AC STEADY-STATE ANALYSIS SINUSOIDAL AND COMPLEX FORCING FUNCTIONS

AC STEADY-STATE ANALYSIS SINUSOIDAL AND COMPLEX FORCING FUNCTIONS

AC STEADY-STATE ANALYSIS 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 SINUSOIDAL AND COMPLEX FORCING FUNCTIONS KVL: L di ( t ) Ri (t ) v (t ) dt In steadystatei (t ) A cos( t ), or i (t ) A1 cos t A2 sin t */ R If the independent sources are sinusoids di */ L (t ) A1 sin t A2 cos t of the same frequency then for any dt variable in the linear circuit the steady state response will be sinusoidal and of ( LA1 RA2 ) sin t ( LA2 RA1 ) cos t the same frequency VM cos t LA1 RA2 0 algebraic problem LA2 RA1 VM To determinethesteadystatesolution we onlyneedto determinetheparametersA1 RV M , A2 LV M 2 2

2 2 R ( L ) R ( L ) B, v (t ) A sin( t ) i SS (t ) B sin( t ) Determining the steady state solution can be accomplished with only algebraic tools FURTHER ANALYSIS OF THE SOLUTION The solutionis i (t ) A1 cos t A2 sin t The appliedvoltageis v (t ) VM cos t For comparisonpurposesone canwritei (t ) A cos( t ) A1 A cos , A2 A sin A1 A A12 A22 , tan RV M LV M , A 2 R 2 (L) 2 R 2 (L) 2 A i (t )

VM R 2 (L) 2 , tan 1 VM R 2 (L) 2 A2 A1 L R cos( t tan 1 L ) R For L 0 thecurrent AL WAYS lagsthe voltage If R 0 (pureinductor)thecurrentlagsthe voltageby 90 SOLVING A SIMPLE ONE LOOP CIRCUIT CAN BE VERY LABORIOUS IF ONE USES SINUSOIDAL EXCITATIONS TO MAKE ANALYSIS SIMPLER ONE RELATES SINUSOIDAL SIGNALS TO COMPLEX NUMBERS. THE ANALYSIS OF STEADY STATE WILL BE CONVERTED TO SOLVING SYSTEMS OF ALGEBRAIC EQUATIONS ... WITH COMPLEX VARIABLES ESSENTIAL IDENTITY: e j cos j sin (Euleridentity) v (t ) VM cos t y (t ) A cos( t ) v (t ) VM sin t y (t ) A sin( t ) * / j (andadd) VM e j t Ae j (t ) Ae j e j t y (t ) If everybody knows the frequency of the sinusoid then one can skip the term exp(jwt) VM Ae j

Example 2 2 R jL R (L ) e IMe j v (t ) VM e j t Assume i (t ) I M e ( j t ) di KVL: L ( t ) Ri (t ) v ( t ) dt di (t ) jI M e ( j t ) dt di L (t ) Ri (t ) jLI M e ( j t ) RI M e ( j t ) dt ( jL R) I M e ( j t ) j ( jL R) I M e e jt ( jL R) I M e j e j t VM e j t VM R j L I M e j */ j L R R j L V ( R j L) I M e j M 2 R (L) 2 IM

VM 2 R (L ) VM R 2 (L ) 2 2 tan 1 e L R tan 1 L R , tan 1 L R v (t ) VM cos t Re{VM e j t } i (t ) Re{I M e ( j t ) } I M cos( t ) C P x jy re j r x 2 y 2 , tan 1 x r cos , y r sin y x 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 ) YM cos( t ) j ( t ) y (t ) YM e SHORTCUT 1 u( t ) U M e Re{U M e j ( t ) } Re{YM e NEW IDEA: U M e j ( t ) U M e j e jt j ( t ) j ( t ) } u U M e j y YM e j SHORTCUTIN NOTATION INSTEADOF WRITING u U M e j WE WRITE u U M ... ANDWE ACCEPTANGLESIN DEGREES U M IS THEPHASORREPRESENTA TIONFORU M cos( t ) u (t ) U M cos( t ) U U M Y YM y (t ) YM cos( t ) SHORTCUT 2: DEVELOP EFFICIENT TOOLS TO DETERMINE THE PHASOR OF THE RESPONSE GIVEN THE INPUT PHASOR(S) Example It is essential to be able to move from sinusoids to phasor representation A cos(t ) A A sin(t ) A 90 V VM 0 v Ve jt I I M

