Chapter 5 Steady-State Sinusoidal Analysis

Chapter 5Steady-State Sinusoidal

Analysis

2008.10

Electrical Engineering and Electronics II Electrical Engineering and Electronics II

Scott

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.

•Main Contents

3. Compute power for steady-state sinusoidal ac circuits.

4. Find Thévenin equivalent circuits for steady-state ac circuits.

5. Determine load impedances for maximum power transfer.

6. Solve balanced three-phase circuits.

•Main Contents

•The importance of steady-state sinusoidal analysis

Electric power transmission and distribution by sinusoidal currents and voltages

Sinusoidal signals in radio communication

All periodic signals are composed of sinusoidal components according to Fourier analysis

5.1 Sinusoidal Currents and Voltages

•Parameters of Sinusoidal Currents and Voltages

•Vm is the peak value, unit is volt

•ω is the angular frequency, unit is radians per second

•f is the frequency， unit is Hertz (Hz) or inverse second.

•θ is the phase angle, unit is radian or degree.

T

2

f 2

90cossin zz

Frequency T

f1

Angular frequency

To uniformity, This textbook expresses sinusoidal functions by using cosine function rather than the sine function. 。

•Parameters of Sinusoidal Currents and Voltages

•Root-Mean-Square Values or Effective Values

dttvT

VT

2

0

rms

1

R

VP

2rms

avg

dttiT

IT

2

0

rms

1

RIP 2rmsavg

Average Power

•RMS Value of a Sinusoid

2rms

mVV

The rms value for a sinusoid is the peak value divided by the square root of two.

However, this is not true for other periodic waveforms such as square waves or triangular waves.

)100cos(100)( ttv

)100cos(100)( ttv A voltage given by is applied to a 50Ω resistance. Find the rms value of the voltage and the average power delivered to the resistance.

70.71V2m

rms

VV

2 2

22

70.71100 W

50

( )( ) 200cos (100 ) W

rmsavg

VP

R

v tp t t

R

Complex Numbers

3 forms of complex numbers

Arithmetic operations of complex numbers

•Rectangular formbaA j

1j

]Re[Aa

]Im[Ab

a

b

Im

Re0

|A|

A

Imaginary unit

Real part

Imaginary part

Magnitude

Angle

Conjugate

22222 baAAba

a

barctan

jA a b A A 2AA A

a

b

Im

Re0

|A|

A

22 baA

a

barctan

|| AA

cos|| Aa

sin|| Ab

Conversion between Rectangular and polar

•Polar form

2 2 arctanb

A a ba

|| AA

•Conversion

baA j cos j sin

cos jsin

A A A

A

j

sinjcos

eA

AA

Euler’s Identity

sinjcosj ea

b

Im

Re0

|A|

A

•Exponential form

Rectangular form

Polar form

Exponential form

baA j

j|| eAA

|| AA

•Three forms

•Arithmetic Operations

baA j dcB jGiven

Identity BA dbca ,

Adding and subtracting

21 jeeBAC

cae 1dbe 2


baA j dcB jGiven

Product

C

BA

eC

eBeABAC

j

jj

BAC

)j(j BAC ee

BAC

B

A

eB

eABAD

j

j

B

ABAD )j(j BAD ee BAD

)j(j BAD eB

AeDD


baA j dcB jGiven

Dividing

•Phasor Definition

111 cos :function Time θtωVtv

1 1 1 peak Phasor: V V

Sinusoidal steady-state analysis is greatly facilitated if currents and voltages are represented as vectors or Phasors.

•rms Phasor or effective phasor.

•In this book, if phasors are not labeled as rms, then they are peak phasors.

5.2 PHASORS

正弦量可以表示为在复平面 complex plane上按逆时针方向旋转的相量的实部 real part。

Angular velocity

•Phasors as Rotating Vector

•Sinusoids can be visualized as the real-axis projection of vectors rotating in the complex plane.

•The phasor for a sinusoid is a ‘snapshot’ of the corresponding rotating vector at t = 0.

θ

•Phase Relationships

•Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ≤180o as the smaller angle between the two phasors.)

•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 .

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



•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.

