Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

18

Power in Circuits

Consider the input impedance of a transmission line circuit, with an

applied voltage

v

(t)

inducing an input current

i

(t)

.

For sinusoidal excitation, we can write



/2 , /2

0

0

( )

cos(

)

( )

cos(

)

v t

V

t

i t

I

t

 









 

where

is the phase difference

between

voltage and current. Note

that

= 0

only when the input impedance is real (purely resistive).

v(t)

i(t)

Z

in

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

19

The time-dependent input power is given by



0 0

0 0

( )

( ) ( )

cos(

) cos(

)

cos( )

cos(2

)

2

P t

v t i t

V I

t

t

V I

t







 





 

The power has two (Fourier) components:

(A) an average value

0 0

cos( )

2

V I

(B) an oscillatory component with frequency

2f

0 0

cos(2

)

2

V I

t

 

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

20

The power flow changes periodically in time with an oscillation like

(B)

about the average value

(A)

. Note that only when

= 0

we have

cos(

)

= 1

, implying that for a resistive impedance the power is

always positive (flowing from generator to load).

When

voltage

and

current

are out of phase, the average value of the

power has lower magnitude than the peak value of the oscillatory
component. Therefore, during portions of the period of oscillation
the power can be negative (flowing from load to generator). This
means that when the power flow is positive, the reactive component
of the input impedance stores energy, which is reflected back to the
generator side when the power flow becomes negative.

For an oscillatory excitation, we are interested in finding the
behavior of the power during one full period, because from this we
can easily obtain the average behavior in time. From the point of
view of power consumption, we are also interested in knowing the
power dissipated by the resistive component of the impedance.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

21

We can also rewrite the time-dependent current as

0

0

0

( )

cos(

)

cos cos

sin sin

i t

I

t

I

t

I

t



 


 



where we have used the trigonometry formula





cos

cos

cos

sin

sin

A

B

A

B

A

B







This result yields an equivalent expression for the power

0

0

0

0

2

0 0

0 0

( )

cos(

)

cos(

) cos( )

cos(

)

sin(

) sin( )

cos( ) cos (

)

sin( ) sin(2

)

2

1

1

P t

V

t I

t

V

t I

t

V I

V I

t

t





















Active (Real

Reactive Power

) Power











Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

22

The

active power

corresponds to the power dissipated by the

resistive

component of the impedance, and it is always positive.

The

reactive power

corresponds to the power stored and then

reflected by the

reactive

component of the impedance. It oscillates

from positive to negative during the period.

Until now we have discussed properties of instantaneous power.
Since we are considering time-harmonic periodic signals, it is very
convenient to consider the time-average power

0

1

( )

( )

T

P t

P t dt

T





where

T = 1 / f

is the period of the oscillation.

We can use either the Fourier or the active/reactive power
formulation to determine the time-average power.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

23

Fourier representation

0 0

0

0 0

0

0 0

1

( )

cos( )

2

1

cos(2

cos

)

2

0

( )

2

T

T

V I

P t

dt

T

V I

t

V

t

T

I

d







 





As one should expect, the time-average power flow is simply given
by the Fourier component corresponding to the average of the
original signal.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

24

Active/Reactive power representation

2

0 0

0

0 0

0

0

0

1

( )

cos( ) cos (

)

1

sin( ) sin(2

)

2

co

0

s( )

2

T

T

P t

V

V

I

t dt

T

V I

t d

I

t

T















This result tells us that the time-average power flow is the average
of the

active power

. The

reactive power

has zero time-average,

since power is stored and completely reflected by the reactive
component of the input impedance during the period of oscillation.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

25

The maximum of the

reactive power

is

{

}

( ) (

)

( )

0 0

0 0

max

max{

sin

sin 2

}

sin

2

2

reac

V I

V I

P

t

=

φ

ω

=

φ

Since the time-average of the reactive power is zero, we often use
the maximum value above as an indication of the reactive power.

The sign of the phase

φ

tells us about the imaginary part of the

impedance (reactance)

φ

> 0

The reactance is inductive

Current is lagging with respect to voltage
Voltage is leading with respect to current

φ

< 0

The reactance is capacitive
Voltage is lagging with respect to current
Current is leading with respect to voltage

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

26

If the net reactance is inductive

V

Z I

R I

j L I





 

I

V

j

L I

R I

Im

Re

Current lags

> 0

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

27

If the net reactance is capacitive

1

V

Z I

R I

j

I

C









- j I /

C

I

V

R I

Im

Re

Voltage lags

< 0

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

28

In many engineering situations, we use the root-mean-square
(r.m.s.) values of quantities. For a given signal

0

( )

cos(

)

v t

V

t





the r.m.s. value is defined as

 

 

 

2

2

2

0

0

0

0

2

2

0

0

0

1

1

cos

cos

1

1

cos

2

2

T

T

rms

V

V

t dt

V

t d t

T

T

V

d

V

















  



This result is valid for sinusoidal signals. Each signal shape
corresponds to a specific coefficient (

peak factor

=

0

/

rms

V

V

) that

allows one to convert directly from peak to r.m.s. values.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

