ac single phase

Characteristics of Sinusoidal

Phasors

Phasor Relationships for R, L and C

Impedance

Parallel and Series Resonance

Examples for Sinusoidal Circuits

Analysis

Single Phase AC

Sinusoidal Steady State Analysis

• Any steady state voltage or current in a linear circuit with a sinusoidal source is a sinusoid– All steady state voltages and currents have the same frequency as

the source• In order to find a steady state voltage or current, all we need to know

is its magnitude and its phase relative to the source (we already know its frequency)

• We do not have to find this differential equation from the circuit, nor do we have to solve it

• Instead, we use the concepts of phasors and complex impedances• Phasors and complex impedances convert problems involving

differential equations into circuit analysis problems

Characteristics of Sinusoids

Outline:1. Time Period: T 2. Frequency: f (Hertz)3. Angular Frequency: (rad/sec)4. Phase angle: Φ5. Amplitude: Vm Im

Characteristics of Sinusoids :

tVv mt sin i

R

+

_

i

R

-

+

v ,i

tt1 t20

Both the polarity and magnitude of voltage are changing.

Radian frequency(Angular frequency): = 2f = 2/T (rad/s ）

Time Period: T — Time necessary to go through one cycle. (s)Frequency: f — Cycles per second. (Hz)

f = 1/T

Amplitude: Vm Im

i = Imsint ， v =Vmsint

v ,i

t 20

Vm , Im


Effective Roof Mean Square (RMS) Value of a Periodic Waveform — is equal to the value of the direct current which is flowing through an R-ohm resistor. It delivers the same average power to the resistor as the periodic current does.

RIRdtiT

T 2

0

21

Effective Value of a Periodic Waveform T

eff dtiT

I0

21

221

22cos1sin1 2

0

2

0

22 mm

Tm

T

meffITI

Tdtt

TItdtI

TI

21

0

2 mT

effVdtv

TV


Phase (angle)

tIi m sin

sin0 mIi

Phase angle

-8

-6

-4

-2

0

2

4

6

8

0 0.01 0.02 0.03 0.04 0.05<0

0


)sin( 1 tVv m )sin( 2 tIi m

Phase difference

2121 )( ttiv

021 — v(t) leads i(t) by (1 - 2), or i(t) lags v(t) by (1 - 2)

221

v, i

t

v

i

21Out of phase

t

v, iv

i

v, i

t

v

i

021 In phase

021 — v(t) lags i(t) by (2 - 1), or i(t) leads v(t) by (2 - 1)


Review

The sinusoidal waves whose phases are compared must:1. Be written as sine waves or cosine waves.2. With positive amplitudes.3. Have the same frequency.

360°—— does not change anything. 90° —— change between sin & cos. 180°—— change between + & -

2sin cos cos3 2

cos sin2


Phase difference

30314sin22201 tv

9030314sin222030314cos22202 ttv 120314sin2220 t

1501203021

30314cos22202 tv

30314cos22202 tv 18030314cos2220 t

210314360cos2220 t

90150314sin2220 t

60314sin2220 t 30603021

Find ?

30314cos22202 tvIf


Phase difference

v, i

t

vi

-/3 /3• ••

3sin tVm

3sin tIm


Outline:1. Complex Numbers 2. Rotating Vector3. Phasors

A sinusoidal voltage/current at a given frequency, is characterized by only two parameters : amplitude and phase

A phasor is a Complex Number which represents magnitude and phase of a sinusoid

Phasors

e.g. voltage response

A sinusoidal v/i

Complex transform

Phasor transform

By knowing angular frequency ω rads/s.

