Zeros of the funcion in Laplace domain

[steve10]

Come on! Life goes on. This is not the end of world, is it? Can't you just let it go? I still remember what you said that I am the kind of person who always wants to have the last words. Well, here I come to give the "last words".

 

[Highlander-SP]

i would like to know how i can represent the e^(-2s) in the nominator....

anybody can help??

 

[steve10]

Yes, I believe a lot of people can help with it and this question is top important. It might affect the world and it might bring us the next generation technology. But before we start enjoying that exciting moment, could you tell us what the hell is "nominator"?

 

[claudiocamera]

Highlander,

e^(-2s) represents just a shift in time domain, It does not affect neither zeros, poles or stability .

If you have any doubts feel free, this is a intenational board, even if you mispell some word there alwas be someone more patient that can realize what you mean. You couldnt careless to bitter coments given by people who lacks manners.

 

[steve10]

What you said above confirms what I said about you before. I am sorry that I cannot say it again here as I have promised Pisoiu. I see nobody can save you as you have been seriously poisoned by your emotion. You refuse the truth and refuse the reality. Simply because I said that point is not a zero, you even don't want to have a try. You said, it's ok to lack basics, as long as you studied. You study nothing.
I am not sure why I have engaged all these useless conversation with you. Honestly, you don't worth me a second.

 

[zorro]

Highlander-SP,

Claudiocamera is right.

F(s) has two finite poles (at 4+j0 and 3+j0), two zeroes at infinity, and a time shift of 2.

Suppose H(s) is a transfer function expressed as a quitient of polynomials in the variable s.
If H(s) has no zero at infinity, then its impulse response includes an impulse at t=0+.
If H(s) has a single zero at infinity, then its impulse response is finite (but not zero) at t=0+.
If H(s) has a double zero at infinity, then its impulse response is zero at t=0+, and its first derivative is finite (but not zero) at t=0+.
… and so on.

Now, consider Hd(s) = exp(-sT)*H(s) . Its impulse response is the same as H(s) but delayed by T.

A quotient of polynomials has a mth-order zero at infinity, where m=degree(denominator)- degree(numerator) [suppose m>=0]. Observe that, if m>=0, the limit of H(s) exists for |s|->Inf, regardless of the path along which s goes to infinity.
The term exp(-sT) cannot be expressed as a quotient of polynomials of finite order. It does not contribute with poles nor zeroes, but with a time delay.
For s->Inf, the limit of exp(-sT) depends of the path: for instance, is 0 for s going to Inf by the positive real axis, is Inf for s going to Inf by the negative real axis, and the limit does not exist (oscillates) if the path is the imaginary axis.


Steve10,

Analyze the problem, and you will see your mistake. See from the first post. Pay attention at the first one, what is called the ‘function’.

Regards

Z

 

[steve10]

Zorro,
Before the discussion, if you have no objection, I would like to get a few things straight:
(1) Please give clear definition of F(s) and H(s) in your post;
(2) Please give the definition of the zero of an analytic function in Complex Analysis;
(3) If you don't think the zero that we are talking comes from the Complex Analysis, then where does it come from? Please define it.

Added after 1 hours 57 minutes:

Zorro,
Are those so demanding? Or you may suggest something similar so that they are easy for both of us. The bottom line is that, when the discussion starts, we should have a clear idea what you are talking about, the function, the definition of the zero, etc. Isn't that fair?
Besides, just for your benefit, if you work in a college, you may want to check with your math people first. Well, it's up to you what you do.

 

[steve10]

Come on! This is really no fun. Look at your post, it won't take a second to write down F(s) and H(s). Besides, I can find a few places in your post mentioning '"zero", which means you have already had it in your mind. I don't think you were talking about something you had no idea what it is, did ya? Come on! Let's have fun. I will then be able to "Analyze the problem" as you suggested.

Ok, let me write them down. You just give your "yes" or "no". Is that ok?

F(s) = 12*exp(-2s) / [(s-4)(s-3)];
H(s) = P(s)/Q(s), where P(s) and Q(s) are polynomials of mth and nth order, respectively.

Are they ok?

Ok, here is the definition of the zero. Because a laplace transform is defined in the complex plane, I'll take the definition of a zero from the Complex Analysis. Suppose f(s) is a complex function and s0 is a point in the complex plane (could be Infinity). If
limit_{s->s0} f(s) = 0

Then we call s0 is a zero or a zero point of f(s).

