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Chapter 5Steady-State Sinusoidal Analysis
1 . Identify the frequency, angular frequency, peak value,rms value, and phase of a sinusoidal signal.
2 . Solve steady-state ac circuits using phasors andcomplex impedances.
3 . Compute power for steady-state ac
4 . Find Th venin and Norton equivalent circuits.
5 . Determine load impedances for maximum powertransfer.
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5. Steady -State Sinus oid al An alysis
* In most circuits, the t r ans ien t respon se (i.e., thecomplimentary solution) decays rapidly to zero, the s teady- s ta te respon se (i.e., the forced response or the particular
solution) persists.* In th is c hapter, we learn eff ic ient method s fo r f indin g th e
s teady-s ta te respon ses for s inu soid al sou rces .
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3
5. Steady -State Sinus oid al An alysis
5 .1 Sinuso idal Currents and Vol tages
5.1.1 Phase and Phase A ng le
* Consider the sinusoidal voltage as shown,
f 2 T
2
T
1 f : ) s (or H zin frequencythe
s,i n period thei s T , 2 T : have we
cycle per by 2 increases angletheSince
degree) i n (usuallyanglephasethei s rad/s in frequencyangular thei s
voltage theof valuePeak thei s V
where t cos V ) t ( v
1 -
m
m
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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage
5.1.1 Phase and Phase A ng le
* We usually use cos ine func t ion to model a sinusoidal signal.In case there is a sine function, we can use the following
conversion:
60 - i s (t) v of anglephasethethat saycan wethus and
60 t 200 cos 10
90 30 t 200 cos 10 t v
30 t 200 sin 10 t v
: example For
90) - cos(z sin(z)
x
x
x
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5
5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage 5.1.2 Roo t-Mean-Squ are (RMS) Values (or Effec tiv e Values )
R I P havewe,dt t i
T
1 I
: as current periodictheof valuerm s thedefinewei f Similarly,
R
V P havewe,dt t v
T
1 V
: as v(t) voltageperiodictheof value(rms) squre- m ean - root thedefiningBy
R / dt ) t ( v T 1 dt
R t v
T 1 dt t p
T 1
T E P
: i s period per poweaverageThe
dt t p E : i s period per delivered energythe
,R
t v t p bygiven i s resistance theto delivered power The
R.resistance ato T period with v(t) voltageperiodicaapplyingConsider
2 rms g av
T
0
2 rms
2 rms
g av
T
0
2 rms
2
T
0
2 T
0
2 T
0
T av
T
0 T
2
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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage
5.1.3 RMS Value of a Sinu so id
m m
rms
m rms m m
rms
2 m
T
0
2 m
T
0
2 2 m
T
0
2 rms
m
I 0.7071 2
I I Similarly,
H z 60155.5V),(V 110V : power l residentia : Note V 0.7071 2
V V
2 sin 2
1 2 T 2 sin
2
1 T
T 2
V
dt 2 t 2 cos 1 T 2
V
dt t cos V T
1 (t)dt v
T
1 V
t
cos V t v
: by given voltagesinusoidal aConsider
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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage Exam ple 5.1 Pow er del ivered to a res is tance by a s inus oidal sou rce
W t 200 cos 100100
W t 100 cos 200 50 t
100 cos 100 R t v t P
W 100 50
) 71 .70 ( R
V P
70.71V 2 / V V
m s 201/f T ,H 50 /2 f
t p an d
P power averagethe,V : F i n d
resistance 50ato applied i s
V t) 100cos(100 v(t) a: Given
2
2 2 2
2 2 rms
g av
m rms
z
av g r ms
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5. Steady -State Sinus oid al Analys is 5.2 Phasors
5.2.1 Definitio n
* A phasor is a vector in complex number plane that represents
the magnitude and phase of a sinusoid.* In ac circuit analysis, voltages and currents are usually
represented as phasors.
