fall 2000ee201phasors and steady-state ac1 phasors a concept of phasors, or rotating vectors, is...

Fall 2000 EE201Phasors and Steady-State AC 1

Phasors

A concept of phasors, or rotating vectors, is used to find the AC steady-state response of linear circuits excited by sinusoidal voltages and/or currents:

x(t) = Xmcos(t + )

That is, x(t) is the real part of the complex quantity Xmej(t + )

Euler’s identity: cos sin , 1je j where j

Then, ( )

( )

cos( ) sin( )

( ) cos( )

j tm m

j tm m

X e X t j t

and x t X t e X e


VectorsThe quantity ej is a unity vector (magnitude =1) pointing at an angle .

Euler’s identity, in essence, allows one to relate polar form representation of the vector to

the rectangular form representation. A shorthand form for representing ej is 1.

cos sinjXe X X jX

j

(Imaginary axis)

(Real axis)

1

sin

cos

The vector Xej =X is stationary and constant. That is, the angle never changes and the magnitude is always X.

(polar form) (rectangular form)(Complex number)


PhasorsA phasor is a rotating vector. The magnitude never changes, but the angle is a function of time. The magnitude of the following phasor is Xm, and the angle is

= t + . The phasor is represented below at times t = 0 and t =/.

The value of Xmcos(t + ) = Re[Xm(t + )] changes depending on the position of the phasor at time t.

( ) ( ) cos( ) sin( )j tm m mX e X t X t j t

rotating at a velocity of radians per second

Xmcos (t + )

Xm

(t = 0)

Xm

Xmcos (t + )

(t = )


Solving circuits in the phasor domain

In general, a circuit containing resistors, inductors, and capacitors and excited by time-varying voltage and current sources must be solved using differential equations. However, if the voltage and current sources are sinusoidal, then algebraic equations can be used to find the AC steady-state response of the circuit. These equations, however, do involve complex quantities and phasors.

Observations:

• The steady-state response of a circuit is the response to an input after the input has been applied a “long time” and the circuit has been allowed to settle down.

• The steady-state value of all voltages and currents in a linear circuit excited by sinusoidal inputs will also be sinusoidal. The voltages and currents will be of the same frequency as the input; the magnitude and phase angle will, in general, be different.

Linear Circuitv(t) = Vmcos(t + v)

i(t) = Imcos(t + i)


Solving circuits in the phasor domain

Since all currents and voltages in the steady-state AC response have the same frequency, a short-hand notation, called phasor notation, is use to represent time-varying sinusoidal signals.

By transforming voltages and signals from the time domain to the phasor domain, we can use algebraic equations and conventional circuit analysis techniques (KVL, KCL, nodal, mesh Thevenin, etc.) to solve for the steady-state voltages and currents in a circuit. In addition to transforming sources from sinusoidal functions of time to phasors, circuit components (R, L , and C) must be transformed, also.

( )cos( ) j tm m mX t e X e X X

time domain representation phasor domain representation


Steady-State AC Voltage and Current Phasors

All steady-state voltages and currents are sinusoidal if the forcing functions (voltage and currents sources) are sinusoidal. These sinusoidal voltages and currents are represented as phasors (rotating vectors).

( )

( )

( ) cos( )

( ) cos( )

v

i

j tm v m m v

j tm i m m i

v t V t e V e V V

i t I t e I e I I

V

v

Vm


+ v(t) -R

i(t)

ResistorsTime Domain

The voltage across a resistor is in phase with the resistor current.

V

v = i

I

Phasor Domain

+ -R

V

I

( ) cos( )

( ) ( ) cos( )

m i m i

m i m i

i t I t I I

v t Ri t RI t V RI

V R I

Justification:


+ v(t) -L

i(t)

InductorsTime Domain

The current through an inductor lags the voltage by 90 degrees.

i I

V

( ) cos( )

( )( ) sin( ) cos( 90 )

( ) cos( 90 ) 90

m i m i

m i m i

m i m i

L

i t I t I I

di tv t L LI t LI t

dt

v t LI t V LI

V j L I jX I

Justification:

Phasor Domain

+ -

I

V

jXL = jL


+ v(t) -C

i(t)

CapacitorsTime Domain Phasor Domain

The current through a capacitor leads the voltage by 90 degrees.