Isn't that simple? Yes, very simple. Ready to roll?

 

[zorro]

Steve10,

Before reading your last post I wrote this answer to your previous one:

(1) Please give clear definition of F(s) and H(s) in your post;

F(s) is the function of the first post of Highlander-SP.
H(s) is any transfer function expressible as a quotient of (finite-order) polynomials in the variable s.

The bottom line is that, when the discussion starts, we should have a clear idea what you are talking about,…

I agree. It seems that the other persons that took part in this discussion had a clear idea about what people were talking about. It seems that you were the only one that misunderstood the subject.

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

Now, I add on that I can't understand your hostility.
OK, that is a zero. Then?

Regards

Z

 

[steve10]

Welcome back, but I don't know what makes you think I am hostile. That sounds funny. Care to explain?

So you think the definitions of F(s) and the zero are ok, huh?

Then, our goal is to decide if Infinity is the zero of F(s)? Am I misunderstanding?

 

[claudiocamera]

Dear Zorro

It seems tha the nervous guy wrote that:

F(s) = 12*exp(-2s) / [(s-4)(s-3)]; if limit_{s->s0} f(s) = 0 Then we call s0 is a zero or a zero point of f(s).

What is limit_{s->infinity} 12*exp(-2s) / [(s-4)(s-3)] ??? If the guy dont want to put aside the exp(-2s) , he can apply L'hoppital rule and find the result.

As far as I am concerned the result is gonna be zero (0 ). So I can't understand his point, he got the result that he is fighting against. Indded he supported the thesis of zero at infinity.

I dont know from what area this guy is, but probably, he have never studied classical filters or root locus analysis and probably he have never heard about zeros at infinity, thats why he might consider this too odd or weird.

I am sure he will reply, as I said before, he always wants to have the last words, I had decided not to answer this topic anylonger and let him with the last word, My mom always tell me to do not discuss with stupid people, and I would take her advice, but your participation changed my mind.

Thanks to you Zorro, Highlander can realize that in this discussion three people ( you , vale and myself) has the same oppinion, against just one that has a different view. Dont worry this same person, struggled in other discussion writing that bilateral laplace transform is useless. I am tired of discuss with a person with so controversial point of views. I also dont know why he is so hostile, in spite of full filling the text with smilling faces, he always try to disqualify other people ( even when they are right)

Thats why I am adreessing this message to Zorro. I want to discuss with people who have manners.

Thanks a lot.

 

[irfan1]

I would like to raise a similar question. What does the zeros of a function in fourier domain mean in the time domain?

 

[steve10]

Ok, if I am not misunderstanding, we can proceed now.

Actually, the next step is pretty easy. We have the function F(s), we have the point s0 = Infinity and we have the definition. Then all we have to do is to feed those two items (F(s) and s0) into the definition to calculate the limit. And then you and I compare the notes to come up with our conclusion.

Oh, well, guess what? While talking to you, I have finished my calculation and my result is:

limit_{s->Infinity}F(s) DOES NOT exist

What's yours?

Added after 21 minutes:

irfan1,
When you say "similar", what do you think the question we are talking about is? Just curious. No hard feeling. Feel free to answer or not to answer.

 

[claudiocamera]

Oh, well, guess what? While talking to you, I have finished my calculation and my result is:

limit_{s->Infinity}F(s) DOES NOT exist

.

Firstly you were not "talking" to anyone, you were "writing". Secondly I would like to make you a question ? Who lacks the basic ? From what a gather limimits with 0/infinity indertemination is solved applying l'hoppital rule. Before go to complex analysis books , would be a great idea studying calculus.

irfan1,
When you say "similar", what do you think the question we are talking about is? Just curious. No hard feeling. Feel free to answer or not to answer.

Why this kind of expressions ? It seens that everybody who dare ask something will receive hostile messages back.

Other questions:

If you are so sure you are right, why the ansiety of answering a post, come back later and answer again ? As you said to highlander you might be thinking that this subject will result in a tech revolution .

If somebody else ask something else, you will answer badly again or you will be more polite ?