)90 ( I and I writecanwe
t sin I t iand t cos I t i for Similarly,
)90 ( V
therefore ,90 t cosV t v toit convert canwe
t sinV t v formtheof is sinusoid the If V :as phasor thedefinewe
t cosV t v formtheof voltagel sinusoidaa For
222111
222111
222
222
222
111
111
I I
V
V
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.1 Definitio n
* Eulers Identity :
* In phasor application:
) (Ae Re ) t Acos(
) t jAsin( ) t Acos( Ae
sin jA cos AAe
sin j cos e l exponentia complex
) t j(
) t j(
j
j
1 1 1 1 1 1
j
jwt
j ) t j(
V V as presented i s ) t cos( V (t) v
voltage sinusoidal theexam ple,F or
) t cos( A AAform phasor i ts sim plyor ) Re(Ae as presented i s )
t
cos( A
result,aAs .Ae term thedeletingby
Ae form theto simplified i s Ae n ,applicatio circuit In
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5. Steady -State Sinus oid al Analys is 5 .2 Phasor s 5.2.4 Phaso r vs . Sinu so ids
* The phasor is simply a snapshot of arotating vector at t=0 .
m
mm
mt j
m
t j
m
m
V
:as defined is v(t) for phasor The
V cosV v(0) 0,t at
and )t ( V eV eV Ret v
:vector rotating a of part real the just s It'
t cosV t v
: shownas voltage al sinusoid the Consider
V
)(Ae Re ) t Acos(
) t jAsin( ) t Acos( Ae
sin jA cos A Ae
j sin cose l exponentiacomplex
) t j(
) t j(
j
j
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.2 Phaso r Sum m ation
90 .9 t cos 54 .14
e 54 .14 Re t v
e e 54 .14 Re t v
e 54 .14 90 .9 54 .14 .5 2 j 33 .14 5 j 50 .2 j 33 .4 10 90 5 30 5 0 10
: number complextheConsider
e 90 5 30 5 0 10 Re t v
: form) polar in (or Identitys Euler' In e e 5 e 5 10 Re
e 5 e 5 e 10 Re t v
e 5 Re e 5 Re e 10 Re
90 t cos 5 30 t cos 5 t cos 10 t v form cosineto functions theal l rewritefirst We
) 90 t cos( 5 ) 60 t sin( 5t) cos( 10v(t)
voltages threeof summation theConsider
) 90 .9 t ( j
t j 90 .9 j
90 .9 j
t j
t
j 90 j 30 j
) 90 t ( j ) 30 t ( j t j
) 90 t ( j ) 30 t ( j t j
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.2 Phaso r Sum m ation
Now w e use Phasor nota t ion to s impl i fy ou r ca lcu lat ion :
Note: In us ing ph asors to add s inuso ids , a ll o f the t erms mu s thave the same f requency .
90.9t cos54.14t v
9.9014.54
90.954.14
5.2 j33.14
5 j50.2 j33.410
905305010
90t cos530t cos5t cos10t v
90t cos560t sin5t cos10t v
321
V
V V V V
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs Exam ple 5.2 Us ing phaso r s t o Add S inuso ids
7 .39t cos97 .29t v7 .3997 .29
14.19 j06 .23
5 j660.814.14 j14.1430104520
3010 and 4520 :are phasors The
(t)v(t)v(t)v Find
60t sin10t v
45t cos20t v that Suppose
s
21 s
21
21 s
2
1
V V V
V V
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.3 Fun dam ental Phaso r Operations
/2V : Root Square
-V
1
V
11 : Reciprocal
-V V
: DivisionV V :tion Multiplica
then ,V ,V If
111
11111
212
1
2
1
212121
222111
V
V
V
V V V
V V
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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.5 Phase Relat ions hip s
.60bylags or 60byleads
:that may sayweand 204 and 403 :as d represente be canThey
20t cos4t v and 40t cos3t v
:voltagestheConsider
1221
21
21
V V V V
V V
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5. Steady -State Sinus oid al Analys is 5 .3 Com plexImpedances
5.3.1 Impedance
* Impedance means com plex res i s t ance .
* The impedance concept is equivalent to stating that capacitors andinductors act as f requency -dependent res is to rs .
* By using impedances, we can solve sinusoidal steady-state circuitwith relatively ease compared to the methods of Chapter 4.
* Except for the fact that we use complex arithmetic, s inus o ida l s t eady- s ta te analys is i s the same as the analys is of res is t ive c i rcu i ts .
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es 5.3.2 Indu ct anc e
form. phasor in law sOhm' is This
Z :have We
90 L L j Z :as inductance the of impedance the Define
L j )90( I 90 L :as voltage Rewrite90byvoltage the lags current the : Note
V I L and )90( I : forms phasor their In
t cos I Ldt
t di Lt v
:inductor an through f low t sin I t i current a Consider
L L L
L
Lm L
mm Lm L
m L
L
m L
I V
I V
V I
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es 5.3.3 Capac itanc e
C C C
C
C m m C
m m C m C
m m C
C
m C
I Z V : have we
90 - C
1
C j
1 Z : as ecapacitanc of im pedancetheDefine
I 90 - C
1 90 I 90 - C
1 V V as voltageRewrite
90 byvoltageleads current the: Note
90 I 90 CV I , V V : forms phasor their In
) 90 t cos( CV ) t sin( CV - dt
dv C (t) i
capacitor aacross ) t cos( V (t) v voltageaConsider
d d l l l d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es
* EL I the IC E m an.* Impedance that are pure imaginary are called reactance .
