Second analysis: +0.5 input reflection coefficient

I got enough email responses to my request to make me believe it might
be worthwhile to put together another analysis.

The analysis I did previously is of limited use because of the total
lack of any loss in the system. It has also caused conceptual
difficulties because of the perfect source's -1 reflection coefficient.
So I'll do another analysis with a more realistic source.

The following analysis will illustrate the startup and progress to
steady state of a lossless, one wavelength, open ended 50 ohm
transmission line as before. But connected to the input end of the line
will be a perfect voltage source in series with a 150 ohm resistance.
This is not a Thevenin equivalent circuit, because it isn't meant to be
an equivalent of anything; it's simply a perfect source and a
resistance, both of which are common idealized linear circuit models.
The only difference between this setup and the one I analyzed earlier is
the addition of the source resistance.

I'm going to make a slight change in notation and call the initial
forward and reverse waves vf1 and vr1 respectively, instead of just vf
and vr as before. I apologize if this causes any confusion.

When we initially connect or turn on the voltage source, the
source/resistance combination sees Z0 looking into the line. So the
voltage at the line input is vs(t) * Z0 / (Z0 + Rs), where vs(t) is the
voltage of the perfect voltage source and Rs is the 150 ohm source
resistance, until the reflection of the original forward wave returns.
So I'm going to specify that

vs(t) = 4 * sin(wt)

so that the initial voltage at the line input is simply sin(wt). At any
point that the forward wave has reached, then,

vf1(t, x) = sin(wt - x)

The open far end of the line has a reflection coefficient of +1, so as
before the returning wave is:

vr1(t, x) = sin(wt + x)

(The change from -x to +x is due to the reversal of direction of
propagation. You'll see this with every reflection.) At any time between
when vf1 reflects from the output end (t = 2*pi/w) and when it reaches
the input end of the line (t = 4*pi/w), the total voltage at any point
the returning wave has reached is

vtot = vf1(t, x) + vr1(t, x) = sin(wt - x) + sin(wt + x) [Eq. 1]

When the returning wave vr1 reaches the input end of the line, it
re-reflects to form a new forward wave vf2. But because of the added 150
ohm resistance, the source reflection coefficient is +0.5 instead of the
-1 of the previous analysis, so

vf2(t, x) = .5 * sin(wt - x)

Just after the reflection, the voltage at the line input will be

vf1(t, 0) + vr1(t, 0) + vf2(t, 0) = sin(wt) + sin(wt) + .5 * sin(wt)
= 2.5 * sin(wt)

So we'll see a 2.5 peak volt sine wave at the input from the time the
first returning wave reaches the source until the next one does, or from
t = 4*pi/w to t = 8*pi/w. And anywhere on the line which vf2 has
reached, we'll see

vtot(t, x) = 1.5 * sin(wt - x) + sin(wt + x)

When vf2 hits the far end and reflects, it creates the second reflected wave

vr2(t, x) = .5 * sin(wt + x)

So now at any point where vr2 has reached we'll have

vtot(t, x) = 1.5 * sin(wt - x) + 1.5 * sin(wt + x) [Eq. 2]

There's something worth pointing out here. Look at the similarity
between Eq. 1, which was the total voltage with only one forward and one
reflected wave, and Eq. 2, which is the total with two forward and two
reflected waves. They're exactly the same except in amplitude -- Eq. 2
vtot is 1.5 times Eq. 1 vtot. Also notice that the equation for vf2 is
exactly the same as for vf1 except a constant, and likewise vr2 and vr1.
Every time a new forward wave is created by reflection from the input
end of the line, a new reflected wave is created by the reflection from
the far end. The reflection coefficient at the far end doesn't change,
so the relationship between each forward wave and the corresponding
reflected wave is the same as for any other forward wave and its
reflection. So the ratio of the sum of all forward waves to the sum of
all reflected waves is the same as for any single forward wave and its
reflection. The relationship between forward and reverse waves
determines the SWR, so this is why the source reflection coefficient
doesn't play any role in determining the line SWR. No matter what it is,
it creates a new pair of waves having the same ratio as every other pair.

Let's look at just one more pair of waves.

vf3 = .5 * .5 * sin(wt - x) = 0.25 * sin(wt - x)
vr3 = .5 * .5 * sin(wt + x) = 0.25 * sin(wt + x)

From the time vf3 is created until vr3 returns to the input, or from t
= 8*pi/w to t = 10*pi/w, at the input end of the line we'll see

v(t, 0) = vf1(t, 0) + vr1(t, 0) + vf2(t, 0) + vr2(t, 0) + vf3(t, 0) =
3.25 * sin(wt)

For the previous round trip period the sine wave amplitude was 2.5 volts
and for this round trip period it's 3.25 volts. Where is this going to
end after an infinite number of reflections?

Well, it had better end at 4 volts, the voltage of the perfect source!
At steady state, the impedance looking into the line is infinite, so the
current from the source is zero. (A comparable analysis of the current
waves would show convergence to itot = 0 at the input.) So there's no
drop across the 150 ohm resistor and the line input voltage equals the
voltage of the source.

