From crose Wed Jan 29 18:38:42 1997
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Date: Wed, 29 Jan 97 18:38:16 EST
From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9701292338.AA03718@MOGLI.rutgers.edu>
To: 330_542
Subject: first assignment
Status: RO


HI Folks,

SO far so good...  no cancellation
yet.  So start and finish the first
assignment (it's a probability review).

I'd like to get latex solutions made up for the other
problem sets as well.  Might someone out there be 
interested in such a task?

Cheers,

Chris Rose

From crose Mon Feb 10 21:59:56 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9702110259.AA19582@MOGLI.rutgers.edu>
To: 330_542
Subject: test
Cc: crose@MOGLI.rutgers.edu
Status: RO


Hi Folks,

This is a test.  If you did not get ths mailin
then you're not on the list.  Please send info
asap if you want to get in on all the discussion.

To start things off, I'd like someone to give the answer
to problem 2.13 (coin weighing) part a.

Then I'd like for a discussion to erupt on whether
the bound is acheivable and when (this latter is a hard problem).

Cheers
Chris Rose

From jlai@ece.rutgers.edu Tue Feb 11 10:00:08 1997
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Date: Tue, 11 Feb 97 9:43:39 EST
From: Jie Lai <jlai@ee.rutgers.edu>
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: test
In-Reply-To: Your message of Mon, 10 Feb 97 21:59:52 EST
Message-Id: <CMM-RU.1.3.855672219.jlai@ee.rutgers.edu>
Status: RO

Hi, Dr. Rose,

I go to the course web page. All the problem set is for 1996. Is that we 
supposed to do? Also, problem 2.13 is not on there. Is this you just add up?
Just start to use this group. Thanks!

Jie Lai

From crose Tue Feb 11 11:49:49 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
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To: jlai@ee.rutgers.edu
Subject: Re: test
Cc: crose
Status: RO

Yup, 2.13 is one I just want you to think about.  And yes
the problem sets are the same as last year.

Cheers

From jlai@liman.Rutgers.EDU Thu Feb 13 10:14:46 1997
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Subject: Prob 2.12
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Status: RO

For prob. 2.12, I got

I(X;Y;Z) = sum{ p(XYZ)log[ (p(XY)p(XZ)p(YZ))/(p(X)p(Y)p(Z)p(XYZ)) ] }

         = sum{ p(XYZ)log[ (p(X|Y)p(X|Z))/(p(X)p(X|YZ)) ] }


It seems a nice SYMMETRY form. From Venn diagram, it also looks ok.
What I donot understand is when and how I(X;Y;Z) can get negative
as the book said. Using the equation from the book:

I(X;Y;Z) = I(X;Y) - I(X;Y|Z)
	 = I(X;Y) - ( I(X;Y,Z) - I(X;Z) )

If X and Y are indep., thus I(X;Y)=0, Y has no mutual info. with X.
It seems to me that I(X;Y,Z) will equal I(X;Z). So I(X;Y;Z)=0. How
come it get negative ?

Please help me get out! Anybody!

Jie

From crose@mogli.rutgers.edu Thu Feb 13 18:13:06 1997
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From: crose@mogli.rutgers.edu (Christopher Rose)
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To: 330_542@MOGLI.rutgers.edu
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Status: RO


QUESTION:

From: Jie Lai <jlai@liman.Rutgers.EDU>
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For prob. 2.12, I got

I(X;Y;Z) = sum{ p(XYZ)log[ (p(XY)p(XZ)p(YZ))/(p(X)p(Y)p(Z)p(XYZ)) ] }
 
         = sum{ p(XYZ)log[ (p(X|Y)p(X|Z))/(p(X)p(X|YZ)) ] }


It seems a nice SYMMETRY form. From Venn diagram, it also looks ok.
What I donot understand is when and how I(X;Y;Z) can get negative
as the book said. Using the equation from the book:

I(X;Y;Z) = I(X;Y) - I(X;Y|Z)
         = I(X;Y) - ( I(X;Y,Z) - I(X;Z) )

If X and Y are indep., thus I(X;Y)=0, Y has no mutual info. with X.
It seems to me that I(X;Y,Z) will equal I(X;Z). So I(X;Y;Z)=0. How
come it get negative ?

