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
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