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