Let's see if we can show this convergence mathematically. The trick is
to take advantage of a simple formula for the sum of an infinite
geometric series.

If F = a + ra + r^2*a + r^3*a, . . . an infinite number of terms

then F = a / (1 - r) if |r| 1. It's a very useful formula. I learned
it in high school algebra, but see that it's in at least one my calculus
and analytic geometry texts. Try it out, if you want, with a pocket
calculator or spreadsheet.

Going back and looking at the analysis, we had

vtot(t, x) = vf1(t, x) + vr1(t, x)

for the first set of waves. The next set was exactly 1/2 the value of
the first set, due to the +0.5 source reflection coefficient. The next
set was 1/2 that value, and so forth. So the first term of the series
(a) is sin(wt - x) + sin(wt - x) and the ratio of terms is 0.5, so

vtot(t, x)(steady state) = (sin(wt - x) + sin(wt + x)) / (1 - 0.5) =
2 * sin(wt - x) + sin(wt + x)

Using the same trig identity as in the earlier analysis,

vtot(t, x)(steady state) = 4 * sin(wt) * cos(x)

This is the correct steady state solution. You can see that it includes
the standing wave envelope cos(x), as it must.

Because superposition applies, we can add up the infinite number of
forward and reverse waves any way we want, and the total will always be
the same. So let's separately add the forward waves and reverse waves.

vf3/vf2 = vf2/vf1 = 0.5, as it is for the ratio of any two successive
forward waves
vr3/vr2 = vr2/vr1 = 0.5, as it is for the ratio of any two successive
reverse waves

So

vf(t, x)(total) = sin(wt - x) / (1 - 0.5) = 2 * sin(wt - x)

and

vr(t, x)(total) = sin(wt + x) / (1 - 0.5) = 2 * sin(wt + x)

These can be used for steady state analysis, as though there is only one
forward and one reverse wave, and the result will be exactly the same as
if the system was analyzed for each of the infinite number of waves
individually. This is almost universally done; the run-up from the
initial state is usually only of academic interest.

If we add the total forward and reverse waves to get the total voltage,
the result is exactly the same as it was when we summed the pairs of
waves to get the total.

Again I ran a SPICE simulation of the circuit, using a 5 wavelength line
for clarity.

http://eznec.com/images/TL2_input.gif is the voltage at the line input.
As predicted, it starts at 1 volt peak, remains there until the first
reflected wave returns to the input end (at t = 10 sec.), then jumps to
2.5 volts. After another round trip, it goes to 3.25.
http://eznec.com/images/TL2_1_sec.gif shows the voltage one wavelength
(1 second) from the source. Here you can see that the voltage is zero
until the initial forward wave arrives at t = 1 sec. From then until the
reflected wave arrives at t = 9 sec., it's one volt peak. From then (t =
9) until the reflected wave re-reflects from the source and returns (t =
11), we have vf1 + vr1 = sin(wt - x) + sin(wt + x). x is 2*pi radians or
360 degrees, so the voltage is simply 2 * sin(wt). And that's what the
plot shows - a sine wave of 2 volts amplitude. Then vf2 gets added in at
t = 11 sec, for a total of sin(wt - x) + sin(wt + x) + 0.5 * sin(wt -
x), or at the observation x point, 2.5 * sin(wt). At t = 19 sec., vr2
arrives and adds another 0.5 * sin(wt) to the total, raising the
amplitude to 3 volts peak. At t = 21 sec., vf3 arrives, adding another
0.25 * sin(wt) for a total amplitude of 3.25 volts. And so forth. As
with the previous analysis, SPICE shows exactly what the analysis predicts.

http://eznec.com/images/TL2_5_sec.gif is the voltage at the open end of
the line. For the first 5 seconds, the voltage is zero because the
initial wave hasn't arrived. At t = 5 sec., the voltage at the end
becomes vf1 + vr1 = 2 * sin(wt). At 15 sec., it becomes vf1 + vr1 + vf2
+ vr2 = 3 * sin(wt). And so forth, just as predicted by the analysis.

Although I've used a line of the convenient length of one wavelength (or
5 wavelengths for the SPICE run), an open circuited end, and a resistive
source, none of these are required. Exactly the same kind of analysis
can be done with complex loads of any values at the line input and
output, and with any line length. But in the general case, returning
waves don't add directly in phase or out of phase with forward waves,
and a wave undergoes some phase shift other than 0 or 180 degrees upon
reflection. So phase angles have to be included in the descriptions of
all voltages. In the general case it's much easier to revert to phasor
notation, but otherwise the analytical process is identical and the
results just as good.

So far I haven't seen any analysis using alternative theories, ideas of
how sources work, or using power waves, which also correctly predict the
voltage at all times and in steady state.

Because there's so much interest in power, I'll calculate the power and
energy at the line input. But I'll put it in a separate posting.

Roy Lewallen, W7EL