l05 kvs sinusoidal steadystate full
TRANSCRIPT
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
1/81
ESc201: IntroductiontoElectronics
SinusoidalSteadystateAnalysis
r. . . r vas avaDept. of Electrical Engineering
IIT Kanpur
1
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
2/81
2 w i r eL o s s i R==
2 .2 2 .2 1K W K V A= 2 .2 2 2 0 1 0K W V A=
2
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
3/81
Communication
20 Hz -20KHz
3
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
4/81
4 1( ) s in ( )
n tf t
=
1,3,5
4
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
5/81
d S in x d v
o s x in xd x= =
c d t=
( 9 0 )S in x d x C o s x S in x= =
v L
d t
=
So as a sinusoidal signal goes through a circuit, it remains a
This makes analysis easier
s nuso
5
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
6/81
VIN
VC
VL6
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
7/81
VIN
VC
VL
Voltage everywhere in the circuit is sinusoidal 7
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
8/81
Transient and Forced Response
8
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
9/81
Transient and Forced Response
9
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
10/81
10
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
11/81
( ) c o s( )m
v t V t = +
11
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
12/81
VM
( ) c o s( )mv t V t = +
12
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
13/81
)604sin(5 otExample-1
What is the amplitude, phase, angular frequency, time period,
frequency?
( ) c o s( )mv t V t = +
c o sv t t=
Amplitude = 5 ; Phase = -150o
Phase in radians:
360o = 2
1 5 02 2 .6 1 8 ra d ia n s
3 6 0
= =
4 /r s =1
4 0 .5T sT
2= = =
z
T
= =13
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
14/81
Example-2 Find the phase difference between the two currents
1 4 sin(377 25 )oi t= + 2 cos t=
c o smx x = +
o o o
1 cos= 1
o o= o=2
sin( 180 ) sinot t = o
cos( 180 ) coso
o
t t =
=
1 2 =
cos( 90 ) sino
t t =
much?14
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
15/81
2 2( ) c o s( )mv t v t = 1 1( ) c o s( 6 0 )o
mv t v t = +
15
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
16/81
Voltage leads current by 90o or lags current by 270o ?
Phase difference is usually considered between -180 to 180o
Add or subtract 360o to bring the phase between -180 to 180o16
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
17/81
1 4 cos(377 65 )oi t=
2 5cos(377 140 )oi t= +
1 2 205o
=
205 360 155o o o = + =
1 ea s 2 y
17
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
18/81
Power dissipation with sinusoidal Voltage
R
( )v tp
R=
18
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
19/81
Power dissipation with sinusoidal Voltage
2
( )v tp=
19
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
20/81
Average
X: x x x ..x1
a v g ix x=
If X is continuous, its average over a time t1
11( )
t
a vx x t d t= 1 0
or per o c s gna s
1 T
=
0
a v gT
20
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
21/81
Average Power21 ( )
Tv t
= 0a v g
T R
2V
We would like to express it like the dc power dissipated in a resisto
2
pR
=2)(
1dttv
T
R
pavg =
21 ( )T
r m sV v t d t =2
r m sV
p =0
This is true for any periodic waveform 21
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
22/81
RMS Value of a Sinusoid
T
2
0
( )r m sV v t d t T
=
( ) c o s( )mv t V t = +
2 1 c o s( 2 2 )c o s ( )2
T Tt
t d t d t
+
+ =
0
10 .5 s in ( 2 2 ) 0 .5TT t T = + =
mr m s
VV =
22
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
23/81
Power dissipation with sinusoidal Voltage
R
( ) c o s( )m
v t V t = +v
2r m s
a v g
V
p = m
r m s
V
V =
2
m
Vp =
( ) c o s ( )i t I t = +
IT 21
2
r m s=
0
( )r m s
I i t d t
T
=232
a v g m=
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
24/81
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
25/81
Performing algebra on sinusoids by representing them as
complex numbers
Strategy
Sinusoidal Complex Perform Algebra onvariables variables Complex variables
Transform complex variable
Back to sinusoid
25
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
26/81
26
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
27/81
27
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
28/81
28
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
29/81
|z|
Polar form:29
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
30/81
Rectangular Polar form
30
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
31/81
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
32/81
?