+ -

-jXC = 1/jCI

V

V

v

I

( ) cos( )

( )( ) sin( ) cos( 90 )

( ) cos( 90 ) 90

m v m v

m v m v

m v m i

C

v t V t V V

dv ti t C CV t CV t

dt

i t CV t I CV

IV jX I

j C

Justification:


Time Domain

R

+ v(t) -

i(t) ( ) ( )v t Ri t

C

+ v(t) -

i(t) ( )( )

dv ti t C

dt

i(t) ( )( )

di tv t L

dt

+ v(t) -

L

Phasor DomainR

V R I

V

I

LV jX I

V

I

jL = jXL

CV jX I

V

I

1/jC = -jXC

Summary


Impedance

In dc circuits we have resistance, R:

I

+V-

I

VR

zz

zz

ivm

m

im

vm

sinZXcosZR

jXRZ

)9090(ZZ

I

V

I

VZ

z > 0 R+jX , an inductive impedance.z < 0 R-jX is a capacitive impedance

In steady-state ac circuits we have impedance, Z:

+ -

imII

vVV m

I

VZ

R = resistance X = reactance


Impedance

In dc circuits :R

1G

1 1

1 1

z

zz

Z Z Z R jX

Y Y G jBZ Z

G and BR X

B is called susceptance

conductance is the reciprocal of resistance

In ac circuits :Z

1Y admittance is the reciprocal of impedance

Parallel and Series Combinations of Impedances

1 2

1 2

1 2

1 1 1 1series n

parallel n

parallel n

Z Z Z Z

Z Z Z Z

Y Y Y Y


Example Draw the frequency-domain network and calculate vo(t). Use a phasor diagram to determine v(t).

vs(t) = 10cos(800t)

+v(t)

-+

vo(t)-

k11.1jLjmH44214.25jLjmH10

14.25jF83.15)800(jjXF83.15 1C

-j25.14 j25.14 1k

1.11k

+

-V

+

-oV

+V=100

-V10 4

442mH


Example (Continued)

-j25.14 j25.14 1k

1.11k

+

-V

+

-oV

+V=100

-V10 4

By Nodal Analysis:

4

4

4

10 10 1010 0

25.14 25.14 1 1.11

1 1 1 1 1 1 110 10 0

25.14 25.14 1 1.11 25.14 25.14 1

1 1 110 10 0

1 1.11 1

o o o o

o

o

V V V VV

j j k j k

V Vj j k j k j j k

V Vk j k k


Example (Continued)

-j25.14 j25.14 1k

1.11k

+

-V

+

-oV

+V=100

-V10 4

By KVL:

oo V010VVV010

)45t800cos(07.7)t(v

97.4407.703.4507.710V03.4507.7V

k10

1

k1

110

k10

1

k11.1j

1

k1

1V

0V01010k1

110

k11.1j

1

k1

1V

o

o

o4

o

Substituting:


Vo = 7.0745

-7.0745

100

V= 10 - Vo = 7.07-45

Finding V using vector addition


E201 Circuits I Phasors

Read chapter 8, Steady-State AC Analysis, pages 399 - 442

HW Assignment #15Chapter 8 – Sections 8.1 thru 8.4Extension Problems: E8.1 thru E8.7

HW Assignment #16Chapter 8 – Sections 8.5 and 8.6Extension Problems: E8.8 thru E8.11End-of- Chapter Problems: P8.27

HW Assignment #17Chapter 8 – Sections 8.7 and 8.8

Extension Problems: E8.12, E.13, E8.14, E8.17End-of- Chapter Problems:P8.36 (45-23.13V)P8.40 (8-90V)P8.44 (8.5-41.73V)P8.46 (vo(t) = 6cos(104t - 23.13)V vL(t) = 8cos(104t + 66.87)V)P8.49 (2.8345A)

fall 2000ee201phasors and steady-state ac1 phasors a concept of phasors, or rotating vectors, is...

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