Robert Hummel wrote:
> >The mismatch of VF has no more effect than the SWR - it all comes out the
> >other end, even though some of it takes a lot of trips back and forth to
> >get there, losing only what is often called the "copper loss," but is
> >really the loss characteristic of the design of each individual cable.
> Can't help but throw in my two cents here...
I'll see your two cents and up you a nickle.
> With the utmost respect to Press (who knows his cable), the idea that "all
> the power goes to the antenna eventually" is one of the most persistent and
> pernicious myths in ham radio. In my mind, it's right up there with "RG-8
> is essentially lossless at HF."
RG-8 in the relatively short lengths used by most amateurs for decently
matched antennas is as lossless as anyone will ever need.
> The idea that an SWR mismatch at an interface doesn't directly cause a loss
> in thru power is valid only in the very special case of steady-state power
> transmission. Amateur radio transmission, however, is not steady-state. It
> is more like impulse transmission, and that makes all the difference.
> Consider a single sharp dit going into a mismatched junction. Part of your
> dit gets transmitted, part gets reflected. At some junction back toward the
> transmitter, lets assume that all the reflected power gets reflected back
> toward the antenna. At the junction, some more of that original dit power
> (now delayed by a coax round-trip) gets radiated. Was that your intent? Did
> you mean to transmit DIT Dit dit .... (echoes continue)?
Consider 300 baud rtty or equivalent cw speed if you can copy that
at worst case you get 300 pulses per second, make it worse by making it
50% duty cycle so each pulse is 1/600th of a second... and since i hate
odd fractions make it even worse at 1msec pulses since that is a nice
round number. now pick a frequency... worst case would be 1.8MHz, but
i like round numbers lets figure it at 1MHz. now, a one 1msec pulse
with a carrier frequency of 1MHz contains 1000 cycles of the carrier.
have a mismatch where 50% of your power is being reflected you will be
within .1% of steady state in 10 cycles leaving 990 steady state ones to
go. double the frequency to get into the 160m band and you have 1990
steady state cycles out of 2000. take that to 10m and it really is
> Now, if you really transmitted a series of dits and dahs or a complex audio
> waveform, all reflected/radiated signals that are not time coincident with
> the original signal contribute to your radiated power, but only as noise.
> The effect is that you are transmitting a noisy signal that's weaker than
> you think.
now, i have shown that a series of normal dits and amateur legal hf data
rates can't do this... but this is an important consideration as has
noted in other applications. consider the data rates seen in video or
high speed digital links. in these the 'echos' of the dits show up as
ghosts on the picture or distorted data bits on the digital stream.
> Consider the analogy of a 4" diameter pipe that goes through an abrupt
> transition to a 2" pipe. At some certain steady, constant pressure on the
> 4" inlet, a measurable amount of water will flow into the pipe and --
> behold! -- the same amount comes out at the 2" end. This is the steady
> state situation.
> But, now consider that you turn on the water briefly, enough to fill the
> entire 4" diameter of the pipe for let's say 1 foot in length. This packet
> of water (neglecting friction the way we neglected loss in the coax
> example) will fly down the 4" pipe like a bullet in a gun barrel. But when
> it hits the 2" abrupt transition, part of the water will be reflected back.
> At the 2" outlet, you will see far less water come out than you put in.
to go back to my discussion above... to fill a piece of coax with a wave
for a 1 foot length.... oh the bad numbers... lets make it a 1 meter
since thats a bit easier to compute. pick a coax with a velocity factor
.66 (standard rg-8 solid poly), this gives a speed of light in the coax
.66*300x10^6 m/s or about 200x10^6 m/s. so a pulse short enough to fill
only 1m of coax would be 1/200x10^6 or about .005 micro seconds long, or
200,000 times shorter than the 1msec pulse considered above.
another way to look at it... a piece of coax long enough to hold that
1msec pulse would be 200km long. its kind of like trying to pump water
bosten to nyc in a garden hose. keep things in scale, rg-8 for hf from
shack to the antenna in your yard is just as good as a garden hose for
from your faucet to your garden... you'll never notice the pressure
but try to use rg-8 or your garden hose to get from boston to nyc and it
falls apart, or try to use rg-8 for microwaves or your garden hose for a
pulsed high pressure washer and see what happens.
> Of course, eventually, all the water may drain out of the pipe. But that
> isn't what you really wanted.
> Repeat this phrase until you believe it: Mismatch loss is REAL loss.
repeat after me...
Mismatch loss is REAL loss, but I'm going to ignore it anyway and have
David Robbins K1TTT (ex KY1H)
firstname.lastname@example.org or email@example.com
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