It's common knowledge that digital technology gives more telephone[4 Khz wide] channels than analog technology - for the same spectrum width. This even with the steep-sided filters and SSB technology.Since the input and output need to be in analog form, I find digital's claim very unintuitive -- almost doubting it.Can someone offer an intuitive explanation ?I guess a 4 HHz bandwidth telco signal contains a lot ofredundancy which is not needed to convey the inteligence ?Speech can even be transmitted as a sequence of 'predeterminedsylables' - what are they called again ?So it seems to me that the improvement over analog methodscan't be acheived merely by 'modeming', which would only addentropy. But once it's in the digital format, standard compression[by removing redundancy] can make for more efficiency ?Thanks for any info,== Chris Glur.

Lot of information is already on line on the topic; Goggle for it.SeeShould get you up to speed so you can search intelligently.----== Posted via - Unlimited-Unrestricted-Secure Usenet News==---- The #1 Newsgroup Service in the World! 120,000+Newsgroups----= East and West-Coast Server Farms - Total Privacy via Encryption =----

How can digital be more spectrum efficient than analog ?

Oct 27, 2006 161 Replies

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(snip)

I used to know the carrier frequencies for up and down stream.

The distinction I was trying to make, which some people seem to find important (though I don't) is between modulated and non-modulated signals. The only one I would count as not modulated is NRZ, where a logic zero is some voltage, and a logic one is a different voltage.

One of the simpler modulation systems is NRZI, where a voltage transition is one, and no transition is zero. I won't argue whether an NRZI signal is analog or digital, only that it is modulated.

A more complicated modulation system, synchronous phase modulation sends one cycle of the carrier at either 0 or 180 degree phase shift. This modulation method, also called Manchester coding, is used by ethernet.

-- glen

Eric,

Forget about the minor technical problems for a minute. The absolute minimum number of code states which is required to realize a given coding gain should be the fundamental issue. I believe this question is answered already; need to review the classic books on that.

Here is a simple reasoning:

In order to decode one bit from the noisy data, we have to process a chunk of data which carries no less then one bit of the joint information. As we are approaching the Shannon's limit, the size of this chunk goes to the infinity.

Coding gain for iterative codes

For many systems, the turnaround delay is a very important issue. Therefore it makes sense to optimize such system not for the maximum capacity, but for the maximum capacity at a given algorithmic delay.

Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

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How can digital be more spectrum efficient than analog ?

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