Time domain

Frequency domain

eR v tComplex form:

cosmv t V t

Phasor form:

j tmv t V e

Angular frequency ω is known in the circuit.

|| mVV

|| mVV

Phasors

Rotating Vector

tIti m sin)(

i

Im

t1

i

t

Im

t

x

y

max

cos sin

sin

j tm m m

j tm m

I e I t jI t

i t I t I I e

A complex coordinates number:

Real value:

i(t1)

Imag

Phasors

Rotating Vector

Vm

x

y

0

)sin( tVv m

Phasors

Complex Numbers

jbaA — Rectangular Coordinates

sincos jAA

jeAA — Polar Coordinates

jeAAjbaA

conversion ： 22 baA

abarctg

jbaeA j cosAa sinAb

|A|

a

b

Real axis

Imaginary axis

jjje j 090sin90cos90

Phasors

Complex Numbers

Arithmetic With Complex Numbers

Addition: A = a + jb, B = c + jd, A + B = (a + c) + j(b + d)

Real Axis

Imaginary Axis

AB

A + B

Phasors

Complex Numbers


Subtraction : A = a + jb, B = c + jd, A - B = (a - c) + j(b - d)

Real Axis

Imaginary Axis

AB

A - B

Phasors

Complex Numbers


Multiplication : A = Am A, B = Bm B

A B = (Am Bm) (A + B)

Division: A = Am A , B = Bm B

A / B = (Am / Bm) (A - B)

Phasors

Phasors

A phasor is a complex number that represents the magnitude and phase of a sinusoid:

tim cos mI

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships between currents and voltages.

Phasors

)sin()cos()( tAjtAeAAe tjtj

)cos(||}Re{ tAAe tj

Complex Exponentials

jeAA

A real-valued sinusoid is the real part of a complex exponential. Complex exponentials make solving for AC steady state an algebraic problem.

Phasors


Outline:I-V Relationship for R, L and C,

Power conversion


v~i relationship for a resistor

_

v

i

R

+

U

I

tItR

VRvi m

m sinsin

tVv m sin

Relationship between RMS:RVI

Wave and Phasor diagrams ： v 、 i

t

v

i

I

V

RVI

Resistor

Suppose

Time domain Frequency domainResistor

With a resistor θ﹦ϕ, v(t) and i(t) are in phase .

)cos()()cos()(

wtItiwtVtv

m

m

IRV

RIVeRIeV

eRIeV

mm

jm

jm

wtjm

wtjm

)()(


PowerResistor

_

v

i

R

+

U

I

P 0

tItVvip mm sinsin tVI mm 2sin

tVI mm 2cos12

tIVIV 2cos

v, i

t

v

i P=IV Tpdt

TP

0

1 T

VIdttVIT 0

2cos11

RVRIIVP

22

• Average Power

• Transient Power

Note: I and V are RMS values.


Resistor

, R=10， Find i and Ptv 314sin311

VVV m 2202

3112

ARVI 22

10220

ti 314sin222 WIVP 484022220


v~i relationshipInductor

dtdiLvv AB

tLIdt

tIdLdtdiLv m

m cossin

90sin tLI m

90sin tVm

tvdt

Li 1

tvdt

Lvdt

L 0

0 11 tvdt

Li

001

tIi m sin Suppose


v~i relationshipInductor

90sin tLIm dtdiLv 90sin tVm

LIV mm Relationship between RMS: LIV

LVI

fLLX L 2

For DC ， f = 0 ， XL = 0.

fX L

v(t) leads i(t) by 90º, or i(t) lags v(t) by 90º


v ~ i relationshipInductor

v, i

t

vi

eL

V

I

LXIjV

Wave and Phasor diagrams ：


PowerInductor

vip tItV mm sin90sin ttIV mm sincos

tIV mm 2sin

2 tVI 2sin

P

t

v, i

t

vi

++

--22max 2

1 LILIW m

2

00 21 LiLidividtW

it Energy stored:

T T

tdtVIT

pdtT

P0 0

02sin11 Average Power

Reactive PowerL

L XVXIIVQ

22 （ Var）


Inductor

L = 10mH ， v = 100sint ， Find iL when f = 50Hz and 50kHz.

14.310105022 3fLX L

Atti

AXVI

L

L

90sin25.22

5.2214.3

2/10050

31401010105022 33fLX L

mAtti

mAXVI

L

Lk

90sin25.22

5.2214.3

2/10050


v ~ i relationshipCapacitor

_

v

i

+

U

I

C

dtdvC

dtdqi

tVv m sinSuppose:

90sincos tCVtCVi mm 90sin tIm

t ttidt

cvidt

cidt

cidt

cv

0

000

1111

i(t) leads v(t) by 90º, or v(t) lags i(t) by 90º

Relationship between RMS:CX

V

C

VCVI

1

fCC

X C 211

For DC ， f = 0 ， XC f

X C1

mm CVI


_

v

i

+

U

I

C tj

m

tjm eCVjdtedVC

dttdvCti

)()(

v(t) = Vm ejt

Represent v(t) and i(t) as phasors:CjX

VVCωjI ==

• The derivative in the relationship between v(t) and i(t) becomes a multiplication by in the relationship between and .• The time-domain differential equation has become the algebraic equation in the frequency-domain.• Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.


