bruce mayer, pe registered electrical & mechanical engineer [email protected]

[email protected] • ENGR-44_Lec-08-2_Impedance.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 43

ImpedanceKCL & KVL



Review → V-I in Phasor Space Resistors

Inductors

Capacitors

RVI

90LVI

90VI C

No Phase Shift

i(t) LAGS

i(t) LEADS



Impedance

The Frequency Domain Analog to Resistance is IMPEDANCE, Z

Since the Phasors V & I Have units of Volts and Amps, Z has units of OHMS

IVZ

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



Impedance cont. Since V & I are

COMPLEX, Then Z is also Complex

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

zivM

M

iM

vM ZIV

IV

||)(IVZ

component REACTive)(component RESISTive

)()(

XR

jXRZ

• Note that the REACTANCE, X, isa function of ω



Impedance cont.2 Thus

The Magnitude and Phase

jXRZ z

RXXRZ

z1

22

tan

Where

z

z

ZXZR

sincos

CjZ

LjZ

Cj

Lj

CL

RZRR

11

IV

IVIV

ImpedanceEq. PhasorElement

Summary Of Passive-Element Impedance

Examine ZC

Cj

Cjjj

CjZC 1

1

CX

CjZ CC

11



KVL & KCL Hold In Phasor Spc

)(1 tv

)(3 tv

)(2 tv)(0 ti )(1 ti )(2 ti )(3 ti

0)()()( 321 tvtvtv :KVL3,2,1,0,)(

0)()()()()(

3210

keIti

titititiktj

Mkk

:KCL3,2,1,)( )( ieVtv itj

Mii

0)( :KVL 321321 tjj

Mj

Mj

M eeVeVeV

0332211 MMM VVV

Phasors! 0321 VVV

1V

3V

2V0I

1I 2I 3I

03210 IIII

Similarly for the SinusoidalCurrents ...



Series & Parallel Impedances Impedances (which have units of Ω)

Combine as do RESISTANCES• The SERIES Case

• The Parallel Case

I 1V

1Z

2V

2ZI

21 ZZZs

k ks ZZ

1Z 2Z

V

I I

V

21

21

ZZZZZ p

k

kp ZZ11



Admittance The Frequency Domain

Analog of CONDUCTANCE is ADMITTANCE• Admittance is Thus

Inverse Impedance

Multiply Denominator by the Complex Conjugate

Find G & B In terms of Resistance, R, and Reactance, X

(Siemens) 1 jBGZ

Y

jXRZY

11

22

22

22

1

XRXB

XRRG

XRjXR

jXRjXR

jXRY

Note that G & R and X & B are NOT Reciprocals

• G CONDUCTance• B SUSCEPTance



Series & Parallel Admittance Admittance

Summarized

Admittances (which have units of Siemens) Combine as do CONDUCTANCES

CjYCj

Z

LjYLjZ

Cj

Lj

CL

GR

YRZRR

1

1

1

1

IV

IV

IV

AdmittanceImpedanceEq. PhasorElement

k

kp YY

The SERIES Case The PARALLEL Case

k ks YY

11



Complex Numbers in MATLAB 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 “*”

>> phiR = 23*pi/180 % 23deg in RadsphiR = 0.4014

>> Z1 = 7 + i*23 % if i or j BEFORE, then need *Z1 = 7.0000 +23.0000i

>> Z2 = 11 - 13jZ2 = 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



Phasors in MATLAB MATLAB Does

NOT Recognize Phasor NOTATION• But it DOES

handle Complex Exponentials

e.g.:

1753

4329

8

7

Z

Z

>> phi7 = -43*pi/180phi7 = -0.7505

>> phi8 = 17*pi/180phi8 = 0.2967

>> Z7 = 29*exp(j*phi7)Z7 = 21.2093 -19.7780i

>> Z8 = -53*exp(j*phi8)Z8 = -50.6842 -15.4957i

>> Zsum = Z7 + Z8Zsum = -29.4749 -35.2737i

>> Zdif = Z7 - Z8Zdif = 71.8934 - 4.2823i

>> Zprod = Z7*Z8Zprod = -1.3814e+003 +6.7378e+002i

>> Zquo = Z7/Z8Zquo = -0.2736 + 0.4739i



Phasors in MATLAB MATLAB Always

returns Complex No.s in RECTANGULAR Form

Can Recover Magnitude & Phase Using Commands• abs(Z)• angle(Z)

