# Derivation Of The Spectrum Due To f(t).d(t - T)?????

• posted

Perhaps you're just kidding, or I misunderstand something. Isn't the Dirac Delta *defined* such that the int_-inf^+inf(delta(t-T)dt = 1.0??? And since the Dirac delta functions value is *defined* as zero everywhere except when t-T = 0, then the lim M->0 int_(T-M)^(T+M) delta(t-T)dt is also equal to 1.0 ???

The partial integral from -inf to (T-M) as lim M->0 remains 0.0 as well as the partial integral from (T+M) to +inf remains 0.0. So the difference between lim M->0 int_(T-M)^(T+M) delta(t-T)dt and int_-inf^+inf delta(t-T)dt is zero (because M never reaches exactly 0).

daestrom

• posted

Sorry, I didn't see your post in time - and then I realised that I posted it in the wrong place with a couple of errors and resubmitted. ah well

Richard

• posted

this is just my expression for integration by parts in general.

I assume this is the line you have problems with. Probably my fault for using the dummy variable 'u', since 'u' had already been used - sorry for the confusion. This is simply an antiderivative of x(t).delta(t-T), not a change of variable from my parts expression. The next line v = 0 if t < T and x(T) if t > T is still correct, and what is important.

Also, I called x(t) f(t) a couple of times by accident - I tried to delete this message from Thunderbird, but it obviously stuck around. I reposted it further up. That'll learn me to proofread. Hope that helps

Richard

• posted

Definition of Integration by Parts Theorem (from "Calculus and Analytic Geometry" 3rd edition):

"If U and V are differentiable functions and int(VdU)dx is an antidervative of VdU, then 'UV - int(VdU)dx' is an antiderivative of UdV."

So.... dV = f(t)delta(t-T) V=int f(t)delta(t-T) = f(T) (a result of the definition of the delta function)

And... U=e^(-st), thus dU=(-s)e^(-st) But the int((-s)e^(-st)) = e^(-st) + C. Not just e^(-st).

int_-inf^+inf (f(t)delta(t-T)e^(-st) = e^(-st)f(T) - int_-inf^+inf f(T)(-s)e^(-st) = f(T)(e^(-st) - (e^(-st) + C)) = Cf(T) where 'C' is the integration constant for the last integration. Only in the particular case of setting C to zero does the result come to zero.

daestrom

• posted

Just kidding :)

absolutely :)

• posted

Not true - what I was getting at in a handwavy manner is that if you make the interval arbitrarily small, because they are continuous, f(t) and exp(-st) approach constant values f(T) and exp(-sT). delta(t) is not continuous, so you can't apply the same principle to this.

Anyway, just in case you misunderstood, this is *not* rigorous, and I would not consider this a proof. I was hoping in a handwavy manner to convince you that David Ullrich's initial proof was correct in the first place.

Richard

• posted

f(t).delta(t-T).exp(-st)dt

f(T).delta(t-T).exp(-sT)dt

Perhaps you missed the "daring insight" ... since we're considering continuous functions, if you make the interval arbitrarily small, exp(-st) and f(t) approach constant values exp(-sT) and f(T) on that interval. As I said, it's not rigorous, I would never submit something like that as a proof, but you should get the idea in a handwavy manner why David Ullrich's proof was correct.

Richard

• posted

Recall that the definition of the Laplacian,

L[f(t)]=integral(-inf,+inf) f(t)*exp(-st)*dt

Recall that by definition an impulse funtion, d(t), is the limit as h->infinity of a rectangle of height h and width 1/h (and thus the area is

1), and centred (occurs) at time t=0. An impulse occuring at time T, is defined by d(t-T).

Let f(t)= g(t)*SUM(n= -inf,+inf) { d(t-nT) }

So the function is a pulse train (or sum there of), each pulse has a strength of g(t) and occurs at every interval of nT. The function g(t) is the sampled signal and the Reiman Sum is used to define when the samples are taken.

The Laplacian of the simple Dirac function h(t)=A*d(t-nT) is equal to the magnitude of the function A multiplied by a the time delay exp(-snT) or,

H(s) = A*exp(-snT)

Now recall our function f(t) which only has non-zero values occuring at intervals of nT. The magnitude of these values are g(t), or more specifically g(nT).

Thus L[f(t)] = F(S) = SUM(n= -inf,+inf) { g(nT)*exp(snT) }

I hope that helps,

Dwayne

Reference: C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., Prentice Hall, Copyright 1999, p. 46-50, 210-211.

• posted

Clueless!