jt di i Ie L (t ) Ri (t ) v dt L( jIe jt ) RIe jt Ve jt In termsof phasorsonehas jLI RI V V I R jL v (t ) 12 cos(377t 425 ) 12 425 y (t ) 18 sin(2513t 4.2 ) 18 85.8 Given f 400 Hz V1 1020 v1 (t ) 10 cos(800 t 20 ) V2 12 60 v2 ( t ) 12 cos(800 t 60 ) Phasors can be combined using the rules of complex algebra The phasor can be obtained using only complex algebra We will develop a phasor representation or the circuit that will eliminate the need of writing the differential equation (V11 )(V2 2 ) V1V2(1 2 ) V11 V1 (1 2 ) V2 2 V2 PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS RESISTORS v ( t ) Ri (t ) VM e ( j t ) RI M e ( j t ) VM e j RI M e j V RIPhasor 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 d ( I M e ( j t ) ) dt jLI M e ( j t ) VM e ( j t ) L Relationship between sinusoids VM e j jLI M e j V jLI Example The relationship between L 20mH , v (t ) 12 cos(377t 20 ). Find i (t ) phasors is algebraic For the geometric view use the result j 190 e j 90 V LI90 377 1220 I ( A) V 1220 L90 V 12 I

I 70 ( A) j L 377 20 10 3 i (t ) The voltage leads the current by 90 deg The current lags the voltage by 90 deg 12 cos(377t 70 ) 3 377 20 10 CAPACITORS I M e ( j t ) C d (VM e ( j t ) ) dt Relationship between sinusoids I M e j jCe j I CV90 I jCV C 100 F , v (t ) 100 cos(314t 15 ). Findi (t ) The relationship between phasors is algebraic In a capacitor the current leads the voltage by 90 deg The voltage lags the current by 90 deg 314 V 10015 I C 190 10015 I jCV I 314 100 10 6 100105 ( A) i (t ) 3.14 cos(314t 105 )( A)

C 150 F , I 3.6 145 , f 60 Hz L 0.05 H , I 4 30 ( A), f 60 Hz Find the voltageacrosstheinductor 2 f 120 V jLI V 120 0.05 190 4 30 V 2460 v (t ) 24 cos(120 60 ) Now an example with capacitors Find the voltage across the capacitor 2 f 120 I jCV V I j C 3.6 145 V 120 150 10 6 190 200 V 235 v (t ) 200 cos(120 t 235 ) 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 Z ( ) R( ) jX ( ) R( ) Resistivecomponent X ( ) Reactivecomponent | Z | R 2 X 2 X z tan 1

R (INPUT) IMPEDANCE V V V Z M v M ( v i ) | Z | z I I M i I M (DRIVING POINT IMPEDANCE) The units of impedance are OHMS Element PhasorEq. Impedance R V RI Z R V jLI Z jL L 1 1 V I Z C jC jC 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 KVL AND KCL HOLD FOR PHASOR REPRESENTATIONS v2 (t ) v1 ( t ) v3 ( t )

i0 (t ) i1 (t ) i2 ( t ) i3 (t ) KVL: v1(t ) v2 (t ) v3 (t ) 0 KCL: i0 (t ) i1 (t ) i2 (t ) i3 (t ) 0 vi (t ) VMie j ( t i ) , i 1,2,3 ik (t ) I Mk e j ( t k ) , k 0,1,2,3 In a similar way, one shows ... KVL: (VM 1e j1 VM 2e j 2 VM 3e j 3 )e jt 0 VM 11 VM 2 2 VM 3 3 0 V1 V2 V3 0 Phasors! V2 V1 V3 I 0 I1 I 2 I 3 0 I0

