[Physltest] [Phys-L] Re: cable number?

[John Denker <jsd@AV8N.COM>]

Anthony Lapinski wrote:
> My colleague has a two-prong BNC cable splitter that reads
> "3.5 dB" on each connector.

If it only has two terminals, it can't be a splitter.

If it's a splitter of the simplest kind, it must have
three terminals, not two:  one of type A (commonly used
as the input) and two of type B (commonly used as the
outputs) ... although the splitting is of course unitary
and indeed reciprocal, so the B-terminals can perfectly
well be used as inputs, so that the A-terminal is the
output, if you want to use the device as a combiner
rather than as a splitter.

> Are these signal losses?

Sure.  More precisely, the official term for this is
_insertion loss_ since it is the loss you get by
inserting the splitter into the circuit.

Obviously it would make sense for each of the two B-
terminals to be labelled 3.5 dB.  If the the two outputs
each get an equal share of the power, then they can get
at most half of the power, i.e. 3.0 dB below the input
level.  In practice you get something like 3.5 not 3.0 dB
of insertion loss due to imperfections such as reflections
due to impedance mismatches here and there.



To answer a question that wasn't asked:

Less obviously, with high probability it would make
sense for all _three_ terminals of the splitter (A,
B1, and B2) to be labelled 3.5 dB.  That's because
practically every three-terminal splitter is built
as a four-terminal splitter (A1, A2, B1, B2) with
the A2 terminal terminated internally.  So the loss
from B1 to A1 is 3.5dB, and the loss from B2 to A1
is the same.

As previously mentioned, the first law of thermodynamics
i.e. conservation of energy tells you that for any
splitter of this kind, there must be at least 3dB
of loss going from A to B1.  That argument does not
however explain why there is a 3dB loss going from
B1 to A  Imagine car traffic splitting at a fork
in the road.

.                       _ _ _ B1
.                     /
.     (west)  A______/          (east)
.                    \
.                     \  _ _ _B2


For traffic flowing west-to-east, you can imagine
every second car taking the upper route, and the
remaining cars taking the lower route, so the
flux drops by 3dB in each case.  For traffic
flowing east-to-west, there is no reason for cars
to disappear, so you would think that there would
be 0dB of loss going from B1 to A.

But it doesn't work that way for EM waves in a
passive device like a splitter.  To see why, you
need to invoke the _second_ law of thermodynamics.
If I terminated each of the three terminals (A, B1,
and B2) with a 50-ohm resistor, the resistor at A
would get twice as hot as the other two resistors,
since it would be getting twice as much black-body
radiation as the B-resistors.

(You can't reach the same conclusion about cars,
because they are spectacularly non-passive and
dissipative ... there are lots of other degrees
of freedom that you would have to account for.)

In any case, the picture you want to have in
mind for any splitter/combiner is this:


.     A1 _ _ _      _ _ _ B1
.             \   /
.              \ /
.              / \
.     A2_ _ _ /   \  _ _ _B2


where the voltages (not powers) are given by the matrix
equation

.           [ vB1 ]    [  cos(th)   sin(th) ]  [ vA1 ]
.           [     ] =  [                    ]  [     ]
.           [ vB2 ]    [ -sin(th)   cos(th) ]  [ vA2 ]

and since the matrix has the form of a rotation matrix
you can be sure that it is unitary without needing to
think about it.  In an ordinary "3dB" splitter/combiner
the mixing angle is th = 45 degrees.  Also useful are
"20dB" splitters, where the angle is smaller, namely
0.1 radian (5.7 degrees).

Again:  The A2 port may be invisible, because it is
internally terminated rather than being brought to
the outside ... but it *must* exist in principle.
Otherwise the device would
  -- violate Liouville's theorem (conservation of
   phase space)
  -- violate the 2nd law of thermodynamics
  -- violate the fluctuation/dissipation theorem
  -- violate the Heisenberg uncertainty principle

Once you have used the 2nd law to convince yourself that
an internally terminated A2 port must exist, you can
verify its existence using the first law.  If you shine
power in the B1 port, only half of it comes out the A1
port, and the rest causes the splitter to get warm.

> His three prong cable reads 3.5 dB, 7 dB, and 3.5 dB.

First of all, I've never seen a "three prong cable".
I hope that was supposed to say three-prong "splitter"
or some such.

Secondly, if those numbers apply to three outputs
on one side (B1, B2, B3) that talk to an un-numbered
input on the other side (A), that's verrry interesting
because it violates conservation of energy.  We have:

    loss         power
    3.5dB       0.446684
    7.0dB       0.199526
    3.5dB       0.446684
               ----------
                1.09289 total

so we get out more power than we put in.  When
you figure in the inevitable internal losses, the
violation is even more spectacular.

Save that splitter!  It is tremendously valuable
as a source of unlimited power for free ... assuming
you believe the markings........

===============================================
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BTW I just got 156 spams in the last ten minutes.  I
normally get 100 to 200 per day, but not all at once!
Gaaack!

I'm mentioning it because they all have forged return
addresses corresponding to phys-l members, so I suspect
other members are going to be hit.  I'll let you know
if I figure out the source(s).
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