電路學 - [第七章] 正弦激勵, 相量與穩態分析
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Transcript of 電路學 - [第七章] 正弦激勵, 相量與穩態分析
Department of Electronic EngineeringNational Taipei University of Technology
•• ( )
• ( )
• ( )
••
Department of Electronic Engineering, NTUT2/41
(I)
•
Department of Electronic Engineering, NTUT
( ) sinmv t V tω=
Vm
t (sec)
T
v(t)
−Vm
2
πω
πω
3
2
πω
2πω
5
2
πω
6
2
πω
0
T
ω
: Vm(volt)
: T (sec)
: f (Hz, cycle/sec)
: (rad/sec)2 fω π=
3/41
(II)
• ( ) ( )sinmv t V tω φ= +
ω
: Vm(volt)
: T (sec)
: f (Hz, cycle/sec)
: (rad/sec)2 fω π=
φ: (rad)檔案中找不到關聯識別碼 rId12 的圖像部分。
Vm
t (sec)
v(t)
−Vm
2
πω
πω
3
2
πω
2πω
5
2
πω
6
2
πω
0
T
φ
1t
Department of Electronic Engineering, NTUT4/41
( )
• vR = iR( ) ( )
• ,
•
R, L, C = + ( ) ( / )
L
div L
dt= 1 t
Cv idtC −∞
= ∫
Department of Electronic Engineering, NTUT5/41
1
• RL i(t)i f (t)
,
KVL+−
L
Ri(t)
( ) ( )cos Vs mv t V tω=
( ) ( ) cosm
di tL Ri t V t
dtω+ =
( ) cos + sinfi t A t B tω ω=
( ) ( ) cosm
di tL Ri t V t
dtω+ =
mRA LB Vω+ = 0LA RBω− + =
( ) ( )sin cos cos sin cosmL A t B t R A t B t V tω ω ω ω ω ω ω− + + + =
, 2 2 2mLV
BR L
ωω
=+2 2 2
mRVA
R Lω=
+( )fi t
( ) ( ) ( )12 2 2 2 2 2 2 2 2
cos sin cos tan cos AR
m m mf m
RV LV V Li t t t t I t
R L R L R L
ω ωω ω ω ω φω ω ω
− = + = − = + + + +
2 2 2
mm
VI
R Lω=
+1tan
L
R
ωφ −= −
( ) ( ) ( )f ni t i t i t= +( ) 1RtL
ni t Ae−=
( )sv t
Department of Electronic Engineering, NTUT6/41
2
• RLC v(t) i(t)
KVL
+−
+
−
5
3R = Ω 5 HL = ( )i t
( )v t1
F25
C =( )sv t
( ) 17cos3sv t t=
( ) ( ) ( ) ( )2
2
1 sd v t dv t v tRv t
dt L dt LC LC+ + =
( ) ( ) ( )2
25 85cos3
d v t dv tRv t t
dt L dt+ + =
( ) ( ) ( ) ( )20cos3 +5sin 3 =5 17 cos 3 2.9 5 17 cos 3 166 Vv t t t t t= − − = −
( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 3 17 3 1760sin 3 15cos3 cos 3 1.33 cos 3 76 A
25 25 5 5
dv t dv ti t C t t t t
dt dt= = = + = − = −
,
( ) 1 2cos3 sin3v t A t A t= +
( ) ( )1 2 1 24 cos3 4 sin 3 85cos3A A t A A t t− + + − − =
1 24 85A A− + = 1 24 0A A− − = 1 20A = −
2 5A =
Department of Electronic Engineering, NTUT7/41
(Complex Number)
•
θa
b
A a jb= +
Re
b
Im
| |A
+A a jb=jA A e Aθ θ= = ∠
[ ]Re cosa A A θ= =
[ ]Im sinb A A θ= =
2 2A a b= + 1tanb
aθ −=
1j = −1 1 1A a jb= + 2 2 2A a jb= +
( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2+A A a jb a jb a a j b b= + + + = + + +
( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2A A a jb a jb a a j b b− = + − + = − + −
• +A a jb= A a jb∗ = −
•
( )( )( ) ( )
21 2 1 1 2 2 1 2 1 2 2 1 1 2
1 2 1 2 1 2 2 1
A A a jb a jb a a ja b ja b j b ba a b b j a b a b
= + + = + + += − + +
( ) ( ) ( )1 2 1 1 2 2 1 2 1 2A A A A A Aθ θ θ θ= ∠ ∠ = ∠ +
•( )( )( )( )
1 1 2 21 1 1 1 2 1 2 1 2 1 22 2 2 2
2 2 2 2 2 2 2 2 2 2 2
a jb a jbA a jb a a b b b a a bj
A a jb a jb a jb a b a b
+ −+ + −= = = ++ + − + +
( )111 2
2 2
AA
A Aθ θ= ∠ −
Department of Electronic Engineering, NTUT8/41
(a+jb) (I)
•
(a) A = 4 + j3
,
A
(b) A = –4 + j3
A θ∠
2 24 3 5A = + = 1 3= tan 36.9
4θ − =
5 36.9A = ∠
1 13 3= tan 180 tan 180 36.9 143.1
4 4θ − − = − = − = −
( )2 24 3 5A = − + =
5 143.1A = ∠
(a)
4
θ = −tan134
| |A =5
Re
3
j
o
o
A
A
jA
9.365
9.3643tan
534
34
1
22
∠=⇒
==⇒
=+=⇒
+=
−θ
Re
j
θ = −180° 3
41tan
−4
| |A =53
( )
oo
oo
A
A
jA
1.14351.143
9.3618043
tan
534
34
1
22
∠=⇒=−=−=⇒
=+−=⇒
+−=
−θ
(b)
Department of Electronic Engineering, NTUT9/41
(a+jb) (II)
(c) A = –4 –j3
( )
o
oo
oo
A
A
jA
9.365
9.361.323
9.3636043tan
534
34
1
22
−∠=⇒
−==
−=−=⇒
=−+=⇒
−=
−θ
j
| |A =5
Re
−3
4
θ = −360° 3
41tan
−3
Re
j
−4
| |A = 5( ) ( )
o
oo
oo
A
A
jA
1.1435
1.1439.216
9.3618043tan
534
34
1
22
−∠=⇒
−==
+=−−=⇒
=−+−=⇒
−−=
−θ
θ = −180° + 3
41tan
A θ∠
( ) ( )2 24 3 5A = − + − =
1 13 3 = tan 180 tan
4 4
180 36.9 216.9 143.1
θ − −− = − −
= + = = −
5 143.1A = ∠ −
(d) A = 4 –j3
( )224 3 5A = + − =
1 13 3 = tan =360 tan
4 4
323.1 36.9
θ − −− −
= = −
5 36.9A = ∠ −
(c)
(d)
Department of Electronic Engineering, NTUT10/41
(I)
•
•
• RL (Complex excitation)(Real excitation)
i(t)
KVL:
cos sinj tm m mV e V t jV tω ω ω= +
cos Re
sin Im
j tm m
j tm m
V t V e
V t V e
ω
ω
ω
ω
=
=
1j t
mv V eω= ( ) [ ]1cos Reg mv t V t vω= =
+−
L
Ri(t)cosmV tω
11+ j t
m
diL Ri V e
dtω=
1j ti Ae ω=
( ) j t j tmj L R Ae V eω ωω + =
-1tan
2 2 2
Lj
m m RV VA e
R j L R L
ω
ω ω−
= =+ +
1tan
1 2 2 2
Lj t
RmVi e
R L
ωω
ω
− − =
+
i1 i(t)
⇒⇒⇒⇒
Department of Electronic Engineering, NTUT11/41
(II)
6 RL
i(t) i1 ( vg(t) v1 )
i1 v1 i f (t)= Re[i1] vg = Re[v1]
1tan
1 2 2 2
Lj t
RmVi e
R L
ωω
ω
− − =
+
( ) [ ]1tan
11 2 2 2 2 2 2
Re Re cos tanL
j tRm mV V L
i t i e tRR L R L
ωω ωωω ω
− − − = = = −
+ +
[ ]11 1Re Re
diL Ri v
dt
+ =
[ ]( ) [ ]( )1 1Re Re cosm
dL i R i V t
dtω+ = ( ) ( ) [ ]1Refi t i t i= =
Department of Electronic Engineering, NTUT12/41
• 11 :
• ?