5.3.4 Resis tanc e
R R
R
R :havewe
R Z asresistanceof impedancethe Define
I V
5 S d S Si id lA l i 5 3C l I d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es
5 S d S Si id lA l i 5 3C l I d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Qu iz - Ex erc is es 5.6, 5.7, 5.8
5 S d S Si id lA l i 5 3C l I d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Qu iz - Ex erc is es 5.6, 5.7, 5.8
5 St d St t Si id lA l i 5 3C l I d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Ad di t ional Example: represen t the c i rcu i t s how n in the
f requency do main us ing im pedances and p hasors .
5 St d St t Si id lA l i 5 3C l I d
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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Ad di t ional Example: represen t the c i rcu i t s how n in the
f requency do main us ing im pedances and p hasors .
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is
5 .4 Circui t A nalys is
* Im pedance c i rcui t analysis is the same as res is t ive c i rcui t analysis, we can directly apply KCL , KVL, nodal analys is ,
m esh analys i s ,
* The above phasor approach can only apply for steady st atewi th s inu so ids o f the same f requency .
0- 0(t)i-(t)i(t)i
:equations KCL
0 0t vt vt v
:equation KVL
321321
321321
I I I
V V V
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exam ple 5.3 S teady-State AC A nalys is of a Ser ies Circu i t* Find the steady-state current, the phasor voltage across each
element, and construct a phasor diagram.
15t 500cos707 .0t i
15707 .0454.141
30100 Z
454.141100 j10050 j150 j100 Z Z R Z
50 jC
1 j Z ,150 j L j Z ,30100
s
C Leq
C L s
V I
V
1054.3515707 .09050
C
1 j
751.106 15707 .090150 L j
157 .7015707 .0100 R
C
L
R
I V
I V
I V
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Exam ple 5.4 Ser ies /Paral le l Com bin at ion of Com plex Impedanc es
* Find the voltage across the capacitor, the phasor currentthrough each element, and construct a phasor diagram
901.09010018010
100 j18010
Z V
1801.0100
18010 RV
1351414.050 j509010
50 j50100 j9010
Z Z
t 1000cos10180t 1000cos10t v
1801050 j50100 j
4571.709010
Z Z Z
50 j50 Z
4571.704501414.0
01 )100 j( 11001
1 Z 1 R1
1 Z
100 jC
1 j Z ,100 j L j Z ,90-10
C
C C
C R
RC L
s
C
RC L
RC sC
RC
C RC
C L s
I
I
V I
V V
V
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exam ple 5.5 Steady -State AC No de-Voltage An alysis* Find the voltage at node 1 using nodal analysis
7 .29t 100cos1.16 t v or 7 .291.16 : for Solve
5.11.0 j2.0 j2 j2.0 j2.0 j1.0
05.15 j10 j
90-2 j5--
10
2node and node1 at KCL Write
11
1
21
21
122
211
V
V
V V
V V
V V V
V V V
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exerc is e 5.11 Steady-Sta te AC Mesh-Current An alys is* Solve for the mesh currents
5 S d S Si id lA l i 5 4Ci i A l i
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le :
)A51.87 - (3t Bcos i(t) when
Land B e min Deter
5 S d S Si id lA l i 5 4Ci i A l i
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le :
5 St d St t Si id lA l i 5 4Ci i tA l i
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is
Ad di t iona l Examp le :
5 St d St t Si id lA l i 5 4Ci i tA l i
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le : Comm erc ia l Ai r l iner Door B r idge Ci rcu i t
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le :
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le :
5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis
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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is
Ad di t iona l Examp le :
5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is
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y y yQuiz Exerc is e 5.10* Find the phasor voltage and phasor current at each element
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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y y5.5 Power in A C Circui t
5.5.1 Voltage, Current an d Im pedanc e
)- where , I - I I
then ,V If :(Note
Z V
I where
I
Z
0V
Z
iscurrent phasor The
(X/R)tan , X R Z where
jX R Z Z impedancethe
0V or t)cos( V v(t) network,the In
ivimvm
vm
mm
mm
1-22
mm
V
V I
V
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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y y5.5.2 Voltage, Curren t and Po w er for a Resist iv e Load
(1) Current is in phase with voltage.(2) Energy flows continuously from
source to load.
rms rms m m
m m
2
m m
m m
m
I V 2
I V P power averageThe
t) cos2 2
1
2
1 ( I V
t cos I V t i t v t p
t cos I t) cos( R
V t i
t cos V t v 0 0,R Z
resistive,pureis load theIf
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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y y5.5.3 Voltage, Curren t and Pow er for an Ind uc tive Lo ad
(1) Current lags the voltage by 90 degree ( ELI )(2) Half of the pow er is pos i t ive , energy is delivered to the inductance
and stored in the magnetic field; the other hal f of the pow er isnegat ive , the inductance returns energy to the source.