Please help me get out! Anybody!

Jie


From kangp@ms.com Thu Feb 13 23:59:04 1997
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Date: Thu, 13 Feb 1997 23:42:47 -0500 (EST)
Message-Id: <9702132342.ZM13477@ms.com>
In-Reply-To: crose@mogli.rutgers.edu (Christopher Rose)
        "" (Feb 13,  5:07pm)
References: <9702132207.AA13552@mogli>
X-Mailer: Z-Mail (3.2.1 10oct95)
To: crose@mogli.rutgers.edu (Christopher Rose)
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Status: RO

I wish I could help ....
I haven't got time to start looking into it yet !!!


On Feb 13,  5:07pm, Christopher Rose wrote:
> Subject:
>
> QUESTION:
>
> From: Jie Lai <jlai@liman.Rutgers.EDU>
> X-Mailer: Mozilla 3.0 (X11; I; SunOS 5.5.1 sun4m)
> Mime-Version: 1.0
> To: ece542@ece.rutgers.edu
> Subject: Prob 2.12
> Content-Type: text/plain; charset=us-ascii
> Content-Transfer-Encoding: 7bit
>
> For prob. 2.12, I got
>
> I(X;Y;Z) = sum{ p(XYZ)log[ (p(XY)p(XZ)p(YZ))/(p(X)p(Y)p(Z)p(XYZ)) ] }
>
>          = sum{ p(XYZ)log[ (p(X|Y)p(X|Z))/(p(X)p(X|YZ)) ] }
>
>
> It seems a nice SYMMETRY form. From Venn diagram, it also looks ok.
> What I donot understand is when and how I(X;Y;Z) can get negative
> as the book said. Using the equation from the book:
>
> I(X;Y;Z) = I(X;Y) - I(X;Y|Z)
>          = I(X;Y) - ( I(X;Y,Z) - I(X;Z) )
>
> If X and Y are indep., thus I(X;Y)=0, Y has no mutual info. with X.
> It seems to me that I(X;Y,Z) will equal I(X;Z). So I(X;Y;Z)=0. How
> come it get negative ?
>
> Please help me get out! Anybody!
>
> Jie
>
>-- End of excerpt from Christopher Rose



From crose Mon Feb 17 13:39:26 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9702171839.AA27090@MOGLI.rutgers.edu>
To: 330_542
Subject: problem 2.12
Status: RO


Since I don't seem to have a record of anyone coming forward
with the question about the negativity of I(X:Y:Z), here's
a solution.

Suppose X and Y are independent. Then as you point out, I(X:Y)=0.

However, just because two variables are independen
does not mean that they are CONDITIONALLY INDEPENDENT.

You know, I think I'm repeating either the solution,
or what one of you posted (Kabir)... let me check a moment...
nope, it looks like I've killed the file!  ARRRRGGGGG!

In any case, consider z=x+y with x and y indep.
Well given z, x and y are no longer independent
so their mututal info is NOT ZERO. This means
that I(X:Y:Z) is negative.

Cheers,

Chris Rose




From jlai@ece.rutgers.edu Tue Feb 18 15:41:35 1997
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This is a MIME-encapsulated message

--MAA24155.856288763/zen.rutgers.edu

The original message was received at Tue, 18 Feb 1997 12:59:22 -0500
from ee.rutgers.edu [128.6.46.13]

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Date: Tue, 18 Feb 97 12:59:47 EST
From: Jie Lai <jlai@ee.rutgers.edu>
To: 330_542@ece
Subject: solution for hw
Message-ID: <CMM-RU.1.3.856288787.jlai@ee.rutgers.edu>

Can we get the solution for HW, just for the first few sets, before
the take-home exam ? So we could learn some techniques on how to 
think, formulate and solve the problems.

Jie

--MAA24155.856288763/zen.rutgers.edu--



From crose Tue Feb 18 17:19:28 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
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To: 330_542
Subject: problem set solutions
Cc: crose
Status: RO


Hi, the solutions to ps 1 and 2 are out on 
the web.  The solutions to ps3 should arrive shortly
(also on the web).

