On Mon, 4 Feb 2008 12:32:39 -0800 (PST)
Keith Dysart wrote:
On Jan 28, 11:30 am, Cecil Moore wrote:
Keith Dysart wrote:
Unfortunately, this is quite wrong. And I continue to
be surprised that you argue that there is a reflection
where there is not an impedance discontinuity.
Since an absence of reflections violates the conservation
of energy principle, there is something wrong with your
assertion and your earlier example was proved to contain
a contradiction.
Psource = Pfor - Pref = Pload
Pfor = Psource + Pref = Pload + Pref
Those equations are true only if reflected energy
does not flow back into the source. I suspect that,
contrary to your assertions, the actual real-world
source presents an infinite impedance to reflected
waves.
Your suspicion is quite incorrect. Real-world generators
which present well defined impedances to the reflected
wave are common.
And your assertion that a lack of reflections will
result in a violation of the conservation of energy
principle is also incorrect.
The equations you present above hold just as well
when there is no reflection at the source.
It might do to analyze a simple example. To simplify
analysis, this example does use ideal components:
there is an ideal 50 ohm transmission line (R=G=0,
uniform 50 ohm impedance along its length), two
ideal resistors (zero tempco, no inductance or
capacitance, just resistance), and an ideal voltage
source (desired voltage is produced regardless of
load impedance, though the external components mean
this source does not need to provide or sink infinite
current).
If you don't think any understanding can be gained
from examples with ideal components, please read
no further.
This simple example has a voltage source in series
with a 50 ohm resistor, driving 45 degrees of 50
ohm line connected to a 150 ohm load.
Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs 45 degrees \ Rl
100 cos(wt) 50 ohm line / 150 ohm
| \ load
| |
+--------------+----------------------+
gnd
The voltage source in series with the 50 ohm
resistor forms a generator.
Vs(t) = 100 cos(wt)
or, in phasor notation
Vs = 100 /_ 0d
where d is degrees or 2*pi/360 when converting to radians
is appropriate.
When the source is turned on, a wave travels down the
line. At the generator terminal (Vg), the forward voltage
is:
Vf.g(t) = 50 cos(wt)
Vf.g = 50 /_ 0d
since the voltage divided evenly between the source
50 ohm resistor and 50 ohm impedance of the line.
At the load, the voltage reflection coefficient is
RCvl = (150-50)/(150+50)
= 0.5
The forward voltage at the load is
Vf.l(t) = 50 cos(wt - 45d)
Vf.l = 50 /_ -45d
The reflected voltage at the load is
Vr.l(t) = 0.5 * 50 cos(wt - 45d)
= 25 cos(wt - 45d)
Vr.l = 25 /_ -45d
The voltage at the load is
Vl = Vf.l + Vr.l
= 50 /_ -45d + 25 /_ -45d
= 75 /_ -45d
Vl(t) = 75 cos(wt - 45d)
At the generator, the voltage reflection coefficient is
RCvg = (50-50)/(50+50)
= 0
so there is no reflection and the system has settled
after one round trip.
The voltage at the generator is, therefore,
Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-45d -45d)
= 50 /_ 0d + 25 /_ -90d
= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)
Whoa! You started with 50 volts applied to the forward line, then froze the wave by finding the time phase of the wave. Mathematically, you want to find the voltage at the generator which is the sum of the terms Vf.g and Vr.g. The sum of those terms is 50 + 0 = 50, not 55.902 /_ -26.565d.
Currents can be similar computed...
If.g(t) = 1 cos(wt)
If.g = 1 /_ 0d
RCil = -(150-50)/(150+50)
= -0.5
If.l(t) = 1 cos(wt - 45d)
If.l = 1 /_ -45d
Ir.l(t) = -0.5 * 1 cos(wt - 45d)
= -0.5 cos(wt - 45d)
Ir.l = -0.5 /_ -45d
Il = If.l + Ir.l
= 1 /_ -45d - 0.5 /_ -45d
= 0.5 /_ -45d
Il(t) = 0.5 cos(wt - 45d)
And just to check, from V = I*R
R = V / I
= Vl(t) / Il(t)
= (75 cos(wt-45d)) / (0.5 cos(wt-45d))
= 150
as expected.
RCig = -(50-50)/(50+50)
= 0
Ig = If.g + Ir.g
= 1 /_ 0d - 0.5 /_ (-45d -45d)
= 1 /_ 0d -0.5 /_ -90d
= 1.118 /_ 26.565d
Ig(t) = 1.118 cos(wt + 26.565d)
The same error here.
