Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1. Engineering 43 ImpedanceKCL & KVL Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Resistors Review → V-I in Phasor Space No Phase Shift • Inductors i(t) LAGS • Capacitors i(t) LEADS

3. Impedance • For each of the passive components, the relationship between the voltage phasor and the current phasor is algebraic (previous sld) • Consider now the general case for an arbitrary 2-terminal element • Since the Phasors V & I Have units of Volts and Amps, Z has units of OHMS • The Frequency Domain Analog to Resistance is IMPEDANCE, Z

4. Since V & I are COMPLEX, Then Z is also Complex Impedance cont. • However, Z IS a COMPLEX NUMBER that can be written in polar or Cartesian form. • In general, its value DOES depend on the Sinusoidal frequency • Impedance is NOT a Phasor • It’s Magnitude and Phase Do Not Change regardless of the Location within The Circuit • Note that the REACTANCE, X, isa function of ω

5. Thus Impedance cont.2 • Summary Of Passive-Element Impedance • The Magnitude and Phase • Examine ZC • Where

6. KVL & KCL Hold In Phasor Spc Similarly for the SinusoidalCurrents ...

7. Impedances (which have units of Ω) Combine as do RESISTANCES The SERIES Case Series & Parallel Impedances • The Parallel Case

8. The Frequency Domain Analog of CONDUCTANCE is ADMITTANCE Admittance is Thus Inverse Impedance Admittance • Multiply Denominator by the Complex Conjugate • G  CONDUCTance • B  SUSCEPTance • Find G & B In terms of Resistance, R, and Reactance, X • Note that G & R and X & B are NOT Reciprocals

9. Admittance Summarized Series & Parallel Admittance • Admittances (which have units of Siemens)Combine as do CONDUCTANCES • The SERIES Case • The PARALLEL Case

10. MATLAB recognizes complex numbers these in these forms Rectangular Exponential Can Use “i” or “j” for √(-1) MATLAB Always returns “i” for √(-1) Sometimes need “*” Complex Numbers in MATLAB >> phiR = 23*pi/180 % 23deg in Rads phiR = 0.4014 >> Z1 = 7 + i*23 % if i or j BEFORE, then need * Z1 = 7.0000 +23.0000i >> Z2 = 11 - 13j Z2 = 11.0000 -13.0000i >> Z3 = 43*exp(j*phiR) % Need * Z3 = 39.5817 +16.8014i >> Z4 = 37*exp(0.61j) Z4 = 30.3270 +21.1961i

11. MATLAB Does NOT Recognize Phasor NOTATION But it DOES handle Complex Exponentials e.g.: Phasors in MATLAB >> phi7 = -43*pi/180 phi7 = -0.7505 >> phi8 = 17*pi/180 phi8 = 0.2967 >> Z7 = 29*exp(j*phi7) Z7 = 21.2093 -19.7780i >> Z8 = -53*exp(j*phi8) Z8 = -50.6842 -15.4957i >> Zsum = Z7 + Z8 Zsum = -29.4749 -35.2737i >> Zdif = Z7 - Z8 Zdif = 71.8934 - 4.2823i >> Zprod = Z7*Z8 Zprod = -1.3814e+003 +6.7378e+002i >> Zquo = Z7/Z8 Zquo = -0.2736 + 0.4739i

12. MATLAB Always returns Complex No.s in RECTANGULAR Form Can Recover Magnitude & Phase Using Commands abs(Z) angle(Z) Phasors in MATLAB >> Zquo = Z7/Z8 Zquo = -0.2736 + 0.4739i >> Asum = abs(Zsum) Asum = 45.9674 >> phi_sum = angle(Zsum) phi_sum = -2.2669 >> phi_sumd = phi_sum*180/pi phi_sumd = -129.8824 >> Aquo = abs(Zquo) Aquo = 0.5472 >> phi_quo = angle(Zquo) phi_quo = 2.0944 >> phi_quod = phi_quo*180/pi phi_quod = 120.0000 >> Zquo_test = Aquo*exp(j*phi_quo) Zquo_test = -0.2736 + 0.4739i

13. BMayer MATLAB Functions MATLAB: a+jb ↔ A∟φ • Example >> Z1r = 13 - 19j Z1r = 13.0000 -19.0000i >> Phasor1 = MagPh(Z1r) Phasor1 = 23.0217 -55.6197 >> Phasor2 = [43 -127] Phasor2 = 43 -127 >> Zr2 = Rectab(Phasor2(1), Phasor2(2)) Zr2 = -25.8780 -34.3413i function Zrectd = Rectab(Mag, phi_deg) % B. Mayer 22Apr09 * ENGR43 % finds for POLAR COMPLEX number Z the Rectangular Equivalet %% note that phi is in DEGREES % a = Mag*cosd(phi_deg); b = Mag*sind(phi_deg); Zrectd = a + j*b function Phasor = MagPh(Zr) % B. Mayer 22Apr09 * ENGR43 % finds for RECTANGULAR COMPLEX number Z %% Magnitude %% Phase Angle in DEGREES Magnitude = abs(Zr); Phase_deg = angle(Zr)*180/pi; Phasor = [Magnitude, Phase_deg];

14. MATLAB Equivalent Functions • Rectangular to Polar • Polar to Rectangular • Both use RADIANS only

15. As Noted Earlier Phasors can be Considered as VECTORS in the Complex Plane See Diagram at Right Imaginary A b a Real Phasor Diagrams • See Next Slide for Review of Vector Addition • Text Diagrams follow the PARALLELOGRAM Method • Phasors Obey the Rules of Vector Arithmetic • Which were orignially Developed for Force Mechanics

