Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Steady-State Sinusoidal Analysis.

Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING

Steady-State Sinusoidal Analysis


Steady-State Sinusoidal Analysis 1. Identify the frequency, angular frequency, peak value, rms

value, and phase of a sinusoidal signal.

2. Solve steady-state ac circuits using phasors and complex

impedances.

3. Compute power for steady-state ac circuits.

4. Find Thévenin and Norton equivalent circuits.

5. Determine load impedances for maximum power transfer.


The sinusoidal function v(t) = VM sin t is plotted (a) versus t and (b) versus t.


The sine wave VM sin ( t + ) leads VM sin t by radian


T

2

f 2

Frequency T

f1

Angular frequency

tt

tt

zz

o

o

o

cos)90sin(

)90cos(sin

)90cos(sin


A graphical representation of the two sinusoids v1 and v2.

The magnitude of each sine function is represented by the length of the corresponding arrow, and the phase angle by the orientation with respect to the positive x axis.

In this diagram, v1 leads v2 by 100o + 30o = 130o, although it could also be argued that v2 leads v1 by 230o.

It is customary, however, to express the phase difference by an angle less than or equal to 180o in magnitude.


4-6

Euler’s identity

tfAtA

f

t

2coscos

2

In Euler expression,

A cos t = Real (A e j t )A sin t = Im( A e j t )

Any sinusoidal function canbe expressed as in Euler form.


The complex forcing function Vm e j ( t + ) produces the complex response Im e j (t + ).

It is a matter of concept to make use of the mathematics of complex number for circuit analysis. (Euler Identity)


The sinusoidal forcing function Vm cos ( t + θ) produces the steady-state response Im cos ( t + φ).

The imaginary sinusoidal forcing function j Vm sin ( t + θ)

produces the imaginary sinusoidal response j Im sin ( t + φ).


Re(Vm e j ( t + ) ) Re(Im e j (t + ))

Im(Vm e j ( t + ) ) Im(Im e j (t + ))


Phasor Definition

111 cos :function Time θtωVtv

tdroppingbye

e

V

j

tj

)Re(

)Re(

:Phasor

)(1

)(1

111

1

1

V

V

V


A phasor diagram showing the sum of

V1 = 6 + j8 V and V2 = 3 – j4 V,

V1 + V2 = 9 + j4 V = Vs

Vs = Ae j θ

A = [9 2 + 4 2]1/2

θ = tan -1 (4/9)

Vs = 9.8524.0o V.


Adding Sinusoids Using Phasors

Step 1: Determine the phasor for each term.

Step 2: Add the phasors using complex arithmetic.

Step 3: Convert the sum to polar form.

Step 4: Write the result as a time function.


Using Phasors to Add Sinusoids

)30cos(10)(

)45cos(20

2

1

ttv

ttv

3010

4520

2

1

V

V


7.39cos97.29 ttvs

7.3906.23

14.19tan,96.29)14.19(06.23

V

122

A

Aes j

7.3997.29

14.1906.235660.814.1414.14

3010452021s

jjj

VVV


Phase Relationships

To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise.

Then when standing at a fixed point,if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ .

Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)


To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval tp between positive peaks of the two waveforms.

Then, the phase angle isθ = (tp/T ) × 360°.

If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t) lags v1(t).




COMPLEX IMPEDANCES

LL Lj IV

90 LLjZ L

LLL Z IV


(b)(a)

(c)

In the phasor domain,

(a) a resistor R is represented by an impedance of the same value;

(b) a capacitor C is represented by an impedance 1/jC;

(c) an inductor L is represented by an impedance jL.


ZcCj

Cj

CVejdt

VedC

dt

dC

Ve

tjtj

tj

1

I

VVI

VI

V

Zc is defined as the impedance of a capacitor

)90(cos,cos,VI tCVithentVvifCjAs

The impedance of a capacitor is 1/jC. It is simply a mathematicalexpression. The physical meaning of the j term is that it will introducea phase shift between the voltage across the capacitor and the currentflowing through the capacitor.


