abt some problem with nyquist's and shannon theory

[hardik]

hello friends,

I need to solve this 2 poblems if anybody can help me out.

Problem 1: A collection of 10 routers is fully connected (a line
between each
pair of routers). How many lines do we need (assuming each is duplex).

Problem 2: If a binary signal is sent over a 4-kHz channel whose
S/N ratio is 30dB, what is the maximum achievable date rate? (you must
check
Shannon and Nyquist and pick the smaller estimate).


I m waiting for ur reply. If I will solve these problems I will post
the answer.

thx,
-hardik

 

[Jim Thompson]

hello friends,

I need to solve this 2 poblems if anybody can help me out.

Problem 1: A collection of 10 routers is fully connected (a line
between each
pair of routers). How many lines do we need (assuming each is duplex).

Problem 2: If a binary signal is sent over a 4-kHz channel whose
S/N ratio is 30dB, what is the maximum achievable date rate? (you must
check
Shannon and Nyquist and pick the smaller estimate).


I m waiting for ur reply. If I will solve these problems I will post
the answer.

thx,
-hardik

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

Go away kid... go flip burgers somewhere.

...Jim Thompson

 

[Ken Smith]

hello friends,

I need to solve this 2 poblems if anybody can help me out.

Problem 1: A collection of 10 routers is fully connected (a line
between each
pair of routers). How many lines do we need (assuming each is duplex).

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".

Problem 2: If a binary signal is sent over a 4-kHz channel whose
S/N ratio is 30dB, what is the maximum achievable date rate? (you must
check
Shannon and Nyquist and pick the smaller estimate).

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

 

[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

 

[Fred Bloggs]

Problem 1: A collection of 10 routers is fully connected (a line
between each
pair of routers). How many lines do we need (assuming each is duplex).

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

Problem 2: If a binary signal is sent over a 4-kHz channel whose
S/N ratio is 30dB, what is the maximum achievable date rate? (you must
check
Shannon and Nyquist and pick the smaller estimate).

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.

 

[Ken Smith]

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.[/QUOTE]

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)

 

[Fred Bloggs]

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.

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)
[/QUOTE]

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.

 

[[email protected]]

Do you mean he should have used R<C?

 

[Fred Bloggs]

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.

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)
[/QUOTE]

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.

 

[Fred Bloggs]

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.

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)
[/QUOTE]

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.

 

[Ken Smith]

[...]

This is easily answered by reference to this piece written by a
pseudo-intellectual pedant:
http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

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.

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)

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.[/QUOTE]


Yes, the homework problem says "binary" signal.

 

[[email protected]]

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.

 

[Ken Smith]

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.

You've got Nyquist standing on his head.

\ / O
! /!\
\!/ <----> !
O / \


1
2 <-----> ---
2

 

[Ken Smith]

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.[/QUOTE]


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?

 

[Fred Bloggs]

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.

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?
[/QUOTE]

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. (<- well known mega-expert),
Prentice Hall 1995, Chapter 9 Communication Over Bandlimited Channels,
Section 9.2.1 Nyquist Criterion for Zero ISI, pg 561.

 

[Keith]

You've got Nyquist standing on his head.

\ / O
! /!\
\!/ <----> !
O / \


1
2 <-----> ---
2

Whew! I thought I was losing it.

 

[Ken Smith]

You've got Nyquist standing on his head.

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

 

[Ken Smith]

Ken Smith 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.

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.

 

[Richard Henry]

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

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

 

[Keith]

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

Poor Harry is dead.