Ig = 1 /_ 0d - 0.5 /_ -90d
= 1 - 0
= 1
Now let us see if all the energy flows balance properly.
The power applied to the load resistor is
Pl(t) = Vl(t) * Il(t)
= 75 cos(wt - 45d) * 0.5 cos(wt - 45d)
Recalling the relation
cos(a) * cos(b) = 0.5(cos(a+b)+cos(a-b))
Pl(t) = 75 * 0.5 * 0.5 (cos(2wt-90d)+cos(0))
= 18.75 cos(2wt-90d) + 18.75 cos(0)
= 18.75 cos(2wt-90d) + 18.75
Since the average of cos is 0, the average power is
Pl.avg = 18.75
To confirm, let us compute using
Pavg = Vrms * Irms * cos(theta)
Pl.avg = (75 * 0.707) * (0.5 * 0.707) * cos(-45d-(-45d))
= 18.75 * 1
= 18.75
Agreement.
Now at the generator end
Pg(t) = Vg(t) * Ig(t)
= 55.902 cos(wt - 26.565d) * 1.118 cos(wt + 26.565d)
= 55.902 * 1.118 * 0.5 (cos(2wt)+cos(-53.130))
= 31.249 cos(2wt) + 31.249 * 0.6
= 31.249 cos(2wt) + 18.75
Pg.avg = 18.75
No. Based on the way I read your CORRECTED sums, the power at the generator Pg should be 50 * 1 = 50.
Good news, the average power into the line is equal to
the average power delivered to the load, as required to
achieve conservation of energy.
Now let us check Pf and Pr
Pf.avg = Vfrms^2 / Zo
= (50*.707)^2 / 50
= 25
Pr.avg = Vrrms^2 / Zo
= (25*.707)^2 / 50
= 6.25
Pnet.avg = Pf.avg - Pr.avg
= 25 - 6.25
= 18.75
Again, as expected, the average energy flow in the line
is equal to the average power into the line and the average
power out of the line.
So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.
It is instructive to continue the cross-checking.
Let us validate the input impedance of the line.
Using a Smith chart, and going towards the source from
a 150 ohm load on a 50 ohm line yields
Zin.smith = 30 - j40
= 50 /_ -53.130
Dividing the voltage by the current at the input to the
line...
Zin = Vg / Ig
= 55.902 /_ -26.565d / 1.118 /_ 26.565d
= 50.002 /_ -53.130
= 30.001 -j 40.001
yields the same result as the Smith chart.
Lastly let's compute the power provided by the source...
Ps(t) = Vs(t) * Is(t)
= 100 cos(wt) * 1.118 cos(wt+26.565)
= 111.8 * 0.5 (cos(2wt+26.565) + cos(-26.565))
= 55.9 cos(2wt + 26.565) + 55.9 * 0.894
= 55.9 cos(2wt + 26.565) + 49.999
Ps.avg = 49.999
and that dissipated in the source resistor
Prs(t) = Vrs(t) * Irs(t)
= (Vs(t)-Vg(t)) * Ig(t)
= (100 cos(wt) - 55.902 cos(wt - 26.565d)) * 1.118 cos(wt +
26.565d)
= 55.902 cos(wt + 26.565d) * 1.118 cos(wt + 26.565d)
= 55.902 * 1.118 * 0.5 (cos(2wt + 53.130) + cos(0))
= 31.249 cos(2wt + 53.130d) + 31.249
Prs.avg = 31.249
The power provided by the source is equal
to the power dissipated in the source resistor and the
power dissipated in the load resistor.
49.999 = 31.249 + 18.75
So all the energy is accounted for, as expected.
When the source impedance is the same as the line
impedance, there are no reflections at the source
and all the energy is properly accounted. There is
no violation of the conservation of energy principle.
...Keith
PS. It is worthwhile to note that there are a large
number (infinite) of source voltage and source
resistor combinations that will produce exactly
the same final conditions on the line. But if
the source resistor is not 50 ohms, then there
will be reflections at the source and it will
take infinite time for the line to settle to
its final state.
With the circuit and conditions presented, the reflected voltage and current return out of phase in time with the source voltage and current. A standing wave condition of 3:1 would always exist throughout the system.
--
73, Roger, W7WKB