Cheers,

From crose Tue Feb 25 18:25:33 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
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To: doolie@eden.rutgers.edu
Subject: Hi
Cc: crose@MOGLI.rutgers.edu
Status: RO


Hi Dong,

Just ask the question on the mailing group first (330_452@mogli.rutgers.edu)
and then set up a time with me if your questions are not answered.
I'm at a meeting on thursday morning, but will be able
to come in for an appointment if you'd like.

Cheers

Chris Rose

From jlai@ece.rutgers.edu Wed Feb 26 17:41:29 1997
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Date: Wed, 26 Feb 97 17:22:19 EST
From: Jie Lai <jlai@ee.rutgers.edu>
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: ready
In-Reply-To: Your message of Mon, 24 Feb 97 10:48:20 EST
Message-Id: <CMM-RU.1.3.856995739.jlai@ee.rutgers.edu>
Status: RO

Hi, Dr. Rose,

Q: For prob. 4, should we compile a program to count the "trans. prob."
of all possible case? I think that means 27*27 possiblities. Do I
understand the question correctly? Also there are 4 paragraphs, not 3.

For paris of letters followed by pairs of letters, there are 27**4
possibilities, most of them are zeros, of course. Is that kind of
prob. count we need to do and then calc. the entropy?

Thanks.

Jie

From crose@mogli.rutgers.edu Wed Feb 26 21:45:29 1997
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From: crose@mogli.rutgers.edu (Christopher Rose)
Message-Id: <9702270213.AA02132@mogli>
To: jlai@ee.rutgers.edu
Subject: Re: ready
Cc: 330_542@MOGLI.rutgers.edu
Status: RO

That's the basic idea.  Yes there are 27**4 possibilities,
but it's pretty obvious from reading the three paragraphs taht there';s
nowhere near those sorts of numbers.  I'm asking for a "rough estimate"
of the transition probs since there's not enough data to
say anything.

Nonetheless, if counting the doubles is too onerous, just count
the relative frequencies of double letter combinations and enter
them in a table.

A computer program IS A GOOD IDEA.  In fact, you can take some text from
anywhere if you'd like (the larger the better) and do the relative frequency
calculations.  So consider this an ammendment to the problem.

Cheers

Chris Rose

From kangp@ms.com Mon Mar  3 10:19:09 1997
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        "Re: ready" (Feb 26,  9:13pm)
References: <9702270213.AA02132@mogli>
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Subject: error ?
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Status: RO

I think there is a typo in question 3.
The transition prob out of state 0 doesn't sum up to 1 !!
Should P     = 0.5 instead of 0.25 ?
	1|0

thanks

From kangp@ms.com Mon Mar  3 10:30:56 1997
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        "Re: ready" (Mar  3,  9:53am)
References: <9703031453.AA05348@mogli>
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Status: RO

Prof Rose,

Prob 2 is tough. Maybe i am too strongly influenced by the transition diagram.

According to the book, not all agglomeration of state X yields a Markov Chain.
So, an answer must exists :)
But I can't see any relation between the transition prob and the past history.

I am not familiar with all classes of Markov (aperiodic, irreducible, etc).

Do I need more knowledge than discussed in the class to solve this problem ?

Perhaps I should solve this problem from a totally different angle ? (i.e. not
straight invocation of the Def of Markov) ??

thanks
/pl

From kangp@ms.com Mon Mar  3 10:33:14 1997
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In-Reply-To: Porlin Kang <kangp>
        "error ?" (Mar  3,  9:59am)
References: <9702270213.AA02132@mogli>  <9703030959.ZM3284@ms.com>
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To: crose@mogli.rutgers.edu (Christopher Rose)
Subject: Re: error ?
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Status: RO

I solved this problem by assuming the typo and
obtain an Entropy Rate as a function of k in part b.