(3) The average po w er is zero , and we say react ive po wer flows backand forth in-between the source and the load.
0 P power averageThe
t sin2 I V t 2 sin 2
I V
t
sin t
cos I V
t i t v t p
t sin I
90 t cos I t i t cos V t v
90 ,90 LZ
,inductance purei s load theIf
rms rms m m
m m
m
m
m
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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y y5.5.4 Voltage, Curren t and Po w er for a Capacit ive Lo ad
(1) Current leads the voltage by 90 degree ( IC E )(2) The average pow er is zero: react ive po wer flows back and forth in-
between the source and the load.(3) Reactive power is negative (positive) for a capacitance
(inductance ).
(4) Reactive power in inductance and in capacitance cancel eachother.
0 P power averageThe
t 2 sin I V t 2 sin 2
I V
t sin t cos I V
t i t v t p
t sin I
90 t cos I t i
t cos V t v
-90 ,90 -
C
1 Z
e,capacitanc pureis load theIf
rms rms m m
m m
m
m
m
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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5.5.5 Pow er Calculat ion for a General (RLC) Lo ad
A ctiv e (Real) load d ue to R
React ive load d ue to L, C
remains.term second theonlyreactive, pure :90
remains;term1st theonlyresistive, pure:0
t 2 sin sin2 I V t 2cos1cos
2 I V
t sint cos sin I V t coscos I V
t cost cos I V t p
t cos I t i
t cosV t v
9090- , Z jX R Z load, RLC general For
mmmm
mm2
mm
mm
m
m
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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5.5.5 Pow er Calculat ion for a General (RLC) Lo ad
t 2 sin sin2 I V
t 2cos1cos2
I V t p
t sint cos sin I V
t coscos I V p(t)t cost cos I V t p
t cos I t i
t cosV t v
mm
mm
mm
2mm
mm
m
m
) t ( 2 sin sin2 I V
) t ( 2cos1 cos2 I V t p
t sin t cos sin I V
t cos cos I V t p
t cos t cos I V t p ) - t cos( I
- , t cos I t i
t cosV t v
vmm
vmm
vvmm
v2
mm
vvmm
vm
ivim
vm
cos I V P have we
,2 / I I and 2 / V V using ,cos2 I V
P
:is P power (real)average the zero,of values average have ))t sin(2( and ))t cos(2( involving terms the Since
rmsrms
mrmsmrmsmm
vv
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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5.5.5 Pow er Calculat ion for a General (RLC) Lo ad
Av erage Power :
Pow er Factor :
Power factor is often stated as percentage, e.g.,90% lagging (i.e., current lags voltage, inductive load)60% leading (i.e., current leads voltage, capacitive load)
Reactiv e Pow er: The last term in power formula is the power flowing back and forth
between the source and the energy-storage elements. Reactivepower is its peak power.Ap paren t Power :
Note: 5kW load is d i fferent f rom 5kVA load .
t 2 sin sin 2
I V t 2 co s 1 cos
2
I V t p m m m m
W cos I V cos 2
I V P rms rms
m m
angle power thecalled i s , cos PF i v
Reactive) Amperes(Volt VAR sinrms I rmsV sin2m I mV Q
2 rms rms 2 2
rms rms 2 2
rms rms 2 2
rms rms
I V sin I V cos I V Q P
Am pere) - (Volt VA I V S
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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5.5.5 Pow er Calculat ion for a General (RLC) Lo ad
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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y y5.5.6 Im pedanc e tr iangle and Pow er Triang le
* The impedance triangle:
* The Power triangle:Apparent power, average (real) power, and reactive power form a
triangle.
