Sinusoidal

Bruce Mayer, PE

BMayer@ChabotCollege.edu

Engineering-43: Engineering Circuit Analysis

1

Bruce Mayer, PE

 SINUSOIDS

• Review basic facts about sinusoidal signals

 SINUSOIDAL AND COMPLEX

FORCING FUNCTIONS

• Behavior of circuits with sinusoidal independent current & voltage sources

• Modeling of sinusoids in terms of complex exponentials

Engineering-43: Engineering Circuit Analysis

2

Bruce Mayer, PE

Outline – AC SS cont.

phasEr

 PHAS O RS

• Representation of complex exponentials as vectors

– Facilitates steady-state analysis of circuits

• Has NOTHING to do with StarTrek

• Generalization of the familiar concepts of

RESISTANCE and CONDUCTANCE to describe AC steady-state circuit operation

Engineering-43: Engineering Circuit Analysis

3

Bruce Mayer, PE

Outline – AC SS cont.2

 PHASOR DIAGRAMS

• Representation of AC voltages and currents as COMPLEX VECTORS

 BASIC AC ANALYSIS USING

KIRCHHOFF’S LAWS

 ANALYSIS TECHNIQUES

• Extension of node, loop, SuperPosition,

Thevenin and other KVL/KCL Linear-

Circuit Analysis techniques

Engineering-43: Engineering Circuit Analysis

4

Bruce Mayer, PE

Sinusoids

 Recall From Trig the

Sine Function

5

Above, The Functional

Relationship x ( t )

X

M sin

Engineering-43: Engineering Circuit Analysis

 t

 Where

• X

M

 “Amplitude” or Peak or Maximum Value

– Typical Units = A or V

• ω 

• ωt 

Sinusoid argument in radians (a pure no.)

 The function Repeats every 2 π; x mathematically

(

 t

2

)

 x (

 t )

Bruce Mayer, PE

Sinusoids cont.

 Now Define the x (

 t

“Period”, T Such That

2

)

 x

( t

T )

 x (

 t )

 From Above Observe

T

2

 

T

2

 and x(t )

 x ( t

T ),

 t (for all t )

6

 Now Can Construct a

DIMENSIONAL (time)

Plot for the sine

Engineering-43: Engineering Circuit Analysis x ( t )

X

M sin

2

 t T

 How often does the

Cycle Repeat?

• Define Next the CYCLIC

FREQUENCY

Bruce Mayer, PE

Sinusoids cont.2

 Now Define the Cyclic

“Frequency”, f f

1

T

2

2

 f

 Describes the Signal

Repetition-Rate in Units of Cycles-Per-Second, or HERTZ (Hz)

• Hz is a Derived SI Unit

 Quick Example

• USA Residential

Electrical Power

Delivered as a 115V rms

,

60Hz, AC sine wave v residence

( t ) v residence

( t )

2

115 V

162 .

6 V sin sin

2

376 .

60

99

 t

 t

 Will Figure Out the

2 term Shortly

• RMS  “ R oot (of the)

M ean S quare”

Engineering-43: Engineering Circuit Analysis

7

Bruce Mayer, PE

Sinusoids cont.3

 Now Consider the

GENERAL Expression for a Sinusoid x ( t )

X

M sin

  t

  

 Where

• θ  “Phase Angle” in

 Graphically, for

POSITIVE θ

" lags by  "

Engineering-43: Engineering Circuit Analysis

8

Bruce Mayer, PE

 Consider Two

Sinusoids with the

SAME Angular x

1 x

2

( t

( t

Frequency

)

)

X

X

M

M

1

2 sin sin

 t t

 Now if

>

• x

1

2 by ( rads or Degrees

 − 

)

• x

2

LAGS x

1 by ( rads or Degrees

 − 

)

Engineering-43: Engineering Circuit Analysis

9

 If

=

, Then The

Signals are IN-PHASE

 If

  

, Then The

Signals are

OUT-of-PHASE

 Phase Angle Typically

Stated in DEGREES, but Radians are acceptable x ( t )

• Both These Forms OK

X

M

X

M sin sin

 

 t t

25 .