On Mar 3,  9:59am, Porlin Kang wrote:
> Subject: error ?
> I think there is a typo in question 3.
> The transition prob out of state 0 doesn't sum up to 1 !!
> Should P     = 0.5 instead of 0.25 ?
> 	1|0
>
> thanks
>-- End of excerpt from Porlin Kang



From crose Mon Mar  3 13:42:28 1997
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Date: Mon, 3 Mar 97 13:42:27 EST
From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9703031842.AA14694@MOGLI.rutgers.edu>
To: crose@mogli.rutgers.edu, kangp@ms.com
Subject: Re:  error ?
Cc: crose@MOGLI.rutgers.edu
Status: RO

I'll check right now....

You're almost correct.  p_0|0 should be 0.75 since
there's no -1 state

Thanks for catching that.

Cheers,

Chris Rose

From crose Mon Mar  3 13:45:02 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9703031844.AA14699@MOGLI.rutgers.edu>
To: crose@mogli.rutgers.edu, kangp@ms.com
Subject: Re: ready
Cc: 330_542@MOGLI.rutgers.edu
Status: RO

Hi Porlin,

I think all you need is the markov property.  You have
to find class(es) of original chains for which
certain subdivisions will be markov as well.

Yes, it can be a tough problem, but if you use
the markov principle as the guide you should be able
to identify at least one structure which always results in a 
new markov chain upon subdivision.

Good luck!

Chris Rose

PS:  And keep thepublic questions coming!

From jlai@ece.rutgers.edu Fri Mar 14 14:36:38 1997
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Date: Fri, 14 Mar 97 14:14:41 EST
From: Jie Lai <jlai@ee.rutgers.edu>
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: grading policy
In-Reply-To: Your message of Fri, 14 Mar 97 13:59:05 EST
Message-Id: <CMM-RU.1.3.858366881.jlai@ee.rutgers.edu>
Status: RO

Hi, Dr. Rose,

Is the channel capacity law we are talking about related to Shannon(?)'s 
original statement, like

      C = W log( 1 + SNR )

for Gaussian channel? How they are related? This will be covered in class, 
isn't it?

Jie

From crose Fri Mar 14 14:40:24 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9703141940.AA01018@MOGLI.rutgers.edu>
To: jlai@ee.rutgers.edu
Subject: Re: grading policy
Cc: crose
Status: RO

We've not gotten there yet, but
it will be covered.  We first have to learn a bit
about continous entropy and taht happens later.

Cheers,

Chris Rose

From kangp@ms.com Fri Mar 14 17:34:37 1997
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Date: Fri, 14 Mar 1997 17:13:14 -0500 (EST)
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In-Reply-To: Christopher Rose <crose@MOGLI.rutgers.edu>
        "grading policy" (Mar 14,  1:59pm)
References: <9703141859.AA00896@MOGLI.rutgers.edu>
X-Mailer: Z-Mail (3.2.1 10oct95)
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: grading policy
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Status: RO

Dr Chris,

I was trying to ask you about this when we meet again.
It seems that the number doesn't add up to the total. Is that intentional (I
meant, does those numbers on each page meant anything ?) ?
On the front page, you wrote 4 numbers that adds up to 130. However, on the
following pages, the numbers were different.

thanks
/pl

On Mar 14,  1:59pm, Christopher Rose wrote:
> Subject: grading policy
> Hi Folks,
>
> I encourage you to look over your exams and the solutions and question
> why you were assigned the grades you were.  When you get out in the
> world and your boss says "good job" or "bad job" you should never just
> take it at face value.  ALWAYS question authority.
>
> In any case, here's the official policy on grading.  If you had a
> problem with a grade I assigned, you must write up your objection and
> submit it to met along with the whole exam.  I will look at the exam
> and make a decision.  Owing to time constraints, however, there will
> only be one such appeals process.  The grade is final after that.  You
> have until monday (the day we come back) to submit your requests for
> regrade.
>
> So hop to it !!!! :)
>
> Cheers,
>
> Chris Rose
>-- End of excerpt from Christopher Rose



From crose Fri Mar 14 17:37:27 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9703142237.AA01307@MOGLI.rutgers.edu>
To: kangp@ms.com
Subject: Re: grading policy
Cc: crose@MOGLI.rutgers.edu
Status: R

Hi Porlin,

Each question was worth 50 points.  I threw numbers
down more or less at random.  However, there should be numbers
which denote the oint totals for each problem somewhere (but
maybe not).  in any case, if you go through and add humbers I 
doubt you'll get the total on the front.  I can take a look
of course if you'd like.  Just drop the exam off
next time you come in (or just wait until monday when we get
back).

