Roy Lewallen wrote:
People who are having trouble with the concept of a -1 voltage
reflection coefficient for a perfect voltage source might benefit from
the following exercise:

Look at my first analysis, where the perfect source was connected
directly to the transmission line. Make no assumptions about the
impedance or reflection coefficient it presents. Then, when the
reflection of the initial forward wave returns, calculate the value the
re-reflected wave must have in order to make the sum of the waves
present, which is the line input voltage, equal to the perfect source
voltage. The voltage of the perfect source can't change, by definition.
The ratio of the re-reflected wave to the returning wave is the voltage
reflection coefficient (since we're dealing with voltage waves).

I'll do it for you:

The forward wave was vf(t, x) = sin(wt - x)
The returning wave was vr(t, x) = sin(wt + x)

The returning wave will strike the input end of the line and create a
new forward wave with value vf2(t, 0) = Gs * vr(t, 0) at the input,
where Gs is the source voltage reflection coefficient.

Before the first forward wave returns, we have only vf(t, 0) = sin(wt)
at the input end of the line. This is of course the source voltage.


After the wave arrives and re-reflects, we have at the input end

vtot(t, 0) = vf(t, 0) + vr(t, 0) + vf2(t, 0)
= vf(t, 0) + vr(t, 0) * (1 + Gs)

This must equal the source voltage, which is the line input voltage, and
cannot change. So plugging in values:

sin(wt) = sin(wt) + sin(wt) * (1 + Gs)

Solving for Gs = Gs = -1


I have made no statement about the "impedance" of the perfect source.
The only thing I've required is that the voltage remains constant, which
is the very definition of a perfect source. You can do a similar
exercise to show that the voltage reflection coefficient of a perfect
current source is +1.

Roy Lewallen, W7EL

I see your example as identical to Keith's example of two wave pulses
traveling in opposite directions.

At the point of interaction, Keith's example has a reflection factor of
1 or zero, depending upon whether the waves bounce or pass. Keith's
example is not a short circuit because two pulses of identical polarity
are interacting so a reflection factor of -1 could never exist.

In your example, the presence of voltage from the ideal source creates
conditions identical to Keith's example for the returning reflected
wave. Accepting this premise, then the reflection factor must be either
1 or zero, depending upon whether the waves bounce or pass.

By assuming that the waves reflect at the ideal source, you proved that
the reflection factor is -1, which is the factor for a short circuit.
This can not be the case, so waves must not reflect at the ideal voltage
source, they must pass.

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)

Someone will certainly say the the vtot(t,0) at the source location is
the source voltage, because it is defined that way. The conditions at
point (t,0) itself is actually unknown (because vf mysteriously appears,
and vr disappears by going off the transmission line), but point (t, 0)
is defined by assumptions. Therefore, at

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)


73, Roger, W7WKB