KVL: 11+ j t
m
diL Ri V e
dtω= 1
j ti Ae ω=
( ) j t j tmj L R Ae V eω ωω + =
-1tan
2 2 2
Lj
m m RV VA e
R j L R L
ω
ω ω−
= =+ +
1tan
1 2 2 2
Lj t
RmVi e
R L
ωω
ω
− − =
+
A
” ”
” ”
j te ω
j te ω
jω
j te ω
1 jω
Department of Electronic Engineering, NTUT13/41
(I)
• (Euler’s formula)
(Phasor)
• rms
• ”−”
( ) ( )cos Re Rej j t j tm mv t V t V e e Veθ ω ωω θ = + = =
jm mV V e Vθ θ= = ∠
2j m
rms rms
VV V e Vθ θ θ= = ∠ = ∠
jm mV e Vθ θ= = ∠V j
m mV V e Vθ θ= = ∠
j te ω
j te ω
1:
Department of Electronic Engineering, NTUT14/41
(II)
I
R
+
=
−V RI
i
R
+
=
−v Ri
I
+
−
i
L
+
−
div L
dt= V j LIω= j Lω
+
−V
1I j CV V I
j Cω
ω= ⇒ =
1
j CωC
+
−v
dvi C
dt=
R R
L j Lω
C1
j Cω
( )
( )
2:
Department of Electronic Engineering, NTUT15/41
(III)
+−
L
Ri(t)cosmV tω +− RI0mV V= ∠
j Lω
j LI R Vω + =1
2 2 2
0tan
Lm mV VV L
IR j L R j L RR
ωω ω ω
−∠= = = ∠ −+ + +
( )1 1tan tan
1
2 2 2 2 2 2 2 2 2Re Re cos tan
L Lj j t
j tR Rm m mV V V Li t e e e t
RR L R L R L
ω ωωω ωω
ω ω ω
− − − − − = = = −
+ + +
2:1:
, !
3:!
!
Department of Electronic Engineering, NTUT16/41
R L C
I
R
+
=
−V RI
(a)
i
R
+
=
−v Ri
t
v i,vi
(b)
i
L
+
−
I
+
−(a) (b)
div L
dt= V j LIω= j Lω
vi
t
v i,
v
i
t
v i,
Department of Electronic Engineering, NTUT
C
+
−v
+
−
I
1
j Cω
dvi C
dt=
(a) (b)
1V I
j Cω=
17/41
(I)
Z (Impedance)
( ) ( )1cosmv t V tω φ= +
( ) ( )2cosmi t I tω φ= + 2mI I φ= ∠
1mV V φ= ∠
( ) ( )1 2= m
m
VVZ Z R jX
I Iθ φ φ= = ∠ ∠ − = + Ω
(a)
+
−
+
−V
I
(b) ( )
( )v t
( )i t
[ ]ImX Z=
[ ]ReR Z=
(Reactance)
(Resistance)
Z ω
ω
( )Z jω( )R R ω=
( )X X ω=
θR
XZ R jX= +
Re
Im2 2Z R X= +
1tanX
Rθ −=
cosR Z θ=
sinX Z θ=
Department of Electronic Engineering, NTUT18/41
(II)
•
• (Admittance)
• Z Y Y
Y = G + jB G = Re[Y] B = Im[Y]
(Susceptance) Y G B
R
VZ R
I= =
LX Lω=L L
VZ jX j L
Iω= = =
1CX
Cω= − 1
C C
VZ jX j
I Cω= = = −
1Y
Z=
( ) ( ) 2 2 2 2
1 1=
R jX R XY j G jB
Z R jX R jX R jX R X R X
−= = = − = ++ + − + +
Department of Electronic Engineering, NTUT19/41
• (a) RL (b)
KVL:
+−
L
Ri( )cos VmV tω
(a)
+− R
j Lω
I0mV ∠
(b)
0L mZ I RI V+ = ∠ ( ) 0mj L R I Vω + = ∠
( ) 1
2 2 2cos tanmV L
i t tRR L
