> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
> Behalf Of Dr. David Kirkby
> Sent: Wednesday, May 17, 2006 9:18 PM
> So this means:
>
> IMD = 0 dB, A=79.0569 V, B = 79.0569 V
> IMD = 5 dB, A=101.203 V, B = 56.9107 V
> IMD = 10 dB,A=120.127 V, B = 37.9873 V
> and so on.
>
>
> As I was saying before, as the IMD decreases, so the signal must go up
> but the IMD voltage must go down.
>
> These numbers are quite different to what Gary got. I think his analysis
> is flawed, but I don't think it is so badly flawed that we should get
> quite different numbers. Anyway, perhaps someone else will look at it.
>
> --
> Dr. David Kirkby BSc MSc PhD CEng MIEE
> Chartered Engineer
----------------------------------------------------------------------------
Hi David,
I see the problem. Your maths are calculating the 30 db down (or 20 db down
in your first example) from one of the two tones power (A) rather than from
PEP even though you say you are using PEP as the reference.
You are coming out with a number 6 db lower than what I come up with, for
the IM voltage level, when I use PEP as the reference.
You are going through a lot of extra math to find the wrong answer.
If you check your answers to find the power contained in each B term you
will find that it is 6 db down from what it should be. 30 db down from 1000
watts should give 1 watt in each B term. Your "B" term is only 1/2 the
voltage it should be and produces .235 watts average.
That in turn makes your voltages too high for the A term (the power in each
tone).
The simplest way to work this out when starting with a given PEP and a known
number of db down for the IM products is to simply find the voltage
contained in each term. Then subtract the voltage contained in the two IM
terms from the PEP (total) voltage. That leaves the voltage contained in the
two original two tone signals. 1/2 that is in each carrier (tone). Or square
that voltage and divide by 50 ohms to find the PEP power contained in just
the two carriers just the way I originally described doing it.
At 50 ohms:
1000 watts PEP = 223.66 volts rms.
30 db down from 1000 watts = 1 watt average power in each IM product.
1 watt = 7.1 volts rms.
223.66 - (2x7.1) = 209.46 volts for the sum of the two carriers. (we
subtract the sum of the two IM products from the PEP voltage)
P= E squared / R.
209.46 x 209.46 = 43873.491
43873.491 / 50 = 877.47 watts PEP generated from the two carriers. (the
useful power out)
The rest of the PEP (122.5 watts) is the result of the two 1 watt average
power distortion products added in to produce the 1000 watts PEP.
The voltage in each carrier (each tone) is 209.46 / 2 = 104.73 volts.
Power = 104.73 x 104.73 = 10968.37.
10968.37 / 50 = 219.37 watts average power in each tone.
Total average power for transmitter = 219.37 + 219.37 = 438.74 watts of
useful power. Plus 2 watts average power of distortion products.
My numbers are slightly different than my original post as I rounded a
little more in the original post.
For those wondering, in order to calculate peak envelope power you must
first add the voltages together of each signal. You can not directly add
powers to find PEP.
73
Gary K4FMX
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