32
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
33/81
33
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
34/81
To add or subtract two complex numbers, convert
the operation
34
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
35/81
35
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
36/81
( ) c o s ( )m
v t V t = +
( )( ) R e ( )j tm
v t V e +=
( ) R e ( c o s( ) s in ( ))m m
v t V t jV t = + + +
( ) c o s( )mv t V t = + R e ( )V t +
mV ( ) c o s ( )mv t V t = +
Phasor36
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
37/81
( ) 3 c o s ( 4 5 )v t t= + 3 4 5
3 c o s( 4 5) 3 s in (4 5 )j+
5 6 0 ( ) 5 c o s( 6 0 )v t t=
37
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
38/81
452045cos20 t
=
38
Complex Impedances
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
39/81
Complex Impedances
,and capacitors can be represented as Complex Impedances
39
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
40/81
40
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
41/81
9 0L MI I = L MV L I =
9 0 9 0L MV L I = +
9 0 9 0V I L =
9 0V I L=
V I j L=
L L LV I Z= LZ j L=
This is l ike ohms law relationship between phasor
voltage and current 41
Example-4
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
42/81
L=0.1HExample 4
i(t)
( ) 2 c o s ( 2 0 0 4 5 )v t t= + 2 0 0 =V rad/s
2 4 5L
V = LL L L
V I j L I j L
= =V
2 4 5 2 4 5 0 .1 4 52 0 2 0 9 0
LI
j = = =
A
( ) 0 .1 c o s( 2 0 0 4 5 )i t t= A
42
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
43/81
=L=0.1H LL
v(t) V
V2 0 0 = rad/s2 0
Lj
=
Carry out analysis with phasors keeping in mind that we canalways transform phasor to the sinusoidal voltage or current
as e case may e.
43
ResistorR
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
44/81
ResistorR
v(t)
c o sMv t t= + ( ) c o s( )Mi t tR
= +
MR
VI
R=
R MV V =
RV
RR
44
L=0 1H R=50Example-5 i(t)
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
45/81
L=0.1H R=50Example 5 i(t)
( ) 2 c o s ( 2 0 0 4 5 )v t t= + 2 0 0 =V rad/s
50j20
L
2 4 5SV = V 45
Example-5
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
46/81
50+j20
p
Zeq
=S
2 4 5 2 4 5 . .5 0 2 0 5 3 .8 5 2 1 .8j
= = =+
( ) 0 .0 3 7 c o s ( 2 0 0 2 3 .2 )i t t= + A
46
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
47/81
50j20
2 4 5SV = V
5 02 4 5
5 0 2 0RV =
+V
47
Capacitor
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
48/81
p
c o sv t V t = +
c
ci C d t=
( ) s in ( )Mi t C V t = +
( ) c o s( 9 0 )oM
i t C V t = + +
9 0C M
I C V = +C MV V =
In a capacitor, current leads voltage by 90048
CapacitorV V
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
49/81
C MV V =
9 0C M
I C V = +
9 0C MI C V =
C CI j C V=
C C CV I Z= CZ jj C C = =
49
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
50/81
50
Example-6
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
51/81
V
51
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
52/81
52
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
53/81
V
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
54/81
1 0 0 3 0V =
0 .7 0 7 1 5I =
54
Example-7
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
55/81
55
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
56/81
V
56
Currents 5 0 5 0R C
Z j=
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
57/81
V
A
57
Example-8
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
58/81
58
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
59/81
59
Example-9
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
60/81
60
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
61/81
61
Example-10
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
62/81
62
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
63/81
63
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
64/81
64
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
65/81
65
sinusoidal sources
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
66/81
66
Example-11
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
67/81
67
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
68/81
68
Power dissipation in RLC Circuits
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
69/81
R 2( )v tp
R=
2
0
1 ( )Ta v g
v tp d tT R
=
v(t)
2r m sV=R
m
V
2r m s
a v g r m sp I R=
2
mr m s
I =69
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
70/81
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
71/81
General Rule ( ) ( )mv t V C o s t =
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
72/81
m o s =
1( ) ( )
T
p v t i t d tT=
For a resistor PF = 1, while for inductor and capacitor it is 0
1 19 0 ; 9 0j L L j = =
is phase difference between voltage and current72
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
73/81
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
74/81
1 0 0 1 4 .1 4 4 5L
V j I= =
4 5 1 3 5 9 0 = + =
2 0 .5R r m sP I R W = = 22 1
2
R rm s RP I R I R= =74
Should a Power company charge a person even though power
consumed is zero?
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
75/81
L v(t)PowerMeter
=a v gRwire
PowerMeter
v(t)L
Power is dissipated and
Rwire
somebody has to pay for it.
75
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
76/81
76
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
77/81
77
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
78/81
2sin | | sin | || |
rms rms rms rms rmsQ V I I Z I I Z
Z = = =
78
Maximum Power Transfer for sinusoidal input
Z R jX= +
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
79/81
Z
ZLVS L L LZ R jX= +L
22 1= =
S
L
VI =
2
2 2
1 SL L
VP R=
2r m s
L L+ + + L L
For maximum load power : X = -X
2
2S
L L
VP R=
LZ Z=
L
Choose RL = R to maximize load power
Maximum power is transferred to the load when load is
complex conjugate of source impedance 79
Maximum Power Transfer for sinusoidal input
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
80/81
conjugate of source impedance
Z Z=
L=
80
Maximum Power Transfer for sinusoidal input when load is
Resistive complex
-
8/12/2019 L05 Kvs Sinusoidal Steadystate Full
81/81
Z
L
Maximum power is transferred to the load whenLR Z=
2 25 0 5 0 7 0 .7 1L
R Z= = + =
81