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!