7

71

Bruce Mayer, PE

1.4

1.2

1.0

0.8

0.6

0.4

   

105 mS

0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

x1 (V or A)

-1.2

x2 (V or A)

-1.4

0.0

0.2

0.4

Sinusoid Phase Difference

1 Period

900 mS

360

1 Period

Out of Phase

* 42°

* 105 mS

0.6

42

For Different Amplitudes,

Measure Phase Difference

• peak-to-peak

• valley-to-valley

• ZeroCross-to-ZeroCross

0.8

1.0

Time (S)

1.2

Engineering-43: Engineering Circuit Analysis

10

PARAMETERS

• T = 900 mS

• f = 1.1111 Hz

• 

• 

• 

1.4

1.6

x2 LAGs x1

Bruce Mayer, PE

1.8

2.0

Useful Trig Identities

11

 To Convert sin↔cos cos

 t

 sin

 t

2 sin

 t

 cos

 t

 To Make a Valid

2

Phase-Angle Difference

Measurement BOTH

Sinusoids MUST have the SAME Frequency &

Trig-Fcn (sin OR cos)

• Useful Phase-Difference

ID’s 

Engineering-43: Engineering Circuit Analysis cos

 t sin

 t

 cos(

 t

 sin(

 t

 

)

 

)

 cos(

)

)

 sin

 cos

 cos

 cos

 cos

 sin

 sin sin

 cos(

 sin(

 

)

)

 cos sin

 cos cos

 sin cos

 sin sin

2

360

(degrees)

180

Bruce Mayer, PE

Example

Phase Angles

 Given Signals v

1

( t )

12 sin( 1000 t v

2

( t )

 

6 cos( 1000 t

60

)

30

)

 Find

• Frequency in Standard

Units of Hz

• Phase Difference

12

 Frequency in radians per second is the

PreFactor for the time variable

• Thus

Engineering-43: Engineering Circuit Analysis

 

1000 S

-1 f ( Hz )

2

159 .

2 Hz

 To find phase angle must express BOTH sinusoids using

• The SAME trigonometric function;

– either sine or cosine

• A POSITIVE amplitude

Bruce Mayer, PE

Example – Phase Angles cont.

 Convert –6V Amplitude to Positive Value Using cos(

)

  cos(

 

180

)

 Then v

2 v

2

 

6 cos( 1000 t

30

)

6 cos( 1000 t

30

 

180

)

 Next Convert cosine to sine using cos(

)

 sin(

 

90

) v

2

6 cos( 1000 t

210

)

6 sin( 1000 t

210

 

90

)

Engineering-43: Engineering Circuit Analysis

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 It’s Poor Form to

Express phase shifts in

Angles >180 ° in

Absolute Value v

2

6

6

6 sin sin

 sin( 1000 t

1000 t

300

)

300

 

360

1000 t

60

 

 So Finally v

1

( t )

12 sin( 1000 t v

2

( t )

6 sin( 1000 t

60

60

)

)

 Thus v

1 by 120 °

2

Bruce Mayer, PE

Sinusoid Phase Difference Example 7.2

12

8

4

0

-4

120°

-8

-12

0.000

0.002

0.004

0.006

Engineering-43: Engineering Circuit Analysis

14 v1 (V) v2 (V)

0.008

0.010

Time (S)

0.012

0.014

0.016

Bruce Mayer, PE

0.018

0.020

Effective or rms Values

 Consider Instantaneous

Power For a Purely i ( t )

R p ( t )

 i

2

( t ) R

15

 Now Define The

EFFECTIVE Value For a Time-Varying Signal as the EQUIVALENT

DC value That Supplies

The SAME AVERAGE

POWER

Engineering-43: Engineering Circuit Analysis

Dissipates this Power as HEAT, the Effective

Value is also called the

HEATING Value for the

Time-Variable Source

• For example

– A Car Coffee Maker

Runs off 12 Vdc, and

Heats the Water in 223s.

– Connect a SawTooth

Source to the coffee

Amplitude for the Same

Time → Effective

Voltage of 12V

Bruce Mayer, PE

rms Values Cont.

