chapter 10 sinusoidal steady-state analysis
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Chapter 10 Sinusoidal Steady-State Analysis . Charles P. Steinmetz (1865-1923), the developer of the mathematical analytical tools for studying ac circuits. Courtesy of General Electric Co. Heinrich R. Hertz (1857-1894). Courtesy of the Institution of Electrical Engineers. cycles/second. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 10
Sinusoidal Steady-State Analysis
Charles P. Steinmetz (1865-1923), the developer of the mathematical analytical tools for studying ac circuits. Courtesy of General Electric Co.
(cos sin )a jb r j
Heinrich R. Hertz (1857-1894). Courtesy of the Institution of Electrical Engineers.
cycles/second Hertz, Hz
Sinusoidal voltage source vs Vm sin(t ).
Sinusoidal Sources
Sinusoidal current source is Im sin(t ).
Amplitude
Phase angle
Angular or radianfrequency = 2f = 2/T
Period = 1/f
Example
Voltage and current of a circuit element.
v
i
The current leads the voltage by radians
circuitelement
+
v_
i
The voltage lags the current by radians OR
Example 10.3-1 3cos32sin(3 10 )
v ti t
Find their phase relationship
sin sin( )2sin(3 180 10 )
t ti t
andsin cos( 90 )
2cos(3 180 10 90 )2cos(3 100 )
i tt
Therefore the current leads the voltage by 100
Triangle for A and B of Eq. 10.3-4, where C .22 BA
Recall
0 cos cos si 3 4n 10. -s fv V t v A t B t
2 2
2 2 2 2cos sinf
A Bv A B t tA B A B
2 2 2 2sin cosB A
A B A B
2 2 cos sis i nco s nA B t t
2 2 cosA B t
1
1
tan ; 0
180 tan ; 0
B AA
B AA
Example 10.3-2 6cos2 8sin 2i t t A B
A
B
1
1
180 tan
8180 tan6
180 53 1
6 0
.
BA
A
10cos(2 126.9 )i t
An RL circuit.
Steady-State Response of an RL circuit
cos ?s m fv V t i
c 10.4-o 1smdiL Ri V tdt
cos sinfi A t B t From #8	
Substitute the assumed solution into 10.4-1 ( sin cos ) ( cos sin ) cosmL A t B t R A t B t V t
mLB RA V
0LA RB Solve for A & B
Coeff. of cosCoeff. of sin
2 2 2 2 2 2andm mRV LVA BR L R L
Steady-State Response of an RL circuit (cont.)
cos sin
cos( )m
i A t B tV tZ
2 2 2Z R L 1tan L
R
Thus the forced (steady-state) response is of the form
cos( )mi I t
mm
VIZ
Complex Exponential Forcing Function
cos
cos( )
s m
m
v V tVi tZ
Input
Response
magnitude phasefrequency
Exponential Signal
cos Re Re
j te m
j ts m m e
v V e
v V t V e v
Note Re a jb a
Complex Exponential Forcing Function (cont.)
s ev v
ee e
diL Ri vdt
try
j tei Ae
( ) j t j tm
jm m
j L R Ae V eV VA e
R j L Z
We get
where1 2 2 2tan andL Z R L
R
Complex Exponential Forcing Function (cont.)
Substituting for Aj j tm
eVi e eZ
We expect
( )
Re Re
Re
Re
cos( )
j j tme
j j tm
j tm
m
Vi i e eZ
V e eZ
V eZ
V tZ
Example2
2 12 12cos3d i di i tdt dt
We replace 312 j tev e
23
2 12 12 j te ee
d i di i edt dt
23 3 3
2, 3 , 9j t j t j te ee
di d ii Ae j Ae Aedt dt
Substituting ie3 3( 9 3 12) 12
12 2 2 453 3
j t j tj Ae e
Aj
Example(cont.)3 ( / 4) 3
(3 / 4)
2 2
2 2
j t j j te
j t
i Ae e e
e
The desired answer for the steady-state current
(3 / 4)Re Re 2 2
2 2 cos(3 45 )
j tei i e
t
Or 2 2 cos(3 )4
t
interchangeable
Using Complex Exponential Excitation to Determine aCircuit’s SS Response to a Sinusoidal Source
Write the excitation as a cosine waveform with a phase angle
Introduce complex excitation
Use the assumed response
Determine the constant A
cos( )s my Y t
( )Re j ts mv V e
( )j tex Ae
jA Be
( ) ( )j t j tex Ae Be
Obtain the solution
The desired response is
( ) Re cos( )ex t x B t
Example 10.5-121H10sin 3s
RLv t
Example 10.5-1(cont.)
