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 ) X M sin t Adimensional plot As function of time X M amplitude or maximum value angular frequency (rads/sec) t argument (radians) T 2 " leads by " Period x (t ) x (t T ), t 1 frequency in Hertz (cycle/sec ) T 2 2 f f " lags by " BASIC TRIGONOMETRY ESSENTIAL IDENTITIES sin( ) sin cos cos sin cos( ) cos cos sin sin sin( ) sin cos( ) cos SOME DERIVED IDENTITIES sin( ) sin cos cos sin cos( ) cos cos sin sin 1 1 sin cos sin( ) sin( ) 2 2 1 1 cos cos cos( ) cos( ) 2 2 APPLICATIO NS cos t sin( t ) 2 sin t cos( t ) 2 cos t cos( t ) sin t sin( t ) RADIANS AND DEGREES 2 radians 360 degrees 180 (rads) (degrees) ACCEPTED EE CONVENTION sin( t ) sin( t 90) 2 LEARNING EXAMPLE cos( t 45) cos( t 45 360) cos( t 45) cos( t 45 180) cos( t ) Leads by 45 degrees Lags by 315 Leads by 225 or lags by 135 LEARNING EXAMPLE v1 (t ) 12 sin(1000t 60), v2 (t ) 6 cos(1000t 30) FIND FREQUENCY AND PHASE ANGLE BETWEEN VOLTAGES Frequency in radians per second is the factor of the time variable f ( Hz) 1000 sec1 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 take care of minus sign with cos( ) cos( 180) 6 cos(1000t 30) 6 cos(1000t 30 180) Change sine into cosine with cos( ) sin( 90) 6 cos(1000t 210) 6 sin(1000t 210 90) We like to have the phase shifts less than 180 in absolute value 6 sin(1000t 300) 6 sin(1000t 60) v1 (t ) 12 sin(1000t 60) (1000t 60) (1000t 60) 120 v2 (t ) 6 sin(1000t 60) (1000t 60) (1000t 60) 120 v1 leads v2 by 120 v2 lags v1 by 120 LEARNING EXTENSION i1 (t ) 2 sin( 377 t 45) i2 ( t ) 0.5 cos(377 t 10) i3 ( t ) 0.25 sin( 377 t 60) i1 leads i2 by_____? i1 leads i3 by_____? cos sin( 90) 0.5 cos(377t 10) 0.5 sin( 377t 10 90) (377t 45) (377 100) 55 sin sin( 180) i1 leads i2 by 55 0.25 sin( 377t 60) 0.25 sin( 377t 60 180) (377t 45) (377t 120) 165 i1 leads i3 by 165 SINUSOIDAL AND COMPLEX FORCING FUNCTIONS Learning Example KVL : L di (t ) Ri (t ) v (t ) dt In steady state i (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 determine the steady state solution we only need to determine the parameters A1 RVM , A2 LVM 2 2 2 2 R ( L ) R ( L ) B, v (t ) Asin( t ) iSS (t ) B sin( t ) Determining the steady state solution can be accomplished with only algebraic tools! FURTHER ANALYSIS OF THE SOLUTION The solution is i (t ) A1 cos t A2 sin t The applied voltage is v (t ) VM cos t For comparison purposes one can write i (t ) A cos( t ) A1 A cos , A2 Asin A1 RVM LVM , A 2 R 2 (L) 2 R 2 (L) 2 A i (t ) A A12 A22 , tan A2 A1 VM 1 L , tan R R 2 (L) 2 VM 1 L cos( t tan ) 2 2 R R (L) For L 0 the current ALWAYS lags the voltage If R 0 (pure inductor) the current lags the voltage by 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 (Euler identity) v (t ) VM cos t y (t ) A cos( t ) v (t ) VM sin t y (t ) A sin( t ) * / j (and add) 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 Learning Example R jL R2 (L )2 e I M e 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 L R ) I M e j e jt ( jL R) I M e j e j t VM e j t VM R jL I M e j */ jL R R jL I M e j VM ( R jL) R 2 (L) 2 IM VM R 2 (L ) 2 VM R 2 (L ) 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 x y 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 ) SHORTCUT 1 u(t ) U M e j ( t ) y(t ) YM e Re{U M e j ( t ) } Re{YM e j ( t ) j ( t ) } U M e j ( t ) U M e j e jt u U M e j y YM e j SHORTCUT IN NOTATION NEW IDEA: INSTEAD OF WRITING u U M e j WE WRITE u U M ... AND WE ACCEPT ANGLES IN DEGREES U M IS THE PHASOR REPRESENTA TION FOR U M cos( t ) u(t ) U M cos( t ) U U M 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) Learning Example Learning Extensions It is essential to be able to move from sinusoids to phasor representation A cos(t ) A A sin(t ) A 90 V V M 0 v Ve jt I I M j t di i Ie L (t ) Ri (t ) v dt L( jIe jt ) RIe jt Ve jt In terms of phasors one has jLI RI V V I R jL The phasor can be obtained using only complex algebra We will develop a phasor representation for the circuit that will eliminate the need of writing the differential equation 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 (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 RI 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 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 Learning 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 ( A) V 1220 I L90 V 12 I I 70( A) j L 377 20 103 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 Learning Example C 100 F , v (t ) 100 cos(314t 15). Find i (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 jCV I C 190 10015 I 314 100 106 100105( A) i (t ) 3.14 cos(314t 105)( A) LEARNING EXTENSIONS L 0.05 H , I 4 30( A), f 60 Hz Find the voltage across the inductor 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 C 150 F , I 3.6 145, f 60 Hz Find the voltage across the inductor 2 f 120 I jCV V V I jC 3.6 145 120 150 106 190 200 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-terminal element Z ( ) R( ) jX ( ) R( ) Resistive component X ( ) Reactive component | 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 Impedance is NOT a phasor but a complex number that can be written in polar or Cartesian form. In general its value depends on the frequency Element Phasor Eq. Impedance R V RI ZR V jLI Z j L L 1 1 V I Z C jC jC 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 VM11 VM 2 2 VM 33 0 V1 V2 V3 0 Phasors! V2 V1 V3 I0 I1 I 2 I3 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 V1 Z1 I V2 I Zs Z1 Z2 Z2 Z1 Z2 V V 1 1 k Zp Zk Z s k Zk LEARNING EXAMPLE I Zp Z1Z 2 Z1 Z 2 f 60 Hz, v (t ) 50 cos( t 30) Compute