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AC STEADY-STATE ANALYSIS
LEARNING GOALS
SINUSOIDS Review basic facts about sinusoidal signals SINUSOIDAL AND COMPLEX FORCING FUNCTIONS Behavior of circuits with sinusoidal independent sources and modeling of sinusoids in terms of complex exponentials PHASORS Representation of complex exponentials as vectors. It facilitates steady-state analysis of circuits. IMPEDANCE AND ADMITANCE Generalization of the familiar concepts of resistance and conductance to describe AC steady state circuit operation PHASOR DIAGRAMS Representation of AC voltages and currents as complex vectors BASIC AC ANALYSIS USING KIRCHHOFF LAWS ANALYSIS TECHNIQUES Extension of node, loop, Thevenin and other techniques
SINUSOIDS
x ( t ) XM sin t argument (radians) angularfrequency (rads/sec) amplitudeormaximum value
t
X M
Adimensional plot As function of time T x ( t ) x ( t T ), t
Period
frequencyinHertz(cycle/sec )
T
f
2 f
"leadsby "
"lagsby "
LEARNING EXAMPLE
cos( t 45 )
cos( t )
Leads by 45 degrees
cos( t 45 )
cos( t 45 180 )
Leads by 225 or lags by 135
cos( t 45 360 )
Lags by 315
LEARNING EXAMPLE FIND FREQUENCYANDPHASEANGLEBETWEEN VOLTAGES v 1 ( t ) 12 sin( 1000 t 60 ), v 2 ( t ) 6 cos( 1000 t 30 ) Frequency in radians per second is the factor of the time variable 1 1000 sec f Hz 159. 2 Hz 2
To find phase angle we must express both sinusoids using the same trigonometric function; either sine or cosine with positive amplitude 6 cos( 1000 t 30 ) 6 cos( 1000 t 30 180 )
takecareof minussignwithcos( ) cos( 180 )
Changesineintocosinewith cos( )sin( 90 )
6 cos( 1000 t 210 ) 6 sin( 1000 t 210 90 ) We like to have the phase shifts less than 180 in absolute value 6 sin( 1000 t 300 ) 6 sin( 1000 t 60 ) ( ) 6 sin( 1000 60 ) ( ) 12 sin( 1000 60 ) 2 1
v t t v t t (^1000 t ^60 ) (^1000 t ^60 )^120 v 1 leads v 2 by 120 ( 1000 t 60 ) ( 1000 t 60 ) 120 v 2 lags v 1 by 120
SINUSOIDAL AND COMPLEX FORCING FUNCTIONS If the independent sources are sinusoids of the same frequency then for any variable in the linear circuit the steady state response will be sinusoidal and of the same frequency v ( t ) A sin( t ) iSS ( t ) B sin( t )
B ,
weonlyneedtodeterminethe parameters To determinethesteadystate solution Learning Example ( t ) Ri ( t ) v ( t ) dt di KVL: L t A t A t dt di i t A t A t i t A t
( ) sin cos ( ) cos sin ( ) cos( ) 1 2 1 2
Insteadystate ,or */ R */ L V t L A RA t L A RA t
M^
cos ( 1 2 )sin ( 2 1 ) cos
L A RA V M
L A RA
2 1
algebraic problem
(^1 ) ( )
( ) R L
LV
A
R L
RV
A
M M
Determining the steady state solution can be accomplished with only algebraic tools
FURTHER ANALYSIS OF THE SOLUTION ( ) cos( ) ( ) cos ( ) 1 cos 2 sin
i t A t v t V t i t A t A t M Forcomparisonpurposesonecan write Theappliedvoltage is The solution is
A 1 A cos , A 2 A sin
(^1 ) ( )
( ) R L
LV
A
R L
RV
A
M M
1 (^22) 2 2 1 ,^ tan A
A
A A A
R
L
R L
V
A
M^
1 2 2 , tan ( )
cos( tan ) ( )
1 2 2 R
L
t R L
V
i t
M^
For L 0 thecurrent ALWAYSlagsthe voltage If R 0 (pureinductor)thecurrentlagsthevoltageby 90
Learning Example j t v t V M e ( ) ( ) ( ) j t Assume i t I M e ( t ) Ri ( t ) v ( t ) dt di KVL: L ( ) ( )
j t t j I M e dt di j j t M j t M j t M j t M j L R I e e j L R I e t Ri t j LI e RI e dt di L
( ) ( ) ( )
j t M j j t j L R I M e e V e
j L R
V
I e j M M
R j L R j L
2 2 ( )
R L
V R j L I e j M M
^
R L R j L R L e
