-------------- Original message --------------
From: "Dr. David Kirkby" <david.kirkby@onetel.net>
> Gary Schafer wrote:
>
> > 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.
>
> But as you say, as soon as you add any IMD, to a signal, whilst keeping
> the signal level constant at 1000 W PEP, the PEP power will exceed 1000
> W due to the distortion.
>
> Are you talking about an AM signal, with a carrier and two sidebands, or
> an SSB one, with no carrier? I assume it is AM.
>
> > The signal should be 223 (not 233 as you noted).
>
> No, see below.
>
> > My attempt is to show how distortion products, however small they may seem,
> > add to the PEP.
>
> Agreed.
>
> > Keeping PEP to 1000 watts would require reducing the signal level as you
> > noted and keeping the IM level the same.
>
> No!! It distorts (pun intended) the figures.
>
> If you run up your amp to something over 1000 W, then decide to turn it
> down to 1000 W (by for example lowing the input drive), whilst keeping
> the IM products the same amount down (-30 dB to take one of your
> examples), then *both* the signal *and* IMD signal must be reduced - not
> just the signal. So the power of the IM products must fall below 1 W
> each or 7 V RMS.
>
> Symbols (P1, P2, f1, f2 ... etc) would make it a lot easier to see. One
> could then do it for any signal levels or IMD you want.
>
> > 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.
>
> I am a bit confused exactly what you are doing, so are not going to work
> it out. But I think your method is flawed.
>
> > 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.
>
> Well this is one of the things you can do any number of ways - it just
> depends on what you want.
>
> As long as you state it, then it does not matter. But as it is, I don't
> think the explanation is clear. Others clearly disagree, as several have
> commented on it being clear. Perhaps it is just me.
>
> > Thanks for the comments.
>
> You are welcome.
>
> > 73
> > Gary K4FMX
> >
> >
> >
>
>
> --
> Dr. David Kirkby BSc MSc PhD CEng MIEE
> Chartered Engineer
Hi David,
Thanks again for your comments.
To accurately portray what I am saying I believe it would take a more complex
math solution than what I am using. Maybe you or someone else can come up with
the proper solution?
In the example being discussed an ssb signal is modulated with two equal tones.
Not an AM signal with carrier. This will create two IM3 products, one above the
two tones and one below the two tones. The -30 db IM products are each equal to
1 watt when the TOTAL output signal is 1000 watts PEP. In other words the IM
products are 30 db down from PEP.
Of course I manipulated the power level of the two signal tones so that the
total power output remained at 1000 watts PEP with each IM product down 30 db.
Each tone = 104.5 volts x 2 = 209 volts = 873.6 watts PEP total for two tones.
Each IM3 product = 7 volts x 2 = 14 volts total for IM3 products.
Total of all = 223 volts = 1000 watts PEP
This scenario will give a total of 223 volts, including the 14 volts from the
addition of the two IM3 products for a total of 1000 watts PEP.
Subtracting the 14 volts from the 223 total volts leaves 209 volts for the two
tone signals. That equates to 873.6 watts PEP for the two tones without the IM3
products added in.
What I am saying is that the 873.6 watts is the effective PEP even though the
amplifier is putting out 1000 watts PEP.
In a practical amplifier of course the IM products and tone levels would vary
together as power levels were changed. But one could manipulate IM products by
changing amplifier tuning as one way of achieving the above scenario.
But this is just an exercise to create an example to relate how the small
amount of IM power creates a much larger peak envelope power.
I am on a different computer for a few days so I don't have a copy of the
original posts.
Any help on the math would be appreciated.
73
Gary K4FMX
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