drick@hplvla.UUCP (drick) (03/08/85)
[ 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 -------------------------------------------------------- If I express any opinions in this note, they are my own. --------------------------------------------------------
leon@hhb.UUCP (Leon Gordon) (03/26/85)
Actually, the delta "functional" can not be represented as
the limit of a sequence of odd functions, but a closely related
function (the "derivative" of the delta "function" can). It
is easy to convince yourself that this function extracts the value
of the derivative of the function it is convoluted with.
that is:
integral[ delta'(t-T) * f(t)]dt = a*f'(0)
where a is a constant factor that I'm
too lazy to figure out at the moment
(some mix of 2's, pi's, and -1's)
By the way, a much clearer and more intuitive way of looking
at delta and related quantities is as a linear functional over
a prescribed function space. Thus delta is the functional
which maps f(x) -> f(0). The properties of delta can be developed
formally without the pain of considering distributions.
leon
{decvax,ihnp4,allegra}!philabs!hhb!leonleon@hhb.UUCP (Leon Gordon) (03/26/85)
P.S. to my last posting:
Another way to look at the delta function is as a
measure over a point set with the property that the measure of
any set containing the origin is 1, and the measure of any set
excluding the origin is zero. Integrals of functions with the
delta measure are then performed in the lebesque sense (summing
the product of a function value times the measure of the set on
which the function has that value. Although delta is a pretty
poor function, it is a very acceptable and well behaved measure.
leon
{decvax,ihnp4,allegra}!philabs!hhb!leonsullivan@harvard.ARPA (John M. Sullivan) (04/05/85)
Using integration by parts shows
$\int \delta' f + \int \delta f' = (\delta f) |^\infty_{-\infty} = 0.$
(I hope those of you who don't have TeX can read this notation anyway.)
Thus the desired constant is -1.
I don't see that distributions are that different from linear
functionals. Viewing $\delta$ as a measure has the disadvantage of
making it hard to define things like its derivative.
--
John M. Sullivan
sullivan@harvard