 For The Resistive Case, i ( t )

Define I eff for the Avg

Power Condition p ( t )

 i

2

( t ) R

R

P av

 2

I eff

R

 The P av

Calc For a

Periodic Signal by Integ

P av

1

T t

0 t

T

0 p ( t ) dt

R

1

T t

0 t

T

0 i

2

( t ) dt

Engineering-43: Engineering Circuit Analysis

16

 If the Current is DC, then i(t) = I dc

, so

P av

R

1

T t

0

0

 t

T

I

2 dc dt

RI

2 dc

1

T t

0 t

0

T

1 ( t ) dt

 2

RI dc

 Now for the Time-

Variable Current i(t) →

I eff

, and, by Definition

P av

 2

RI eff

 2

RI dc

P av

Bruce Mayer, PE

rms Values cont.2

 In the Pwr Eqn

P av

R

1

T t

0  t

T

0 i

2

( t ) dt

 2

RI dc

 2

RI eff

 Equating the 1 st & 3 rd

Expression for P av find

I eff

T

1 t

0

0

 t

T i

2

( t ) dt

 This Expression Holds for ANY Periodic Signal

Engineering-43: Engineering Circuit Analysis

17

 Examine the Eqn for I eff and notice it is

Determined by

• Taking the Square R OOT of the time-averaged, or

M EAN, S QUARE of the

Current

 In Engineering This

Operation is given the

Short-hand notation of

“rms”

 So

I eff

I rms

Bruce Mayer, PE

RMS Value for Sinusoid

 Find RMS Value for sinusoidal current i

• Note the Period T

2

I

M

 cos

  t

  

 Sub Above into RMS Eqn:

I rms

 

1

T

T

0

 Use Trig ID:

I

2

M cos

2 cos

 

2

 t

1

2

1

2

 dt

 cos

1 2

Engineering-43: Engineering Circuit Analysis

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Bruce Mayer, PE

RMS Value for Sinusoid

 Sub cos 2 Trig ID into RMS integral

I rms

I

M

1

T

T

0

1

2

1

2 cos

2

 t

2

 

 dt

 1 2

I rms

 Integrating Term by Term

I

M

1

T

T

0

1

2 dt

1

T

T

0

1

2 cos

2

 t

2

  dt

 1 2

Engineering-43: Engineering Circuit Analysis

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Bruce Mayer, PE

RMS Value for Sinusoid

0

 Rearranging a bit

I rms

I

M

1

2

1

T

0

T

1 dt

1

2

1

T

T

0 cos

2

 t

2

  dt

 1

 But the integral of a sinusoid over ONE

PERIOD is ZERO, so the 2 nd Term goes

2

I rms to Zero leaving

2

I

M

1

2

1

T

T

0

1 dt

 1

I

M

1

2 T

 t



1



T

0

1 2

I

M

1

2 T

T

1

1 2

I

M

2

Engineering-43: Engineering Circuit Analysis

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Bruce Mayer, PE

Sinusoidal rms Alternative

 For a Sinusoidal Source

Driving a Complex

( Z = R + jX) Load

P av

1

2

V

M

2

, res

R

 If the Load is Purely

Resistive

P av

1 V

2

M

1

2

RI

M

2 R 2

 Now, By the “effective”

P av

V

M

2

1

2

V

2

2

M

V

2 dc

R

 2

V rms

R

V rms

V

M

V eff

R

2

R

2

V rms

2

0 .

707 V

M

1

2

2

RI

M , res

 Similarly for the rms

P av

Current

1

2

I

M

R

2

2

I

M

2

2

I rms

2

I dc

R

 2

I eff

R

 2

I rms

R

I rms

I

M

2

0 .

707 I

M

Engineering-43: Engineering Circuit Analysis

21

Bruce Mayer, PE

Sinusoidal rms Values cont

 In General for a

Sinusoidal Quantity u ( t )

U

M cos(

 t

 

) and the effective value is

U eff

U rms

U

M

2

 For the General,

P

1

2

V

M

I

M cos(

 v

  i

)

22

V

M

I

M cos(

 v

  i

2 2

 By the rms Definitions

Engineering-43: Engineering Circuit Analysis

)

P av

V rms

I rms cos(

  v

 i

)

 Thus the Power to a

Calculated using These

Quantities as Measured at the SOURCE

• Using a True-rms DMM

– The rms Voltage

– The rms Current

• Using an Oscilloscope and “Current Shunt”

– The Phase Angle

Difference

Bruce Mayer, PE

Example

 rms Voltage

 Given Voltage

Waveform Find the rms Voltage Value

T

 Find The Period

• T = 4 s

 Derive A Math Model for the Voltage WaveForm

Engineering-43: Engineering Circuit Analysis

23

 During the 2s Rise Calc the slope

• m = [4V/2s] = 2 V/s

 Thus The Math Model for the First Complete

Period v

 

 0

2 t

2

0

 t t

 Use the rms Integral

4

2

U rms

T

1 t

0 t

T

0 u

2

( t ) dt

Bruce Mayer, PE

Example

 rms Voltage cont.

