SJTU1 Chapter 9 Sinusoids and Phasors. SJTU2 Sinusoids A sinusoid is a signal that has the form of the sine or cosine function.

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Chapter 9

Sinusoids and Phasors

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Sinusoids

A sinusoid is a signal that has the form of the sine or cosine function.

anglephase

umentt

frequencyangular

amplitudeVmwhere

tVv m

arg

)cos(

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)cos( tVv m

t

fT

22

radians/second

(rad/s)

f is in hertz(Hz)

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)cos()(

)cos()(

222

111

tVtv

tVtv

m

m

Phase difference:

byvlagsv

byvleadsv

phaseinarevandv

phaseofoutarevandvif

tt

210

210

210

210

)()(

21

21

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Complex Number

form lexponentia

form sinusoidal

formpolar

formr rectangula

jrez

jrsinrcosz

rz

jyxz

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Phasor

a phasor is a complex number representing the amplitude and phase angle of a sinusoidal voltage or current.

Eq.(8-1)

and

Eq. (8-2)

Eq.(8-3)

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When Eq.(8-2) is applied to the general sinusoid we obtain

Eq.(8-4)

The phasor V is written as

Eq.(8-5)

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Fig. 8-1 shows a graphical representation commonly called a phasor diagram.

Fig. 8-1: Phasor diagram

Two features of the phasor concept need emphasis:

1. Phasors are written in boldface type like V or I1 to distinguish them from signal waveforms such as v(t) and i1(t).

2. A phasor is determined by amplitude and phase angle and does not contain any information about the frequency of the sinusoid.

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In summary, given a sinusoidal signal                      , the corresponding phasor representation is            . Conversely, given the phasor            , the corresponding sinusoid is found by multiplying the phasor by        and reversing the steps in Eq.(8-4) as follows:

Eq.(8-6)

)cos()( tVtv m VmVTime domain representation

Phase-domain representation

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Properties of Phasors

• additive property

Eq.(8-7)

Eq.(8-8)

Eq.(8-9)

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• derivative property

Eq.(8-10)

Vjdt

dv Time domain representation

Phase-domain representation

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• Integral property

Time domain representation

Phase-domain representation

j

Vvdt

The differences between v(t) and V:

1. V(t) is the instantaneous or time-domain representation, while V is the frequency or phasor-domain representation.

2. V(t) is a real signal which is time dependent, while V is just a supposed value to simplify the analysis

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The complex exponential is sometimes called a rotating phasor, and the phasor V is viewed as a snapshot of the situation at t=0.

Fig. 8-2: Complex exponential

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+ j

+ r e a l

- r e a l

- j

t

V m

= 0

= 9 0 o r / 2

= - 9 0 o r - / 2

= 1 8 0 o r

15 10 5 0 5 10 1515

13.5

12

10.5

9

7.5

6

4.5

3

1.5

010V rms ac s igna l a t 0.5 Hz

voltage in volts

angular freq

uency times

time in radian

s

0

12 .566

t n

14.14214.142 v real t( ) n

0 5 10 1515

12

9

6

3

0

3

6

9

12

1510V rms ac s igna l a t 0.5 Hz

angular frequency times time in radians

Voltage in v

olts

14 .142

14.142

v imag t n

12.5660 t( )n

)sin(

is axis )(imaginary j on thephasor

rotating theof projection The

tVv mima g

)cos(

is axis real on thephasor

rotating theof projection The

tVv mrea l

case particular In this

5.02cos102cos ttfVv m

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EXAMPLE 8-1(a) Construct the phasors for the following signals:

(b) Use the additive property of phasors and the phasors

found in (a) to find v(t)=v1(t)+v2(t).

SOLUTION

(a) The phasor representations of v(t)=v1(t)+ v2(t) are

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(b) The two sinusoids have the same frequent so the additive property of phasors can be used to obtain their sum:

The waveform corresponding to this phasor sum is

V1

V2

1

j

V

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EXAMPLE 8-2 (a) Construct the phasors representing the following signals:

(b) Use the additive property of phasors and the phasors found i

n (a) to find the sum of these waveforms.SOLUTION:

(a) The phasor representation of the three sinusoidal currents are

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(b) The currents have the same frequency, so the additive property of phasors applies. The phasor representing the sum of these current is

Fig. 8-4

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EXAMPLE 8-3Use the derivative property of phasors to find the time derivative of v(t)=15 cos(200t-30° ).