1 20cos 45 Vv t t

2 10sin 60 Vv t t

1 20 45 V V 2 10 30 V V

•Example:

1 2Find ?sv v v

Solution:


1 2

20 45 10 30

14.14 j14.14 8.660 j5

23.06 j19.14

29.97 39.7 V

sV V V

29.97cos 39.7 Vsv t t


•Exercise

•Answers

1

1

2

1. ( ) 10cos( ) 10sin( )

2. ( ) 10cos( 30 ) 5sin( 30 )

3. ( ) 20sin( 90 ) 15cos( 60 )

o o

o o

v t t t

i t t t

i t t t

1

1

2

1. ( ) 14.14cos( 45 )

2. ( ) 11.18cos( 3.44 )

3. ( ) 30.4cos( 25.3 )

o

o

o

v t t

i t t

i t t


5.3 COMPLEX IMPEDANCES

•By using phasors to represent sinusoidal voltages and currents, we can solve sinusoidal steady-state circuit problems with relative ease.

•Sinusoidal steady-state circuit analysis is virtually the same as the analysis of resistive circuits except for using complex arithmetic.

•Sinusoidal response of resistance• Relationship between voltage and current

Uu

i

I

0 t

Ru

icos( )mi I t

cos( )

cos( )m

m

v Ri RI t

V t

rms rmsV RI

rms m

rms m

V VR

I I

then：

• 2.Phasor relation and diagram

Substituting for current and voltage phasors

Rv

i

IV

O0

RV

I

mV RI RI

rms rms rmsV RI RI

rms

rms

V VR

I I

Phasor diagram

cos( )mi I t

mI I

rmsrmsII

•Sinusoidal response of resistance

Plot the phasor diagram

IV

00m mI I I

00m mV RI RI

VR

I phasors expression in

Ohm’s Law


p vi

2 cos 2 cosrms rmsV t I t

22 cos cos 2rms rms rms rms rms rmsV I t V I V I t

No doubt, p≥0,Resistance always absorbs energy.

• Power

Instantaneous power


0t

vi

p

VrmsIrms

2VrmsIrms

• Power

22 cos (1 cos 2 )rms rms rms rmsp V I t V I t


Average Power

——Average value of instantaneous power in a period

0 0

1 1(1 cos 2 )

T T

rms rmsP pdt V I t dtT T

2

2 rmsrms rms rms

VV I I R

R

Average Power represents the consumed power, is also named as

Real power.

• Power


u、 i have the same frequency， u leads i by 90o

i

u π 2π0 ωt

•Sinusoidal response of Inductance• Relationship between voltage and current

sinmi I t

0cos sin( 90 )m m

div L LI t V t

dt

rms rmsV LI m mV LI

2rms mL

rms m

U UX L fL

I I

DefinitionReactance电抗－感抗

sinmi I t 0sin( 90 )mv V t

• Effective Value

•Sinusoidal response of Inductance

cosmi I t

0cos( 90 )mv V t

AII 00

i

v L

jωL

I

V

0 090 0m L m L m LV V jX I jX I jX I

j jL

VZ X L

I

Phasor diagram：

I

LV jX I

• Phasor relations


Complex impedance

• Power

sinmi I t

sin sin( 90 )L m mp p vi V I t t

sin cos

sin 2m m

rms rms

V I t t

V I t

0sin( 90 )mv V t

0t

ui

pVrmsIrms


0

ti

UI

0

t

p

释放能量

吸收能量

u

释放能量

吸收能量

i

u

i

u

i

u

i

u

——magnitude of magnetic field increases， inductance absorbs energy

——magnitude of magnetic field decreases， inductance supplys energy

increases 0

( ) 0, increasesL

dii

dtp t W

，，

decreases 0

( ) 0, decreasesL

dii

dtp t W

，，

• Power


0

0

1

1sin 2 0

T

T

rms rms

P pdtT

V I tdtT

•Inductance is an energy-storage element rather than an energy-consuming element

Average PowerAverage Power

• Power


The power flows back and forth to inductances and capacitances is called reactive power Q, it is the peak instantaneous power。

22Lrms

Lrms Lrms L LrmsL

VQ V I X I

X Unit: Var ( 乏 )

Reactive PowerReactive Power

Reactive power is important because it causes power dissipation in the lines and transformers of a power distribution system. Specific Charge is executed by electric-power companies for reactive power.