 Calc the rms Voltage

T

V rms

1

4

4

0 v

2

  dt

 Numerically

1

4

2

0

( 2 t )

2 dt

1

4

4

2

0 dt

0

V rms



1

3 t

3



0

2

8

3

( V )

1 .

633 V

Engineering-43: Engineering Circuit Analysis

24

Bruce Mayer, PE

Example

Average Power

 Given Current

Waveform Thru a 10 Ω

Resistor , then Find the

Average Power

 Find The Period

• T = 8 s

 Apply The rms Eqns

U rms

T

1 t

0 t

T

0 u

2

( t ) dt P av

I

2 rms

R

 The “squared” Version

2

I rms

1

8 s

2

0 s

4

2 dt

 

6

4 s s

 

2 dt

8 A

2

 Then the Power

P av

 2

I rms

R

8 A

2 

10

 

80 W

Engineering-43: Engineering Circuit Analysis

25

Bruce Mayer, PE

Sinusoidal Forcing Functions

 Consider the Arbitrary

LINEAR Ckt at Right.

 If the independent source is a sinusoid of constant frequency then for ANY variable in the LINEAR circuit the

Response will be

SINUSOIDAL and of the SAME

FREQUENCY

26

 Mathematically

Engineering-43: Engineering Circuit Analysis v ( t ) i

SS

( t )

V

M

 sin(

A sin(

 t t

 

 

)

)

 Thus to Find i ss

(t), Need

ONLY to Determine

Parameters A &

Bruce Mayer, PE

Example

RL Single Loop

 Given Simple Ckt Find i(t) in Steady State

 Write KVL for Single

Loop v

Ri ( t )

L di i ( t

)

A cos(

 t

 

) dt using cos(x

 i ( t ) i ( t ) i ( t )

 y)

 cosx

 cosy sinx

A

A cos cos

 cos

 cos

 t t

A

 sin

A

 sin sin

 t sin

A

1 cos

 t

A

2 sin

 t

 siny

 t di

( t )

 

A

1

 sin

 t

A

2

 cos dt

Engineering-43: Engineering Circuit Analysis

 t

27

 Sub Into ODE and

Rearrange

V

M

(

 cos

L

A

1

 t

( L

A

2

RA

2

RA

1

)

) sin cos

 t t

Bruce Mayer, PE

Example

RL Single Loop cont

 Recall the Expanded

V

M cos

(

L

A

1

 t

ODE

RA

2

) sin

 t

( L

A

2

RA

1

) cos

 t

 Equating the sin & cos

PreFactors Yields

L

A

1

L

A

2

RA

2

RA

1

V

M

0

 Recognize as an

ALGEBRAIC Relation for 2 Unkwns in 2 Eqns

Engineering-43: Engineering Circuit Analysis

28

 Solving for

Constants A

A

1

R

2

RV

(

M

L )

2

,

1

 Found A

1

& A

2

ONLY Algebra and A

A

2

R

2

2

LV

M

(

L )

2 using

• This is Good

Bruce Mayer, PE

Example

RL Loop cont.2

 Using A

1

& A

2 i the i(t) Soln

( t )

A

1 cos

 t

State

A

2 sin

 t

A

1

R

2

RV

M

(

L )

2

 Also the Source-V v ( t )

V

M cos

 t

A

2

R

2

LV

M

(

L )

2

 Would Like Soln in

Form

 Comparing Soln to

Desired form → use sum-formula Trig ID

A cos( x

 i ( t )

A cos(

 t

 

) y )

A cos x

 cos y

 If in the ID

=x, and

ωt = y, then

A cos

 

A

1

R

2

RV

(

M

L )

2

A sin

 

A

2

R

2

LV

M

(

L )

2

A sin x

 sin y

Engineering-43: Engineering Circuit Analysis

29

Bruce Mayer, PE

Example

RL Loop cont.3

 Dividing These Eqns

Find

A sin

 

L tan

  cos

 

A

 Now

A cos

  

A sin

 

2

A

2

 cos

2

  sin

2

A

2

R

30

 Find A to be

A

R

2

V

M

(

L )

2

 Subbing for A &

 in

Solution Eqn

Engineering-43: Engineering Circuit Analysis i ( t )

R

2

V

M

(

L )

2 cos

 t

 tan

1

L

R

 Elegant Final Result,

But VERY Tedious Calc for a SIMPLE Ckt 

• Not Good

Bruce Mayer, PE

Complex Exponential Form

 Solving a Simple,

One-Loop Circuit Can

Be Very Tedious for

Sinusoidal Excitations

 To make the analysis simpler relate sinusoidal signals to COMPLEX

NUMBERS.

