Best real world example is an eme capable station blowing a
relay that is rated for 2kw+, the reflected power turns into
a LOT of heat! It also effects coax & hardline cables by
creating Hot spots!
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...
> 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."
> 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)?
> 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.
> 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.
> 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.
> * Robert L. Hummel *
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