10sin 3 10cos(3 90 )sv t t (3 90 )10 j t
ev e
ee e
diL Ri vdt
(3 90 )j t
ei Ae
(3 90 ) (3 90 ) (3 90 )3 2 103 2 10
10 102 3 4 9
j t j t j t
j
j Ae Ae ej A A
A ej
1 3tan 56.32
Example 10.5-1(cont.)
56.3 (3 90 ) (3 146.3 )10 1013 13
j j t j tei e e e
The solution is
The actual response is
10Re cos(3 146.3 ) A13ei i t
The Phasor Concept
A sinusoidal current or voltage at a given frequency is characterized by its amplitude and phase angle.
( )( ) Re
cos( )
j t
m
m
I
i t I e
t
Magnitude Phase angleThus we may write
( )( ) Re j j tmi t I e e
unchanged
The Phasor Concept(cont.)
A phasor is a complex number that represents the magnitude and phase of a sinusoid.
( )jm mI e I Iphasor
The Phasor Concept may be used when the circuit is linear , in steady state, and all independent sources aresinusoidal and have the same frequency.
A real sinusoidal current
( )( ) cos( ) Re j tm m
jm m
i t I t I e
I e I
I phasor notation
The Transformation
( )( ) cos( ) Re j tm my t Y t Y e
jm mY e Y Y
Time domain
Frequency domain
Transformation
The Transformation (cont.)
( ) 5sin(100 120 )5cos(100 30 )
i t tt
5 30 I
Time domain
Frequency domain
Transformation
Example10.6-2s
diL Ri vdt
( )cos( ) Re j ts m mv V t V e
( )cos( ) Re j tm mi I t I e
Substitute into 10.6-2( ) ( )( ) j t j t
m m mj LI RI e V e Suppress j te
( ) j jm mR I e V ej L I V
( )j L R
VI
( 200 200)0
283 45
45283
m
m
jV
V
VI
Example (cont.)
R 200 , L 2 H, 100 /rad s
( ) cos(100 45 ) A283
mVi t t
Phasor Relationship for R, L, and C Elements
v Ri
Time domain
Frequency domain
R orR
VV I I
Resistor
Voltage and current are in phase
Time domain Frequency domain
div Ldt
j L orj L
VV I I
Inductor
Voltage leads current by 90
Time domain Frequency domain
j C orj C
II V V
dvi Cdt
Capacitor
Voltage lags current by 90
Impedance and Admittance
Impedance is defined as the ratio of the phasor voltage to the phasor current.
m m
m m
V VI I
VZI
Ohm’s law in phasor notation
magnitude
phase
Z
or jZe R jX Z Zpolar exponential rectangular
Graphical representation of impedance
Z Z2 2R X Z
1tan XR
Resistor
Inductor
Capacitor
RZj LZ
1j C
Z
R
L
C
Admittance is defined as the reciprocal of impedance.
1 1
Y YZ Z
In rectangular form
2 2
1 1 R jX G jBR jX R X
Y
Z
conductance
susceptanceResistor
Inductor
Capacitor
1GR
Y
1j L
Y
j CY
L
C
G
Kirchhoff’s Law using Phasors
KVL 1 2 3 0n V V V V
KCL 1 2 3 0n I I I I
Both Kirchhoff’s Laws hold in the frequency domain.
and so all the techniques developed for resistive circuits hold
Superposition Thevenin &Norton Equivalent Circuits Source Transformation Node & Mesh Analysis etc.
Impedances in series
1 2 3eq n Z Z Z Z Z
Admittances in parallel
1 2 3eq n Y Y Y Y Y
Example 10.9-1 R = 9 , L = 10 mH, C = 1 mF i = ?
KVL2 3 sR I Z I Z I V
2
3
11 10
j L j
jj C
Z
Z (9 1 10) sj j I V100 0 7.86 45
(9 9) 9 2 45or 7.86cos(100 45 ) A
s
ji t
VI
Example 10.9-2 v = ?
