abt some problem with nyquist's and shannon theory

And how many CEU's will I get for doing YOUR homework?

Go away kid... go flip burgers somewhere.

...Jim Thompson

This one has an obvious math answer which is in fact wrong. If you've ever tried wiring up a number of routers or anything else, you will know the right answer is more than the simple one.

You need to know the following added information to get the real answer.

What day of the week is it and what is the time?

How far is it to the store to buy extra cables?

What are the hours of said store?

Did you buy those cheap cables in the bin?

BTW: the first version of this problem involved "speaking tubes".

In this one, it depends if you are in marketing or engineering.

Bwahahahaha ;-)

...Jim Thompson

I take "duplex" to mean full-duplex so single cable comm. Then conceptually label the routers 1,2,...,10, and the cabling would go as (1,2),(1,3),...,(1,10),(2,3),(2,4),....,(9,10) where (m,n) denotes the cable between routers mn. This is 9+8+...+1=(1/2)*9*10=45 cables.

This is easily answered by reference to this piece written by a pseudo-intellectual pedant:

C= BW x Log2(1+S/N)= 4e3*Log(1+1000)/Log(2)=40Kbps for Shannon

Nyquist's Theorem applies to the equivalent bit rates of noiseless channels:

C = 2*W log base2( L ) bits/sec o where 2W is 2 times the highest frequency contained in the noiseless channel, and o where L = number of discrete levels (e.g., binary = two levels, 0 and 1)

C=8e3*log2(2)= 8Kbps

Despite the claims of others, real life is just this simple.

In article , Fred Bloggs wrote: [...]

Except the answer you gave is wrong. Go back and re-read his homework question very carefully and you will find the mistake quickly, (I assume based on my past experience with you)

The Shannon answer is right for the case where you have the freedom to use multi-level encoding or a symbol set greater than two. The OP states "binary signal," so that Shannon does not apply in this case. The Nyquist answer is an assumption based on two level detection at the receiver.

Do you mean he should have used R

You must be saying that because the analog bw of the channel is 4K then the this will only support a data rate that represents 2 samples or 2Kbps.

The 8Kbps is right for the Nyquist answer, derived from zero ISI considerations, when you forget about the fact that this requires pulses of the form sinc(2piWt)- and that is binary in a strict sense. Anything else does not refer to Nyquist's original theorem.

Yes, the homework problem says "binary" signal.

Huh? The Nyquist Theorem say that a channel of bw=H=4Khz using V=2 states has a maximum capacity of 2H*log2(V) = 8Kbps. Doesn't "binary" in the question simply mean 2 states? I think Fred's first answer is correct.

-- Joe Legris

You've got Nyquist standing on his head.

\\ / O ! /!\\ \\!/ ! O / \\

In article , Fred Bloggs wrote: [...]

Since part of the sinc() function must appear before Nyquist was born, I suggest you re-cast that in minimum phase not zero phase.

BTW: Isn't this just a way to sneak a multilevel encoding in?

Nyquist was working with the noiseless telegraph channel with brickwall bandlimiting to find maximum pulse rate with zero intersymbol interference, or the equivalent thereof in his day, 1920's or something. The ideal solution is pulses of the form sinc(2piWt) at a rate of 2W is the maximum possible. This was easily extended to the Log2(L) multiplier term for multilevel. As usual there are an infinity of extensions to this with ordinary digital pulses, raised cosine, and others, but they are non-closed form, parameterized with BER against types of noises and noise power, and they are not called Nyquist's Theorem. This comes right out of Digital Communication Techniques, Signal Design and Detection, Simon, M.K., Hinedi, S.M., Lindsey, W.C. (

Whew! I thought I was losing it.

Oooops, I miss read your post. Yes, the highest frequency needed to do a bit stream is 1/2 the bit rate.

I think you may have misunderstood the crack about the sinc() function. You can also shift the phase of the higher frequencies later to make a minimum phase signal. This signal has the exact same bandwidth as the sinc() but doesn't require non-zero values before Nyquists birth.

OK, now you've got him lying on his back.

Poor Harry is dead.