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Re: [Amps] How intermod limits your PEP

To: "'Dr. David Kirkby'" <david.kirkby@onetel.net>,<amps@contesting.com>
Subject: Re: [Amps] How intermod limits your PEP
From: "Gary Schafer" <garyschafer@comcast.net>
Reply-to: garyschafer@comcast.net
Date: Tue, 16 May 2006 21:42:47 -0400
List-post: <mailto:amps@contesting.com>

> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
> Behalf Of Dr. David Kirkby
> Sent: Tuesday, May 16, 2006 8:15 PM
> To: amps@contesting.com
> Subject: Re: [Amps] How intermod limits your PEP
> 
> Gary Schafer wrote:
> > HOW INTERMOD LIMITS YOUR PEP.
> >
> >
> >
> > Some interesting notes on PEP and intermod.
> 
> 
> I believe your conclusions are invalid.
> 
> > With 1000 watts PEP output and 3rd order IM products down -30 db from
> PEP
> > (-24 db from from one tone of a two tone test ) the useful PEP output is
> > only 873.6 watts. The other 126 watts PEP is wasted power.
> 
> I don't think so - see below.
> 
> > At -20 db 3rd order products from PEP (-14 db from 1 of two tones) the
> > useful PEP output power is only 636 watts.
> >
> >
> >
> > Here?s how it works:
> >
> > P = E squared / R.
> >
> > 1000 watts PEP into 50 ohms is around 223 volts at 50 ohms.
> >
> > -30 db down from 1000 watts is 1 watt. 1 watt is about 7 volts at 50
> ohms.
> >
> >
> >
> > With a two tone test signal there are two signals that are 6 db down
> from
> > the PEP power. 6 db down is ½ the voltage so each of those signals will
> have
> > 111.5 volts. (111.5 volts is =  ~ 250 watts average power in each
> signal.)
> >
> >
> >
> > To find PEP all voltages of all parts of the signal must first be added
> > together. Then squared and divided by 50 ohms. Thus if we add 111.5 +
> 111.5
> > we get 223 volts. That squared and divided by 50 ohms gives us the 1000
> > watts PEP.
> >
> >
> >
> > But we also have our 3rd order distortion products present and it is not
> > just one signal that makes it up but two! We have 1 watt (-30db) signal
> on
> > the high side of the original two tones and we have another 1 watt (-
> 30db)
> > product on the low side of the original two tones. Each of those 1 watt
> > signals contains 7 volts of signal.
> >
> >
> >
> > When we figure PEP we need to add all the voltages together first. So
> with 7
> > volts in each that makes 14 volts of distortion products.
> >
> >
> >
> > 223 volts in the two tone signals plus 14 volts in the two 3rd order IM
> > signals gives us 237 volts. Square that and divide by 50 ohms gives us
> 1123
> > watts PEP.
> >
> >
> >
> > If we limit the amplifier power output to 1000 watts PEP (maybe that?s
> all
> > it is capable of) we first subtract the distortion products to find
> useful
> > PEP output.
> 
> You need to keep them in the same ratio.
> 
> > Since our amplifier is limited to 1000 watts PEP output we must subtract
> the
> > 14 volts from the 223 volts that make up the 1000 watt signal. That
> leaves
> > 209 volts for both of the two tone signals, as high as we can run them.
> > Square 209 and divide by 50 ohms leaves us with 873 watts PEP of useful
> > power for our 1000 watt amplifier.
> 
> So if I am not mistaken, your example keeps to 1000 W PEP by
> 
> 1) Reducing the useful signal
> 2) Keep the IM products at the same level
> 
> That means you have increased the relative level of the distortion. They
> are no longer in the same ratio as before.
> 
> Previously the ratio of signal to IM was
> 
> 233/7, with PEP = 1123 W
> 
> But now it is 209/7 with PEP = 1000 W.
> 
> The error will be small for low values of distortion, but will increase
> with increasing distortion.
> --
> Dr. David Kirkby BSc MSc PhD CEng MIEE
> Chartered Engineer

Hi David,

Yes I am in error, thank you, but only on the examples where the PEP is
greater than 1000 watts (1123 watt example) I believe.
The signal should be 223 (not 233 as you noted).

My attempt is to show how distortion products, however small they may seem,
add to the PEP.

Keeping PEP to 1000 watts would require reducing the signal level as you
noted and keeping the IM level the same. That gives the same number of db
down from PEP for the distortion products.
My attempt was referencing distortion products to PEP and not to signal
level, which I think makes the 1000 watt display correct.

My first example of PEP being greater than 1000 watts would indeed give a
ratio where the distortion products would be better than 30 db below PEP but
not by much. It would be 30.5 db rather than 30 db if I am now doing that
right.

It would probably be easier to just relate distortion products to the level
of one of the signals of the two tone signal rather than to PEP and show how
PEP changes.

Thanks for the comments.
73
Gary  K4FMX


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