Wpower_solutions - Solutions Set 7 Power Flow on...

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 )

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