equivalent impedance and current 120 , V 5030, Z R 25 ZR R Z L jL ZC 1 jC 1 j120 50 106 Z L j 7.54, ZC j53.05 Z L j120 20 103 , ZC Z s Z R Z L ZC 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) LEARNING ASSESSMENT FIND i (t ) 377 Z R 20 Z L j 377 40 103 j15.08 V 120(60 90) ZC Zeq ZC || ( Z R Z L ) I j j53.05 377 50 106 Zeq 30.5616 + j 4.9714 30.9639.239 V 120 30 3.876 39.924( A) Z eq 30.9639.239 Parallel Combinatio n of Admittanc es (COMPLEX) ADMITTANCE Y p Yk 1 G jB (Siemens) Z G conductanc e B Suceptanc e Y k YR 0.1S 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 L C V RI V jLI 1 V I jC ZR Z jL Z 1 jC 1 j1( S ) j1 Y p 0.1 j1( S ) Series Combinatio n of Admittanc es 1 1 Ys k Yk 0.1S Element Phasor Eq. Impedance R YC Admittance 1 Y G R 1 Y jL Y jC 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 LEARNING EXAMPLE VS 6045(V ) LEARNING EXTENSION FIND Y p , I Y p YR YL 0.5 j 0.25 Y p 0.5 j 0.5 j1 0.25 0.75 j 0.5( S ) Y p 0.901433.69( S ) 2 j4 2 j4 Yp 0.5 j 0.25( S ) 2 j4 j8 I Y pV 0.901433.69 1020 I Y pV (0.5 j 0.25) 6045( A) I 9.01453.79( A) Zp I 0.559 26.565 6045( A) I 33.5418.435( A) LEARNING EXAMPLE SERIES-PARALLEL REDUCTIONS 1 2 j4 2 2 j 4 (2) (4) 2 1 4 j2 Y34 4 j2 20 Y2 Z3 4 j 2 Y4 j 0.25 j 0.5 j 0.25 Z 4 1 / Y4 j 4 1 ( j 2) 1 j2 1 Z1 1 j 0.5 Z1 1 j 0.5 Z1 1 (0.5) 2 Z1 0.8 j 0.4() Z4 j 4 ( j 2) 8 j4 j2 j2 Y2 0.1 j 0.2( S ) Y34 0.2 j 0.1 Z2 2 j 6 j 2 2 j 4 Z34 4 j 2 Z 234 Y234 0.3 j 0.1( S ) Z 234 1 1 0.3 j 0.1 Y234 0.3 j 0.1 0.1 Z 2 Z 34 3 j1 Z 2 Z 34 Zeq Z1 Z234 3.8 j 0.6 3.8478.973 LEARNING EXTENSION FIND THE IMPEDANCE ZT 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 Z2 2 j 2 ( R P ) Z2 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 LEARNING EXAMPLE SKETCH THE PHASOR DIAGRAM FOR THE CIRCUIT Any one variable can be chosen as reference. For this case select the voltage V KCL : I S V V jCV R jL (capacitiv e) | I L || IC | | I L || IC | IC jCV IL V jl 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 1 the current as reference VC I j C VS VR VL VC 1. DRAW ALL THE PHASORS | VL || VC | DIAGRAM WITH REFERENCE VS 12 290 VL 18135(V ) Read values from diagram! I 345( A) VR 1245(V ) (Pythagora s) VC 6 45 FIND THE FREQUENCY AT WHICH v (t ) AND i (t ) LEARNING BY DOING v (t ) ARE IN PHASE i.e., the phasors for i (t ), v (t ) are co - lineal 1 V I jLI RI jC C L PHASOR DIAGRAM R jLI 1 I jC V RI V and I are co - lineal iff jL 2 Notice that I was chosen as reference 1 I jLI RI jC I 1 1 0 2 jC LC 1 9 4 10 3 . 