2rmsrms22rmsrms22rmsrms22rmsrms
I V sin I V cos I V Q P
Ampere)-(Volt VA I V S
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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47
y y5.5.6 Im pedanc e tr iangle and Pow er Triang le
reactance theacross voltagethei s V whereX
V Q
resistance theacrtoss voltagethei s V whereR
V
P : Also
load) capacitive 0,X load; inductive0,(X
X I sin 2
I V Q power reactivetheSimilarly,
R I R 2
I ) Z I V : (Note
Z
R
2
I V cos
2
I V P power averageThe
Z
X sin ,Z
R cos ,X j R Z Z
Xrms
2 Xrms
Rrms
2 Rrms
2 rms
m m
2 rms
2 m
m m
m m m m
R
V
R
1
2
V
R
1
Z
R
2
V
Z
R
2
V
Z
R
2
I V P
2 Rrms
2 Rm
2
22m
2
2mmm
t 2 sin sin2 I V
t 2cos1 cos2 I V
t p mmmm
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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48
y yExam ple 5.6 AC Pow er Calcula t ion
VAR5.045 sin1.0071.7
sin I V Q power reactiveThe
W 5.045cos1.0071.7
cos I V P : power The
A1.02
1414.0
2 I
V 071.7 2
10
2V
45 )135( 90
sourcethe fromtaken power reactiveand power the Find (1)
rmsrms , s
rmsrms , s
rms
srms , s
iv
I
V
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t E l 5 6 ACP C l l i
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49
Exam ple 5.6 AC Pow er Calcula t ion
.resistancethebyabsorbed is sourcethebydelivedrd power theof All
0 P ,0 P
0isinductanceand ecapacitancthebyabsorbed power thecourse,Of
W 5.010021.0 R
2 R I P
:resistancethetodelivered power (real)The(4)
)QQQ: Note( VAR5.0 )100( 2
1.0 X I Q
:capacitor thetodelivered power reactiveThe(3)
VAR0.1 )100( )1.0( X I Q
:inductor thetodelivered power reactiveThe(2)
C L
22
R2rms , R R
C L
2
C 2
rms ,C C
2 L
2rms L
I
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5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t
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51
Ad dit ional Example:
5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t
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52
Ad dit ional Example:
5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t
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53
Ad dit ional Example:
5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t
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54
Ad dit ional Example:
5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t
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55
Ad dit ional Example:
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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56
Exam ple 5.7 Using Pow er Tr ianglesFind power, reactive power, and power factor for the source
and the phasor currents as shown.
We first find the power and reactive power for each load, thensum over to obtain the power and reactive power for thesource.
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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57
Exam ple 5.7 Using Pow er Tr iangles
kVAR559.3101.5660.8QQQ
kW 1055 P P P
: sourcethebydelivered power reactiveand power The
kVAR101.5Q
57 .45tan5000tan P Q
57 .45 )7 .0arccos( B,load For
kVAR660.8500010
P I V Q
kW 5 )5.0( 10cos I V P
5.0cos:havewe A,load For
B A
B A
B
B B B
B
224
2 A
2 Armsrms A
4 A Armsrms A
A
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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58
Exam ple 5.7 Using Pow er Tr iangles
59.4915 ,59.49 )59.19( 30
A15 I 2 I
A61.10 /V I V I :current effectiveThekV 12 / V :voltage sourceeffectiveThe
kVA61.10Q P I V S : power apparent The
leading.94.21%or 9421.0 cos : factor power The
59.19Q/P tan :angle power The
kVAR559.3QQQ ,kW 10 P P P
:by sourcedelivered power reactiveand power The
ivi
rmsm
rmsrmsrmsrms
rms
22rmsrms
1
B A B A
I I
I
V
v
i
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5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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60
Exam ple 5.8 Pow er-Facto r Correct ionA 50kW load operates from a 60-Hz 10kV-rm s line with a power
factor of 60% lagging. Compute the capacitance that mustbe placed in parallel with the load to achieve a 90% laggingpower factor.
F 126 .1
2356 377
1
X
1C
0.377 60 2
2356 4245010
QV
X
kVAR45.42QQQ
kVAR22.24 tan P Q84.25 )9.0( cos
kVAR67 .66 tan P Q
13.53 )6 .0( cos
C
24
C
2
rmsC
LnewC
newnew
1new
L
1 L
66.67kVAQ L
50kW P
22.42kVAQ new
0.6 PF 50kW
5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t
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61
Qu iz - Exerc ise 5.12 Pow er in A C Ci rcu i t s.
5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton
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62
5.6 Thevenin and Norton Equivalent Circui ts
* A two terminal circuit composed of sinusoidal sources (of thesame frequency), resistances, capacitances, andinductances can be simplified to Thevenin or Nortonequivalent c i rcu i t .