16. Parallelogram Rule For Vector Addition Examine Top & Bottom ofThe Parallelogram Triangle Rule ForVector Addition Vector Addition isCommutative Vector Subtraction →Reverse Direction ofThe Subtrahend C B C B Vector Addition

17. For The Ckt at Right, Draw the Phasor Diagrams as a function of Frequency First Write KCL Example  Phasor Diagram • That is, we Can Select ONE Phasor to have a ZERO Phase Angle • In this Case Choose V • Next Examine Frequency Sensitivity of the Admittances • Now we can Select ANY Phasor Quantity, I or V, as the BaseLine

18. The KCL Example  Phasor Diagram cont • This Eqn Shows That as ω increases • YL DEcreases • YC INcreases • Now Rewrite KCL using Phasor Notation • Examining the Phase Angles Shows that in the Complex Plane • IR Points RIGHT • IL Points DOWN • IC Points UP • As ω Increases, IC begins to dominate IL

19. Case-I: ω=Med so That YL YC Example  Phasor Diagram cont.2 • Case-III: ω=Hi so That • YC 2YL • The Circuit is Basically CAPACITIVE • Case-II: ω=Low so That • YL 2YC • The Circuit is Basically INDUCTIVE

20. KCL & KVL for AC Analysis • Simple-Circuit Analysis • AC Version of Ohm’s Law → V = IZ • Rules for Combining Z and/or Y • KCL & KVL • Current and/or Voltage Dividers • More Complex Circuits • Nodal Analysis • Loop or Mesh Analysis • SuperPosition

21. Methods of AC Analysis cont. • More Complex Circuits • Thevenin’s Theorem • Norton’s Theorem • Numerical Techniques • MATLAB • SPICE

22. For The Ckt At Right, Find VS if Example • Then I2 by Ohm • Solution Plan: GND at Bot, then Find in Order • I3 → V1 → I2 → I1 → VS • I3 First by Ohm • Then I1 by KCL • Then V1 by Ohm

23. Then VS by Ohm & KVL Example cont. • Then Zeq • Note That we haveI1 and VS • Thus can find the Circuit’s Equivalent Impedance

24. For The Ckt at Right Find IO Use Node Analysis Specifically a SuperNode that Encompasses The V-Src  KCL at SN Nodal Analysis for AC Circuits • The Relation For IO • And the SuperNode Constraint • In SuperNode KCL Sub for V1

25. Solving for For V2 Nodal Analysis cont. • The Complex Arithmetic • Recall • Or

26. Same Ckt, But Different Approach to Find IO Note: IO = –I3 Constraint: I1 = –2A0 Loop Analysis for AC Circuits • The Loop Eqns • Simplify Loop2 & Loop3 • Solution is I3 = –IO • Recall I1 = –2A0 • Two Eqns In Two Unknowns

27. Isolating I3 Loop Analysis cont • Then The Solution • The Next Step is to Solve the 3 Eqns for I2 andI3 • So Then Note • Could also use a SuperMesh to Avoid the Current Source

28. = Recall Source SuperPosition + • Circuit With Current Source Set To Zero • OPEN Ckt • Circuit with Voltage Source set to Zero • SHORT Ckt • By Linearity

29. Same Ckt, But Use Source SuperPosition to Find IO Deactivate V-Source AC Ckt Source SuperPosition • The Reduced Ckt • Combine The Parallel Impedances

30. Find I-Src Contribution to IO by I-Divider AC Source SuperPosition cont. • The V-Src Contribution by V-Divider • Now Deactivate the I-Source (open it)

31. Sub for Z” AC Source SuperPosition cont.2 • The Total Response • Finally SuperPose the Response Components

32. When Sources of Differing FREQUENCIES excite a ckt then we MUST use SuperPosition for every set of sources with NON-EQUAL FREQUENCIES An Example Multiple Frequencies • We Can Denote the Sources as Phasors • But canNOT COMBINE them due to DIFFERENT frequencies

33. Must Use SuperPosition for EACH Different ω V1 first (ω = 10 r/s) Multiple Frequencies cont.1 • V2 next (ω = 20 r/s) • The Frequency-1 Domain Phasor-Diagram

34. The Frequency-2 Domain Phasor-Diagram Multiple Frequencies cont.2 • Recover the Time Domain Currents • Finally SuperPose • Note the MINUS sign from CW-current assumed-Positive

35. 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 Resistance → Impedance Analogy Source Transformation

36. Same Ckt, But Use Source Transformationto Find IO Start With I-Src Source Transformation • Then the Reduced Circuit • Next Combine the VoltageSources And Xform

37. The Reduced Ckt Source Transformation cont • Now Combine theSeries-ParallelImpedances • The Reduced Ckt • IO by I-Divider

38. WhiteBoard Work • Let’s Work This Nice Problem to Find VO

39. All Done for Today CharlesProteusSteinmetz Delveloper of Phasor Analysis

42. WhiteBoard Work • Let’s Work this Nice Problem • See Next Slide for Phasor Diagrams

43. P8.29 Phasor Diagrams • Tip-To-Tail Phasor (Vector) Addition

44. WhiteBoard Work • Let’s Work Some Phasor Problems