162 10 (rad / s) 3 6 10 10 f 5.033 103 Hz 2 LEARNING EXTENSION Draw a phasor diagram illustrating all voltages and currents j4 4 90 I 445 2 j4 4.472 63.435 I1 3.57818.435( A) I1 I2 Current divider 1 20 I 445 2 j4 4.472 63.435 I 2 1.789108.435 Simpler than I 2 I I1 V 2 I1 7.15618.435(V ) DRAW PHASORS. ALL ARE KNOWN. NO NEED TO SELECT A REFERENCE BASIC ANALYSIS USING KIRCHHOFF’S LAWS PROBLEM SOLVING STRATEGY For relatively simple circuits use Ohm' s law for AC analysis;i.e., V IZ The rules for combining Z and Y KCL and KVL Current and voltage divider For more complex circuits use Node analysis Loop analysis Superposit ion Source transforma tion Thevenin' s and Norton' s theorems LEARNING EXAMPLE COMPUTE ALL THE VOLTAGES AND CURRENTS Compute I1 Use current divider for I2 , I3 Ohm' s law for V1 , V2 V1 690 I 2 Zeq 4 ( j 6 || 8 j 4) V1 16.2678.42(V ) 24 j 48 32 j8 24 j 48 Z eq 4 8 j2 8 j2 V2 7.2815(V ) 56 j 56 79.19645 9.60430.964() 8 j2 8.24614.036 V 2460 I1 S 2.49829.036( A) Z eq 9.60430.964 Z eq j6 690 I1 2.49829.036( A) 8 j2 8.24614.036 8 j4 8.944 26.565 I2 I1 2.49829.036( A) 8 j2 8.24614.036 I3 I1 2.529.06 I 2 2.71 11.58 V2 4 90 I3 I3 1.82105 LEARNING EXTENSION IF VO 845, COMPUTE VS THE PLAN... COMPUTE I3 COMPUTE V1 COMPUTE I2 , I1 COMPUTE VS VO ( A) 445( A) 2 V1 (2 j 2) I3 8 45 445 I3 V1 11.3140(V ) I2 V1 11.3140 5.657 90( A) j2 290 I1 I 2 I3 5.657 90 445 I1 j5.657 (2.828 j 2.828)( A) I1 2.828 j 2.829( A) VS 2 I1 V1 2(2.828 j 2.829) 11.3140 VS 16.97 j5.658(V ) VS 17.888 18.439 ANALYSIS TECHNIQUES PURPOSE: TO REVIEW ALL CIRCUIT ANALYSIS TOOLS DEVELOPED FOR RESISTIVE CIRCUITS; I.E., NODE AND LOOP ANALYSIS, SOURCE SUPERPOSITION, SOURCE TRANSFORMATION, THEVENIN’S AND NORTON’S THEOREMS. COMPUTE I0 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 I0 V2 ( A) 1 (1 j1) (1 j1)(1 j1) (1 j1) 2(1 j1) 6 (1 j1)(1 j1) 1 j1 V2 4 8 j2 1 j V2 (4 j )(1 j ) 2 3 5 I 0 j ( A) 2 2 I0 2.92 30.96 NEXT: LOOP ANALYSIS 2. LOOP ANALYSIS ONE COULD ALSO USE THE SUPERMESH TECHNIQUE SOURCE IS NOT SHARED AND Io IS DEFINED BY ONE LOOP CURRENT LOOP 1 : I1 20 I2 I 0 I3 LOOP 2 : (1 j )( I1 I 2 ) 60 (1 j )( I 2 I3 ) 0 LOOP 3 : (1 j )( I 2 I3 ) I3 0 CONSTRAINT : I1 I 2 20 SUPERMESH : (1 j ) I1 60 ( I 2 I 3 ) 0 MUST FIND I3 2 I 2 (1 j ) I3 6 (1 j )(2) (1 j ) I 2 (2 j ) I3 0 (1 j) /* (1 j ) /* ( 2) 2(2 j ) I3 (1 j )(8 2 j ) 5 3 10 6 j I j ( A) I3 0 2 2 4 2 MESH 3 : ( I 3 I 2 ) (1 j ) I 3 0 I 0 I 2 I3 NEXT: SOURCE SUPERPOSITION Circuit with voltage source set to zero (SHORT CIRCUITED) SOURCE SUPERPOSITION I I L2 1 L = V 1 L + 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 VL2 3. SOURCE SUPERPOSITION I 0' 10( A) Z ' (1 j ) || (1 j ) (1 j )(1 j ) 1 (1 j ) (1 j ) COULD USE SOURCE TRANSFORMATION TO COMPUTE I"0 V1" 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 Z" Z " 1 || (1 j ) NEXT: SOURCE TRANSFORMATION Source transformation is a good tool to reduce complexity in a circuit ... WHEN IT CAN BE APPLIED!! “ideal sources” are not good models for real behavior of sources A real battery does not produce infinite current when short-circuited ZV + - RV VS a ZI a