I1 I2 I3 The components will be represented by their impedances and the relationships will be entirely algebraic!! SPECIAL APPLICATION: IMPEDANCES CAN BE COMBINED USING THE SAME RULES DEVELOPED FOR RESISTORS I I V1 V2 Z1 Z2 I I Zs Z1 Z2 Z2 V Z1 V 1 1 k Zp Zk

Z s k Z k LEARNING EXAMPLE Zp Z1Z 2 Z1 Z 2 f 60 Hz , v (t ) 50 cos( t 30 ) Computeequivalent impedance andcurrent 120 , V 5030 , Z R 25 Z R R Z L j L ZC 1 j C 1 j120 50 10 6 Z L j 7.54, Z C j 53.05 Z L j120 20 10 3 , Z C Z s Z R Z L Z C 25 j 45.51 I V 5030 5030 ( A) ( A)

Z s 25 j 45.51 51.93 61.22 I 0.9691.22 ( A) i (t ) 0.96 cos(120 t 91.22 )( A) (COMPLEX) ADMITTANCE 1 G jB (Siemens) Z G conductanc e B Suceptanc e Y 1 1 R jX R jX 2 Z R jX R jX R X 2 R R2 X 2 X B 2 R X2 G ParallelCombinatio n of Admittanc es Y p Yk k YR 0.1S YC

1 j1( S ) j1 Y p 0.1 j1( S ) SeriesCombinatio n of Admittanc es 1 1 Ys k Yk Element PhasorEq. Impedance Admittance 1 R V RI Z R Y G R 1 Z jL Y V jLI L jL 1 V I 1 C Z Y jC jC jC 0.1S j 0.1S 1 1 1

Ys 0.1 j 0.1 10 j10 (0.1)( j 0.1) 0.1 j 0.1 0.1 j 0.1 0.1 j 0.1 1 10 j10 Ys 10 j10 200 Ys 0.05 j 0.05 S Ys FIND THE IMPEDANCEZT Z1 4 j 6 j 4 Z1 4 j 2 Y12 Y1 Y2 ( R P ) Z1 4.47226.565 Y1 0.224 26.565 ( P R)Y1 0.200 j 0.100 Y12 Y1 Y2 0.45 j 0.35 ( R P )Y12 0.570 37.875 Z12 1.75437.875 ( P R) Z12 1.384 j1.077 Z 2 2 j 2 ( R P ) Z 2 2.82845 Y2 0.354 45 ( P R)Y2 0.250 j 0.250 1 4 j2 ZT 2 (1.384 j1077) 3.383 j1.077 Y1 2 4 j 2 ( 4) ( 2) 2 1 2 j2

Y2 2 2 j 2 (2) (2) 2 1 1 0.45 j 0.35 Z12 Y12 0.45 j 0.35 0.325 1 Z12 Y12 PHASOR DIAGRAMS Display all relevant phasors on a common reference frame Very useful to visualize phase relationships among variables. Especially if some variable, like the frequency, can change SKETCH THE PHASOR DIAGRAM FOR THE CIRCUIT Any one variable can be chosen as reference. For this case select the voltage V V V KCL: I S jCV R jL (capacitiv e) | I L || I C | | I L || I C | I C jCV IL V j l

INDUCTIVE CASE CAPACITIVE CASE (inductive ) LEARNING EXAMPLE DO THE PHASOR DIAGRAM FOR THE CIRCUIT 377( s 1 ) 2. PUT KNOWN NUMERICAL VALUES | VL VC || VR | VR RI VL jLI It is convenient to select DIAGRAMWITHREFERENCE VS 12 290 1 the current as reference VL 18135 (V ) VC I j C VS VR VL VC 1. DRAW ALL THE PHASORS | VL || VC | Read values from diagram! I 345 ( A) VR 1245 (V ) (Pythagora s) VC 6 45

