Dr. David Kirkby wrote:
Sorry, I made a few errors. Here are some corrections.
> V4 = B Sin( 2 Pu f4 t + phi4)
That should have been Pi, not Pu.
> Define the IMD (in dB) as
>
> 10 Log_base_10(P1/P3)
>
> Since P is proportional to V^2, it follows that
>
> IMD = 10 log_base_10(V1^2 / V1^2)
That should have been
IMD = 10 log_base_10(V1^2 / V3^2)
> = 20 log_base 10(V1/V2)
Oops, another mess up, that should have been
20 log_base 10(V1/V3)
>
> The PEP, which is the average power at the peak of the waveform occurs
> when the 4 voltages V1, V2, V3 and V4 are all in phase.
>
> The PEP is (2A + 2B)^2 / 2 R.
Perhaps I should have written that as (2A + 2B)^2 / (2 R), to avoid any
confusion.
> for anyone interested, but basically I solve for the voltages A and B,
> subject to these two equations
>
> PEP == (2 A + 2 B)^2 / (2 R)
> IMD == 20 Log[10, A/B]
I've not no negative sign. Whether you call the IM product 20 dB or -20
dB is more a matter of convention than anything else. I have not really
used the convention, but the maths is right (I think).
> Let Mathematica crunch it.
>
> sparrow /export/home/drkirkby % math
> Mathematica 5.2 for Sun Solaris (UltraSPARC)
> Copyright 1988-2005 Wolfram Research, Inc.
> -- Motif graphics initialized --
BTW, that is over $3000 on my Sun, and $1880 on a Windows or Linux PC,
so is not cheap, but as I said, you can get a 14-day demo.
> In[1]:= Solve[{PEP == (2 A + 2 B)^2 / (2 R), IMD == 20 Log[10, A/B]},{A,B}]
> Lets compute the power in these two signals.
>
> P1=P3=158.112%2/ (50 * 2) = 249.994 W
Ooops, that should be P1=P2=249.994W.
P3 and P4 are vitually zero.
--
Dr. David Kirkby BSc MSc PhD CEng MIEE
Chartered Engineer
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