RI b IS b THE MODELS ARE EQUIVALENT S WHEN RV RI R Z I Z VZ VS RI S S IZ SV Improved model Improved model for voltage source for current source Source Transformationcan be used to determine the Thevenin or Norton Equivalent... BUT THERE MAY BE MORE EFFICIENT TECHNIQUES 4. SOURCE TRANSFORMATION IS 8 2j 1 j Z (1 j ) || (1 j ) 1 V ' 8 2 j Now a voltage to current transformation NEXT: THEVENIN I0 I S 4 j (4 j )(1 j ) 5 3 j 2 1 j (1 j )(1 j ) 2 THEVENIN’S EQUIVALENCE THEOREM LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A ZTH RTH vTH i a vO b _ i a LINEAR CIRCUIT vO _ LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B b PART B PART A Phasor Thevenin Equivalent Circuit for PART A vTH Thevenin Equivalent Source RTH Thevenin Equivalent 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 8 2j I0 53j ( A) 2 NEXT: NORTON NORTON’S EQUIVALENCE THEOREM LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A i a vO b _ LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B Phasors ZN iN RN i a LINEAR CIRCUIT vO _ b PART B PART A Norton Equivalent Circuit for PART A iN Thevenin Equivalent Source RNZ N Thevenin Equivalent Resistance Impedance 6. NORTON ANALYSIS ZTH (1 j ) || (1 j ) 1 I0 Possible techniques: loops, source transformation, superposition BY SUPERPOSTI ON 60 8 2 j I SC 20 ( A) 1 j 1 j I SC 4 j (4 j )(1 j ) 5 3 j 2 1 j (1 j )(1 j ) 2 LEARNING EXAMPLE FIND V0 USING NODES, LOOPS, THEVENIN, NORTON WHY SKIP SUPERPOSITION AND TRANSFORMATION? Supernode constraint : V1 V3 120 KCL @ Supernode V V0 V3 V2 V1 V2 V3 40 3 0 1 1 j j KCL@V2 V2 V1 V V3 2I x 2 0 j 1 NODES KCL@ V0 V0 V0 V3 40 0 V3 2V0 4 1 1 V1 V3 12 Controllin g variable V1 2V0 16 V3 V0 Ix 1 V3 V0 V0 4 j (V2 2V0 16) 2(V0 4) (V2 2V0 4) 0 Notice choice of ground j (V2 2V0 16) (V2 2V0 4) (V0 4) j (2V0 4) 4 8 4j Adding : V0 1 2 j LOOP ANALYSIS MESH CURRENTS DETERMINED BY SOURCES I 2 40 I3 2 I x I3 2( I 4 4) MESH 1 : jI1 120 1( I1 I 3 ) 0 MESH 4 : 1( I 4 I 2 ) 1 I 4 j ( I 4 I 3 ) 0 CONTROLLING VARIABLE : I x I 4 I 2 VARIABLE OF INTEREST : V0 1 I 4 (V ) MESH CURRENTS ARE ACCEPTABLE I 4 4 I 4 j ( I 4 2( I 4 4)) 0 ( 2 j ) I 4 ( 4 8 j ) I 4 V0 84j 1 2 j 48 j j 2 j j Alternative procedure to compute Thevenin impedance: 1. Set to zero all INDEPENDENT sources 2. Apply an external probe THEVENIN ZTH I "x FOR OPEN CIRCUIT VOLTAGE Vtest I "x KVL Vtest I "x jI "x ZTH 1 j () I x' 40 ZTH 1 j 2I x 80 VOC 4 8 j (V ) V0 1 (4 8 j )(V ) 2 j Supernode constraint V1 V3 12 V1 V3 120 KCL@ Supernode V3 V3 V3 V2 V1 V2 40 0 / j 1 j 1 j NORTON V2 V3 V2 V1 0 / ( j ) 1 j V Controllin g Variable : I x''' 3 1 KCL@ V2 : 2 I X''' I SC 2 jV3 j (V2 V3 ) (V2 V3 12) 0 (1 j )V2 (1 3 j )V3 12 (1 j )V3 jV3 jV2 (V3 12) V2 4 j I x''' V3 ( A) 1 I SC I x''' 4 USE NODES (1 j )V2 2 jV3 12 4 j 4j 48j (1 j )V3 4 j V3 I SC 