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Lecture 15Lecture 15Sinusoidal Signals, Complex
N b PhNumbers, Phasors
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Steady-State Sinusoidal Analysis1. Identify the frequency, angular frequency,
peak value rms value and phase of apeak value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and complex impedancesphasors and complex impedances.
3 Compute power for steady state ac3. Compute power for steady-state ac circuits.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Steady-State Sinusoidal Analysisy y
4. Find Thvenin and Norton equivalent circuits.
5. Determine load impedances for maximum power transfermaximum power transfer.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Sinusoidal Currents and Voltagesg)cos()( += tVtv m
Vm is the peak value
is the angular frequency in radians per q y psecond
i h h l is the phase angle
T is the period
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
T is the period
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)cos()( += tVtv m
)()( mHambley mixes units; t in radians, in degrees
Frequency
==sec
1 cyclesHzT
f
2 radians
A f
=secT
Angular frequency
f 2=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
( ) ( )o90cossin = zz
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Root Mean Square (RMS) Valuesq ( )
tVtV m )cos()( += The average value of tVtV m )cos()( + cos(t+) is zero. This is the squared qversion of the signal, and its mean value is .
tVTR
dtR
tVT
pdtT
PT
m
Tm
T
Average0
22
0
22
0
)(cos11)(cos11
+=+==
RVrms
2000
=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
R
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Root Mean Square (RMS) Valuesq ( )
( )dttvT
VT
2rms
1 = ( )dttiTIT
2rms
1 =( )T 0rms ( )T 0rms VP
2rms
avg = RIP 2rmsavg =Ravg rmsavg
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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RMS Value of a Sinusoid
V2rmsmVV =
The rms value for a sinusoid is the peak l di id d b th t f t Thivalue divided by the square root of two. This
is not true for other periodic waveforms such as square waves or triangular waves.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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AC Power in the US
60-Hz 115-V power is a RMS value for60-Hz 115-V power is a RMS value for the voltage. The peak voltage is:
VVV RMSm 16321152 ===
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Power Delivered to a 50 Resistor
)100cos(100)( ttv = 100
50)71.71( 22 W
RvP rmsavg ===
100
m
m
vv
Vv
==
50)100(cos100)()(
50222 t
Rtvtp
R==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
2RMSv = )100(cos200 2 t=
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Phasor Definition
( ) ( )111 cos :function Time tVtv +=( ) ( )111111 :Phasor V =V
rePhas rePhas
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Phasor Arithmetic66.85
2310
2110)60sin(10)60cos(1060101 jjjZ +=+=+== ooo
54.354.3225
225)45sin(5)45cos(54552 jjjZ +=+=+== ooo
10550455601021 ZZ == ooo
1524556010
2
1
ZZ =
= ooo
2.1254.854.354.366.8521 jjjZZ +=+++=+
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
12.546.154.354.366.8521 jjjZZ +=+=
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Eulers Formula
+= )sin()cos(j je
=+=
)sin()cos()sin()cos(
j
j
je
je
=)I ()i (
)Re()cos(j
je
=+==
1)(sin)(cos
)Im()sin(22j
j
e
e
= 1)()(
je
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Eulers Formula
)sin()cos( += j je1)sin()cos(
)sin()cos(
=+=+=
j
j
je
je
11lim....028759045235367182818284.2 +==n
e
....264389793238461415926535.3
1lim....028759045235367182818284.2
= +
n n
e
1
1
==
jej
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Adding Sinusoids Using Phasorsg g
Step 1: Determine the phasor for each termStep 1: Determine the phasor for each term.
Step 2: Add the phasors using complex arithmetic.Step 2: Add the phasors using complex arithmetic.
Step 3: Convert the sum to polar form.Step 3: Convert the sum to polar form.