• The Analysis Of the

Converted To Solving

Systems Of Algebraic

Equations ...

Identity (Appendix A) e j

  cos

  j sin

• Where j

 

1

 Note:

• The Euler Relation can

Be Proved Using Taylor’s

Series (Power Series)

Expansion of e j

Engineering-43: Engineering Circuit Analysis

31

Bruce Mayer, PE

Complex Exponential cont

 Now in the Euler

Identity, Let

   t

 So e j

 t  cos

 t

 j sin

 t

 Next Multiply by a

V

M e

Constant Amplitude, V

M j

 t 

V

M cos

 t

 jV

M sin

 t

32

 Separate Function into Re al and Im aginary

Parts

Engineering-43: Engineering Circuit Analysis

Re

Im

V

M

V

M e e j

 t j

 t

V

M

V

M cos

 t sin

 t v

 Notice That if

V

M cos

 t v

Then

Re

V

M e j

 t

V

M cos

 t

 Now Recall that

LINEAR Circuits Obey

SUPERPOSITION

Bruce Mayer, PE

Complex Exponential cont.2

 Consider at Right The

Linear Ckt with Two

Driving Sources

 By KVL The Total V-src v

 

Applied to the Circuit

 v

1

 

 v

2

 

V

M cos

 t

 jV

M sin

 t

V

M e j

 t

 Now by SuperPosition

The Current Response to the Applied Sources

Engineering-43: Engineering Circuit Analysis

33 v

1

V

M cos

 t

General

Linear

Circuit v

2

 jV

M sin

 t i

     

I

M

1 cos

  t

2

  

I

M e j

  t

  

 jI

M sin

  t

  

 This Suggests That the…

Bruce Mayer, PE

Complex Exponential cont.3

I

M

 Application of the

Complex SOURCE will

Result in a Complex

RESPONSE From

Which The REAL

(desired) Response

Can be RECOVERED;

That is cos

  t

  

Re

I

M e j

  t

  

 Thus the To find the

Response a COMPLEX

Source Can Be applied.

Engineering-43: Engineering Circuit Analysis

34 v

1

V

M cos

 t v

2

 jV

M sin

 t

General

Linear

Circuit

 Then The Desired

Response can Be

RECOVERED By Taking the REAL Part of the

COMPLEX Response at the end of the analysis

Bruce Mayer, PE

Realizability

 We can NOT Build

Physical (REAL)

Sources that Include

IMAGINARY Outputs

35

V

M e j

 t jV

M sin

 t

 We CAN, However,

BUILD These

V

M cos

 t

V

M sin

 t

Engineering-43: Engineering Circuit Analysis v

1

V

M cos

 t

General

Linear

Circuit v

2

 jV

M sin

 t

 We can also NOT invalidate Superposition if we multiply a REAL

Source by ANY

CONSTANT Including “j”

 Thus Superposition Holds, mathematically, for

V

M e j

 t 

V

M cos

 t

 jV

M sin

 t

Bruce Mayer, PE

Example

RL Single Loop

36

COMPLEX forcing

Function, and Recover the REAL Response at

The End of the Analysis

• Let v ( t )

V

M e j

 t

 In a Linear Ckt, No

Circuit Element Can

Change The Driving

Frequency, but They

May induce a Phase

Shift Relative to the

Driving Sinusoid

Engineering-43: Engineering Circuit Analysis v ( t )

V

M e j

 t

 Thus Assume Current i ( t )

Response of the Form

I

M e j

  t

  

I

M e j

 e j

 t

 Then The KVL Eqn v ( t )

Ri ( t )

L di dt

( t )

Bruce Mayer, PE

Example – RL Single Loop cont.

