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From: drick@hplvla.UUCP (drick)
Newsgroups: net.math
Subject: Interview Q revisited
Message-ID: <7700003@hplvle.UUCP>
Date: Thu, 7-Mar-85 22:25:00 EST
Article-I.D.: hplvle.7700003
Posted: Thu Mar  7 22:25:00 1985
Date-Received: Tue, 12-Mar-85 09:24:07 EST
Organization: HP Loveland Instrument Division
Lines: 112

[ take this, bug! ]

Here is the followup on the interview question I posted a couple of 
weeks ago.  I should note that this was the most "mathematical"
question the interviewee was asked.  Most of the other questions
dealt with electronics.


>Recently, I interviewed someone who claimed to have a lot of
>coursework in communication theory.  The recent discussion of
>Dirac's delta function (function used advisedly) reminded me
>of a problem I gave this interviewee, to wit:
>
>The Fourier Transform is defined as:
>
>F{f(t)} = F(w) = Integral[-inf,inf]: f(t)e^(-jwt)dt.
>[ j is the square root of -1, w is usually omega ]
>
>a.  Prove that F{ d(f(t))/dt } = jwF(w).
>    (What assumptions are necessary?)

Normally, I don't ask proofs, but I wanted the interviewee to
have this theorem available, and he *did* have a master's 
degree...

Integrate by parts.  To get the unwanted term to drop out, one 
can assume that f(t) has finite extent in time, or make other
more complicated assumptions.

>b.  If f(t) is the function sketched below [here I substitute the
>    definition because graphics work poorly on the net], find F(w).
>    (Hint: use the result of part a.)
>
>    f(t) =  -1/T,   -5T/2 < t < -T
>             2/T,      -T < t <  T
>            -1/T,       T < t < 5T/2
>               0,       else.
>

Naturally, a candidate who doesn't see the implication of the
above theorem can just brute force the integration here.  A
nicer way is to differentiate f(t) to get:

fprime(t) = (-1/T) * delta( t + 5T/2)  
             + (3/T) * delta( t + T )
             - (3/T) * delta( t - T )
             (1/T) * delta( t - 5T/2)

Taking the transform of fprime(t) is very simple because of the
"sifting" property of the delta function.  (All students of
communication theory know this property.)  After dividing through
by jw, and changing exponentials into sines, one obtains:

F{f(t)} =  6 sin(wT)  -  5 sin(5wT/2)
           ---------     ------------
              wT          (5/2)wT

Any student of communication theory would recognize the form
sin(x)/x as a "sinc" function.

>c.  What happens to F(w) in the limit as T approaches zero?  What does
>    this suggest about f(t)?

Applying L'Hopital's rule to evaluate the limit yields unity.
But all students of communication theory know that unity is
the transform of a delta function!  Thus, f(t) becomes a delta
function in the limit as T goes to zero.

Most engineering students know how to use the delta function, but
not how it is rigorously defined.  Usually, they are shown only one
function that converges to a delta; this question was designed not
only to discover the applicant's facility with simple mathematics 
and familiarity with an important "function" of the engineering
trade, but to teach him a little bit more about that "function",
namely, that there are *lots* of things that converge to a delta
function.

Incidently, I chose f(t) to be an even function.  One can also 
find odd functions that converge to more or less the same thing.
I say "more or less," because I think a delta defined as a limit
of odd functions might have subtly different properties.  Can
someone on the net with a better grounding in the theory of 
distributions enlighten me?

This problem is more elementary than many that are posted to
this notes group, therefore, I expected to get flamed for
cluttering up the net with trivia.  Instead, I got flamed
for _asking_the_interviewee_a_hard_question_!  Come now.
I already *knew* he could do easy questions -- else he 
wouldn't have gotten the interview to begin with.  What I wanted
to find out was how he approached *hard* problems.  That's
the only kind of problem we can expect to make money by solving
around here (HP Loveland makes 5-1/2 digit voltmeters.  Noone
will pay us to make 3-1/2 digit voltmeters because they can
build their own. :-) ). 

To the respondent who thought this was discriminatory:  Of course
it is!  Why interview people at all if you're not trying to get
good ones?  I can honestly say that I respect *every* engineer I
work with here -- and I aim to keep it that way!

Cheers,

David L. Rick
Hewlett-Packard Company
Loveland Instrument Division
...!hplabs!hplvla!hplvle!drick
 

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If I express any opinions in this note, they are my own.
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