ωωω
− = − +
1
2 2 2
0tanm mV V L
IR j L RR L
ωω ω
−∠ = = ∠ − + +
Z j L Rω= + 0mVVI
Z R j Lω∠= =
+
Department of Electronic Engineering, NTUT20/41
3
• (a) v(t) i(t) 1. (a) (b)
2. (c)
3. (c)
10 0V = ∠ 2 rad secω =
( )Z 2 L j L jω= = Ω ( )11 CZ j j
Cω= − = − Ω
1 1.5 2Z j= + ( )( )2
1 10.5 0.5
1 1
jZ j
j
−= = −
− +
( ) ( )0.283sin 2 81.9 Vv t t= −
1 2
10 0 10 04 36.9
2.5 36.9TI Z Z
∠ ∠= = = ∠ −+ ∠
( )1 4 36.92.83 8.1 A
1 1 2 45TI I
j
∠ −= = = ∠− ∠ −
( ) ( ) ( )2.83sin 2 8.1 Ai t t= +
+−
+
−( )10sin2 Vt
1.5 Ω 1 H
0.5 F 1 Ω( )Ti t ( )i t
( )v t
(a)
+−
+
−10 0∠ 1 Ω
2 j Ω
1 j− Ω
1.5 Ω
TII
V
(b)
+−
+
−
TI
10 0∠ V
1Z
2Z
(c)
,
,
,
( ) ( )2
1 2
0.5 0.510 0 10 0
1.5 2 0.5 0.5
Z jV
Z Z j j
−= × ∠ = × ∠+ + + −
0.283 81.9 V= ∠ −
Department of Electronic Engineering, NTUT21/41
4
• (a) i(t)
1. (a) (b)
a KCL:
+−
a
12
18F
+
−v1
i t( )
4Ω( )10cos4 At v1
(a)
+−
a
12
−j2 Ω
+
−V1
I
4Ω( )10 0 A∠ V1
(b)
( )1 1
21
48
C
jZ j j
Cω= − = − = − Ω
4rad secω =10 0I = ∠
1 1
12 10 02
V VI
j
−+ = ∠
− ( )1 1 10 0I j+ = ∠ ( )10 0
7.07 45 A1 1
Ij
∠= = ∠ −+
( ) ( )( )7.07cos 4 45 Ai t t= −
, ,
Department of Electronic Engineering, NTUT22/41
(Phasor Diagram)
• (Vector)
•
Vg= VR + VL + VC
(b) VR , VL , VC
V|VL| |VC|
+−
Vg
R
+ −VR + −VL+
−VC
1j Cω
j Lω
VL VC+
VR
(a) RLC I
(a) RLC
0I I= ∠
I I I
(b)
VR
VL
VC
(a)
VL
VC (c) VR + VL + VC = Vg
Department of Electronic Engineering, NTUT23/41
Department of Electronic Engineering, NTUT24/41
5 −
• (a) i(t)
1. (a) (b)
2. V V + 3000I
KVL:
4.
3. KCL:
(a)
+−
+−
1 k
2Ω
2 kΩ1
F5
µ( )3000 Vi t
( ) ( )3000 v t i t+
2 kΩ
1 F
5µ
( ) ( )4cos5000 Vsv t t=
( )sv t
I V
+−
+−
1 k
2Ω 3000V I+
3000I
4 0sV = ∠ ( )21 2 k
5j− Ω ( )2 1 kj− Ω
(b)
( ) ( )( ) ( )( )4 3000
01 2 2 1 103103 1 2 1032 5
V V V I
jj
− ++ + =−−
( )4
1103
2
VI
−=
324 10 53.1 A=24 53.1 mAI − °= × ∠ ∠
( ) ( )24cos 5000 53.1 mAi t t °= +
Department of Electronic Engineering, NTUT25/41
( )i t ( )v t
6 −
• (a) i1(t) i2(t)
1. ω = 2 rad/s : (b)
(b)
+− 1I 2I
3 Ω 2 j Ω 1 Ω
1 j Ω
2 j− Ω1 Ω18 0sV = ∠
2.
( ) 1 24 2 18 0j I I °+ − = ∠
( )1 22 1 0I j I− + − =
2 2 0 AI °= ∠1 4.47 26.6 AI °= ∠ −
3.
