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On Mon, 1 Jun 2009 07:01:27 -0700 (PDT), GreenXenon
Your FM receiver antenna picks up about 2 or 3 femtovolts from the
airwaves. That ends up getting tuned in as your stereo media source.
quoted:
The SI derived unit for voltage is the volt.
1 volt is equal to 1.0E+15 femtovolt.
Now stop being a cross posting retard when you make these stupid queries
that you are even too dumb to do a simple research hunt for.
Your FM receiver antenna picks up about 2 or 3 femtovolts from the
airwaves. That ends up getting tuned in as your stereo media source.
quoted:
The SI derived unit for voltage is the volt.
1 volt is equal to 1.0E+15 femtovolt.
Now stop being a cross posting retard when you make these stupid queries
that you are even too dumb to do a simple research hunt for.
- Re: Frequency-to-voltage and visa versa at 1.602 =?UTF-8?Q?=C3=97?= 10^-19 volt.
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>:
>> e02d32b9-b1f7-47e9-90f2-5af9088f0...@b1g2000vbc.googlegroups.com...
>> wrote:
>> > f104efd2-6a60-4dc7-9cb2-6eb697a4b...@g19g2000vbi.googlegroups.com...
>> > ...
>> > | In my hypothetical device the input of a signal that has a frequency
>> > | of A Hz and a peak-to-peak amplitude of B volts will result in the
>> > | output of a signal that has a frequency of B Hz and a peak-to-peak
>> > | amplitude of A x [1.602 × 10^-19 volts].
>> > ...
>> > Even if the charge is quantified, a potential difference is not.
>> | Ok.
>> > Why should a pp amplitude be a multiple of 1.602 × 10^-19 volts?
>> | Because 1 electron has a charge of 1.602 × 10^-19 coulomb.
>> If you are ok with the first statement, you cannot agree the second.
>Well, the voltage is not exactly 1.602 × 10^-19 but close. Right?
>> Let a capacitor C retaining a charge Q. The potential difference is
>> U=Q/C.
>> As C is not quantified, U is not quantified. By changing the plates
>> distance, you can adjust C to get any value for U including
>> non-multiples of 1.602 × 10^-19 volts. You can have U = 2.1*10-19 V or
>> anything else.
>> | There are several applications I can think of for the aforementioned
>> | device:
>> | 1. Transmitting/recording too high a frequency signal on a medium that
>> | does not have the bandwidth required to handle the high-frequency
>> | 2. Transmitting/recording too large and amplitude signal on a medium
>> | that does not have the dynamic range required to handle the large
>> | amplitude
>> | 3. Generating a higher-frequency signal from a bunch of lower-
>> | frequency signals
>> It is theorically feasable. For example, you could do an A/D conversion
>> of what you are interested in in your input signal and code the digital
>> data with the amplitude of the transmitted signal. Using steps of 10^-18
>> volts, a 0->1 V signal could carry in one shot a near 60 bits word!
>> Using a 1 Mhz pass-band, you could transmit at least a 60 Mb/s signal.
>> Physically it's not possible. The problem is less in the A/D conversion
>> than in the noise which avoids to recover the information when a too
>> small step is used for the discrimination. Even if no noise was added to
>> the signal during its transmission, it is a limit to the initial A/D
>> conversion. An "exact" frequency-to-amplitude and visa versa, or
>> anything else, cannot have an arbitrarily accuracy due to the noise. The
>> noise independantly of its origin (thermal, quantum...) is the only
>> limit to signal transmission. See Shannon.
>If 1.602 × 10^-19 volt is too small, then what is the smallest
>physically-possible voltage that can be detected or processed given
>the state of today's technology?