Cheers,

Chris Rose

From kangp@ms.com Wed Apr  9 13:46:46 1997
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To: crose@MOGLI.rutgers.edu
Subject: 2 questions
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Status: RO

Dr Rose,

On page 99 of the text,
there is a statement that says :

(roughly in the middle)
o The two longest codewords differ ....
  ....
    ...
  ....
    Now we can exchange the longest lenght codewords so the two lowest prob
  are associated with 2 siblings on the tree. This does not change the expected
  length ....


My question is : how could the reassigment of probability with the codewords
not change the expected length ? Shouldn't it be decreased ?


------------------

Somehow we skipped the proof of the Kraft's infinite source Uniquely decodable.
Did you mentioned that you don't like the proof in the book ?

thanks
/pl

From crose Wed Apr  9 17:59:53 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9704092159.AA00648@MOGLI.rutgers.edu>
To: 330_542
Subject: questions
Cc: crose
Status: R

Dr Rose,

On page 99 of the text,
there is a statement that says :

(roughly in the middle)
o The two longest codewords differ ....
  ....
    ...
  ....
    Now we can exchange the longest lenght codewords so the two lowest prob
  are associated with 2 siblings on the tree. This does not change the expected
  length ....
 

My question is : how could the reassigment of probability with the codewords
not change the expected length ? Shouldn't it be decreased ?


------------------

Somehow we skipped the proof of the Kraft's infinite source Uniquely decodable.
Did you mentioned that you don't like the proof in the book ?

thanks
*****************

Question 1: the two longest codewords are siblings which
means they are the same length.  Therefore swapping them
changes nothing.

Question 2: The proof is really simple, but it makes me uncomfortable
because you're taking limits when you're not sure they exist.  To
clean is simple. We just show that if the limit approached a number
larger than 1, then there must exist some N for which the finite sum
is greater than 1 (an impossibility owing to Kraft) and we're done.

From crose Fri Apr 11 13:11:30 1997
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9704111711.AA02962@MOGLI.rutgers.edu>
To: 330_542
Subject: questions
Cc: crose@MOGLI.rutgers.edu
Status: R


Prof. Rose,

How could we explain 188, eqn 8.11 ?
H(Y) = H(Y,E) where event E is {Y=e}.

Seems like this equation is not used in the proof though.

*******************

Remember that H(Y,E) = H(Y)+H(E|Y).

Now what is H(E|Y) where E is the event
that Y=e?

And yes it is used in the proof since equation 8.12
is a direct application.

Cheers

From kangp@ms.com Fri Apr 11 13:24:30 1997
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Date: Fri, 11 Apr 1997 12:58:07 -0400 (EDT)
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In-Reply-To: Christopher Rose <crose@MOGLI.rutgers.edu>
        "questions" (Apr 11,  1:11pm)
References: <9704111711.AA02962@MOGLI.rutgers.edu>
X-Mailer: Z-Mail (3.2.1 10oct95)
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: questions
Mime-Version: 1.0
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Status: R

Prof,

I guess my intended question was :
H(Y) = H(Y,E) ?

thanks
/pl

On Apr 11,  1:11pm, Christopher Rose wrote:
> Subject: questions
>
> Prof. Rose,
>
> How could we explain 188, eqn 8.11 ?
> H(Y) = H(Y,E) where event E is {Y=e}.
>
> Seems like this equation is not used in the proof though.
>
> *******************
>
> Remember that H(Y,E) = H(Y)+H(E|Y).
>
> Now what is H(E|Y) where E is the event
> that Y=e?
>
> And yes it is used in the proof since equation 8.12
> is a direct application.
>
> Cheers
>-- End of excerpt from Christopher Rose



From crose Fri Apr 11 14:39:59 1997
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Status: R

