In the clear light of day, reread all the words below. What you've done is
take my fairly clear and correct statements, assumed that I said something I
did not or assume I did not know something I did know and then take issue
with your assumptions. I respectfully suggest that in the future you
concentrate more on helping the original requestor instead of taking issue
with someone else who is trying to help.
Almost all of what we learn we learn by analogy. In some people that
learning is short-circuited by their fear -- I don't know if it's fear of
being wrong or ignorant or what -- so that when they hear the first part of
the analogy, their internal response is, "I know that!" and the new
knowledge is lost.
Guard against that.
73, Mike N4NT
----- Original Message -----
From: "Michael Tope" <W4EF@dellroy.com>
To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>
Sent: Saturday, July 30, 2005 2:07 AM
Subject: Re: [TenTec] Filters, amps and IMD, etc
----- Original Message -----
From: "Mike Hyder -N4NT-" <n4nt_m_o_hyder@charter.net>
> Hi Mike--
>
> As I said, I was over my head. Please help me understand something you
> said
> in your last paragraph, to wit: "If you drop the input signal tones levels
> by 10dB, the 3rd order products will drop 30dB for a net change in
> relative
> IMD level of 20dB (e.g. IMD changes from -20dBc to -40dBc as you drop the
> tone levels 10dB from the amplifiers 1dB gain compression point)."
>
> Please compare that with what I said: "For this reason, a 10dB attenuation
> in the line will attenuate the desired signal by that much but will
> attenuate the intermodulation distortion products by much more" and
> explain
> where my thinking differs from yours.
>
> 73, Mike N4NT
>
Perhaps it is just the way you worded your statement, Mike, but I
took what you said to mean that an attenuator actually reduces IMD
regardless of where it is in the RF chain. An attenuator can't reduce
IMD products that are already present at its input any more than
it can reduce the fundamental signal. It will reduce the IMD at its
input by an amount equal to the amount that it reduces the
fundamental. If the IMD is down 40dB relative to the fundamental
signal at the input of an attenuator, it will still be down 40dB relative
to the fundamental at the output of the attenuator. An attenuator can,
however, reduce the fundamental signal level hitting a downstream
non-linear stage thereby reducing the amount of IMD which that
stage produces. In that case (non-linear stage downstream from the
attenuator) the amount of IMD produced by the offending non-linear
stage will be reduced in greater proportion than reduction in the
level of the fundamental signal (~3 for 1 proportion in the case of
the 3rd order IMD products).
You also said the following:
"Amplification works in the opposite way. It brings up one
signal but brings up the IMD products more than it does the
one signal. In addition, if the amplifier is not perfectly linear, it
serves to accentuate the harmonic products and worsen the
problem (see paragraph 1 above)."
The wording of this statement implies that IMD products are
created even in perfectly linear amplifiers (e.g. your wording
implies that problem is always there, and just gets worse when
the amplifier is non-linear). In point of fact a pefectly linear
amplifier (if you could find one), doesn't generate any IMD
products. Any IMD already present at the input of a linear
amplifier will be amplified by the same amount as the fundamental
signal. Of course, any real amplifier will have some non-linearity.
If you increase the signal level to a typical amplifier that has some
inherent non-linearity, the IMD products will grow by an amount
greater than the increase in input signal level (~3 for 1 in the case
of the 3rd order product). This is where the concept of intercept
point comes from.
If you keep increasing the signal level to a real amplifier (e.g. any
real amplifier will have some inherent non-linearity), since the 3rd
order IMD products grow faster than the increase in fundamental
signal level, eventually the level of the 3rd order products present
at the amplifier output catch up or "intercept" the level of the
fundamental at the output. Actually, however, a typical amplifier
doesn't ever get there as its output will saturate just before it
reaches the 3rd order intercept point, so generally the intercept
point is a mathematical extrapolation from the slopes of
fundamental signal level vs. input power and the 3rd order
product level vs. input power curves.
In any event, perhaps this is what you meant to say, but it
just wasn't clear to me from your wording.
73 de Mike, W4EF................................
> ----- Original Message -----
> From: "Michael Tope" <W4EF@dellroy.com>
> To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>;
> <k4qo@earthlink.net>
> Sent: Friday, July 29, 2005 10:34 AM
> Subject: Re: [TenTec] Filters, amps and IMD, etc
>
>
>
> ----- Original Message -----
> From: "Mike Hyder -N4NT-" <n4nt_m_o_hyder@charter.net>
>
>> IMD is the product of two or more signals. 3rd order IMD is a signal's
>> fundamental mixing with the second harmonic of another signal...
>>
>> For this reason, a 10dB attenuation in the line will attenuate the
>> desired
>> signal by that much but will attenuate the intermodulation distortion
>> products by much more.
>
> That isn't quite right, Mike. This only happens when you put the
> attenuator in front of a non-linear amplifier. In this case, the
> attenuator
> will appear to reduce IMD products more than the fundamental signal
> is, but what is really happening is that you are reducing the drive to the
> non-linear amplifier stage so that it doesn't produce as much IMD.
> If you put a 10dB attenuator after the stage which produces the IMD
> distortion, it will knock the signal down and the distortion products
> by the same amount, 10dB. In other words, if the IMD products are
> already present, the attenuator will knock them down by the same
> amount as the desired signal (the attenuator can't tell the difference
> between the desired signal and the distortion products).
>
>>
>> Amplification works in the opposite way. It brings up one signal but
>> brings
>> up the IMD products more than it does the one signal. In addition, if
>> the
>> amplifier is not perfectly linear, it serves to accentuate the harmonic
>> products and worsen the problem (see paragraph 1 above).
>>
> A perfectly linear amplifier by definition will amplify distortion
> products
> by the same amount as the fundamental signal (amplifiers can't tell the
> difference between the signal we want and the signal we don't want). Most
> amplifier, are somewhat non-linear so they typically add IMD products.
> The amount of IMD that they add depends on how hard you drive the
> amplifier. At the amplifiers 1dB gain compression point, an amplifier
> will typically add 3rd order IMD products that are approximately 20dB
> down from the fundamental tones. If you drop the input signal tones levels
> by 10dB, the 3rd order products will drop 30dB for a net change in
> relative IMD level of 20dB (e.g. IMD changes from -20dBc to -40dBc
> as you drop the tone levels 10dB from the amplifiers 1dB gain compression
> point).
>
> 73 de Mike, W4EF.......................
>
>> I have just exceeded my knowledge.
>>
>> 73, Mike N4NT
>>
>> ----- Original Message -----
>> From: "Randy K4QO" <k4qo@earthlink.net>
>> To: <tentec@contesting.com>
>> Sent: Thursday, July 28, 2005 11:50 PM
>> Subject: [TenTec] Filters, amps and IMD, etc
>>
>>
>> If the amp that kicks in at the tighter bandwidths is the problem, why
>> not just adjust the gain of that amp down a bit. Wouldn't that improve
>> the IMD without resorting to 3rd party filters? Or is it that ANY
>> additional amp stages, regardless of gain causes the IMD to get worse?
>>
>> Iquiring minds wanna know!
>>
>> 73,
>> Randy
>> K4QO
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