 

[steve10]

There are half of a dozen of people who want to leave their home t - world (earth) and to travel to the s - world, which is a remote planet. When landing, they are accidentally thrown into a small place which looks like a small town in the t-world. They start commenting,"wow, the s-world has been exagerated, and it turns out to be such a small place like the backyard of my house in the t - world." Suddenly, a local gentleman pops out from nowhere and offer his help as a tour guide, but the half of a dozen of playboys refuse,"We don't need your help. Everything here is so similar to the ones in our t - world. Look at that big chimney ... ". "No, that is not a chimney", the poor local man tries to correct, "that is the place through which people are launched to anywhere in the universe." "Don't play that dirty trick, peasant. The six of us all agree that is a chimney, while you are only one ...."

Here, I urge you guys honestly ask yourself the following questions in s - domain and see how much you know about the s - domain:

(1) Did you know that s=1 is not a zero of (s-1)^(4/3), while x=1 is a zero of (x-1)^(4/3)? s=1 is a branch of (s-1)^(4/3).

(2) Did you know that, if f(s) has the first order derivative, then it has infinite many order derivatives?

(3) Did you know that, for any point s0, f(s) has the first order derivative at s0 <=> the integration of f(s) along any closed curve aound s0 is equal to zero? Can you find anything in the real function that the differentiability is decided by integrability?

(4) Did you know that sin(s) can take any value? It means, for example, you can find a complex number s0 such that sin(s0) = 1000000000000!.

(5) Did you know that exp(-2s) can take any value (except 0) you designate? (Picard theorem)

(6) If you treat w=f(s) as a map from the complex s -plane to complex w - plane, did you know that the map will retain the angle of any two straight lines?

(7) If you write f(s) = u(x,y) + i V(x,y), where u and v are real, then both u and v are harmonic functions?


I cannot list them all ...

Zorro,
Don't bother about that limit.
It's really not my business whatever you guys think and you guys do.


irfan1,
I asked you that question because, at the very beginning, I thought they are interested in something in t - domain, exactly like you do. Gradually, I realize that, even though they keep mentioning t - domain, they are really not interested in anything there. Isn't that funny? My very first a couple of posts showed it. Finally, I got used to it.

 

[irfan1]

My point is the following: just a single point in frequency domain or s-domain has no meaning in the time domain or t-domain. That is simply because the transformation is not one to one or we are considering the integral transformations. Each point in t-domain has a contribution from each point in the s-domain.

 

[claudiocamera]

I will not tell any fair tale, neither quote math theorems. Fortunately the discussion was enriched by guys who seem to be willing to discuss appropriatelly.

I see no practical reasons to stabilish relationships between t and s or jw domain in every cases you face, since transformation in signals analysis is mostly used to make things that are hard to analyse in time domain easy to analyse in frequency domain and vice versa. Nevertheless; quite often, in order to undestand some points and realized how useful is transformation tool, it is necessary to compare the two domains, for instance: multiplication in frequency -> correlation in time and as was the case: time shift -> phase shift. this kind of approach not necessarily try to suggest that it should take in consideration correspondency one to one in every other aspect such for example zeros and poles.

The question raised by irfan1 is very interesting , I cant remeber trying to connect zeros or poles in frequency domain to t domain point by point, probably because the natural purpose of transformation tool that I quoted, despite this fact the question raised a good point .

Now considering a practical approach from laplace functions:

A butterworth filter has an equivalent Laplace function of H(s)H(-s) = 1 / [ 1 +(s/jwc)^2k] where wc is the cutoff frequency and k is the filter's order.
The question is: What are the zeros of a butterworth fileter ?

I think the answer should be connected with we are discussing.

Another remark, I was wondering about the topic's title "Zeros of the funcion in Laplace domain" probably was taken from a question in a text book ( well , this only the original poster can confirms) It seems to be the kind of question that wants to indroduce some concept to be explored in applications, what should be this concept ? just a hint.

Regards.

 

[steve10]

I'd like to take this chance to introduce a nice theorem in Complex Analysis which has been widely used.

For any function F(s), if all the isolated singularities can be surrounded by a closed curve C, then the integral of F(s) along C is equal to zero.

