"Jeff" wrote in
.com:

Owen and Cecil are right: the source (transmitter) has no effect
whatever on the VSWR on the line.

That isn't just an assertion - it is part of the bedrock transmission
line theory. Owen referred to "reputable textbooks", one of which
would surely be 'Theory and Problems of Transmission Lines' by R A
Chipman [1]. This book gains a lot of its reputation from its very
complete mathematical development of the theory, showing all the
detailed working.

I am sorry but you are not correct, I have not read Chipman so I
cannot comment on his analysis or your interpretation of his results,
but my understanding , practical experiments and CAD analysis would
lead me to disagree.

If we take the situation where the source is matched (50ohms) to the
5.35 wavelength transmission line (lossless to simplify things) with a
100ohm load, I agree that the vswr is 4:1, unchanging with frequency.

Plotted on a Smith Chart when swept against frequency this gives a
circle centred on 1 (50ohms) with a radius of 4. i.e. on a constant
VSWR circle.

Now if we change the source impedance to 100ohms and repeat the same
sweep and re-plot, keeping the chart normalized to 50 ohms, the circle
moves on the resistance axis, still with a radius of 4 and now passing
though 2 (100 ohms) resistive. The centre moves to about 0.6 (30ohms).
It then becomes obvious that the locus of the circle is NOT a constant
VSWR against frequency.

You will come to the same conclusion if you normalize the chart to 100
ohms, the new source impedance and re-plot.

The coax is acting as an impedance transformer, causing a shift along
the resistance axis.

Looking at it another way, the vswr changes sinusoidally with
frequency, in our example, between 2:1 and 8:1. (The same as the Smith
chart plot with a circle of radius 4 centred at about 0.6).


If you are asserting that VSWR on a real or even theoretical line varies
sinudoidally with displacement, it is time to go back to basics. You need
some time with a reputable text book.

Owen