Not quite right:
Intermodulation products are extraneous signals generated by mixing two or
more signals among themselves: they appear at very definite frequencies in
the spectrum and are perfectly correlated with the signals that participated
in their generation.
Cross modulation on the other hand is a mechanism, whereby the information
present on one carrier appears on another carrier in the passband of the
amplifier.
Intermodulation products are multiplicative, not additive: if the first
stage amplifier generates IM products of, say -6o dBc , a second identical
amplifier will generate its own IM products PLUS the Im of the first stage
multiplied by the first stage gain SQUARED. That stems from the higher IM
products of the second stage, driven by the higher sinal: the output of the
first stage.
Alex 4Z5KS
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of Dr. David Kirkby
Sent: Saturday, April 17, 2010 11:53 AM
To: 'AMPS'
Subject: Re: [Amps] A tale of two IMs What happens?
Gary Schafer wrote:
> I worked in the cable TV industry for a short time a long while ago. I
> dug out my old CATV handbook and they give a pretty good explanation
> on intermod although they call it "cross modulation". I quickly looked
> over the part on IM and this is what I came away with:
>
> They say that the easiest way to look at it is to use percentages of
> distortion rather than db. In other words if we have two cascaded amps
> of the same type and they each have .1% IM products then the result
> out of the second amp will be .2%. Adding a 3rd amp of the same type
> also with .1% IM would yield a total of .3% IM out of the 3rd amp.
> So it seems that the IM products directly add together.
> They go on to say that with 2 amplifiers that gives an increase of 3
> db in total IM over a single amp.
>
> Now the amps in this case were tube amps of the "distributed" type,
> which makes them very broad band and they hold their phase shift
> constant. They refer to them as "well behaved" amplifiers.
>
> With "not so well behaved" amplifiers the total IM output is a
> different story. They show a typical dip in IM products in the mid to
> 3/4 output range. So IM products are higher at around 1/3 power, dip
> down at around 3/4 power and are highest at full output.
> I suspect that most of that would come from phase shifts in the amps
> being "not well behaved amplifiers".
>
> The not so well behaved amplifier scenario is probably similar to
> amateur exciter/amp setups. In a post I made on this a few days ago I
> mentioned that in most amplifiers there is a dip in the IM products
> mid way in 3rd order and a dip in 5th order at a higher level and then
> the IM products increasing again at full output.
> According to Orr these IM products can be pretty closely predicted
> mathematically from the tube curves.
>
> I think that Collins was trying to reduce the total IM products from
> the
> 30S1/KWM2 system when they recommended a specific length interconnect
> cable for the RF input to the amp. They make mention of the concern
> for the phase of the signal at the plate of the driver and the phase
> of the signal at the plate of the amp. I am guessing that the attempt
> was to have some amount of cancellation of the total IM products by this.
>
> 73
> Gary K4FMX
That's interesting. At first I thought this seemed silly, but then I thought
about it, and I believe it may be semi-true in the case of CATV, but not
true at all in the case of amateur radio.
In the case of amateur radio the purpose of the amplifier is to produce a
bigger signal, rather than simply overcome the losses in the cables. For
CATV purposes, the amplifiers many not added to produce a bigger signal, but
just to overcome losses. That is a VERY different case.
If you have to get a signal over a very long distance using cables, then the
losses of the cables obviously become significant, and you might need to
have many amplifiers in series. So the source is connected to the load like
this:
source -> CL -> A -> CL -> A -> CL -> A -> CL -> A -> load
where:
CL = 20 dB of cable loss
A = 20 dB gain amplifier
In the above case, the 80 dB dB gain just overcomes the 80 dB cable loss. So
the main signal at the load is just at the power same level as the main
signal at the source. In this case, I can see how the IM level would
deteriorate rapidly.
At each amplifier, the level of distortion increases, but the level of the
main signal is kept the same.
In the case of an amateur radio power amp, the distortion gets bigger, but
so too does the signal. Since the IMD is given as a ratio, it stay much more
constant.
So I don't believe this 3 dB applies in the context of the amateur radio
amplifier stuck on the end of a rig to get a bigger signal. But I do believe
it could (approximately at least) apply in the case of multiple amplifiers
in series which just have enough gain to overcome the losses.
Dave, G8WRB
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