1 tan 2 2 ( ) R L j M M e R L
V
I e
1 tan 2 2 ( )
R
L
R L
V
I
M M
1 2 2 , tan ( )
( ) Re{ } cos( ) ( ) cos Re{ } ( )
i t I e I t v t V t V e M j t M j t M M
cos , sin , tan 2 2 1 x r y r y x r x y x jy re
C P
j
PHASORS ESSENTIAL CONDITION ALL INDEPENDENT SOURCES ARE SINUSOIDS OF THE SAME FREQUENCY BECAUSE OF SOURCE SUPERPOSITION ONE CAN CONSIDER A SINGLE SOURCE
u ( t ) U M cos( t )
THE STEADY STATE RESPONSE OF ANY CIRCUIT VARIABLE WILL BE OF THE FORM
y ( t ) Y M cos( t )
SHORTCUT 1 ( ) ( ) ( ) ( )^ ^ j t u t U e y t YM e j t M Re{ } Re{ } ( ) (^ ^ ) j t U e YM e j t M NEW IDEA: j j t M j t U M e U e e ( ) j M j u U M e y Y e ANDWEACCEPTANGLESIN DEGREES
INSTEADOFWRITING WE WRITE
SHORTCUTIN NOTATION
M j u U M e u U
U M IS THEPHASORREPRESENTATIONFOR UM cos( t )
u ( t ) U M cos( t ) U UM Y YM y ( t )Re{ YM cos( t )}
SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF THE RESPONSE GIVEN THE INPUT PHASOR(S)
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS RESISTORS ( ) ( )
j t M j t V M e RI e v t Ri t V RI V e RI e j M j M
Phasor representation for a resistor Phasors are complex numbers. The resistor model has a geometric interpretation The voltage and current phasors are colineal In terms of the sinusoidal signals this geometric representation implies that the two sinusoids are “in phase”
INDUCTORS (^ )
( ) ( ) j t M j t M I e dt d V e L ( )
j t j LI M e
V j LI
The relationship between phasors is algebraic 90 1 90 j j e For the geometric view use the result
V LI 90
The voltage leads the current by 90 deg The current lags the voltage by 90 deg j M j V M e j LI e Learning Example L 20 mH , v ( t ) 12 cos( 377 t 20 ). Find i ( t ) j L
V
I
V
A
L
I
3
I A
cos( 377 70 ) 377 20 10
3
i t t Relationship between sinusoids
LEARNING EXTENSIONS Findthevoltageacrossthe inductor L 0. 05 H , I 4 30 ( A ), f 60 Hz
2 f 120
V j LI
V 120 0. 05 1 90 4 30
V 24 60
v ( t ) 24 cos( 120 60 )
Findthevoltageacrossthe inductor
C 150 F , I 3. 6 145 , f 60 Hz
2 f 120
j C
I
I j CV V
120 150 10 1 90
6
V
V
cos( 120 235 )
v ( t ) t
Now an example with capacitors
IMPEDANCE AND ADMITTANCE For each of the passive components the relationship between the voltage phasor and the current phasor is algebraic. We now generalize for an arbitrary 2-termina element v i z M M M i M v Z I
V
I
V
I
V
Z
(INPUT) IMPEDANCE (DRIVING POINT IMPEDANCE) The units of impedance are OHMS j C Z Z j L I j C V V j LI C L R V RI Z R 1 1 Element PhasorEq. Impedance mpedance is NOT a phasor but a complex umber that can be written in polar or Cartesian form. In general its value depends n the frequency Reactive component Resistive component
X
R
Z R jX R
X
Z R X
z 1 2 2 tan
SPECIAL APPLICATION: IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED FOR RESISTORS I V 1 Z 1 V 2 Z 2 I Zs Z 1 Z 2 Z 1 (^) 2 Z
V
I I
V
1 2 1 2
Z Z
Z Z
Z p
Zs (^) kZk k k Z p Z
LEARNING EXAMPLE Computeequivalentimpedanceand current
f 60 Hz , v ( t ) 50 cos( t 30 )
6 3 120 50 10
j Z j Z
V Z
L C R Z (^) L j 7. 54 , ZC j 53. 05 Z (^) s ZR ZL ZC 25 j 45. 51 ( ) 25 45. 51
A
Z j
V
I
s
A
I 0. 96 91. 22 ( A ) i ( t ) 0. 96 cos( 120 t 91. 22 )( A )
ZR R
ZL j L
j C
ZC
LEARNING ASSESSMENT FIND i ( t )
V 120 ( 60 90 )
ZR 20
377 40 10 15. 08 3 ZL j j
- 05 377 50 10 6 j j ZC
Zeq Z (^) C || ( ZR ZL ) Zeq 30. 5616 + j 4. 9714 30. 963 9. 239
- 876 39. 924 ( )
- 963 9. 239
120 30
A Z V I
eq