V IwCj



v, i

t

vi

I

V

CXIjV

Wave and Phasor diagrams ：


PowerCapacitor

Average Power: P = 0

Reactive PowerC

C XVXIIVQ

22 （ Var）

90sinsin tItVvip mm tVItIV mm 2sin2sin2

P

t

v, i

t

vi

++

--

Energy stored:

t vv

CvCvdvdtdtdvCvvidtW

0 0

2

0 21

22max 2

1 CVCVW m


Capacitor

Suppose C=20F ， AC source v=100sint ， Find XC and I for f = 50Hz, 50kHz 。

1592

11Hz50fCC

Xf c

A44.02

c

m

c XV

XVI

159.02

11KHz50fCC

Xf c

A4402

c

m

c XV

XVI


Review (v – i Relationship)

Time domain Frequency domain

iRv IRV

ICj

V 1

ILjV dtdiLvL

dtdvCiC C

X C 1

LX L ,

,

, v and i are in phase.

, v leads i by 90°.

, v lags i by 90°.

R

C

L


Summary: R ： RX R 0

L ： ffLLX L 22 iv

C： ffccX C

12

11

2 iv

IXV

Frequency characteristics of an Ideal Inductor and Capacitor: A capacitor is an open circuit to DC currents; A Inductor is a short circuit to DC currents.


Impedance (Z)

Outline:Complex currents and voltages.ImpedancePhasor Diagrams

• AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:

ZIV

Complex voltage ， Complex current ， Complex Impedance

vmj

m VeVV v

imj

m IeII i

ZeZeIV

IVZ jj

m

m iv )(

‘Z’ is called impedancemeasured in ohms ()

Impedance (Z)

Complex Impedance

ZeZeIV

IVZ jj

m

m iv )(

Complex impedance describes the relationship between the voltage across an element (expressed as a phasor) and the current through the element (expressed as a phasor).

Impedance is a complex number and is not a phasor (why?).

Impedance depends on frequency.

Impedance (Z)

Complex Impedance

ZR = R ＝ 0; or ZR = R 0

Resistor——The impedance is R

cj

c jXCje

CZ

21

)2

( iv

or 901

CZC

Capacitor——The impedance is 1/jωC

Lj

L jXLjLeZ 2

)2

( iv

or 90 LZL

Inductor——The impedance is jωL

Impedance (Z)

Complex ImpedanceImpedance in series/parallel can be combined as resistors.

_

U

U

Z1 +

Z2 Zn

I

n

kkn ZZZZZ

121 ...

_

In

Zn

+

U

I

Z2 Z1

n

k kn ZZZZZ 121

11...111

21

12

21

21 ZZ

ZIIZZ

ZII

Current divider:

n

kk

ii

Z

ZVV

1

Voltage divider:

Impedance (Z)

Complex Impedance

_

+

V

I

1I Z1

Z2 Z

2121

2

2121

21

21

1

2

21

11

ZZZZZZZVI

ZZZZZZZZV

ZZZ

VI

ZZZII

Impedance (Z)

Complex Impedance

Phasors and complex impedance allow us to use Ohm’s law with complex numbers to compute current from voltage and voltage from current

20kW+

-1mF10V 0 VC

+

-

w = 377Find VC

• How do we find VC?• First compute impedances for resistor and capacitor:

ZR = 20kW = 20kW 0 ZC = 1/j (377 *1mF) = 2.65kW -90

Impedance (Z)

Complex Impedance

20kW+

-1mF10V 0 VC

+

-

w = 377Find VC

20kW 0

+

-2.65kW -9010V 0 VC

+

-

Now use the voltage divider to find VC:

46.82 V31.154.717.20

9065.20 10VCV

)0209065.2

9065.2(010

kk

kVVC

Impedance (Z)

Impedance allows us to use the same solution techniquesfor AC steady state as we use for DC steady state.