 Taking the 1 st Time di dt

Derivative for the

Assumed Solution d dt

I

M e j

  t

  

 j

I

M e

 Then the Right-Handv ( t ) j

  t

  

V

M e j

 t

 Then the KVL Eqn

Side (RHS) of the KVL

( j

L

R ) I

M e j

 e j

 t 

V

M e j

 t

 di

L ( t

L dt

 j

I

)

M

(

( j

L j

L

Ri ( t ) j

  t

   e

R

R

)

)

I

I

M

M e e

 j j

 e

R

  t

   j

 t

I

M e j

  t

  

 Canceling e j

 t and

Solving for I

M e j

I

M e j

 

V

M j

L

R

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

37

Example – RL Single Loop cont.2

 Clear Denominator of

The Imaginary

Component By

Multiplying by the

Complex Conjugate

I

M e j

 

R

V

M

 j

L

R

R

 j

L j

L

I

M e j

 

V

M

R

2

( R

 j

L )

(

L )

2

 Then the Response in

Rectangular Form: a+jb

Engineering-43: Engineering Circuit Analysis

38 v ( t )

V

M e j

 t

R

2

V

M

R

(

L )

2

R

2

V

L

M

(

L )

2

 a

 b

 Next A & θ

Bruce Mayer, PE

Example – RL Single Loop cont.3

 First A

A

 a

2  b

2 

V

M

R

 

V

M

L

R

2 

(

L )

2

2

2

R

2

V

M

 

2

L

2

39

 And θ

  tan

 tan

1 tan

1

1

L R

 

L R

Engineering-43: Engineering Circuit Analysis v

R

R

( t

2

2

)

V

M

R

(

L )

V

V

M

(

L

L )

M

2

2 e

 j

 a b t

 A &

Correspondence with Assumed Soln

• A → I

M

• θ → 

 Cast Solution into

Assumed Form

Bruce Mayer, PE

Example – RL Single Loop cont.4

I

M

 The Complex

Exponential Soln e j

  e j 

 tan

1

L

R



R

2

V

M

 

2 v ( t )

V

M e j

 t

 Where

I

M

R

2

V

M

 

2

,

   tan

1

L

R

 Finally RECOVER the

DESIRED Soln By

Taking the REAL Part of the Response

 Recall Assumed Soln i

( t

I

)

M

I

M cos

 e

 j

 t e

 j

 t

I j

M e sin j

  t

  

 t

   

Engineering-43: Engineering Circuit Analysis

40

Bruce Mayer, PE

Example – RL Single Loop cont.5

 By Superposition v ( t )

V

M e j

 t v ( t

)

 i ( t )

V

M

 cos

Re{ I

M t e j

Re{ V

M

  t

  

} i ( t )

I

M cos(

 t

 

) e j

 t

} i

 

 Explicitly

R

2

V

M

(

L )

2 cos

 SAME as Before tan

1

L

R

41

Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE

Phasor Notation Imaginary

 If ALL dependent

Quantities In a Circuit

(ALL i’s & v’s) Have The

SAME FREQUENCY,

Then They differ only by

Magnitude and Phase x ( t )

• That is, With Reference to the Complex-Plane

Diagram at Right, The dependent Variable

Takes the form

Ae j

  t

   

  e j

 t b

A

 a

Real

 Borrowing Notation from

Vector Mechanics The

Frequency PreFactor Can

Be Written in the

Shorthand “Phasor” Form

Ae j

 

A

 

Engineering-43: Engineering Circuit Analysis

42

Bruce Mayer, PE

Phasor Characteristics

 Since in the Euler Reln

The REAL part of the expression is a

COSINE, Need to express any SINE

Function as an

Equivalent CoSine

A

• Turn into Cos Phasor cos(

 t

 

)

A

   and the Trig ID

A sin(

)

A cos(

 

90

)

So

A sin(

 t

 

)

A

   

90

Engineering-43: Engineering Circuit Analysis

43

 Examples v ( t )

12 cos( 377 t

425

)

12

 

425

 y ( t )

18 sin( 2513 t

4 .

2

)

18 cos( 2513 t

4 .

2

 

90

)

18

 

85 .

