjxe jx +=Recall:( )[ ]
[ ][ ]
φ
ω
ωφ
φωφω
jm
tj
tjjm
tjmm
eVV
eV
eeVeVtVtV
⋅=
⋅=
⋅⋅=
⋅=+= +
~
~Re
ReRe)cos()(
Where is called a Phasor
Phasors
φjm eVV ⋅=~
What does the Phasor tell us about the sinusoidal voltage or current
Phasors
φjm eVV ⋅=~
What does the Phasor tell us about the sinusoidal voltage or current
For example if a Phasor voltage is ojeV 305~ ⋅=
and the frequency was known as kHzf 5=could you tell me what the voltage waveform was in time?
PhasorsFor example if a Phasor voltage is
ojeV 305~ ⋅=and the frequency was known as kHzf 5=could you tell me what the voltage waveform was in time?
)3050002cos(5)( ottv +⋅= π
ojeV 305~ ⋅= kHzf 5=
Phasor Transform of Sinusoidal Sources
)cos()( φω += tVtV m
Time-Domain Phasor-Domain
φjm eVV ⋅=~
Phasor Transform of Sinusoidal Sources
)cos()( φω += tIti m
Time-Domain Phasor-Domain
φjm eII ⋅=~
Phasor Transform of Sinusoidal Sources: Examples
)3
1000cos(10)( π+= ttV
Time-Domain Phasor-Domain
310~ πjeV ⋅=
Phasor Transform of Sinusoidal Sources: Examples
Time-Domain Phasor-Domain
)40000,10cos(5.1)( otti +=ojeI 405.1~ ⋅=
Phasor Transform of Sinusoidal Sources: Examples
)4
1000sin(7)( π−= ttV
Time-Domain Phasor-Domain
?~ =V
Inverse Phasor Transform
?)( =tV
Time-Domain Phasor-Domain
58~ πjeV ⋅=
kHzf 5=
Inverse Phasor Transform
)5
50002cos(8)( ππ +⋅⋅= ttV
Time-Domain Phasor-Domain
58~ πjeV ⋅=
kHzf 5=
Inverse Phasor Transform
?)( =tV
Time-Domain Phasor-Domain
jV 15~ =MHzf 1=
Inverse Phasor Transform
)2
102cos(15)( 6 ππ +⋅=tV
Time-Domain Phasor-Domain
215
15~
πje
jV
⋅=
=
MHzf 1=
Phasors with Circuit Components
=== tjtjtjjm e
RVeIeeIti I ωωωφ~
Re~ReRe)(
R)(ti tj
tjjm
vm
eV
eeV
tVtvv
ω
ωφ
φω
~Re
Re
)cos()(
=
=
+=
= tjtj eRVeI ωω~
Re~Re
Phasors with Circuit Components
R)(ti tj
tjjm
eV
eeVtv v
ω
ωφ
~Re
Re)(
=
=
= tjtj eRVeI ωω~
Re~Re
= tjtj eRVeI ωω~
Re~Re
Phasors with Circuit Components
R)(ti tj
tjjm
eV
eeVtv v
ω
ωφ
~Re
Re)(
=
=
= tjtj eRVeI ωω~
Re~Re
RVI~~ =
IVR ~~
=
Phasors with Circuit Components
R tjeVtv ω~Re)( =
IVR ~~
=
The ratio of the phasor voltage to phasor current is called the impedance (symbol Z)
tjeIti ω~Re)( =
Phasors with Circuit Components
IVRZ ~~
==
The ratio of the phasor voltage to phasor current is called the impedance (Z)
IVZ ~~
=
In general For a resistor
C
[ ] tjtjtj eCjVeVdtdCeI
dttdvCti
ωωω ω ⋅⋅==
=
~Re~Re~Re
)()(
tjeVtv ω~Re)( = tjeIti ω~Re)( =
Phasors with Circuit Components
C
tjtj
tjtj
eCjVeI
eCjVeIωω
ωω
ω
ω
⋅⋅=
⋅⋅=~Re~Re
~Re~Re
tjeVtv ω~Re)( = tjeIti ω~Re)( =
Phasors with Circuit Components
C
CjVI ω⋅= ~~
tjeVtv ω~Re)( = tjeIti ω~Re)( =
CjIVZ
ω1
~~==
Phasors with Circuit Components
C tjeVtv ω~Re)( =
tjeIti ω~Re)( =
CjIVZ
ω1
~~==I
VZ ~~
=
In general For a capacitor
Phasors with Circuit Components
L tjeVtv ω~Re)( =
tjeIti ω~Re)( =
[ ]
⋅==
=
∫
∫tjtjtj eV
LjdteV
LeI
dttvL
ti
ωωω
ω~1Re~Re1~Re
)(1)(
Phasors with Circuit Components
L tjeVtv ω~Re)( =
tjeIti ω~Re)( =
⋅=
⋅=
tjtj
tjtj
eVLj
eI
eVLj
eI
ωω
ωω
ω
ω
~1Re~Re
~1Re~Re
Phasors with Circuit Components
L tjeVtv ω~Re)( =
tjeIti ω~Re)( =
⋅= tjtj eVLj
eI ωω
ω~1Re~Re
VLj
I ~1~ω
= LjIVZ ω== ~~
Phasors with Circuit Components
L tjeVtv ω~Re)( =
tjeIti ω~Re)( =
LjIVZ ω== ~~
IVZ ~~
=
In general For an inductor
Phasors with Circuit Components
Phasors with Circuit ComponentsSummary
Impedance, Z, Ω
Resistor
Inductor
Capacitor
LjZ ω=
CjZ
ω1
=
RZ =
Phasors with Circuit ComponentsSummary
Impedance,Z, Ω
Admittance,Y, sieman
Resistor
Inductor
Capacitor
LjIVZ ω== ~~
CjIVZ
ω1
~~==
RIVZ == ~~
RVIY 1~~==
LjVIY
ω1
~~==
CjVIY ω== ~~
Combining Impedances
1 mH
0.5 mH
1 µF
10 ΩeqZ
Convert all of the components to their impedance values and determine the equivalent impedance of the entire network
kHzf 10=
1 mH
0.5 mH
1 µF
Z=10 ΩeqZ
Convert all of the components to their impedance values and determine the equivalent impedance of the entire network
kHzf 10=
jjCj
Z 9.15101000,102
116 −=
⋅⋅⋅== −πω
jjLjZ 8.62101000,102 3 =⋅⋅⋅== −πω
jj
LjZ
4.31105.0000,102 3
=⋅⋅⋅=
=−π
ω
Combining Impedances
62.8j Ω
31.4j Ω
-15.9j Ω
10 ΩeqZ
Convert all of the components to their impedance values and determine the equivalent impedance of the entire network
kHzf 10=
jjjjjZeq 8.4908.99.158.62
4.31104.3110
+=−++⋅
=
Combining Impedances
62.8j Ω
31.4j Ω