BASIC ANALYSIS USING KIRCHHOFFS LAWS PROBLEM SOLVING STRATEGY For relatively simple circuits use Ohm's law for ACanalysis;i.e., V IZ Therulesfor combiningZ andY KCL ANDKVL Currentandvoltagedivider For more complex circuits use Node analysis Loop analysis Superposition Thevenin' s and Norton' s theorems MATLAB PSPICE ANALYSIS TECHNIQUES PURPOSE: TO REVIEW ALL CIRCUIT ANALYSIS TOOLS DEVELOPED FOR RESISTIVE CIRCUITS; I.E., NODE AND LOOP ANALYSIS, SOURCE SUPERPOSITION, SOURCE TRANSFORMATION, THEVENINS AND NORTONS THEOREMS. COMPUTEI 0 V2 60 V 20 V2 2 0 1 j1 1 j1 1 1 6 V2 1 2 1 j1 1 j1 1 j1 V2

1. NODE ANALYSIS V1 V V 20 2 2 0 1 j1 1 1 j1 V1 V2 60 V I 0 2 ( A) 1 (1 j1) (1 j1)(1 j1) (1 j1) 2(1 j1) 6 (1 j1)(1 j1) 1 j1 V2 4 8 j 2 1 j V2 (4 j )(1 j ) 2 3 5 I 0 j ( A) 2 2 I 0 2.92 30.96 Circuit with voltage source set to zero (SHORT CIRCUITED) SOURCE SUPERPOSITION I

I L2 1 L = 1 L V + VL2 Circuit with current source set to zero(OPEN) Due to the linearity of the models we must have I L I L1 I L2 VL VL1 VL2 Principle of Source Superposition The approach will be useful if solving the two circuits is simpler, or more convenient, than solving a circuit with two sources We can have any combination of sources. And we can partition any way we find convenient 3. SOURCE SUPERPOSITION I 0' 10 ( A) Z ' (1 j ) || (1 j ) (1 j )(1 j ) 1 (1 j ) (1 j ) COULDUSESOURCETRANSFORMA TION TO COMPUTEI "0 V1"

Z" Z " 1 || (1 j ) Z" Z" " " 60 (V ) I 0 " 60 ( A) Z 1 j Z 1 j 1 j 1 j I 0" 6 2 j ( 1 j ) 3 j I 0" 6 ( A) 1 j 6 6 " 1 j I j ( A) 0 2 j 4 4 5 3 I 0 I 0' I 0" j ( A)

2 2 THEVENINS EQUIVALENCE THEOREM L IN E A R C IR C U IT M a y co n ta in in d ep en d en t a n d d ep en d en t so u rces w ith th e ir c o n tr o llin g va ria b les PART A ZTH b _ RTH vTH a vO i i a L IN E A R C IR C U IT vO _ L IN E A R C IR C U IT M a y co n ta in in d ep en d en t a n d

d ep en d en t so u rces w ith th e ir co n tro llin g va ria b les PART B b PART B PART A Phasor Thevenin Equivalent Circuit for PART A vTH TheveninEquivalent Source RTH TheveninEquivalent Resistance Impedance 5. THEVENIN ANALYSIS Voltage Divider VOC 10 6 j 1 j (8 2 j ) 2 (1 j ) (1 j ) ZTH (1 j ) || (1 j ) 1 82j

5 3j I0 ( A) 2 EXAMPLE Find the current i(t) in steady state The sources have different frequencies! For phasor analysis MUST use source superpositio Frequency domain SOURCE 2: FREQUENCY 20r/s Principle of superposition LEARNING BY DESIGN USING PASSIVE COMPONENTS TO CREATE GAINS LARGER THAN ON PRODUCE A GAIN=10 AT 1KhZ WHEN R=100 2 LC 1 L 1.59mH C 15.9 F

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