Best advice I've heard on this subject so far.

- posted
18 years ago

Best advice I've heard on this subject so far.

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- posted
18 years ago

nope, you've got it all wrong bean. obviously you're incompetent. well we knew that from the start. need better than that if you're going to stay here.

- posted
18 years ago

in article snipped-for-privacy@individual.net, Nimrod at snipped-for-privacy@me.com wrote on 12/03/2004 02:27:

i found a few old postings where ***he*** says he's class of '72 (not 2002).

same middle name, same first name, same last name. education in mathematics. lot'sa coincidences for someone on this side of the pond. i'm sure you UKers know him better.

i picked that up.

i know that. another reason i thought it was the young guy in the photo is that his arguments are really neophyte (his "integration-by-parts" and his concepts surrounding bandlimited sampling are indicative of someone who hasn't really done it) and his means of argument are very immature. not reflective of someone in their 50s.

it's exactly what i did.

i agree.

r b-j

- posted
18 years ago

That evaluation is the one by which all electronic engineers are taught.

- posted
18 years ago

You really need to grow up.

This is an international forum and not the staff room of the infants's school in which you apparently teach.

You're changing your tune, once again.

The discussion was about int f(t).d(t-T), now your equivocating about int f(t) only.

Integration and differentiation of products is not done by simple substitution.

- posted
18 years ago

The anti-derivative of

(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))

is pretty continuous in the classical sense, and indeed, in any sense you care to present.

- posted
18 years ago

Quack? Honk?

- posted
18 years ago

There are some historical consistencies with the email addresses with which he's been associated that put him as an experienced ringer (a person who, with other folks, pulls the ropes to toll peals and carols from church bell towers), with email traffic to related lists back to at least 1998. He was upsetting some of those folks with his views, too, and his account of his historical geographic locations is consistent with where he is now.

So I think the young guy in the website you found is unlikely to be Airy. The employment history issues also seem to indicate the same.

He seems to be interested in and experienced with music, which is something he has in common with you and I, though. ;)

Apologies to those of you who've dealt with this for way too long. He's still a little new to us at comp.dsp, but we've had very similar experiences with other people.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

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- posted
18 years ago

For any nonzero eps, yes. But for any nonzero eps, that function is not the delta function.

- posted
18 years ago

in article snipped-for-privacy@news.west.cox.net, Eric Jacobsen at snipped-for-privacy@ieee.org wrote on 12/03/2004 13:43:

i've come around to that conclusion. it's just that with an identical first, middle, and last name and some formal study in electrical engineering and math and being in the UK, that just seemed to be too many coincidences for me initially.

"***very*** similar"? i think Beanie is in a league of his own. his trolling mastery ***way*** outstrips eBob.

r b-j

- posted
18 years ago

: He appears to be unemployable and becomes fixated on those who have won out : arguements against him. said NIMROD hiding in the shadows thowing stones from a safe distance

: : If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. said NIMROD hiding in the shadows thowing stones from a safe distance

- posted
18 years ago

It's hard to decide. You're just trolling, or you're really unable to follow simple arguments.

Hint: The "f" in the statement from me you quote below is not the same as the f that we've been talking about. If I'd realized it was going to cause confusion I would have used a different letter...

David C. Ullrich

- posted
18 years ago

Uh, yes it is. If that function were a delta function you'd have a point.

David C. Ullrich

- posted
18 years ago

Take ARB with you.

- posted
18 years ago

If you integrate it over the entire range from -oo to +oo, then yes, the value is f(T). BUT, you can't integrate -oo^+oo f(t).d(t-T) separately from the '(-s)e^(-st)dt' term in the second term of your 'integration by parts'.

The second term of your 'integration by parts' is "-int_-oo^+oo F(t)(-s)e^(-st) dt". That is ***not*** the same thing as "- (int_-oo^+oo F(t)) (int_-oo^+oo (-s)e^(-st)dt) " You can't integrate the two factors separately, so you can't 'pull' the f(T) 'out front' just yet.

