I'm another communications engineer (satellite industry) and I've been attacking the claims of IAUS in America Online's investing forum. Here's an old post I did about the mathematics of the Shannon limit and DWM.
Paul McGinnis / TRADER@cup.portal.com / PaulMcG@aol.com
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I should warn everyone first that this post involves a little math. It uses the Shannon-Hartley theorem on digital modulation, for unencoded systems. [For further information, look at the college textbook, "An Introduction to Communication Systems", Allan Hambley, Computer Science Press, 1990, pages 409-410.] Since the numbers are so absurd, most calculators will not be able to perform this calculation, but I used the EG-02 program for the Macintosh. Even then, I had to use a sum of logarithms (base 10) to get the results, since EG-02 blew up with the absurd numbers.
IAS published their "DWM Test Results" (in the promotional packet they send out) and claimed that they could transmit 1.2 billion bytes per second over a 100 KHz bandwidth. Since a byte has 8 bits, this would be 9.6 billion bits per second or 9.6 Gbps (the Giga means Giga or 10 to the power of 9). I had to use the sum of the logarithmic values (in 1.2 Gbps increments) to calculate the total power required.
A widely used formula for this the Eb/No ratio, is given by the digital modulation formula below. (NOTE: ^ means "to the power of", / means divide by, and * means multiply by. For example, 2^5 would mean 2 to the power of 5 which equals 32.)
Eb
-- = [2^(fb/B) - 1] * B/fb
No
(in dB or decibels, a measurement of signal power, the formula would be)
Eb
-- = 10*log([2^(fb/B) - 1] * B/fb)
No
In this equation, fb is the bit rate (in bps [bits per second), and B is the bandwidth in Hertz. In the case of the DWM claims, the values would be:
fb = 9.6 Gbps or 9.6 * 10^9 or 9,600,000,000
B = 100 KHz (Kilohertz) or 1 * 10^5 or 100,000
This means that the DWM fb/B ratio would be 9.6 * 10^4 or 96,000 . Now, before we go any further, it should be pointed out that the most advanced digital modulation methods available today, such as 32-ary PSK (Phase Shift Keying) have a fb/B ratio around 5.
Crunching through the numbers I found that just having 1.2 Gbps (1/8 of the DWM claim) would require a Eb/No (signal to noise ratio) of 1.9088277 * 10^3608 (this is the raw number, not the decibels). Yes, that means approximately 1.9 followed 3,607 zeros. There isn't enough energy in our entire solar system to provide the power needed by an amplifier with this kind of output. Crunching the numbers further revealed that the total Eb/No would be 288,662 dB, which is extremely absurd. Just tell your local engineer you want an amplifier with an Eb/No of 288,662 dB and see what their reaction is! |