• All the analysis techniques we have learned for the linear circuits are applicable to compute phasors– KCL & KVL– node analysis / loop analysis– Superposition– Thevenin equivalents / Norton equivalents– source exchange

• The only difference is that now complex numbers are used.

Complex Impedance

Impedance (Z)

Kirchhoff’s Laws

KCL and KVL hold as well in phasor domain.

KVL ： 01

n

kkv vk- Transient voltage of the #k branch

01

n

kkV

KCL: 01

n

kki

01

n

kkI

ik- Transient current of the #k branch

Impedance (Z)

Admittance

• I = YV, Y is called admittance, the reciprocal of impedance, measured in Siemens (S)

• Resistor:– The admittance is 1/R

• Inductor:– The admittance is 1/jL

• Capacitor:– The admittance is jC

Impedance (Z)

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships between currents and voltages.

2mA 40

–

1mF VC

+

–

1kW VR

+

+

–

V

I = 2mA 40, VR = 2V 40 VC = 5.31V -50, V = 5.67V -29.37

Real Axis

Imaginary Axis

VR

VC

V

Impedance (Z)

Parallel and Series Resonance

Outline:RLC Circuit,

Series Resonance

Parallel Resonance

v

vR

vL

vC

CLR vvvv

CLR VVVV Phasor

I

V

LV

CV

RVIZ

XRI

XXRI

IXIXIR

VVVV

CL

CL

CLR

22

22

22

22

)(

)()(

)(

）CL XXX (

22 XRZ 22 )1(c

LR

(2nd Order RLC Circuit )Series RLC Circuit

Parallel and Series Resonance :

22 XRZ 22 )1(c

LR

IZVVVV CLR 22 )(

Z X = XL-XC

R

V

RVCLX VVV R

XXV

VV

CL

R

CL

1

1

tan

-tan

Phase difference:

XL>XC >0 ， v leads i by — Inductance Circuit

XL<XC <0 ， v lags i by — Capacitance Circuit

XL=XC =0 ， v and i in phase — Resistors Circuit

Series RLC Circuit


CLR VVVV CL XIjXIjRI

ZIjXRIXXjRI CL )()]((

)( CL XXjRIVZ

ZjXRZ22 )( CL XXRZ

RXX CL 1tan

iv

v

vR

vL

vC

Series RLC Circuit


Series Resonance (2nd Order RLC Circuit )

CLR VVVV CL XIjXIjRI R

XXarctgV

VVarctg CL

R

CL

CLCL VVLC

XXWhen 1,

VVR 0and —— Series Resonance

Resonance condition

I

LV

CV

VVR

LCfor

LC

211

00

f0 f

X

CfX C 2

1

fLX L 2

Resonant frequency


Series Resonance

RV

ZVIRXXRZ CL

00

220 )(•

Zmin ； when V = constant, I = Imax= I0

RXX CL RIXIXI CL 000 VVV CL

• Quality factor Q,

RX

RX

VV

VVQ CLCL

CLCL VVLC

XX )1(

Resonance condition:

When,


Parallel RLC Circuit

V

I

LI CI

)(

1/11

222222 LRLCj

LRR

CjLjRLjR

LjR

CjLjRCjLjR

Y

Parallel Resonance

Parallel Resonance frequencyL

CRLC

2

0 11

LXR In generally )2

1( 0LC

f

LC1

0

0)( 222

LR

LCWhen

2220 LRRY

,

In phase withV I

VL

RC

CLR

RVL

LCR

RVLR

RVVYII

22222

0200 1Zmax Imin:



V

I

LI CI

VLCj

LVj

LjRVIL

00

1

VLCjVCjIC 0

0|||||| 0 III CL Z .

RCRLQ

0

0 1

0IjQI L

0IjQIC

•Quality factor Q,

0000 YY

YY

II

IIQ CLLC



Review

For sinusoidal circuit ， Series ： 21 vvv 21 VVV

21 iii 21 III ？

Two Simple Methods: Phasor Diagrams and Complex Numbers

Parallel :


ac single phase

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