ECE 303 - Sum2014
Solutions Set 7:
Power Flow on Transmission Lines
1
7-1.
Let the voltage across a 100 Ω resistor be
v
(
t
) = 5 cos(2
π
1000
t
) and let the current be
i
(
t
).
(a) Compute and plot the current
i
(
t
) over 0
≤
t
≤
3 msec. (b) Compute and plot the instantaneous
power
p
(
t
) =
v
(
t
)
i
(
t
) over 0
≤
t
≤
3 msec. (c) Compute
P
avg
, the time-average power, using the
time-domain integral given by eq. (3) in the notes.
DETAILED SOLUTION:
(a) Compute the current
i
(
t
)
:
The current is easily computed using Ohm’s Law:
i
(
t
) =
v
(
t
)
R
=
5 cos(2
π
1000
t
)
100
(
a
)
Simplifying (
a
) then gives the desired result :
i
(
t
)
=
0
.
05 cos(2
π
1000
t
)
(
b
)
(b) Compute the instantaneous power
p
(
t
) =
v
(
t
)
i
(
t
)
:
The instantaneous power is given by the product of current and voltage :
p
(
t
) =
v
(
t
)
i
(
t
)
(
c
)
Substitute into equation (
c
) the voltage given in the problem statement and the current from (
b
) :
p
(
t
) = 5 cos(2
π
1000
t
)
×
0
.
05 cos(2
π
1000
t
)
(
d
)
Performing the multiplication in (
d
) then gives the desired result :
p
(
t
) = 0
.
25 cos
2
(2
π
1000
t
)
(
e
)
(c) Compute
P
avg
, the time-average power, using the time-domain integral.
From the notes, the time-domain integral definition of the time-average power is given by
P
avg
=
1
T
p
integraldisplay
T
p
t
=0
p
(
t
)
dt
(
f
)
The period
T
p
is given by the inverse of the frequency of the sinusoid. With
f
= 1000 Hz from the
problem statement, we therefore have
T
p
=
1
f
=
1
1000
= 0
.
001
(
g
)
Substituting (
e
) and (
g
) into (
f
) then gives the following integral :
P
avg
=
1
0
.
001
integraldisplay
.
001
t
=0
0
.
25 cos
2
(2
π
1000
t
)
dt
= 250
integraldisplay
.
001
t
=0
cos
2
(2
π
1000
t
)
dt
(
h
)
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ECE 303 - Sum2014
Solutions Set 7:
Power Flow on Transmission Lines
2
Prob. 7-1 (cont.)
To evaluate (
h
), use the following trigonometric identity :
cos
2
α
=
1
2
+
1
2
cos2
α
(
i
)
Use (
i
) in (
h
) with
α
= 2
π
1000
t
. This gives
P
avg
= 125
integraldisplay
.
001
t
=0
bracketleftBig
1 + cos(2
π
2000
t
)
bracketrightBig
dt
(
j
)
Expanding (
j
) into the two integrals then gives
P
avg
= 125
integraldisplay
.
001
t
=0
dt
+
125
integraldisplay
.
001
t
=0
cos(2
π
2000
t
)
dt
(
k
)
The first integral in (
k
) evaluates to 0.001 and the second integral (containing the cosine) in (
k
)
evaluates to zero. This provides the following answer for the time-average power :
P
avg
=
0
.
125
(
l
)
ECE 303 - Sum2014
Solutions Set 7:
Power Flow on Transmission Lines
3
7-7.
A lossy transmission line operating at
f
0
= 200 KHz has per kilometer propagation constant
γ
= 0
.
8 +
j
5. The characteristic impedance
Z
0
= 226
-
j
35 Ω. A transmitter voltage waveform
v
0
(
t
) =
A
cos(2
πf
0
t
) enters the line at
y
= 0
+
. The time-average power received at a distance
700 meters from the transmitter must be at least 10 milliwatts. Compute the minimum value of
A
which will provide this power.
DETAILED SOLUTION:
Here is the constraint expressed in watts as a mathematical relation:
P
avg
≥
0
.
01
(
a
)
The expression for
P
avg
has been found from the notes to be
P
avg
=
A
2
e
-
2
αx
2
|
Z
0
|
cos
φ
(
b
)
where
|
Z
0
|
and
φ
are obtained from the polar form for
Z
0
:
Z
0
=
|
Z
0
|
e
j φ
= 228
.
7
e
-
j
0
.
154
(
c
)
It should also be recognized that since
α
is given in km
-
1
units then the distance
x
must be in km
units. Therefore, we need to use
x
= 0
.
7 km
(
d
)
in this problem. Now substitute the expression for
P
avg
from (
b
) into the left side of (
a
), and
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- Summer '08
- ALEXANDER
- Volt, Transmission line, Impedance matching, Telegrapher's equations, Characteristic impedance, Solutions Set
-
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