5.6.1 Thevenin Equ ivalent Circu its
* The Thevenin impedance can also beobtained by zeroing sources.
5.6.2 Nort on Equ ivalent Circu its
sc
t
sc
oct
oct
Z I V
I V
V V
scn
sc
t
sc
oct I
Z
I I
V
I
V
5. Steady -State Sinus oid al Analys is 5 .6 Thevenin and No rton
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63
Exam ple 5.9 Thevenin and Norton Eq uivalents
V 90100
4571.7045414.1
Z I
A45414.1 9011
A01100
0100100V
terminalsat circuit t Apply shor
50 j50
4571.704501414.0
1 )100 j( 11001
1 Z
sources zeroing by Z f ind We
t sct
s R sc
s R
t
t
V
I I I
I
5. Steady -State Sinus oid al Analys is 5 .6 Thevenin and No rton
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64
Ad dit ional Example:
5. Steady -State Sinus oid al Analys is 5 .6 Thevenin and No rton
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5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton E l 5 10 M i P T f
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67
Exam ple 5.10 Maxim um Pow er Trans fer
W 25 )50( 2
1 R I P
90150 j5050 j50
90100 Z Z
50 j50 Z Z
when power output max.havewill we(a)
50 j50 Z Since
2
load 2
arms
load t
t a
*t load
t
V I
W 71.2071.70
2
7653.0 R I P
50.67 765.050.2266 .130
90100
71.7050 j50
90100
Z Z
71.7050 j50 Z R
resistive,betohasload the If (b)
2
load 2
brms
load t
t b
t load
V I
5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 14 dE i 5 15
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68
Quiz Exerc is e 5.14 and Ex ercis e 5.15
5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 14
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69
Quiz Exerc is e 5.14
5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 15
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Quiz Exerc is e 5.15
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5. Steady -State Sinu so idal An alysis SUMMARY
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5. Steady -State Sinu so idal An alysis SUMMARY
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Chapter 6
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Chapter 6Frequency Response, Bode Plots,
and Resonance1. State the fundamental concepts of Fourier analysis.2 . Determine the output of a filter for a given input
consisting of sinusoidal components using the
filter
s transfer function.3 . Use circuit analysis to determine the transferfunctions of simple circuits.
4 . Draw first-order lowpass or highpass filter circuitsand sketch their transfer functions.
5.
Understand decibels, logarithmic frequency scales,and Bode plots.6 . Calculate parameters for series and parallel
resonant circuits.
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6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.1 Fourier A nalys is
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77
y
* All real -wo r ld s igna ls a re sums of s inu so ida l com ponents having various frequencies, amplitudes, and phases.
* The square wave is a special example:
* Most of the real-world signals areconfined to finite range of frequency.
* It is important to learn how the circuits
respond to components havingdifferent frequencies.
.. .t ) s in(5 5 4A
t ) s in(3 3 4A
t ) s in (
4A(t ) v 0 0 0 sq
T 2 where0
6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.2 Filt ers
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* Filters process the sinusoidal components of an input signaldifferently depending of the frequency of each component.
Often, the goal of the f i l ter i s to re ta in the com pon ents incer ta in f requency ranges and reject com pon ents in o therf requency ranges .
6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.3 Fil ters and Transfer Fun ctio ns
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79
* Since the impedances of inductances and capacitanceschange with frequency, RLC c i rcu i t s p rov ide one w ay torealize electri cal fil ters .
* The t ransfer fun ct ion of a two-port filter is defined as:
90- f 2
190-C 1 Z ,90 fL290 L Z C L
phaset h eis- H(f)
magni tudetheisH(f) w h e r e
H(f) H(f)) f H
in o u t
in
o u t
VV
V
V)(
6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns Exam ple 6.1 Us ing Transfe r Func t ion to F ind Outpu t
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80
For the transfer functions shown, find the output signal,given the input: )40t 2000cos( 2 )t ( vin
in
out 303 )1000( H
Hz 1000 f is signal input theof f requencyThe
V
V
402inV 706 402303* )1000( H inout V V )702000cos(6)( t t vout
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6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns Exam ple 6.2 Mul t i -inpu t com po nents , Superpos i t ion Pr inc iple
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82
)70t 4000cos( )t 2000( sco2 )t ( vin
)10t 4000cos( 2 )30t 2000cos( 6 )t ( vout
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Ideal Filters
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83
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters 6.2 First Ord er Low -Pass Fil ters
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84
A low -pass f i l ter is designed to p ass low-f requencycom ponents and re jec t h igh- f requency c om ponents . Inother words, for low frequencies, the output magnitude isnearly the same as the input; while for high frequencies, theoutput magnitude is much less than the input.