i

v C

sinmv V t

0

0

sin( 90 )

sin( 90 )

m

m

dvi C C U t

dt

I t

Current leads voltage by 90°

2m m mI CV fCV u

i 20

•Sinusoidal response of capacitance• Relationship between voltage and current

i

v C

2rms rms rmsI CV fCV

2m m mI CV fCV

1 1

2rms m

Crms m

V UX

I I C fC

Definition

Reactance电抗－容抗

• Relationship between voltage and current

•Sinusoidal response of capacitance

sinmv V t0sin( 90 )mi I t A 090mI I A

00mV V V

i

v C

I

U -jωC

• Phasor diagram

U

CX

UjI

0

0

0j j

90 jm m m

Cm m m

V V VUX

I I II


0sin sin( 90 )

sin 2m m

rms rms

p vi V tI t

V I t

0t

ui

pUrmsIrms

•PowerPower•Instantaneous power


0

t

UI

0

t

p

释放能量

吸收能量

ui

释放能

量

吸收能量

i

u

i

u

i

u

i

u

increases increases

0 ( ) 0

u Wc

dup t

dt

，，

，

——capacitance charges，it absorbs energy

——capacitance discharges，it supplys energy

decreases decreases

0 ( ) 0

u Wc

dup t

dt

，，

，

•PowerPower


0 0

1 1sin 2 0

T T

rms rmsP pdt V I tdtT T

Average PowerAverage Power

•PowerPower

•Capacitance is an energy-storage element rather than an energy-consuming element.


• Assuming the same current acts on the inductance and capacitance respectively, the initial phase is 0， then

sinmi I t sin( 90 )C mv V t

2C Crms Crms Crms CQ V I I X

0sin sin( 90 )

sin 2m m

rms rms

p vi V tI t

V I t

Reactive PowerReactive Power•PowerPower


It is given that the frequency of sinusoidal source is 50Hz, the rms or effective value is 10V, capacitor is 25μF, determine the rms value of current. If the frequency is 5000Hz, then what is the rms value of current now？

i

u C

when f =50Hz

6

1 1127.4

2 2 3.14 50 (25 10 )CX fC

mAAX

UI

C

rms

rms78078.0

4.127

10

When f =5000Hz

6

1 11.274

2 2 3.14 5000 (25 10 )CX fC

AX

UI

C

rms

rms8.7

274.1

10

The higher frequency is under fixed rms value of voltage, the bigger the rms value of current flowing through capacitor.

Solution

•Summaryele

ment

circuitRelationship

between u and i

　 Ohm’s Law

Complex impedance

Phasors diagram

R

L

C

R

i

v v Ri

div L

dt

dvi C

dt

IRU

IjXU L

IjXU C

I

UR

I

UjXC

I

UjX L

U

I

U

I

U

I

i

v C

i

v L

•Inductance

•Capacitance

In phase

•Resistance

•True or False1.1. 220 sin( 45 )Vv ω t

e V 45220V ？)A30(sin24 tω ？

RMS e A j304I3.3. ComplexComplex

j45

)A60(sin10 tωi ？Peak ValuePeak Value

RMS 100VV ？RMS e j15100 V

V ？ minusminus

2. RMS 10 60 AI4.4.

RMS V 100 15V

•Complex Impedances

Complex Impedance

Two-terminal circuit of

zero sources

+

-V

I

ii

( )vv

V ΨV VZ Ψ Ψ Z

I I Ψ I

MagnitudeV

ZI

Angle ivΨ Ψ

j jLL L

L

VZ L X

I

:L

:C

RR

R

VZ R

I

:R | | 0RZ R

| |2LZ L

1| |

2CZ C

1j

jC

C CC

VZ X

I C

•Complex Impedances of R L C

•The Laws of Phasor Form

(KCL)： 0i 0I

(KVL)： 0v 0V

v iR V IZ

Voltage-DividerVoltage-Divider1

11 2

ZV V

Z Z

ZZeqeq ==ZZ11++ZZ22Equivalent ImpedanceEquivalent Impedance

22

1 2

ZV V

Z Z

ZZ11

ZZ22

++

––

++

++

–– ––

V1V

2V

I

ZZ

++

––

V

I

•Complex Impedances in Series

IZZ

ZI

21

21

ZZ

++

––

V

I

V

I

ZZ11 ZZ22

++

––

1I 2I

1 2

1 1 1

eqZ Z Z

IZZ

ZI

21

12 Current-DividerCurrent-Divider

Equivalent ImpedanceEquivalent Impedance

•Complex Impedances in Parallel

Chapter 5 Steady-State Sinusoidal Analysis

Documents

sinusoidal voltages

sinusoidal components

peak phasors

phase circuits

steadystate ac circuits

peak value

complex impedancesby

complex arithmetic