1 j 1 j I SC (4 8 j ) j 8 4j (1 j ) j 1 j Now we can draw the Norton Equivalent circuit ... NORTON’S EQUIVALENT CIRCUIT ZTH I SC V0 (1) I 0 (V ) 1 j 8 4 j (V ) Current Divider 2 j 1 j EQUIVALENCE OF SOLUTIONS 12 4 j (8 4 j )(1 j ) 3 j (1 2 j )(1 j ) Using Norton’s method V0 Using Thevenin’s 48j j V0 j 2 j Using Node and Loop methods V0 84j 1 2 j LEARNING EXTENSION COMPUTE V0 USE THEVENIN USE NODAL ANALYSIS V1 4j VOC V1 1230 V1 V1 V1 V0 0 / 2 j 2 1 j2 j 3 4 j 4 j (2 6 j ) 2 j2 2 6 j 40 3 1 || j 2 j2 1230 1230 2 (1 || j 2) 2(1 2 j ) 2 j 24120 12120 26j 1 3 j ZTH 2 || 1 || j 2 VOC V0 V1 V0 0 V1 (1 j )V0 j 1 ZTH j (V1 1230) 2 jV1 V1 2(V1 V0 ) 0 2V0 (1 2 2 j j )(1 j )V0 j1230 (2 (1 3 j )(1 j ))V0 190 1230 12120 12120 V0 2.1275(V ) 44j 5.6645 j1 VOC 1 + - V0 V0 ZTH 1 VOC 1 j LEARNING EXTENSION COMPUTE V0 USING MESH ANALYSIS V1 USING NODES V1 240 V 290 1 0 2 22j 2 V0 V1 22j USING SOURCE SUPERPOSITION 2 240 222j CONSTRAINT I1 I 2 290 I1 I 2 2 j V0V SUPERMESH 240 2 I1 2 jI 2 2 I 2 0 V0I 2 2( I 2 2 j ) (2 2 j ) I 2 24 (4 2 j ) I 2 24 4 j V0 2 I 2 24 4 j 24.339.46 10.8636.03 2 j 2.24 26.57 2 290 42j V0 V0V V0I LEARNING EXTENSION COMPUTE V0 2V V0" V0 V0' V0" 1. USING SUPERPOSITION 2 || (2 2 j ) V2 V1 V1 2 || (2 2 j ) (120) j 2 (2 || 2 2 j ) V0' V0' 2 V1 22j (2 j ) || (2 2 j ) 240 2 (2 j || (2 2 j ) V0" 2 V2 22j 2. USE SOURCE TRANSFORMATION 2j Z I eq I1 2 V0 I eq 120 6 90 12 6 j I1 V0 2I1 j 2 120 2 I1 2 j 2 6 90 Z 2 || j 2 Z I eq Z 22j V0 USE NORTON’S THEOREM j2 I1 I SC ZTH 2 || j 2 ZTH I1 2 V0 ZTH I SC ZTH 2 2 j V0 2I1 120 6 90 I SC LEARNING 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 USING MATLAB MATLAB recognizes complex numbers in rectangular representation. It does NOT recognize Phasors Unless previously re-defined, MATLAB recognizes “i” or “j” as imaginary units » z2=3+4j z2 = 3.0000 + 4.0000i » z1=4+6i Phasors Rectangula r z 1045 » a=45; % angle in degrees » ar=a*pi/180, %convert degrees to radians ar = 0.7854 » m=10; %define magnitude » x=m*cos(ar); %real part x = 7.0711 » y=m*sin(ar); %imaginary part y = 7.0711 » z=x+i*y z = 7.0711 + 7.0711i z = 7.0711 + 7.0711i; » mp=abs(z); %compute magnitude z1 = mp = 10 4.0000 + 6.0000i » arr=angle(z); %compute angle in RADIANS In its output MATLAB always uses “i” for arr = 0.7854 the imaginary unit » adeg=arr*180/pi; %convert to degres adeg = 45 x=real(z) x= 7.0711 y=imag(z) y= 7.70711 LEARNING EXAMPLE COMPUTE ALL NODE VOLTAGES V1 1230 V1 (1 j 0.5 j )V2 jV3 0.5 jV5 0 jV2 ( j 0.5 1)V3 0.5V5 0 0.5V1 (0.5 1 j )V4 V5 0 0.5 jV2 0.5V3 V4 (1 0.5 j 0.5 j )V5 245 V1 1230 0 0 0 0 1 V1 1230 1 1 j 0. 5 V j1 0 j 0. 5 0 2 j1 1.5 j1 0 0. 5 0 V3 0 0.5 0 0 1.5 j1 1 V4 0 2 45 0 j 0 . 5 0 . 5 1 1 . 5 j 0 . 