Just to clarify - the "Shannon's Law" which Harold is talking about is
the Shannon-Hartley Channel Capacity theorem, which is related to but
not the same as the Shannon-Nyquist sampling theorem I mentioned :)

Sean


On Tue, Nov 15, 2011 at 11:33 PM, Harold Hallikainen
<harold@hallikainen.org> wrote:
>
>> 1. Forgive me for asking, but what's stopping me from using a 2.4GHz VCO
>> and sending in a 100Mhz (100Mbit/s) data stream into the input pin and
>> receiving it on the other end (via FSK)? There just /has/ to be some
>> gotcha, because it seems simple enough. I know I should be able to send =
in
>> at least a 1Hz data stream and see it reliably, but where does the data
>> speed limit come in, other than the effects of nearing carrier wave
>> frequency?
>
> Sidebands... You can use trig identities to show the sidebands for AM at
> carrier +/- the modulating frequency. For FM, there are an infinite numbe=
r
> of sidebands, at +/- n* the modulating frequency. You use Bessel function=
s
> to determine the amplitude of each sideband, but you can estimate that th=
e
> majority of the power is within +/- F+d where F is the highest modulating
> frequency and d is the deviation. So, for US broadcast FM, which, for
> stereo has a highest modulating frequency of 53kHz and a deviation of
> +/-75kH, the majority of the sidebands are within +/- 75+53kHz or
> +/-128kHz of the carrier. For FSK, you're doing FM with a square wave. So=
,
> it consists of the fundamental (1/2 bit rate) plus a few odd harmonics,
> (up to 3 or 5). So... that's why you can't run infinite bitrate in a
> non-infinite channel.
>
> This also gets back to Shannon's law which (summarizing) says that the
> number of bits per second you can send down a channel is proportional to
> the bandwidth of the channel and proportional to the signal to noise rati=
o
> of the channel. FM is the "original spread spectrum" where, by using a
> larger bandwidth than the information bandwidth you get a better signal t=
o
> noise ratio on the demodulated output than you get on the RF channel.
> You're trading noise for bandwidth.
>
> Looking at this like at 56kbps modem, which uses pulse amplitude
> modulation, let's say you send 8,000 pulses per second down a phone line
> (which happens to be the sampling rate for telephone ADCs). If the pulse
> has 256 different levels you can encode 8 bits in that one pulse (8 bits
> per baud). But if noise makes it so you can't reasonably distinguish
> between those 256 levels, you have to run fewer levels (like 7 bits per
> baud for 128 levels). As noise increases, you get less bits per baud.
>
> So... it's a tradeoff. Bitrate throughput is proportional to channel widt=
h
> and channel s/n.
>
> Welcome to the wonderful world of RF!
>
> Harold
> WA6FDN
> Former broadcast engineer
>
>
>
>
>
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