>> Zquo = Z7/Z8Zquo = -0.2736 + 0.4739i

>> Asum = abs(Zsum)Asum = 45.9674

>> phi_sum = angle(Zsum)phi_sum = -2.2669

>> phi_sumd = phi_sum*180/piphi_sumd = -129.8824

>> Aquo = abs(Zquo)Aquo = 0.5472

>> phi_quo = angle(Zquo)phi_quo = 2.0944

>> phi_quod = phi_quo*180/piphi_quod = 120.0000

>> Zquo_test = Aquo*exp(j*phi_quo)Zquo_test = -0.2736 + 0.4739i



MATLAB: a+jb ↔ A∟φ BMayer MATLAB

Functionsfunction 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 DEGREESMagnitude = abs(Zr);Phase_deg = angle(Zr)*180/pi;Phasor = [Magnitude, Phase_deg];

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



MATLAB Equivalent Functions

Rectangular to Polar Polar to Rectangular

Both use RADIANS only



Phasor Diagrams As Noted Earlier

Phasors can be Considered as VECTORS in the Complex Plane • See Diagram at Right

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

A

b

a Real

Imaginary



Vector Addition Parallelogram Rule

For Vector Addition Examine Top & Bottom of

The Parallelogram• Triangle Rule For

Vector Addition• Vector Addition is

Commutative• Vector Subtraction →

Reverse Direction ofThe Subtrahend

B

B

C

C

QPQP

PQQP



Example Phasor Diagram For The Ckt at Right,

Draw the Phasor Diagrams as a function of Frequency

First Write KCL

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

k kS

S

S

Y

CjLjR

CjLjR

sAdmittance

11

VI

VI

VVVI



Example Phasor Diagram cont

The KCL

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

This Eqn Shows That as ω increases• YL DEcreases• YC INcreases

Now Rewrite KCL using Phasor Notation

VVVI CjLjRS

901;9011as

90900

jj

CVL

VR

VM

MMS

I



Example Phasor Diagram cont.2

Case-I: ω=Med so That• YL YC

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

VI CjC

LjL VI

|||| CL II

|||| CL II

0 LC II

RS II



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



Methods of AC Analysis cont. More Complex Circuits

• Thevenin’s Theorem• Norton’s Theorem• Numerical Techniques

– MATLAB– SPICE



Example For The Ckt At Right,

Find VS if Then I2 by Ohm

Solution Plan: GND at Bot, then Find in Order• I3 → V1 → I2 → I1 → VS

I3 First by Ohm

458VoV

AVO

45423VI

Then V1 by Ohm 454458)22( 31 IV j

)(0314.111 VV

)(90657.5902

0314.112

12 AV

j

VI

Then I1 by KCL 45490657.5321 III

))(828.2828.2(657.51 Ajj I

)(829.2828.21 AjI



Example cont. Then VS by Ohm & KVL

Then Zeq Note That we have

I1 and VS

Thus can find the Circuit’s Equivalent Impedance

0314.11)829.2828.2(22 11 jS VIV

)(658.597.16 VjS V

439.18888.17 VSV

1IVZ S

eq

56.26472.4

00.200.4829.2828.2

439.18888.17

eq

eq

eq

jAj

V

Z

Z

Z

eqZ



Nodal Analysis for AC Circuits For The Ckt at Right

Find IO

Use Node Analysis• Specifically a SuperNode

that Encompasses The V-Src KCL at SN

The Relation For IO

And the SuperNode Constraint

0111

0211

221

jj

VVV

0606

21

12

VVVV

or

)(1

2 AOVI

011

0211

06 22

2

jj

VVV

In SuperNode KCL Sub for V1



Nodal Analysis cont. Solving for For V2

Recall

The Complex Arithmetic

12 V

OVI

Or

1162

1111

111

2 jjj

V

116)11(2

)11)(11()11()11)(11()11(

2 jj

jjjjjj

V

281

42 j

j

V

23

25

41

128

2 jjj

V

)(23

25 AjO

I

96.3092.2 AOI



Loop Analysis for AC Circuits Same Ckt, But Different

Approach to Find IO

Note: IO = –I3

Constraint: I1 = –2A0

Simplify Loop2 & Loop3 The Loop Eqns

Two Eqns In Two Unknowns