For example, if F(s) = 12*exp(-2s) / [(s-4)(s-3)], obviously, you can easily see two singularities, s=3 and s=4 (poles). If you think it's hard to see what the INFINITY is, then you can make your assumtion. For example, you assume that the INFINITY is a zero, then s=3 and s=4 are all the singularities. Therefore, if you make a closed curve C surounding the two points (s=3 and s=4), the integral of F(s) along C should be zero.
Let's just do a simple calculation. According to the residue theorem, the integral of F(s) along C is 2*Pi*i times of the sum of the residues of F(s) at s=3 and s=4.
Sum of the Residues of F(s) (at s=3 and s=4) = 2*Pi*i *(limit_{s->3} F(s)*(s-3) + limit_{s->4} F(s)*(s-4))
=2*Pi*i* (-12*exp(-6) + 12 *exp(- 8 ))
which is NOT zero. Apparently, it indicates that your assumtion that INFINITY is a zero is wrong.

Actually, from the final result, you can see that the exponential function really messes things up.

 

[claudiocamera]

What all of us are observing here is an example of someone stucked in pure theory and unable to cross the border to practical approach.
It is known and was written by me and by Zorro that exp(-2s) has no effect in the zeros and poles of the transfer function , I believe this same approach was also realized by vale. I have followed other discussion in which these guys took place and I could note that they have knowledge of signals and systems and DSP, just as myself.

All those who studied these subject is tired to know the shift in time laplace transform property wich reflect a phase shift in frequency. It is very obvious to us that exp(-2s) is a shift in phase in frequency domain, thats why in analysing zeros and poles there is no difference from work to the functions F(s) = 12*exp(-2s) / [(s-4)(s-3)], or H(s)= 12 / [(s-4)(s-3)], definetely these functions has the same set of zeros and poles , there is no doubt of that and there is no nice complex analysis theorem that states otherwise.

I presented the butterworth filter function because it is apractical approach of the topic discussed, the zeros of this functions is at infinity, as are the zeros of chebshev filters and as are the zeros of any function F(s)= H(s)/P(s) where m is the order of H(s) ; n is the order of P(s) and n>m, thus F(s) will have ( n-m) zeros at infinity. ( it was already quoted by zorro, and I reinforce it here), anyone who studied root locus analysis had to learn this, and there is no nice theorem from complex analysis that states otherwise.

The issue here is that there are guys who are used to signals and systems analysys discussing with someone that probably is doing something wrong in complex analisys, once the theories are linked and so cannot colapse. Honnestly I will not review, residues, cauchy and other stuffs, I dont need this to know that the exponetial dont mess anything up, it just shift frequency phase, what correspond to a shift in time, despite someone dislike to references in time domais, it is essential to quote it here.

I have no doubt in this topic and I will avoid returns to it. If someone insist in have a distorted view It is not up to me to change his mind, for those who have been following the topic, i believe there are already plenty of discussion, the poin of views are well defined. Thanks for the attention I will move on.

Regards

 

[steve10]

Complex Analysis is relatively old special subject of mathemtics. It has all its concepts perfectly defined, like zeros, poles, singularities, integral, residue, .... In other words, it's very unusual if someone has to appeal to something out of it to prove something in it. Here, we got a situation. In order to prove a point is or is not a zero, one has to go back and forth from t to s. It sounds really funny to me. It sounds like when you ask someone: "please tell me if the whale is a mammal or not." He says:"hold on! I will first go to the moon to collect evidence." He then makes a trip to the moon and brings back a load of rocks:"Look! All the rocks are red! It means the whale is not a mammal." We have a perfect system on the earth, while the moon exists based on our earth. Why do we have to appeal to the moon to verify something on the earth? ... jesus. Complex Analysis has a lot of applications, and the laplace transform is just one of them. Complex Analysis exsits independently of anything else, while laplace transform exists on top of Complex Analysis.


This is a mathematics&physics board, but it's not a engineering board. It's very natural people talk in mathemative theories, and it is NOT UNUSUAL people also talk about something else. What is really UNUSUAL is that if you are blamed to be too theoretical ... can you believe it? That makes me think ... assume there are three forums for elemntary school students, middle school students and college students, respectively. The number of the college student is much lower than that of elementary school students. Someday, the elementary school students all flood into the forum for the college students and, at the beginning, the college students are all happy with the sudden increase of the posts. But very quickly, their posts are getting fussed:"what are those roach-like symbols, big boy? Get out here!" "Well, those are my six day's and seven night's hard work ...", the poor college boy scrambles for reasons. "Come on! Out! .... Come back here! ... your glasses."

Added after 13 minutes:

It's really fun, isn't it?