8

 Phasors Combine As the Complex Polar

Exponentials that they are

( V

1

 

1

)( V

2

 

2

)

V

1

V

2

(

1

 

2

)

V

1

V

2

1

2

V

V

2

1

(

1

 

2

)

Bruce Mayer, PE

Example

RL Single Loop

44

 As Before Sub a

Complex Source for the

Real Source

 The Form of the

Responding Current

I

I

M

  i

I

M e j

 e j

 t

• Phasor Variables

Denoted as BOLDFACE

CAPITAL Letters

 Recall The KVL

L di

( t )

Ri ( t )

 dt

Engineering-43: Engineering Circuit Analysis v

V v

V

M

0

V

M e j

 t

 For the Complex

( j

L

Quantities

R ) I

M e j

 e j

 t

V

M e j

 t

 Then The KVL Eqn in j

L I the Phasor Domain

R I

V

 as I

I

M e j

I

R

V j

L

Bruce Mayer, PE

Example

RL Single Loop cont.

 To recover the desired

Time Domain Solution

I

Substitute

V

V

M

0

I

M

   i

 v

Ve j

 t

I

M e j

 e j

 t

45

 Then by Superposition i

Take

 

Re

 

I

M e j

 e

I

Re I cos j

  t

  

M

 e

 t

  

M

 j

 t

Engineering-43: Engineering Circuit Analysis

 This is A LOT Easier

Than Previous Methods

 The Solution Process in the Frequency Domain

Entailed Only Simple

Algebraic Operations on the Phasors

Bruce Mayer, PE

Resistors in Frequency Domain

 The v-i Reln for R v if

( t i

)

 

Ri ( t

I

M

) cos

  t

  

RI

M

 e

( j

 t

 

) 

V

M e

( j

 t

 

)

Then

V

M

  

RI

M

 

46

 Thus the Frequency

Domain Relationship for

Resistors

V

R I

Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE

Resistors in

-Land cont.

 Phasors are complex numbers. The Resistor

Model Has A Geometric

Interpretation

 In the Complex-Plane

The Current & Voltage

Are CoLineal

• i.e., Resistors induce

NO Phase Shift

Between the Source and the Response

• Thus resistor voltage and current sinusoids are said to be “IN PHASE”

Engineering-43: Engineering Circuit Analysis

47

R

→ IN Phase

Bruce Mayer, PE

Inductors in Frequency Domain

 The v-i Reln for L

V

M e

 j

 j (

 t

  v

)

LI

M e j

 j

 t d

L

  dt

 i

I

M e j

  t

  i

 or

V

M e j

 v

 j

LI

M e j

 i

 Thus the Frequency

Domain Relationship for

Inductors

V

 j

L I

Engineering-43: Engineering Circuit Analysis

48

Bruce Mayer, PE

Inductors in

-Land cont.

 The relationship between phasors is algebraic.

j

• To Examine This Reln

Note That

 

1

1

90

  e j 90

 Therefore the Current and Voltage are OUT of

PHASE by 90 °

• Plotting the Current and

Voltage Vectors in the

Complex Plane

49

 Thus V

 j

L I

V

M e j

LI v

M

 

LI

  M e

  e j

 i j

 i j

  i

90

V

M e j

 v

 

LI

V

 

L I

Engineering-43: Engineering Circuit Analysis

90

M e

Bruce Mayer, PE

Inductors in

-Land cont.2

 In the Time Domain

L → current LAGS

 Phase Relationship

Descriptions

• The VOLTAGE LEADS the current by 90 °

• The CURRENT LAGS the voltage by 90 °

Engineering-43: Engineering Circuit Analysis

50

 Short Example

Given

L

20 mH , v ( t )

12 cos( 377 t

20

)

Find i ( t )

 

377

V

12

20

I

V j

L so with

I

12 j

 

 

L V

1

90

20

A

90

I

377

12

20

10

3

 

70

( A )

 In the Time Domain i ( t )

1 .

593 A

 cos( 377 t

70

)

Bruce Mayer, PE

Capacitors in Frequency Domain

 The v-i Reln for C

I

M e

( j

 j

 t

 

CV

M i

) e

 j

  d

C t

  dt

 v

V

M e j

  t

  v

 or

I

M e j

 i

 j

CV

M e j

 v

 Thus the Frequency

Domain Relationship for

I

Capacitors

 j

C V

Engineering-43: Engineering Circuit Analysis

51

Bruce Mayer, PE

Capacitors in

-Land cont.

 The relationship between phasors is algebraic.