KCL
10 010 10 10 10j j
V V V
0.1 (0.05 0.05) 0.1 1010 63.3 18.4
0.158 18.4or 63.3cos(1000 18.4 ) V
j j
v t
V V V
V
Node Voltage & Mesh Current using Phasors
va = ? vb = ?
cosC 100 μF, L 5 mH
1000 rad/s
s mi I t
11 10j
j C Z
21 1 1 (1 )5 5
jj L
Y
3 10Z
KCL at node a
1 3
a a bs
V V V IZ Z
KCL at node b
3 2
0b a b
V V VZ Z
Rearranging1 3 3
3 2 3
( ) ( )( ) ( ) 0
a b s
a b
Y Y V Y V I
Y V Y Y V
1 3 3
3 2 3
( )0
matrix
a s
b
Y Y Y V IY Y Y V
Y
Admittance matrix
If Im = 10 A and 0s mI I
Using Cramer’s rule to solve for Va
Therefore the steady state voltage va is
100(3 2 )4
100(3 2 )(4 )17
100 (10 11 )17
87.5 47.7
aj
jj j
j
V
87.5cos(1000 47.7 ) Vav t
Example 10.10-1 v = ?
10cosC 10 F, L 0.5 H
10 rad/s
sv tm
use supernode conceptas in #4
3 3
51 10
5 5
L
C
L
j L j
jj C
R j
Z
Z
Z Z
Example 10.10-1 (cont.)1
1
22
33
1 110
1 1 1 (1 )10
1 1 (5 5)50
C
R
jR
j
Y
YZ
YZ
KCL at supernode 1 2 3 1( ) 10 ( ) 0s s Y V V Y V Y V Y V V
Rearranging
1 2 3 1 3 1 1 3
1 1 3
1 2 3 1 3
( 10 ) ( 10 )( 10 )
( 10 )
s
s
Y Y Y Y Y V Y Y Y VY Y Y VV
Y Y Y Y Y
1 1 3
1 2 3 1 3
( 10 )10 0( 10 )10 10 63.42 5
jj
Y Y YVY Y Y Y Y
Example 10.10-1 (cont.)
Therefore the steady state voltage v is
10 cos(10 63.4 ) V5
v t
Example 10.10-2 i1 = ?
10 2 cos( 45 )C 5 mF, L 30 mH
100 rad/s
sv t
31 2
10 2 45 10 10
L
C
s
j L j
jj C
j
Z
Z
V
Example 10.10-2 (cont.)
KVL at mesh 1 & 2
1 2
1 2
(3 3) 3(3 3) ( 3 2) 0
sj jj j j
I I VI I
Using Cramer’s rule to solve for I1
1(10 10)j j
I
where is the determinant(3 3)( ) 3(3- 3) 6 12j j j j j
1(10 10) 1.05 71.6
6 12j j
j
I
Superposition, Thevenin & Norton Equivalentsand Source Transformations
Example 10.11-1 i = ?10cos10 V3 A
C 10 mF, L 1.5 H
s
s
v ti
10 03 0
s
s
VI
Consider the response to the voltage source acting alone = i1
Example 10.11-2 (cont.)
1 5s
pj L
VI
Z
5(1 ) and L 15Cp
C
R jR
ZZZ
Substitute
110 0
5 15 (5 5)10 10 45
10 10 200
j j
j
I
Consider the response to the current source acting alone = i2
Example 10.11-2 (cont.)
0
210 (3) 2 A15
I
Using the principle of superposition
0.71cos(10 45 ) 2 Ai t
Source Transformations
V I
I V
Example 10.11-2 IS = ?
10cos( 45 ) V100 rad/s
sv t
10 1010 45
s
s
j
ZV
10 45200 4510 0200
s
I
Example 10.11-3 Thevenin’s equivalent circuit
?
1
2
1 jj
ZZ
1
1 2
2 2 451
OC s
t
V I ZZ Z Z
Example 10.11-4 Thevenin’s equivalent circuit
10 20 03 80 0
s
OC
V I
V V V
10 4 ( 10 40)40 10
O
t
j jj
V I V I
Z
Example 10.11-4 Norton’s equivalent circuit
?
1 23
1 2t
Z ZZ Z
Z Z1 2 2
2 2 3
( ) ( )( ) ( ) 0
SC s
SC
Z Z I Z I V
Z I Z Z I
Phasor Diagrams
A Phasor Diagram is a graphical representation of phasorsand their relationship on the complex plane.
Take I as a reference phasor
0I I
The voltage phasors are
090
90
R
L
C
R RIj L LI
j IC C
V IV I
V I
Phasor Diagrams (cont.)
KVLs R L C V V V V
For a given L and C there will be a frequency that
2
1
1 1or
L C
LC
LC LC
V V
Resonant frequency
s RV VResonance
Summary
Sinusoidal Sources Steady-State Response of an RL Circuit for Sinusoidal Forcing FunctionComplex Exponential Forcing Function The Phasor Concept Impedance and Admittance Electrical Circuit Laws using Phasors