   ----- Transcript of session follows -----
550 335_542... User unknown
550 log... User unknown

   ----- Unsent message follows -----
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From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9704111839.AA03112@MOGLI.rutgers.edu>
To: kangp@ms.com
Subject: Re: questions
Cc: 335_542

Please reread my statements AND question.  I think
you'll find you answer there.
 Cheers,

***************
I guess my intended question was :
H(Y) = H(Y,E) ?

thanks
/pl

On Apr 11,  1:11pm, Christopher Rose wrote:
> Subject: questions
>
> Prof. Rose,
>
> How could we explain 188, eqn 8.11 ?
> H(Y) = H(Y,E) where event E is {Y=e}.
>
> Seems like this equation is not used in the proof though.
>
> *******************
>
> Remember that H(Y,E) = H(Y)+H(E|Y).
>
> Now what is H(E|Y) where E is the event
> that Y=e?
>
> And yes it is used in the proof since equation 8.12
> is a direct application.
>
> Cheers
>-- End of excerpt from Christopher Rose



From crose Fri Apr 11 14:44:08 1997
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Date: Fri, 11 Apr 97 14:44:02 EDT
From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9704111844.AA03142@MOGLI.rutgers.edu>
To: 330_542
Subject: mistake in address
Status: R


Sent this to the wrong adress.  sorry folks..
_______________
Please reread my statements AND question.  I think
you'll find you answer there.
 Cheers,

***************
I guess my intended question was :
H(Y) = H(Y,E) ?

thanks
/pl

On Apr 11,  1:11pm, Christopher Rose wrote:
> Subject: questions
>
> Prof. Rose,
>
> How could we explain 188, eqn 8.11 ?
> H(Y) = H(Y,E) where event E is {Y=e}.
>
> Seems like this equation is not used in the proof though.
>
> *******************
>
> Remember that H(Y,E) = H(Y)+H(E|Y).
>
> Now what is H(E|Y) where E is the event
> that Y=e?
>
> And yes it is used in the proof since equation 8.12
> is a direct application.
>
> Cheers
>-- End of excerpt from Christopher Rose



From kangp@ms.com Fri Apr 11 14:56:40 1997
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Date: Fri, 11 Apr 1997 14:20:14 -0400 (EDT)
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In-Reply-To: Christopher Rose <crose@MOGLI.rutgers.edu>
        "Re: questions" (Apr 11,  2:39pm)
References: <9704111839.AA03112@MOGLI.rutgers.edu>
X-Mailer: Z-Mail (3.2.1 10oct95)
To: Christopher Rose <crose@MOGLI.rutgers.edu>
Subject: Re: questions
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Status: R

of course ...
H(E|Y) = 0 given event E is {Y=e}.
So, H(Y) = H(Y,E)

thanks
/pl

On Apr 11,  2:39pm, Christopher Rose wrote:
> Subject: Re: questions
> Please reread my statements AND question.  I think
> you'll find you answer there.
>  Cheers,
>
> ***************
> I guess my intended question was :
> H(Y) = H(Y,E) ?
>
> thanks
> /pl
>
> On Apr 11,  1:11pm, Christopher Rose wrote:
> > Subject: questions
> >
> > Prof. Rose,
> >
> > How could we explain 188, eqn 8.11 ?
> > H(Y) = H(Y,E) where event E is {Y=e}.
> >
> > Seems like this equation is not used in the proof though.
> >
> > *******************
> >
> > Remember that H(Y,E) = H(Y)+H(E|Y).
> >
> > Now what is H(E|Y) where E is the event
> > that Y=e?
> >
> > And yes it is used in the proof since equation 8.12
> > is a direct application.
> >
> > Cheers
> >-- End of excerpt from Christopher Rose
>
>
>-- End of excerpt from Christopher Rose



From crose Fri Apr 11 14:30:37 1997
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From: crose (Christopher Rose)
Message-Id: <9704111830.AA10416@qbeast.rutgers.edu>
To: kangp@ms.com
Subject: Re: questions
Cc: crose
Status: R

Yup!