6.2.1 Transfer Fun ctio n
fC j21 R
have we ,V phasor ahaving l sinusoidaais signal input the shown,as
f ilter pass-loworder - f irst theConsider
in
in
V I
RC f j21 fC j21 R fC j21 inin V V I V
fC j21
out
fRC 2 j11 H(f)
in
out
V V
frequency power" -half " the
frequency,break" " the RC 21
f defineWe B
) f f j( 11
H(f) B
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters 6.2.2 Magn itud e and Phase Plots of th e Trans fer Fun ctio n
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85
)(11
)( B f f j
f H 2)(1
1)(
B f f f H )arctan()(
B f f
f H
Power Half V P since ,V 2
1V ,2
1 H(f ) , f f As
90 H(f )also
rejected,components f requency-high 0 H(f ) , f f As
0 H(f )also passed,components f requency-low 1 ) f ( H 0, f As
2rmsrmsinrmsout B
B
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Exam ple 6.3 Calcula t ion of RC Low -pass Outpu t
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86
1000 f 0,5
100, f 0,5 10, f 0,5
)t 2000cos( 5 )t 200cos( 5 )t 20cos( 5 )t ( v
33in
22in11in
in
V
V V
) f f ( j1
1 ) f ( H
B
Hz RC
f B 10010*10*)21000(*21
21
6
71.59950.0 )10( H
457071.0 )100( H 29.840995.0 )1000( H
71.5975.4 )10( H 1in1out V V
)71.5t 20cos( 975.4 )t ( v 1out
45535.3 )100( H 2in2out V V )45t 200cos( 535.3 )t ( v 2out
29.844975.0 )1000( H 3in3out V V
)29.84t 2000cos( 4975.0 )t ( v 3out
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Exam ple 6.3 Calcula t ion of RC Low -pass Outpu t
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87
1000 f 0,5 100, f 0,5 10, f 0,5
)t 2000cos( 5 )t 200cos( 5 )t 20cos( 5 )t ( v
33in
22in11in
in
V V V
(t)v )29.84t 2000cos( 4975.0 (t)v )45t 200cos( 535.3 (t)v )71.5t 20cos( 975.4 )t ( v
3out
2out
1out out
6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Quiz Exercis e 6.4: A no ther First -Order Lo w -Pass Fil ter
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88
This i s a lso a low-pass f i l ter
L R/2 f where
) j(f /f 1
1 H(f )
is functiontransfer thethat Show
B
Bin
out
V
V
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6. Frequ ency Respo nse 6.3 Decibels and th e Casc ade Con nectio n 6.3.2 Casc ade two -Port Netw ork s
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90
2in
2out
1in
1out
1out
2out
1in
1out
1in
2out
in
out
) f ( H V V
V
V
V
V
V
V
V
V
V
V
)()()( 21 f H f H f H
dB2dB1dB ) f ( H ) f ( H ) f ( H
6. Frequenc y Respo ns e 6 .4 Bod e Plots 6.4 Bod e Plots
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91
2)(1
1)(
B f f f H
6. Frequ ency Respo nse 6 .4 Bo de Plots 6.4 Bo de Plots
ffb k
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92
2)(1
1)(
B f f f H
2
B
dB
) f f ( 1
1log 20 ) f ( H
BdB B
B
f f
log 20 ) f ( H f f For
dB0 H(f) f f For
B f f
f H arctan)(
90 H(f ) , f 10 f For
0 H(f ) /10, f f For
B
B
B f frequencybreak
6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters 6.5 First-Order Hig h-Pass Fil ters
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93
6.5.1 Transfer Fu nc tion
)(1)(
)( B
B
in
out
f f j f f j
V V
f H
RC f B 2
1
2 B B
f f 1
f f ) f H(
B f f
f H arctan90)(
6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters 6.5.2 Bo de Plo ts
2
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94
21)(
B
B
f f
f f f H
21log10log20)(
B BdB f
f f f
f H
0 f H , f 10 f For
90 f H /10, f f For
0 f H , f f For
f f log 20 ) f ( H , f f For
B
B
dB B
BdB B
6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters Exerc is e 6.13 An oth er First -Order High-Pass Fil ter
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95
L R/2 f where
) f f ( j1 ) f f ( j
V V ) f ( H
:iscircuit theof
functiontransfer thethat Show
B
B
B
in
out
6. Frequ ency Respo nse 6.5 First -Order Filters First-Order Lo w -Pass Fil ters
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96
First-Order High -Pass Fil ters
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6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.1 Reson ant Circ uits (Secon d Ord er)
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98
C f 21
L f 2
f requency" resonant " thecalled is f