29

The

peak factor

for

sinusoidal signals

is

0

2

1.4142

rms

V

V





For a

symmetric triangular signal

the peak factor is

0

3

1.732

rms

V

V





For a

symmetric square signal

the peak factor is simply

0

1

rms

V

V



V

0

t

t

V

0

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

30

For a

non-sinusoidal periodic signal

, we can determine the r.m.s.

value by using a very important theorem of vector spaces. If we
decompose the non-sinusoidal signal into its

Fourier components

 

1

2

3

( )

( )

( )

( )

av

k

k

V t

V

V t

V t

V t

V

t













then

 







 



2

2

2

( )

k

k

k

k

rms

rms

rms

V t

V t

V

t





So, the r.m.s. value of the signal is computed as

 

 

 

2

2

2

2

1

2

3

rms

av

rms

rms

rms

V

V

V

V

V











Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

31

So far, we have used

peak values

for the amplitude of voltage and

current. In terms of

r.m.s. values

, the time-average power for a

sinusoidal signal is

0

0

( )

cos( )

cos( )

2

2

rms rms

V

I

P t

V

I







Finally, we can relate the time-average power to the phasors of
voltage and current. Since









0

0

0

0

( )

cos(

)

Re

exp(

)

( )

cos(

)

Re

exp(

) exp(

)

v t

V

t

V

j t

i t

I

t

I

j

j t



 





 






we have phasors

0

0

exp(

)

V

V

I

I

j







Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

32

The time-average power in terms of phasors is given by

 





*

0 0

0 0

1

1

( )

Re

Re

exp(

)

2

2

cos( )

2

P t

V I

V I

j

V I









Note that one must always use the complex conjugate of the phasor
current to obtain the time-average power. It is important to
remember this when voltage and current are expressed as
functions of each other. Only when the impedance is purely

resistive,

I = I* = I

0

since

= 0

.

Also, note that the time-average power is always a real positive
quantity and that it

is not

the phasor of the time-dependent power.

It is a common mistake to think so.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

33

Now, we consider power flow including explicitly the generator, to
understand in which conditions

maximum power transfer

to a load

can take place.

 

*

1

1

Re

2

R

in

G

G

R

in

G

G

R

in

in in

Z

V

V

Z

Z

I

V

Z

Z

P

V

I













Z

G

V

G

V

in

I

in

Generator

Z

R

Load

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

34

As a first case, we examine resistive impedances

G

G

R

R

Z

R

Z

R





Voltage and current are

in phase

at the input. The time-average

power dissipated by the load is

*

2

2

1

1

( )

2

1

2

(

)

R

G

G

G

R

G

R

R

G

G

R

R

P t

V

V

R

R

R

R

R

V

R

R













Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

35

To find the

load

resistance that maximizes power transfer to the

load for a given generator we impose

( )

0

R

d P t

dR





from which we obtain

2

2

4

0

(

)

(

)

2

(

)

0

(

)

(

)

2

0

R

R

G

R

G

R

R

G

R

G

R

G

R

R

R

G

d

R

dR

R

R

R

R

R

R

R

R

R

R

R

R

R

R









































We conclude that for maximum power transfer the load resistance
must be identical to the generator resistance.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

36

Let’s consider now

complex impedances

R

R

R

G

G

G

Z

R

jX

Z

R

jX









For maximum power transfer, generator and load impedances must
be complex conjugate of each other

*

R

G

R

G

R

G

Z

Z

R

R

X

X







 

This can be easily understood by considering that,

to maximize

the

active

power

supplied to the load,

voltage

and

current

of the

generator should remain

in phase

. If the reactances of generator

and load are opposite and cancel each other along the path of the
current, the generator will only see a resistance. Voltage and
current will be in phase with maximum power delivered to the load.

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

37

The

total time-average power

supplied by the

generator

in

conditions of maximum power transfer is

 

*

2

2

1

1

1

1

1

Re

2

2

2

4

tot

G

in

G

G

R

R

P

V I

V

V

R

R









The

time-average power

supplied to the

load

is

 

*

*

*

2

2

1

1

1

Re

Re

2

2

1

1

1

Re

2

4

8

R

in

in in

G

G

G

R

G

R

R

R

G

G

R

R

Z

P

V

I

V

V

Z

Z

Z

Z

R

jX

V

V

R

R







































!















!

Transmission Lines

© Amanogawa, 2000 – Digital Maestro Series

38

The

power dissipated

by the

internal generator impedance

is





*

2

2

2

1

Re (

)

2

1

1

1

1

1

1

4

8

8

G

G

in

in

G

G

G

R

R

R

P

V

V

I

V

V

V

R

R

R













We conclude that, in conditions of

maximum power transfer,

only

half

of the

total active power

supplied by the

generator

is actually

used by the

load

. The generator impedance dissipates the

remaining half of the available active power.

This may seem a disappointing result, but it is the best one can do
for a real generator with a given internal impedance!