( )2 2sin 2 Ai t t=
( ) ( )1 4.47sin 2 26.6 Ai t t °= −
Department of Electronic Engineering, NTUT26/41
+−
3 Ω
(a)
1 H 1 Ω
1 Ω
( ) ( )18sin 2 Vsv t t=
( )sv t ( )1i t
1H
21
F4
( )2i t
7 −
• (a)
1. (b) I i(t) I = I1 + I2
2. I1 I2 (c) (d)
3. (c)
+−
+−
5 Ω
(a)
4 Ω1 H( )i t
1F
20( )1sv t
( ) ( )1 20sin 2 Vsv t t= ( ) ( )2 10sin 2 Vsv t t=
( )2sv t
+−
+−
5 Ω 4 Ω2 j Ω
20 0∠
10 j− Ω10 0∠
I
(b)
+−
5 Ω 4 Ω2 j Ω
20 0∠
10 j− Ω
1I
(c)
+−
5 Ω 4 Ω2 j Ω
10 j− Ω10 0∠
2I
(d)
( ) ( )1
10 4 2 10 4 25 5 10
10 4 2 4 8
j j j jZ
j j j
− + − += + = + = Ω
− + + −
11
20 0 20 02 0 A
10I
Z
° °°∠ ∠= = = ∠
Department of Electronic Engineering, NTUT27/41
7 − ( )
3. (d)
:
4. I
( )2
10 54 2 4 2 4 2 8
5 10
jZ j j j
j
−= + + = + + − =
− +−
5 Ω 4 Ω2 j Ω
10 j− Ω10 0∠
2I
(d)
3I
( )3
10 0 50 A
8 4I
∠= = ∠
( )2 3
10 2 1 5153.4 A
5 10 2 2
j jI I
j
− − += − = = ∠−
( )1 2
2 12 0 1 0.5 1.12 26.6 A
2
jI I I j
− += + = ∠ + = + = ∠
( ) ( ) ( )1.12sin 2 26.6 Ai t t= +
Department of Electronic Engineering, NTUT28/41
•Zth
( ) Zth
: V Z Ioc th sc=
a
b
a
b
Zth
Eth Voc= +−
a
b
a
b
IN Isc= Zth
Department of Electronic Engineering, NTUT29/41
8 −
• (a) a-b
1. Voc
a
b
+−
4 Ω 10 j Ω
6 Ω
I
4 j− Ω
(a)
( )6 0 V∠
a
b
+−
4 Ω 10 j Ω
4 j− Ω
(b)
( )6 0 V∠
+
−
ocV
a
b
4 Ω 10 j Ω
4 j− Ω
(c)
thZ
46 0 3 3 4.24 45 V
4 4oc
jV j
j° °−= × ∠ = − = ∠ −
−
2. Zth
( )4 410 2 8 8.246 75.96
4 4th
jZ j j
j°−
= + = + Ω = ∠ Ω−
4.24 45 4.24 450.375 90 A
2 8 6 11.3 45I
j
° °°
°
∠ − ∠ −= = = ∠ −+ + ∠
3. a
b
+−
2 8 j+ Ω
6 Ω
I
4.24 45 V°∠ −
Department of Electronic Engineering, NTUT30/41
9 −
• (a) a-b I
1. Voc Zth
2. 6Ω
a
b
+−
4 Ω 10 j Ω
6 Ω
I
4 j− Ω
(a)
( )6 0 V∠
4.24 450.515 120.96 A
8.246 75.96oc
th
VI
Z
°°
°
∠ −= = = ∠ −∠
( )2 80.515 120.96 0.375 90 A
2 8 6
jI
j° °+= ∠ − = ∠ −
+ +
b
a
2 8 j+ Ω 6 Ω
I
0.515 120.96 A°∠ −
Department of Electronic Engineering, NTUT31/41
(AC Steady-State Power) (I)
• ( ) ( ) ( ) ( ) ( )1 1cos cosm mp t v t i t V t I tω φ ω φ= ⋅ = + × +
( ) ( ) ( )1 20 0cos cos
T T
m mt tW p t dt V t I t dtω φ ω φ
= == = + × +∫ ∫
( ) ( )1 2 1 2
0
1 1 1sin 2 cos cos
2 2 2
T
m m m m
t
V I t t V I Tω φ φ φ φ θω =
= − + + + − =
( )1 2
1 1cos cos cos cos cos
2 2 2 2m m
m m m m rms rms
V IWP V I V I V I S
Tφ φ θ θ θ θ= = − = = = =
( ) ( )1 2 1 20
1cos 2 cos
2
T
m mtV I t dtω φ φ φ φ
== + + + − ∫
(((( ))))
Peak RMS ( )
cosθ : Power Factor (PF)
( )1 2θ φ φ= −( )
S:
+
−v(t)
i(t)
