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