From seupark@paul.rutgers.edu Mon Apr 14 19:36:16 1997
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Subject: questions ?
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Status: R

Dear Professor,

I have some questions.

1) On page 193, eqn 8.29, --where I(.) is the indicator function.
 ; What is the indicatior function ? Could you give me some examples of
  indicator function ?

2) On page 198, Theorem 8.7.1(The channel coding theorem) 
  -- there exists a sequence of (2nR, n) codes with - - 
  ; a sequence of (2nR, n) codes, it means a code ? Is it right ?
    In the proof, it is eqn 8.62. Do I understand rightly ?

3) On page 227, Theoren 9.2.3  -- , to first order in the exponent.
  ; What is the "first order in the exponent" ? 
    Could you explain it with some examples ?

4) On page 239, eqn 10.2, power constraint.
  ; I don't understand xi(codeword)2. 
    If the code is the code of example 5.1.1(page 79),
    What is xi2 in this case ?
    In example, C(2) = 10, (10)2 = ? = 100 ?   

Thanks.

From crose Mon Apr 14 19:47:13 1997
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Date: Mon, 14 Apr 97 19:47:10 EDT
From: Christopher Rose <crose>
Full-Name: Christopher Rose
Message-Id: <9704142347.AA06945@MOGLI.rutgers.edu>
To: seupark@paul.rutgers.edu
Subject: Re:  questions ?
Cc: 330_542
Status: R

1) and indicator function is 1 when the "indicated condition"
is true and zero otherwise.

2) A sequence of codes is what is meant.  You're looking
for codes which may vary with n.  That is, there's a different
code for each n.  HOwever, the coding procedure (typical coding)
does not change.

3) FIrst order in exponent means to first order in n.

4) x_i(w)^2 is the i^th power level used
in sequence of n channel uses.  Remember that for continuous
channels we're using continous levels as opposed to binary symbols.
Thus, the example on page 79, which is for discrete sources
is not applicable.  Think of it this way.  For the continuous
channel we're sending a signal during a channel use (say on an interval
(0,T)).  We can sent a large signal (with large power) and correspondingly
large x_i or we can send a smaller signal.  That is, we're sending
actual levels over the channel.

The only place where discreteness comes in is that we're
using sequences of these powers to code discrete messages.
Specifically, we have 2^(nR) messages which we're trying to code
with a sequence of n levels (x_i).

Cheers

From oekelly@winlab.rutgers.edu Mon Apr 14 23:24:29 1997
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Date: Mon, 14 Apr 1997 22:58:13 -0400
From: oekelly@winlab.rutgers.edu (Owen Ernest Kelly)
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To: crose@MOGLI
Subject: Re: talk
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Status: R

The best I can say "huge". She seems to really really have the big picture.
Her basic point was that

source coding + channel coding + modulation 

should all be designed *jointly*.  I could have told you THAT, but it seems
overwhelmingly difficult.  She has just given it the college try anyway and come
away a winner.  Her talk examined when you can get away with isolated designs
and when you can't.  Very impressive.  If I had an NSF career award to give away
I wouldn't hesitate to give it to her.

Another amazing result was that if you have both power control and rate control, 
then the proper way to combat fading is to adjust your *rate* and leave power 
constant. To be more formal, if you compare capacity curves for 

i) shannon capacity
ii) adjustable rate, adjustable power
iii) adjustable rate, fixed power
iv) fixed rate, adjustable power

then iii) is much closer to i) than iv). That is, you take really big capacity 
hit if you fix the rate. 

-Owen

From jlai@ece.rutgers.edu Mon Apr 28 11:58:50 1997
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Date: Mon, 28 Apr 97 11:28:00 EDT
From: Jie Lai <jlai@ece.rutgers.edu>
To: 330_542@mogli.rutgers.edu
Subject: dual code Cd ?
Message-Id: <CMM-RU.1.5.862241280.jlai@ece.rutgers.edu>
Status: R

Dr. Rose,

On page 76 of handout, there are sth. about linear code C and its dual
code Cd. My question is, what is the defination of dual code Cd, how
is it got from C, is there any relations between their generator and
parity check matrix ?

Thanks.

Jie