The 'definition' is part of the Dirac function. The value is zero for all values < 0 and all values > 0 and undefined at zero. lim_m->0 [int_-oo^(0-m)delta(t) dt] = 0 and lim_m->0 [int_(0+m)^+oo delta(t) dt] = 0 and lim_m->0[int_(0-m)^(0+m) delta(t) dt] = 1.0

What you can do is evaluate the integral in steps between -oo and +oo like this...

-int_-oo^+oo F(t)(-s)e^(-st) dt = - lim_m->0 [ int_-oo^(T-m) F(t)(-s)e^(-st) dt + int_(T+m)^+oo F(t)(-s)e^(-st) dt ]

Since F(t) = 0 for all values of t < T, and F(t) = f(T) for all values of t > T ...

-int_-oo^+oo F(t)(-s)e^(-st) dt = - lim_m->0 [ 0 + int_(T+m)^+oo f(T) (-s)e^(-st) dt ]

Since this last integral is over the range of T+m to +oo, and over that specific range F(t) is a constant, here we can 'pull it out front'.

= - f(T) lim_m->0[ int_(T+m)^+oo (-s)e^(-st) dt] = - f(T) [ e^(-s(+oo)) - e^(-sT) ]

As long as s > 0 e^(-s(+oo)) = zero so...

= f(T) [e^(-sT)]

daestrom

- posted
18 years ago

But the integral you need for your integration by parts is...

int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt

Not just

int_-oo^+oo f(t)d(t-T) dt

And you can't 'factor' integrals

int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt is ***not*** equal to

See my other post for one method of evaluating the term properly.

daestrom

- posted
18 years ago

That much is true. BUT, it is only true for ***some*** definite integrals such as from -oo to +oo. It is ***not*** true for all definite integrals, such as from -oo to zero (assuming T > 0).

As X moves from -oo to +oo, the value of int_-oo^X f(t)d(t-T) dt starts out as a horizontal line at zero, then at T, it is discontinous and 'jumps' to f(T). From X=T to +oo it is again a horizontal line, but with a value now of f(T).

There ***is*** some information given by the definition of the delta function though that can be useful. For example, the following definite integrals can be found from the definition.

Because the Dirac delta function is zero for all values < 0...

For all positive values of m, lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0

And because the Dirac delta function is zero for all values > 0...

For all positive values of m, lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0

And finally, since given the two conditions above and the part of the definition that says int_-oo^+oo d(t) dt = 1 ...

For all positive values of m, lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1

In point of fact, these three definite integrals can be considered the

Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isn't much use in solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current context, G(t) = (-s)e^(-st) ). You can't 'factor' the integral into two parts like this.

int_-oo^+oo f(t)d(t-T)G(t) dt is ***NOT*** equal to [int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]

But that is what you are in effect doing when you say... int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo (-s)e^(-st) dt

One thing you ***can*** do is... int_-oo^+oo f(t)d(t-T)G(t) dt = int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt Where 'a' is any arbitrary real number.

Then, if we set things up so that 'a' has a value approaching 'T' from the negative in the first integral, and from the positive in the second integral, we can make some progress. So define our 'a' as (T-m) where 'm' is positive and => 0 as a limit in the first integral and (T+m) in the second. Substituting this into our problem, we get...

int_-oo^+oo f(t)d(t-T)G(t) dt = lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo f(t)d(t-T)G(t) dt ]

Now, over the definite integral of -oo to (T-m), the value of int_-oo^(T-m) f(t)d(t-T) dt is always zero so the first integral is zero. Over the definite integral of (T+m) to +oo, the value of int_(T+m)^+oo f(t)d(t-T) dt is always f(T) so for ***THIS PARTICULAR DEFINITE INTEGRAL***, it can be factored. So over these two particular definite integrals, you can factor out the term. But note that it has different values over these two particular definite integrals.

int_-oo^+oo f(t)d(t-T)G(t) dt = lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t) dt ]

And, given G(t) = (-s)e^(-st)...

int_-oo^+oo f(t)d(t-T)G(t) dt = f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]

The integral of (-s)e^(-st) is easy...

int_-oo^+oo f(t)d(t-T)G(t) dt = f(T) { e^(-oo) - e^(-sT) } = f(T) { 0 - e^(-sT) }=

-f(T)e^(-sT)

Substituting into the original integration by parts... int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))

And since Uint(V) is zero... int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)

daestrom

- posted
18 years ago

That much is true. BUT, it is only true for ***some*** definite integrals such as from -oo to +oo. It is ***not*** true for all definite integrals, such as from -oo to zero (assuming T > 0).

