Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Outline – AC Steady State
 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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
 IMPEDANCE AND ADMITANCE
• 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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Sinusoids
 Recall From Trig the
Sine Function
5
 For the RADIAN Plot
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 
Radian, or Angular,
Frequency in rads/sec
• ωt 
Sinusoid argument in radians (a pure no.)
 The function Repeats every 2 π; x mathematically
(
 t

2

)
 x (
 t )
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Sinusoids cont.3
 Now Consider the
GENERAL Expression for a Sinusoid x ( t )

X
M sin
  t
  
 Where
• θ  “Phase Angle” in
Radians
 Graphically, for
POSITIVE θ
" leads by  "
" lags by  "
Engineering-43: Engineering Circuit Analysis
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Leading, Lagging, In-Phase
 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
LEADS x
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

1.4
1.2
1.0
0.8
0.6
0.4
   
105 mS
0.2
0.0
-0.2
-0.4
x1 LEADs x2
-0.6
-0.8
-1.0
x1 (V or A)
-1.2
x2 (V or A)
-1.4
0.0
0.2
file =Sinusoid_Lead-Lag_Plot_0311.xls
0.4
Sinusoid Phase Difference

1 Period
900 mS

360

1 Period
Out of Phase
* 0.733 rads
* 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
• 
= 6.981 rad/s
• 
= 1.257 rads = 72°
• 
= 0.5236 rads = 30°
1.4
1.6
x2 LAGs x1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt
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
 
)
 
)
 Additional Relations sin(
 cos(




)

)

 sin
 cos
 cos
 cos

 cos

 sin
 sin sin
 cos(
 sin(


 

)
)

 cos sin

 cos cos



 sin cos

 sin sin



2

 radians

360

(degrees)


180
 rads

(rads)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
13
 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 °
LEADS v
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Sinusoid Phase Difference Example 7.2
12
8
4
0
-4
120°
-8
-12
0.000
0.002
file =Sinusoid_Lead-Lag_Plot_0311.xls
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt
0.018
0.020

Effective or rms Values
 Consider Instantaneous
Power For a Purely i ( t )
Resistive Load
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
 Since a Resistive Load
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
Maker and Adjust the
Amplitude for the Same
Time → Effective
Voltage of 12V
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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”
Definition for a R Load
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
Complex-Load Case av

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
Reactive Load Can be
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
STEADY-STATE
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
 In Steady State Expect
)

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
Steady State Will Be
Converted To Solving
Systems Of Algebraic
Equations ...
 Start with Euler’s
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Example

RL Single Loop
36
 This Time, Start with a
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Capacitors in

-Land cont.2
 In the Time Domain
C → current LEADS
 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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

All Done for Today
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

WhiteBoard Work
 Let’s Work Text
Problem 8.5
i(t)
+ v(t)
_
2

Engineering-43: Engineering Circuit Analysis
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
 ad
  bc



 jd j
2
 ad bd
 n
 m

Ae j
 
De j
 
ADe 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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
 ad



 j 2 bd m c
2  j

  cd d
2 n m

 ac
 bd c
2

 j
2 bc
 ad
 d
 The Exponential Form is
Cleaner
• See Next Slide
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt

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

 
)
 The Real Steady State
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
BMayer@ChabotCollege.edu • ENGR-43_Lec-08-1_AC-Steady-State.ppt