L05 Kvs Sinusoidal Steadystate Full

8/12/2019 L05 Kvs Sinusoidal Steadystate Full

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ESc201: IntroductiontoElectronics

SinusoidalSteadystateAnalysis

r. . . r vas avaDept. of Electrical Engineering

IIT Kanpur

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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=

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Communication

20 Hz -20KHz

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4 1( ) s in ( )

n tf t

=

1,3,5

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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

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VIN

VC

VL6


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VIN

VC

VL

Voltage everywhere in the circuit is sinusoidal 7


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Transient and Forced Response

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Transient and Forced Response

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( ) c o s( )m

v t V t = +

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VM

( ) c o s( )mv t V t = +

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)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


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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


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2 2( ) c o s( )mv t v t = 1 1( ) c o s( 6 0 )o

mv t v t = +

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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


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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

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Power dissipation with sinusoidal Voltage

R

( )v tp

R=

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2

( )v tp=

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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

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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


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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 =

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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=


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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

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|z|

Polar form:29


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Rectangular Polar form

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?

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To add or subtract two complex numbers, convert

the operation

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( ) 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


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( ) 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=

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452045cos20 t

=

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Complex Impedances


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Complex Impedances

,and capacitors can be represented as Complex Impedances

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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


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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

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=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.

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ResistorR


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ResistorR

v(t)

c o sMv t t= + ( ) c o s( )Mi t tR

= +

MR

VI

R=

R MV V =

RV

RR

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L=0 1H R=50Example-5 i(t)


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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


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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

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50j20

2 4 5SV = V

5 02 4 5

5 0 2 0RV =

+V

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Capacitor


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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


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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 = =

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50

Example-6


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V

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V


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1 0 0 3 0V =

0 .7 0 7 1 5I =

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Example-7


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V

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Currents 5 0 5 0R C

Z j=


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V

A

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Example-8


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Example-9


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Example-10


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sinusoidal sources


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Example-11


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Power dissipation in RLC Circuits


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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


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General Rule ( ) ( )mv t V C o s t =


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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


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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?


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L v(t)PowerMeter

=a v gRwire

PowerMeter

v(t)L

Power is dissipated and

Rwire

somebody has to pay for it.

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2sin | | sin | || |

rms rms rms rms rmsQ V I I Z I I Z

Z = = =

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Maximum Power Transfer for sinusoidal input

Z R jX= +


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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


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conjugate of source impedance

Z Z=

L=

80

Maximum Power Transfer for sinusoidal input when load is

Resistive complex


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Z

L

Maximum power is transferred to the load whenLR Z=

2 25 0 5 0 7 0 .7 1L

R Z= = + =

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L05 Kvs Sinusoidal Steadystate Full