Department of Electronic Engineering, NTUT32/41
(II)
V I ( )
(a)
( )( )
1 1
2 2
V
Arms
rms
V V V
I I I
φ φφ φ
= ∠ = ∠ = ∠ = ∠
1cos cos
2 m mP VI V Iθ θ= =
cosR
Zθ = ( ) 2cos
RP VI Z I I I R
Zθ
= = =
I+
−V
Z
+
−v(t)
i(t)
j
θ
(a)
Z R jX Z θ= + = ∠
Z
[ ]Re Z R=
[ ]Im Z X=θ Z
Department of Electronic Engineering, NTUT33/41
10
• (a)
(a)
+−
( )i t 10 Ω
0.5 H( )sv t
( ) ( )40sin 2 Vsv t t=
+
−
10 Ω
10 j Ω( )400 V
2sV = ∠
(b)
I
400 V
2V °= ∠
( )( )10 20 0.5 10 10Z R j L j jω= + = + = +
10 2 45 , 45θ° °= ∠ Ω =
( )40 2 02 45 A
10 2 45I
°°
°
∠= = ∠ −∠
( )40cos 2 cos45 40W
2rms rmsP V I θ ° = = =
Department of Electronic Engineering, NTUT34/41
(Complex Power)
• S = P + jQ S VA P WQ VAR
(a)
QS
θP
j
QS
θPj
(b)
cosP S θ=( ):
sinQ S θ=( ):
:( )cosθ:sinθ
( ) ( )rms rmsrms
V VVI I
Z Z Z
φ φ θ φ θθ
∠= = = ∠ − = ∠ −∠
0θ >
0θ <( ) ( )rms rmsI I Iφ θ θ φ∗ = ∠ − − = ∠ −
rms rmsS S V I VIθ θ ∗= ∠ = ∠ =
( )rms rms rms rmsVI V I V I Sφ θ φ θ θ∗ = ∠ ⋅ ∠ − = ∠ = ∠
ReP VI∗ = ImQ VI∗ = S VI∗=
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R L C
• R θ = 0°
=
• ( )
θ +90° – 90°
( )
( )
cos0 1PF = =2
2 rmsrms rms rms
VP S V I I R
R= = = =
( )cos 90 0PF = ± =
j Lω 1j
Cω−
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R
Z
j
θ
−XC
(a)
90 0θ− < <
CZ R jX= − 0 90θ< <
LZ R jX= +
I
Vθ
(b)
RC RL
jXL
Z
R
θ
(c)
I
Vθ
(d)
cosθ θ RC RL( )
I V
I V
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11
•
(1).
(2).
(3).
(4).
(5).
(6).
(7).
10 15Z °= ∠ Ω 50 0V °= ∠
50 05 15 A
10 15
VI
Z
°°
°
∠= = = ∠ −∠
5 15 AI ∗ °= ∠
( )( )50 0 5 15 250 15S VI∗ ° ° °= = ∠ ∠ = ∠
( )250 VAS S= =
15θ °=
cos15 0.966PF °= =
( )cos 250cos15 241.5 WP S θ °= = =
( )sin 250sin15 64.7 VARQ S θ °= = =
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• ZL Zth θL = –θth
ZL ()
•
Ia
b
+−
( Eth )
2 22 2
max 2 4th th
L L L th thth th
E EP I R I R R
R R
= = = =
( )L thZ Z=
L thθ θ= − L thZ Z∗=
thE
th th thZ Z θ= ∠
L L LZ Z θ= ∠
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12
•W
( )( )500 0 2000 90/ / 485.07 14.04 470.58 117.68
500 2000th LZ R X jj
° °°
∠ ∠= = = ∠ = +
+
470.58 117.68L L CZ R jX j= − = − Ω
2 2
max
1207.65W
4 4 470.58th
th
EP
R= = =
×
a
b
RL= ?
X = ?XL= 2kΩ
R = 05. kΩ
RMS+−120 0thE = ∠
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•
•
Department of Electronic Engineering, NTUT
j te ω
j te ω
jωj te ω
jω
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