-15.9j Ω
10 ΩeqZ
Convert all of the components to their impedance values and determine the equivalent impedance of the entire network
kHzf 10=
Ω==+= ⋅⋅ ojjeq eejZ 7.7939.1 6.506.508.4908.9
Combining Impedances
Solving Sinusoidal Steady State Circuits
STEP #1: Transform all sources to the phasor domain using the phasor transform
Solving Sinusoidal Steady State CircuitsSTEP #2: Transform all components (i.e. resistors, inductors and capacitors) into their complex impedance values.
Solving Sinusoidal Steady State Circuits
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
I~
Solving Sinusoidal Steady State Circuits
STEP #4: Transform phasor results back into the time domain.
tjeIti ω~Re)( =
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
Find vc(t)
)(tvc
Solving Sinusoidal Steady State CircuitsExamples
STEP #1 and #2: Transform all sources to the phasor domain using the phasor transform and transform all components to their impedances
)cos()( tAtv ω= CR
)(tvc
Time-Domain Phasor-Domain
0~ jAeV = Cjω1
R
cV~
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
Time-Domain Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
Solve this circuit like you would if it was only just filled with resistors and DC sources.
Solving Sinusoidal Steady State CircuitsExamples
Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
CjR
CjVVc
ω
ω1
1~~
+=Voltage divider
Solving Sinusoidal Steady State CircuitsExamples
Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
11
1
1~~
+⋅=⋅
+=
CRjA
CjCj
CjR
CjVVc ωωω
ω
ωVoltage divider
Solving Sinusoidal Steady State CircuitsExamples
Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
Voltage divider)
1(tan22
0
1
)(11
~RCj
j
c
eRC
eACRjAV ω
ωω −
⋅+
⋅=
+=
Solving Sinusoidal Steady State CircuitsExamples
Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.
Voltage divider)
1(tan
2
1
)(1~ RCj
c eRC
AVω
ω
−−⋅
+=
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
Time-Domain Phasor-Domain
0~ jAeV = Cjω1
R
cV~
STEP #4: Transform phasor results back into the time domain.
)1
(tan
2
1
)(1~ RCj
c eRC
AVω
ω
−−⋅
+=( )
))(tancos(1
)( 1
2RCt
RC
Atvc ωωω
−−+
=
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Lets look at this solution at different frequencies.
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Lets look at this solution at different frequencies.
1. At DC (ω=0) what does the capacitor voltage look like?
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Lets look at this solution at different frequencies.
1. At DC (ω=0) what does the capacitor voltage look like?
( )ARCt
RC
Atvc =⋅−⋅⋅+
= − ))0(tan0cos(01
)( 1
2 Does this make sense?
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Lets look at this solution at different frequencies.1. At very very high frequencies (ω=infinity) what does the capacitor voltage look like?
Solving Sinusoidal Steady State CircuitsExamples
)cos()( tAtv ω= CR
)(tvc
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Lets look at this solution at different frequencies.1. At very very high frequencies (ω=infinity) what does the capacitor voltage look like?
( )0))(tancos(
1)( 1
2=⋅∞−⋅∞
⋅∞+= − RCt
RC
Atvc Does this make sense?
Solving Sinusoidal Steady State CircuitsExamples
Lets now plot the magnitude of the capacitor sinusoid as a function of frequency
ω
( )21 RC
A
ω+
A
0 0 RC20
As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.
RC10
Solving Sinusoidal Steady State CircuitsExamples
Lets now plot the magnitude of the capacitor sinusoid as a function of frequency
ω
( )21 RC
A
ω+
A
0 0 RC20
As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.
RC10
THIS PLOT IS CALLED A MAGNITUDE FREQUENCY RESPONSE PLOT
Solving Sinusoidal Steady State CircuitsExamples
As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.
( )))(tancos(
1)( 1
2RCt
RC
Atvc ωωω
−−+
=
Solving Sinusoidal Steady State CircuitsExamples
Lets now plot the phase angle of the capacitor sinusoid as a function of frequency ω
)(tan 1 RCω−−
o90−0 RC20
As the frequency increases the phase of the output voltage (vc(t)) goes from 0 degrees to -90 degrees.
RC10
THIS PLOT IS CALLED THE PHASE FREQUENCY RESPONSE PLOT
o0
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