5 V 5 V2 V1 V2 V3 V2 V5 0 1 j1 j2 V3 V2 V3 V5 V3 0 j1 2 1 V4 V1 V4 V5 V4 0 2 1 j1 V V4 V5 V2 V5 V3 V5 245 5 0 1 j2 2 j1 YV I R V Y 1I R %example8p18 %define the RHS vector. ir=zeros(5,1); %initialize and define non zero values ir(1)=12*cos(30*pi/180)+j*12*sin(30*pi/180); ir(5)=2*cos(pi/4)+j*2*sin(pi/4), %echo the vector %now define the matrix y=[1,0,0,0,0; %first row -1,1+0.5j,-j,0,0.5j; %second row 0,-j,1.5+j,0,-0.5; %third row -0.5,0,0,1.5+j,-1; %fourth row 0,0.5i,-0.5,-1,1.5+0.5i] %last row and do echo v=y\ir %solve equations and echo the answer ir = 10.3923 + 6.0000i 0 0 Echo of RHS 0 1.4142 + 1.4142i y = Columns 1 through 4 1.0000 0 0 -1.0000 1.0000 + 0.5000i 0 - 1.0000i 0 0 - 1.0000i 1.5000 + 1.0000i -0.5000 0 0 0 0 + 0.5000i -0.5000 Column 5 0 Echo of Matrix 0 + 0.5000i -0.5000 -1.0000 1.5000 + 0.5000i Echo of Answer v = 10.3923 7.0766 1.4038 3.7661 3.4151 + + + - 6.0000i 2.1580i 2.5561i 2.9621i 3.6771i 0 0 0 1.5000 + 1.0000i -1.0000 LEARNING EXAMPLE FIND THE CURRENT I_o 6 meshes and two current sources 5 non-reference nodes and two voltage sources… slight advantage for nodes NODE EQUATIONS MATRIX FORM MATLAB COMMANDS LINEAR EQUATION ANSWER AC PSPICE ANALYSIS Circuit ready to be simulated Select and place components L R Ground set, meters specified VAC C Wire and set correct attributes **** AC ANALYSIS TEMPERATURE = 27.000 DEG C ************************************************************************ ****** Results in output file FREQ VM($N_0003) VP($N_0003) 6.000E+01 2.651E+00 -3.854E+01 **** 05/20/01 09:03:41 *********** Evaluation PSpice (Nov 1999) ************** * C:\ECEWork\IrwinPPT\ACSteadyStateAnalysis\Sec7p9Demo.sch **** AC ANALYSIS TEMPERATURE = 27.000 DEG C ************************************************************************ ****** FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2) 2.998E-03 5.146E+01 LEARNING APPLICATION NOISE REJECTION Noise has much higher frequency (700kHz) than signal. Find a way to ‘block’ high frequencies Impedance X should have low (zero) value at low frequencies and very high at noise frequency Reduce amplitude of noise by 10 EXAMPLE A GENERAL IMPEDANCE CONVERTER (GIC) VA VC VIN (ideal OpAmp) @A: @C: @E: Suitable choices of impedances Permit to create any desired equivale A 1kOhm resistor coverts to 1 H equivalent inductance!! LEARNING BY DESIGN USING PASSIVE COMPONENTS TO CREATE GAINS LARGER THAN ONE PRODUCE A GAIN=10 AT 1KhZ WHEN R=100 2 LC 1 L 1.59mH C 15.9 F LEARNING BY DESIGN PASSIVE SUMMING CIRCUIT - BIAS T NETWORK vO ( t ) 2.5 2.5cos t , 2 f ; f 1GHz B should have zero impedance for DC and block high frequencie PROPOSED SOLUTION A should block DC and have very low impedance at 1GHz AT DC THE CAPACITOR IS ALWAYS OPEN CIRCUIT BUT AT 1GHz ONE WOULD NEED INFINITE INDUCTANCE. JUST MAKE THE IMPEDANCES VERY DIFFERENT PROPOSE C 1; L 10k; 2 109 end vO ( t ) 2.5 2.50025cos 2 109 t