f requencytheresistive, purelyis(f ) Z When
00
0
s
LC 2
1 f 0
)CR f 1/2( L/R f 2Q :resistancetheto frequencyresonant theat
inductancetheof reactancetheof ratiotheasdefined is factor qualityThe
00 s
f f
f f
jQ1 R ) f ( Z :asrewrittenbecanimpedancethe Now 00
s s
fC 2
1 j R fL2 j ) f ( Z
is sourcetheby seenimpedanceThe
s
6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.1 Reson ant Circ uits (Secon d Ord er)
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99
LC 2
1 f frequencyresonant With 0
)CR f 1/2( L/R f 2Q factor qualityand
00 s
f f
f f
jQ1 R 00
s fC 21
j R fL2 j ) f ( Z
is sourcetheby seenimpedanceThe
s
6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.2 Series Reson ant Circ uits as B and-Pass Fil ter
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100
f f
f f
jQ R f Z s s0
0
1)(
-
RV
) f f f f ( jQ1 R
f Z 00 s s
s
s V V I
) f f f f ( jQ1 R
00 s
s R
V I V
) f f f f ( jQ11
00 s s
R
V
V
rejected others pass, f naer freq.
Filter Pass- Band
) f f
- f f
( Q1 f H
0
-1/2
20
0
2 s
s
R
V
V
6. Frequenc y Respo ns e 6 .6 Series Reson ances 6.6.2 Series Reson ant Circu its as B and -Pass Fil ter
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101
s0 L H /Q f f f B
:asdefined iswidth- Band
L H f and f :asdenoted
21/ f H at occurs
: f requency power -half The
2 B
- f f ,2 B
f f 0 L0 H
-1/2
20
0
2 s
s
R ) f f
- f f
( Q1 f H V
V
6. Frequ ency Respo nse 6 .6 Series Reson ances Exam ple 6.5 Ser ies Resonant Circu i t
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102
Hz 1000 LC 2
1 f : frequency Resonant 0
10 R
L f 2Q: Factor Quality 0 s
Hz 10010
1000Q f
B:width- Band s
0
950Hz 2
B- f f , Hz 1050
2 B
f f 0 L0 H
6. Frequ ency Respo nse 6 .6 Series Reson ances Exam ple 6.5 Ser ies Resonant Circu i t
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103
j1000-C f -J/2 Z
1000 j L f 2 j Z
1000Hz f f requencyresonant At
0c
0 L
0
100 Z Z R Z c L s
001.0100
01 Z s
sV I
01001.0100 R R I V
9010001.01000 j Z L L I V
9010001.01000 j Z C C I V
6. Frequenc y Respo ns e 6.7 Parallel Reso nan ces
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104
6.7 Paral lel Reso nan ce
LC 2
1 f frequencyresonant With 0
CR) f (2 L) f R/(2Q factor qualityand 00 p
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8 Ideal and Secon d-Order Fil ters
6 8 1IdealFilt ers
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105
6.8.1 Ideal Filt ers
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.1 Ideal Filt ers
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106
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.2 Secon d-Order L ow -Pass Fil ter
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107
) f f f f ( jQ1 ) f f ( jQ
V V
) f ( H 00 s
0 s
in
out
LC 21 f 0
R L f 2
Q 0 s
CR f 21
0
1Qchoose
passban d theinconstant elyapproximat beto gainthe
want we f ilter,adesign In
s
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.2 Seco nd -Order High -Pass Fil ter
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6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters Exam ple 6.7 Fi l ter DesignD ig d d filt ith L 50 H th t
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111
Design a second-order filter with L=50mH that passescomponents higher in frequency than 1kHz , rejectscomponents lower than 1kHz .
We need a high-pass filter.To obtain a approximatelyconstant transfer functionIn the pass-band, we choose
LC 2
1 f since
1kHz f select and 1Q
0
0 s
F 507 .0 L f )2(
1C havewe
202
1.314Q
L f 2 R and
s
0
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters * The Popu lar Sallen-Key Fil ters
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112
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters
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* High er-ord er Fi l ters u s ing Cascade of 2 n d -ord er Fil ters
6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters
* High er-ord er Fi l ters u s ing Cascade of 2 n d -ord er Fil ters
8/10/2019 Ac Sinusoidal Ckt
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