Really? And what exactly is your supporting evidence?

A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing?

• posted

It is precisely because of this that he does it. His pathology requires that he annoy people.

Bob

• posted

It is interesting the size of "the amount of posts".

If I am wrong in what I say, it would be a simple matter to ignore me and what I posit.

That so many people, people who would seek to represent themselves as authorities, respond with emotional postings that do not address the points raised, seems to suggest that I am not wrong, and that they are embarrassed about a basic failing in their professed knowledge and so feel threatened.

"the amount of posts", posts that are non-technical and of an undesirable ad hominem style has reached such a scale that I have binned them all this morning.

This leaves just one point that has not yet been addressed, and that is, that any mathematical analysis of a real system must, to be respectable, deal with measurements of that system.

There is no system of which I am aware that has sampling pulses whose measurements match those of the Diracian Impulse...

1. Their amplitudes do not approach infinity.

1. Their areas do not approach unity.

2. In any case, the operation of f(t).d(t - T) is not defined unless under an integral sign, and cannot be evaluated unless under that integral sign.

1. The evaluation of the spectrum of the Diracian relies on it being over all time, -oo^+oo. It is improper, therefore, to attempt to evaluate other operations using the Diracian with a reduced domain, and yet still rely on the spectrum derivation.

• posted

TIA

• posted

Classic crackpot. If nobody says you're wrong you must be right. But if a large number of people say you're wrong it follows you must be right as well.

Things must be very pleasant in your little world.

Btw, s>It is interesting the size of "the amount of posts".

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David C. Ullrich

• posted

You claim to be a mathematician of 20 years' standing.

I suggest that you do not try to be a teacher of mathematics.

Mathematics is an abstract subject for which your students must be relaxed in mind in order to be able to take on abstractions. Your emotive and insulting posts will disrupt such a state of mind. In my time as a part-time tutor of mathematics to adults I came across this time and time again - people who have been put off mathematics, not by the subject per se, but by the aggression and personal attacks addressed to them by their teachers.

Shame on you.

(I notice that you do not address the mathematical points below)

• posted

This is the definition of the Laplace Transform. The Laplacian is something else entirely: the Laplacian of the function u(x,y) is u_xx + u_yy (using subscripts for partial derivatives).

(It's also at the very least a non-standard definition of the Laplace transform - the usual definition uses an integral from 0 to infinity.)

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David C. Ullrich

• posted

I suggest that if you wish to gain any respect and be seen as a potential mathematics professional that you drop your wont for making personal remarks and that you concentrate on the matters in hand.

You seem to have no experience of acadaemic debate, and I can tell you that Argumentum Ad Hominem, which is your wont, has no place in reputable acadaemic institutions.

Do us all a favour, Mr.Ullrich and grow up, or else leave this forum.

• posted

In my time I came across classes in which the range of mathematical abilities varied from the incompetent to the brilliant. I can't recall a case of anyone being put off by the teacher - but plenty that were put off by teaching methods. You may, of course, be referring to your own case - which is not necessarily generally true.

• posted

But you yourself have deleted whole posts - some of which were long and detailed and which dealt with the points you raised - because you perceived something you defined as 'childish' early on. If you believe that to be a mature approach to debate, then you are in a very small minority.

You have been repeatedly told that the way to deal with (what you perceive as) 'childishness' is to ignore it and deal with the substance. This (positive) behaviour reinforces the message that negativity is not the means to achieve a satisfactory outcome. Unfortunately, your approach merely reinforces negative behaviour and is ultimately self-defeating.

FUs set to remove uk.radio.amateur, where Bean's approach to discussion has been discussed ad nauseum.

• posted

I suggest that if you wish to gain anyone's respect in this forum, you learn to quit top posting like a newbie, quit trying to impress everyone with your amazingly brilliant but trivial technobabble, and learn not to get all pissy the first time someone begs to differ with you.

• posted

Guffaw. Every once in a while I get advice like this here on sci.math. Curiously, it always comes from someone who's totally wrong about some bit of mathematics, who's insisting he's right and refusing to accept simple explanations of why he's wrong.

Been teaching math for 20 years, and no situation even remotely like that has ever come up in the classroom.

Guffaw again. The mathematical points have been addressed quite completely. Below I refer to a post where I do address a new mathematical point (that you're simply doing the basic calculus wrong), which post you _say_ that you deleted unread.

Which is of course another classic crackpot technique: ignore substantive objections and then claim they don't exist (or repeatedly ignore substantive objections and then whine about the fact that they're not repeated in every reply.)

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David C. Ullrich

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