As X moves from -oo to +oo, the value of int_-oo^X f(t)d(t-T) dt starts out as a horizontal line at zero, then at T, it is discontinous and 'jumps' to f(T). From X=T to +oo it is again a horizontal line, but with a value now of f(T).

There ***is*** some information given by the definition of the delta function though that can be useful. For example, the following definite integrals can be found from the definition.

Because the Dirac delta function is zero for all values < 0...

For all positive values of m, lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0

And because the Dirac delta function is zero for all values > 0...

For all positive values of m, lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0

And finally, since given the two conditions above and the part of the definition that says int_-oo^+oo d(t) dt = 1 ...

For all positive values of m, lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1

In point of fact, these three definite integrals can be considered the

Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isn't much use in solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current context, G(t) = (-s)e^(-st) ). You can not 'factor' the integral into two parts like this.

int_-oo^+oo f(t)d(t-T)G(t) dt is ***NOT*** equal to [int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]

But that is what you are in effect doing when you say... int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo (-s)e^(-st) dt

One thing you ***can*** do is... int_-oo^+oo f(t)d(t-T)G(t) dt = int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt Where 'a' is any arbitrary real number.

Then, if we set things up so that 'a' has a value approaching 'T' from the negative in the first integral, and from the positive in the second integral, we can make some progress. So define our 'a' as (T-m) where 'm' is positive and => 0 as a limit in the first integral and (T+m) in the second. Substituting this into our problem, we get...

int_-oo^+oo f(t)d(t-T)G(t) dt = lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo f(t)d(t-T)G(t) dt ]

Now, over the definite integral of -oo to (T-m), the value of int_-oo^(T-m) f(t)d(t-T) dt is always zero so the first integral is zero. Over the definite integral of (T+m) to +oo, the value of int_(T+m)^+oo f(t)d(t-T) dt is always f(T) so for ***THIS PARTICULAR DEFINITE INTEGRAL***, it can be factored. So over these two particular definite integrals, you can factor out the term. But note that it has different values over these two particular definite integrals.

int_-oo^+oo f(t)d(t-T)G(t) dt = lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t) dt ]

And, given G(t) = (-s)e^(-st)...

int_-oo^+oo f(t)d(t-T)G(t) dt = f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]

The integral of (-s)e^(-st) is easy...

int_-oo^+oo f(t)d(t-T)G(t) dt = f(T) { e^(-oo) - e^(-sT) } = f(T) { 0 - e^(-sT) }=

-f(T)e^(-sT)

Substituting into the original integration by parts... int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))

And since Uint(V) is zero... int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)

daestrom P.S. Sorry to all if this post shows up twice. My ISP's news-server seems to have misplaced the first copy so I'm posting again directly to google-groups.

- posted
18 years ago

Whatever he is I don' believe he is a troll! He is a very mixed up person and can be very vindictive and vicious at times. I "kill-filed" him a long time ago but unfortunately others quote him in their posts and reply to the rubbish he puts out,. Only by totally ignoring him will he eventually leave the newsgroups that he presently infests.

Peter

- posted
18 years ago

I am not a troll, and neither am I mixed up, vindictive or vicious.

The person speaking below has a particularly nasty attitude problem which manifests itself in his behaving like a 5 year old in these NG, an example of which can be seen below. Notice that is is a gratuitous contribution unrelated to any other matter.

Peter Day is kill-filing me because he made a fool of himself by uttering repeated abusive remarks to which I did not rise; I chastised him for his foolish childishness and now he is sulking in a kill file.

Stupid boy!

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