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zeptomole

Member Since 2004

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Tuesday Sep 07, 2004

Sep 7, 2004
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Let's take an incoming analog transmission. Any kind you like, over any medium. Myself, I prefer light signals, converted into electrical impulses by fields of rod/cone wavelength/intensity sensors, transmitted along an optic nerve.

Take your signal and realize that, as it travels over its imperfect medium, it will become corrupted in some way. Assuming that an incoming signal can be compared versus a reference signal, you can measure the deviation from the reference and call the resulting difference 'noise'.

Thus you get a signal-to-noise ratio. The rigorous definition involves measuring both the reference signal and the noise in microvolts; a simple conversion results in a ratio measured in decibels. SNR = 20 Log10(Vs/Vn), I believe. You may leave rigor behind and simply apply the logarithm to your own signal measurements.

The logarithm is then simply a translation, a simplifier. If your measured noise is ever greater than your measured signal, the logarithm will cast the resulting calculation into the negative. If your signal is greater than the noise, your calculation will be positive.

Back to the task at hand. I am measuring the analog delights of my visual senses, light focused and constrained by both naturally-occuring and artificially-produced lenses, with its quantum payloads continually emptied out onto the surfaces of my retina.

Note that there is no reference signal for this. Note that this phonomenon, this lack of an absolute reference, has been much discussed; I'll invoke Descartes and be done with the matter. Your own investigations into this are encouraged but it is tangential to our current discussion.

As there is no reference signal, if a ratio is to be measured, a number for the noise must be estimated. Luckily, a substitute for noise exists in this case. An unreliable memory informs me that signals along the optic nerve peak at roughly 100 microvolts, or 0.1 millivolts. If this signal is measured in millivolts, let us use a seemingly unrelated number - blood/alcohol level - as noise.

Thus : we end up with the following equation. D = Log10((2Von)/BA), where Von is the strength of the signal along the optic nerve measured in millivolts, and BA is your blood-alcohol level. The two is a constant applied for cleanliness of the result. Assuming that your optic nerve is continually excited, result D will only be negative when your blood-alcohol level climbs above 0.2 percent*. Thus we end up with a simple identifier, the sign of D, which states whether or not we are extremely drunk.

Through a steady and thorough application of Seagram's Seven on the lining of my alimentary canal, I achieved a negative D yesterday.



*Note that when your optic signals die down, the ratio falls into the negatives quite quickly. This explains the phenomenon of why you feel so drunk when you close your eyes - the optic signal drops magnitudes, yet the blood-alcohol level stays level, resulting in a highly negative ratio.

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