The phasor for the sinusoid is V=15∠-30 °  . According to the derivative property, the phasor representing the dv/dt is found by multiplying V by j .

SOLUTION:

The sinusoid corresponding to the phasor jV is

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Device Constraints in Phasor Form

Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.

Resistor:

Re

jIm

I

V

0

IV

mm RIV

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Device Constraints in Phasor Form

Inductor:

Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.

90IV

mm LIV

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Device Constraints in Phasor Form

Capacitor:

Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.

90VI

mm CVI

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Connection Constraints in Phasor Form

KVL in time domain

Kirchhoff's laws in phasor form (in frequency domain)

KVL: The algebraic sum of phasor voltages around a loop is zero.

KCL: The algebraic sum of phasor currents at a node is zero.

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The Impedance Concept

The IV constraints are all of the form

V=ZI or Z= V/I Eq.(8-16)

where Z is called the impedance of the element

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms()

reactance. theis ZImX and resistance theis ZReR where jXRZ

The impedance is inductive when X is positive

is capacitive when X is negative

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The Impedance Concept

sin,cos

tan, where 122

ZXZRandR

XXRZ

ZZ

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EXAMPLE 8-5

Fig. 8-5

The circuit in Fig. 8-5 is operating in the sinusoidal steady state with                           and                     . Find the impedance of the elements in the rectangular box.

SOLUTION:

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37.90.278/RVI L23

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The Admittance Concept

The admittance Y is the reciprocal of impedance, measured in siemens (S)

V

I

ZY

1

Y=G+jB

Where G=Re Y is called conductance and B=Im Y is called the susceptance

2222,

1

XR

XB

XR

RG

jXRjBG

How get Y=G+jB from Z=R+jX ?

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CjYcapacitor

LjYinductor

GR

Yresistor

C

L

R

:

1:

1:

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Basic Circuit Analysis with Phasors

Step 1: The circuit is transformed into the phasor domain by representing the input and response sinusoids as phasor and the passive circuit elements by their impedances.

Step 2: Standard algebraic circuit techniques are applied to solve the phasor domain circuit for the desired unknown phasor responses.

Step 3: The phasor responses are inverse transformed back into time-domain sinusoids to obtain the response waveforms.

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Series Equivalence And Voltage Division

where R is the real part and X is the imaginary part

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EXAMPLE 8-6

Fig. 8-8

The circuit in Fig. 8 - 8 is operating in the sinusoidal steady state with

(a) Transform the circuit into the phasor domain. (b) Solve for the phasor current I. (c) Solve for the phasor voltage across each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c)

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SOLUTION:

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PARALLEL EQUIVALENCE AND

CURRENT DIVISION

Rest of the circuit

Y1Y1 Y2 YN

I

VI1 I2 I3

phasor version of the current division principle

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EXAMPLE 8-9

Fig. 8-13

The circuit in Fig. 8-13 is operating in the sinusoidal steady state with iS(t)=50cos2000t mA.(a) Transform the circuit into the phasor domain. (b) Solve for the phasor voltage V.(c) Solve for the phasor current through each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c).

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SOLUTION:

(a) The phasor representing the input source current is Is=0.050°∠  A. The impedances of the three passive elements are

Fig. 8-14

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And the voltage across the parallel circuit is

The sinusoidal steady-state waveforms corresponding to the phasors in (b) and (c) are

The current through each parallel branch is

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EXAMPLE 8-10

Fig. 8-15

 Find the steady-state currents i(t), and iC(t) in the circuit of Fig. 8-15 (for Vs=100cos2000t V, L=250mH, C=0.5  F, and R=3k ).

SOLUTION:

Vs=100 0°∠

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Y←→ TRANSFORMATIONS△

The equations for the to Y △transformation are

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The equations for a Y-to- transformation are △

when Z1=Z2=Z3=ZY or ZA=ZB=ZC=ZN.

ZY=ZN /3 and ZN =3ZY balanced conditions

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EXAMPLE 8-12Use a  to Y transformation to solve for the phasor current I△ X in Fig. 8-18.

Fig. 8-18

SOLUTION:

ABC to Y△

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