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No EE input in this story at least WRT docsis. Too many HFC plants with enormous capital and labor investment in bidirectional amps which boil down to amplify

by VLM 12y ago

No EE input in this story at least WRT docsis. Too many HFC plants with enormous capital and labor investment in bidirectional amps which boil down to amplify 30 MHz and down and send upstream that direction while amplify 50 MHz and up and set downstream the opposite direction. Maybe in all new construction or retrofit or after a hurricane rebuild... Maybe. The biggest problem I see is Shannons Law will not be denied, and if the DSP to pull this off burns more watts than just transmitting a couple more milliwatts to get a better SNR you can spend on higher speed... What I'm getting at is if you want a TX RX ratio of 50:50 then merely increasing TX power by a cheap 3 DB will give you enough SNR margin to double your speed, making it possible to run "full speed" while listening half the time. The real world is never quite that simple. Then again real DSP processing is neither free or low power. (Let me expound on Shannon's Law... If you double your transmitted data by doubling your power to double your SNR at the same bit error rate for less power than perhaps a couple watts of DSP, then you're better off doubling your power. For, say, wifi, or bluetooth, or cellphones. Or if its cheaper to spend a couple watts on DSP than to double your tx power, maybe for satellites, then the DSP obviously wins. I'm thinking low power wifi is perhaps the worst case scenario for this new tech despite the implications in the article...)

nardi 12y ago
People always react to new RF technologies with "SHANNON'S LAW!!!" and ignore the fact that Shannon's Law refers to channels, not frequencies. If you create a new channel through active interference, you get more bandwidth.
- wirelessjunkie 12y ago
  That's absolutely right. By isolating the transmitter and receiver, you are able to create an independent channel for the receiver, which can operate independently of the transmit channel, while each of those channels obey Shannon's law. Another correction -- Shannon's law is logarithmic -- C (bit/sec/Hz) = logbase2(1+SNR) ~= logbase2(SNR) for high SNR. Thus a 3dB increase in SNR is a 1 bit increase in capacity. Clearly, there is a low ROI on increasing tx power beyond a certain point.
mnw21cam 12y ago
> Let me expound on Shannon's Law... If you double your transmitted data by doubling your power to double your SNR That's not how the Shannon-Hartley Theorem goes. http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem C = B log(1 + S/N) So doubling S (the signal) doesn't double C (the channel capacity). The improvement will always be less than this. If the signal to noise ratio is already fairly good, then the improvement of doubling the signal can be very small.

by VLM 12y ago

nardi 12y ago

People always react to new RF technologies with "SHANNON'S LAW!!!" and ignore the fact that Shannon's Law refers to channels, not frequencies. If you create a new channel through active interference, you get more bandwidth.

wirelessjunkie 12y ago
That's absolutely right. By isolating the transmitter and receiver, you are able to create an independent channel for the receiver, which can operate independently of the transmit channel, while each of those channels obey Shannon's law. Another correction -- Shannon's law is logarithmic -- C (bit/sec/Hz) = logbase2(1+SNR) ~= logbase2(SNR) for high SNR. Thus a 3dB increase in SNR is a 1 bit increase in capacity. Clearly, there is a low ROI on increasing tx power beyond a certain point.

mnw21cam 12y ago

> Let me expound on Shannon's Law... If you double your transmitted data by doubling your power to double your SNR That's not how the Shannon-Hartley Theorem goes. http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem C = B log(1 + S/N) So doubling S (the signal) doesn't double C (the channel capacity). The improvement will always be less than this. If the signal to noise ratio is already fairly good, then the improvement of doubling the signal can be very small.