• Recall j

 

1

1

90

  e j 90

 Therefore the Voltage and Current are OUT of

PHASE by 90 °

• Plotting the Current and

Voltage Vectors in the

Complex Plane

 Thus

I

M e j

 i

 j

CV

M e j

 v

 e

 j 90

 

CV

M

CV

M e j

  v

I

 

C V

90

 e j

 v

90

Engineering-43: Engineering Circuit Analysis

52

Bruce Mayer, PE

Capacitors in

-Land cont.2

 In the Time Domain

 Phase Relationship

Descriptions

• The CURRENT LEADS the voltage by 90 °

• The VOLTAGE LAGS the current by 90 °

Engineering-43: Engineering Circuit Analysis

53

 Short Example

Given

C

100

F , v ( t )

100 cos( 314 t

15

)

Find i ( t )

 

314

I

V

100

15

 j

C V

I with j

1

90

 

C

1

90

 

100

15

I

314

100

10

6 

100

105

( A )

 In the Time Domain i ( t )

3 .

14 A

 cos( 314 t

105

)

Bruce Mayer, PE

All Done for Today

Square

rms voltage = 0.707 peak voltage peak voltage = 1.414 rms voltage average voltage = 0.637 peak voltage

Engineering-43: Engineering Circuit Analysis

54 rms voltage = 1.11 average voltage peak voltage = 1.57 average voltage average voltage = 0.9 rms voltage

Bruce Mayer, PE

WhiteBoard Work

 Let’s Work Text

Problem 8.5

i(t)

+ v(t)

_

2

Engineering-43: Engineering Circuit Analysis

55

Bruce Mayer, PE

Complex No.s

Bruce Mayer, PE

BMayer@ChabotCollege.edu

Engineering-43: Engineering Circuit Analysis

56

Bruce Mayer, PE

57

Complex Numbers Reviewed

Imaginary

 Consider a General

Complex Number n

 a

 jb

 This Can Be thought of as a VECTOR in the

Complex Plane b

 Where

A

 a

Real

 This Vector Can be

Expressed in Polar j

 

1

 j

 j

 

1

(exponential) Form Thru the Euler Identity a

 jb

Ae j

A (cos

  j sin

Engineering-43: Engineering Circuit Analysis

)

 Then from the Vector Plot

A

 a

2  b

2

  tan

1 b a

Bruce Mayer, PE

Complex Number Arithmetic

 Consider Two Complex n

Numbers

 a

 jb

Ae j

 m

 c

 jd

De j

 The PRODUCT n•m n

 m ac

 ac

 j

 a

 jb

 c

 bc bd

  bc

 jd j

2

 n

 m

Ae j

 

De j

 

    

 The SUM, Σ, and

DIFFERENCE,

, for these numbers n

 m

 n

 m

 a a

 c c

 

  b b

 d d

Engineering-43: Engineering Circuit Analysis

58

 Complex DIVISION is

Painfully Tedious

• See Next Slide

Bruce Mayer, PE

Complex Number Division

 For the Quotient n/m in

Rectangular Form n m

 a

 c

 jb jd

 The Generally accepted

Form of a Complex

Quotient Does NOT contain Complex or

Imaginary

DENOMINATORS

Engineering-43: Engineering Circuit Analysis

59

 Use the Complex

CONJUGATE to Clear the Complex

Denominator n m

 a c

 jb jd

 c c

 jd jd n

 ac

 j

 bc dc

 j 2 bd m c

2  j

  cd d

2 n m

 ac

 bd c

2

 j

2 bc

 d

 The Exponential Form is

Cleaner

• See Next Slide

Bruce Mayer, PE

Complex Number Division cont.

 For the Quotient n/m in Exponential Form n m

Ae j

 j

De

A

D

 e j

    

 However Must Still Calculate the Magnitudes A & D...

Engineering-43: Engineering Circuit Analysis

60

Bruce Mayer, PE

Phasor Notation cont.

61

 Because of source superposition one can consider as a SINGLE source, a System That contains REAL and

IMAGINARY

Components u ( t )

U

M cos(

Re

U

M e j

  e j

 t t

 

)

Response Of Any

Circuit Variable Will Be

Of The Form

Engineering-43: Engineering Circuit Analysis y ( t )

Y

M cos(

 t

 

)

 Or by SuperPosition

Re{ U

M e j

  e j

 t

}

Re{ Y

M e j

  e j

 t

}

 Since e j

 t is COMMON To all Terms we can work with ONLY the PreFactor that contains Magnitude u ( t )

U and Phase info; so t

 

)

M cos(

U

 y ( t )

U

M

  

Re{ Y }

Y

Y

M

Y

M

 cos(

 t

 

)

Bruce Mayer, PE