MM

MM

CVI

tCVithentVvifCjAs

)90(cos,cos,VI


L

tjtj

tj

ZLjLj

LIejdt

IedL

dt

dL

Ie

I

VIV

IV

I

ZL is defined as the impedance of an inductor

.cos),90(cos

)90(cos,cos,IV

tLIvandtIior

tLIvthentIiifCjAs

The impedance of a inductor is jL. It is simply a mathematicalexpression. The physical meaning of the j term is that it will introducea phase shift between the voltage across the inductor and the currentflowing through the inductor.


MMMM

MM

LIVtLIvandtIior

tLIvthentIiifCjAs

,cos),90(cos

)90(cos,cos,IV



90

90111

LLjZ

Z

CCj

CjZ

Z

L

LLL

C

CCC

IV

IV

RR RIV

Complex Impedance in Phasor Notation



Vm Im


Kirchhoff’s Laws in Phasor Form

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.

The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.


Circuit Analysis Using Phasors and Impedances

1. Replace the time descriptions of the voltage and current sources with the corresponding phasors.

(All of the sources must have the same frequency.)

2. Replace inductances by their complex impedances

ZL= jωL. Replace capacitances by their complex

impedances ZC = 1/(jωC). Resistances remain the same as their

resistances.


3. Analyze the circuit using any of the techniques studied earlier performing the calculations with complex arithmetic.


)105500cos(35.35)(,10535.35

159035.3515707.09050I50

)75500cos(05.106)(,7505.106

159005.10615707.090150I150

)15500cos(7.70)(,157.70 I100

15707.0

4530707.0454.141

30100I

454.141100100

)50150(100

ttv

jV

ttv

jV

ttvV

Z

V

j

jZ

L

C

L

L

RR

total

total



5050

4571.704501414.0

01

01.001.0

1

01.001.0

100

1

100

1

)100/(1100/1

1

/1/1

1

2

j

jZ

jj

j

j

j

jjAs

jZcRZ

RC

RC


V1000cos10)1351000(cos10)(

135104571.70

4571.704510

5050

4571.709010

5050100

4571.709010

)(

tttv

jjj

divisionvoltageZZ

ZVsVc

c

RCL

RC


)1351000(cos414.0)(

135414.04571.70

9010

5050

9010

5050100

9010

tti

jjj

ZZ

VsI

RCL


A)451000(cos1.0)(

451.090100

13510

100

13510

)1351000(cos1.0)(

1351.0100

13510

tti

jZc

VI

tti

R

VI

R

CC

R

CR



Solve by nodal analysis

)2(5.105.1510

)1(2902510

122

211

eqj

VV

j

V

eqjj

VVV


V)74.29100cos(1.16

74.291.16

43.6369.331.1643.632236.0

69.336.3

2.01.0

23

23)2.01.0(

)2(2)1(

5.11.02.0

)2(

22.0)2.01.0(

1,

1

1122.02.01.0

)1(

)2(5.105.1510

)1(2902510

1

1

1

1

21

21

211

122211

tv

j

jV

jVj

eqeqbyVSolving

VjVj

eqFrom

jVjVj

jj

jj

j

j

jjAsjVjVjV

eqFrom

eqj

VV

j

Veqj

j

VVV


Vs= - j10, ZL=jL=j(0.5×500)=j250

Use mesh analysis,


)135500cos(7)(

1357250)135028.0(

V)45500cos(7)(

457

90250)135028.0(

A)135500cos(028.0

135028.0

4590028.04533.353

9010

250250

10

0)250(250)10(

0

ttv

RIV

ttv

ZIV

ti

I

j

jI

jIIj

VVVs

R

R

L

LL

ZR



5-j50 j200


5-j50 j200


100

j100 -j200

j100


100

j100 -j200

j100





AC Power Calculations

cosrmsrmsIVP

cosPF

iv

sinrmsrmsIVQ


rmsrmspower apparent IV

2rmsrms

22 IVQP

RIP 2rms

XIQ 2rms

R

VP R

2rms

X

VQ X

2rms










THÉVENIN EQUIVALENT CIRCUITS


The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.

ocVV t

We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.


The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

scsc

oc

I

V

I

V ttZ

scII n






Maximum Power TransferIf the load can take on any complex

value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.



Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Steady-State Sinusoidal Analysis.

Documents

v1 v2

function vm e j t

j term

complex response im

phase angle

v2 lags v1

sinusoids v1

sum of v1