0))(1(06))(1( :2 LOOP

3221 IIjIIj

01)()1( :3 LOOP 332 IIIj)2)(1(6)1(2

)1(6)1(2 :L2

32

132

jIjIIjIjI

0)2()1( :L3 32 IjIj Solution is I3 = –IO

Recall I1 = –2A0



Loop Analysis cont Isolating I3

The Next Step is to Solve the 3 Eqns for I2 and I3

So Then Note

Then The Solution

)28)(1()2(2)1( 32 jjjj I

4610

3j

I )(23

25

0 Aj I

2I

Could also use a SuperMesh to Avoid the Current Source

0)1()(1 :3 MESH0)(106)1( :SUPERMESH

02 :CONSTRAINT

323

321

12

IIIIII

II

jj

32 III O



Recall Source SuperPosition

1LI

1LV

2LI

2LV= +

Circuit With Current Source Set To Zero• OPEN Ckt

Circuit with Voltage Source set to Zero• SHORT Ckt By Linearity

2121LLLLLL VVVIII



AC Ckt Source SuperPosition Same Ckt, But Use

Source SuperPosition to Find IO

Deactivate V-Source

The Reduced Ckt

Combine The Parallel Impedances 1

)1()1()1)(1()1(||)1('

jjjjjjZ



AC Source SuperPosition cont. Find I-Src Contribution

to IO by I-Divider

The V-Src Contribution by V-Divider

Now Deactivate the I-Source (open it)

1'Z

)(0111

102'0 A

I

)1(||1" jZ

jjjZ

1111)1(||1"

)(061"

""

1 VjZ

Z

V

)(061

1 "

""

1" A

jZZ

O

VI



AC Source SuperPosition cont.2 Sub for Z”

The Total Response

Finally SuperPose the Response Components

)(61

21

21

"0 A

jjjjj

I

63)1(

1"0 jj

j

I

)(46

46"

0 AjI

)(23

25

0 Aj I

j

23

231"

0'00 III



Multiple Frequencies When Sources of Differing

FREQUENCIES excite a ckt then we MUST use SuperPosition for every set of sources with NON-EQUAL FREQUENCIES

An Example

1V 2V

We Can Denote the Sources as Phasors 1050&0100 21 VV VV

But canNOT COMBINE them due to DIFFERENT frequencies



Multiple Frequencies cont.1 Must Use

SuperPosition for EACH Different ω

V1 first (ω = 10 r/s)

The Frequency-1 Domain Phasor-Diagram

1V

11010, jZL

12020, jZL

V2 next (ω = 20 r/s)



Multiple Frequencies cont.2 The Frequency-2

Domain Phasor-Diagram

Recover the Time Domain Currents

2V

43.7320cos24.24510cos07.7"' tAtAtititi– Note the MINUS sign from CW-current assumed-Positive

Finally SuperPose



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 Resistance → Impedance Analogy

+-

Im pro ved m odelfo r vo ltage source

Im proved m o delfo r current source

SVVR

SI

IRa

b

a

bSS

IV

RIVRRR

WHEN SEQUIVALENT AREMODELS THEVZ IZ

SS

IV

ZZZZ

IV



Source Transformation Same Ckt, But Use

Source Transformationto Find IO

Start With I-Src Then the Reduced Circuit

jV 28'

Next Combine the VoltageSources And Xform j

jZSeries

SS

128'VI



Source Transformation cont The Reduced Ckt Now Combine the

Series-ParallelImpedances

The Reduced Ckt

jj

S

1

28I

1

111)1(||)1(

2

jjjjjZ p

pZ

235 j

O

I

IO by I-Divider

235 j

jj

jjjj

SO

1114

14

21II



WhiteBoard Work

Let’s Work This Nice Problem to Find VO



All Done for Today

CharlesProteus

Steinmetz

Delveloper of Phasor Analysis



WhiteBoard Work

20

6m H 10

i s( t)

3 .33 F

i 1 ( t) i 2 ( t)

See Next Slide for Phasor Diagrams

Let’s Work this Nice Problem

13.8100

13.85000cos100mA

tmAtiSI



P8.29 Phasor Diagrams

13.8100 mASI

41.273.143 1 mAI

44.10998.84 2 mAI

11.88.99 mASI

41.273.143 1 mAI

44.10998.84 2 mAI

Tip-To-Tail Phasor (Vector) Addition



WhiteBoard Work

z 10m H 500F

2 4

Let’s Work Some Phasor Problems

bruce mayer, pe registered electrical & mechanical engineer [email protected]

Documents