Step 4: Write the result as a time function.p
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Using Phasors to Add Sinusoids( ) ( )o45cos201 = ttv ( ) ( )( )
o60sin102 += ttv ( )( )ooo
3010
9060cos10 += t( )o30cos10 = to4520V o45201 =V
o3010VELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
30102 =V
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Using Phasors to Add Sinusoidsg
45201 = oV)45sin(20)45cos(20
1
j += oo
14.1414.142220
2220 jj ==
30102 = oVj
5668110310
)30sin(10)30cos(10 += oo
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
jj 566.82110
2310 ==
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Using Phasors to Add Sinusoids
566.814.1414.1421s
+=+= VVV
jj
g
14.1980.22 = jjj
8.29)14.19()80.22( 22
=+=Vs
o01.408.2214.19
8.2214.19)( 1 =
== TanTan s
o01.408.29s =V
( ) ( )ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
( ) ( )o01.40cos8.29 = ttvs
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Exercise 5.4
)90cos(10)cos(10
)sin(10)cos(10)(1o+=
+=tt
tttv
[ ][ ]00110
)90sin()90cos()0sin()0cos(109010010
++=+++=
+=
jjj
V
[ ][ ]110
00110=
++j
j
45)1(11)(
14.142101010
1
22
o=====+=
TanTan V
4514.141
o=V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
)45cos(14.14)(1o= ttv
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Exercise 5.4
6053010
)60cos(5)30cos(10
)30sin(5)30cos(10)(
oo
oo
oo
+=++=
+++=tt
ttti
V[ ] [ ]
)23
215
21
2310
)60sin()60cos(5)30sin()30cos(106053010
+
+=
+++=+=
jj
jjV
67.02.1133.45.2566.8
)2222
+=++=
jjj
jj
67.067.0
2.11)67.0()2.11(
1
22
o=+=V
42.32.11
42.32.11
67.02.11
67.0)( 1
o
o
==
== TanTan
V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
)42.3cos(2.11)( o+= tti
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Exercise 5.4
6015020)60cos(15)cos(20
)60cos(15)90sin(20)(2o
oo
+=
++=tt
ttti
V[ ] [ ][ ] 31
)60sin()60cos(15)0sin()0cos(206015020
+++=
+=jj
V
[ ]135.27135.720
23
21150120
=+=
++=
jj
jj
1313
4.30)13()5.27( 22
=+=V
3.255.27
135.27
13)( 1 =
== TanTan
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
)3.25cos(4.30)(2o= tti
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Phase RelationshipsSinusoids can be visualized as the real-visualized as the realaxis projection of vectors rotating in thevectors rotating in the complex plane. The phasor for a sinusoidphasor for a sinusoid is a snapshot of the corresponding rotatingcorresponding rotating vector at t = 0.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Phase RelationshipsTo determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. Then when standing at aThen when standing at a fixed point, if V1 arrives first followed by V2 after a rotation of we say that Vrotation of , we say that V1leads V2 by . Alternatively, we could say that V2 lags V1by . (Usually, we take as the smaller angle between the two phasors.)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
wo p so s.)
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Phase RelationshipsTo determine phase relationships betweenrelationships between sinusoids from their plots versus time, find the shortest time interval tpbetween positive peaks of th t f Ththe two waveforms. Then, the phase angle is = (tp/T ) 360 If the peak of v1(t) 360 . If the peak of v1(t)occurs first, we say that v1(t)leads v2(t) or that v2(t) lags
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
2( ) 2( ) gv1(t).
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Complex Impedances-Inductorp p
mL titi += )sin()( m
LL
mL
tLidt
tdiLtv
titi
+== )cos()()()s ()(
i
dt
I = 90LLmL
mL
LLjZLjLLi
iIIV
I
===
=
90)90(
90
L LLjZ == 90
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
LLL Z IV =
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Complex Impedances InductorComplex Impedances-Inductor
LL Lj IV = o90== LLjZL jL
Z IV LLL Z IV =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Complex Impedances-Capacitor
Z IV
p p p
CCC Z IV =o90111 ===
CCjCjZC CCjC
RIV RR RIV =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Impedances-Resistorp
RR RIV =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Exercise 5.5
o
o
)30(10)(
)30cos(10)(1 =tt
ttv
oo
o
)45cos(10)45sin(10)(
)30cos(10)(
3
2
=+=+=
tttv
ttv
o30101 =V V1 lags V2 by 60 Find the phasor for each voltage and state phase relationship between them
o
o
4510
30102
1
+=
V
V1 2
V2 leads V3 by 75 V l d V b 15 45103 =V V1 leads V3 by 15
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Exercise 5.6
250)200cos(100
==L
HLtv Find the impedance of the inductance, the phasor current and the phasor
voltage. Draw the phasor diagram.
905050)250)(200(0100
25.0
=
=L
jHjLjZ
HLV
voltage. Draw the phasor diagram.
90290500100
905050)25.0)(200(
===
====L
L
L
Z
jHjLjZVI
9050LL Z
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Exercise 5.7
100)200cos(100
==c
FCtv
Find the impedance of the capacitor, the phasor current and the phasor voltage. Draw the phasor diagram.
1110100
100
=
=C
j
FCV
voltage. Draw the phasor diagram.
9020100
9050102)10100)(200(
11126
=====
C
C xj
xj
Cj
CjZ
V
90290500100 +=
==C
CC Z
VI
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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Exercise 5.8
50)200cos(100)(
==R
Rttv Find the impedance of the resistor, the
phasor current and the phasor voltage. Draw the phasor diagram.
0500100
50
=
=R
RZ
RV
Draw the phasor diagram.
020100050
=====
CR
R RZVI 02
050RR Z
I
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.
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