Zeros of the funcion in Laplace domain

[Highlander-SP]

ok..
i know that e^(-2s) in Laplace domain represents a delay in time domain, but i'd like to see how matlab draw the poles / zeroes diagram from the function:

F(s) = 12*exp(-2s) / [(s-4)(s-3)]

how can i represent this full function in the matlab program?

tks

 

[zorro]

Hi,

Steve10,

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

The explanation is in your own posts.

Steve10, you misunderstood at the beginning the question posed by Highlander-SP: you mistook what was “the function”, you confused whether the transfer function was f(t) or F(s). This is not grave. Perhaps you are not familiar with some terms of circuit analysis employed in electrical engineering. This is not grave. The point is that you don’t have reason for being so provocative and treat other people as they were ingorant. It’s funny for you? Not for me. It is pathetic.

Now, let’s go back to the question about zeroes.

F(s) has an essential singularity at Infinity because of the term exp(-2s). I agree with that.
But, for circuit analysis, F(s) is better specified saying that it is composed by the product (the cascade) of two transfer functions:
1) a function that has poles at s=4 and s=3, and unity gain at d.c.;
2) a function that, in time domain, represents a pure time delay of 2.
The first has the two poles (and in consequence two zeroes, that are at Inf).
The second does not contribute with poles nor zeroes to F(s), but with the essential singularity at Inf.

For circuit analysis purposes (response in frequency domain, transient response, stablity analysis, etc.) it is not useful to describe F(s) as a function with an essential singularity at Inf. Saying that, the only conclusion would be that standard methods of circuit analysis could not be applied.
Instead, describing F(s) in the other way, those methods are appropriate.

For instance, the doble zero at Inf says the behaviour of F(s) at “high frequencies” (in the imaginary axis): for frequencies high enough, the amplitude of the transfer function decays at 40 dB/decade, and the phase is +/-Pi. The additional term exp(-sT) doesn’t modify the amplitude response, and adds a linear phase term.


Irfan1,

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

Irfan1, if a transfer function has a zero at s0, then when the input (in time domain) is x(t)=exp(s0*t), the output y(t) is zero. Note that this is true for x(t) coming from t=-Inf to t=Inf, i.e. x(t) is not exp(s0*t)*u(t) where u(t) is the step function. (The symbol “*” stands for multiplication.)

If the input is x0(t)=exp(s0*t)*u(t), then y(t) contains a transient response that is a linear combination of functions of the form exp(p_n*t)*u(t), where p_n are the poles of the circuit. These are the “natural responses” of the circuit. But there is no a term exp(s0*t)*u(t) in y(t) if s0 is a zero.

If the input is x0(t)=exp(s1*t)*u(t) and s1 is not a zero of the transfer function, then there is also a term exp(s1*t)*u(t) in y(t). This is the forced response. It is also called steady-state response, because it is the response that is observed at the output once the transient response extinguished but the term exp(s1*t)*u(t) in y(t) did not. This happens if s1 has real part >=0. The term “steady state” comes from the use of signals exp(j*w*t+theta) [or sin(j*w*t+theta)], that don’t decay with time.
Note that the transient response extinguishes only if the circuit is stable (it has not poles with real part >=0).

In the above sentence, the expression “the transient response extinguished” means that it is small enough that it is negligible compared with the steady state response, because, rigorously speaking, it decays asymptotically to 0 but without vanishing completely. [OK, in practice it becomes unnoticeable when it is weaker than noise, or it disappears when the circuit is destroyed, but let’s leave this facts out of our mathematical model for the moment].

Regards

Z

 

[irfan1]

Dear zorro,
While I agree with some of your comments, I would still want to make a remark about your analysis. To me, it seems that s0 enters to the equation as a parameter rather than a variable in the s domain (in your analysis). I beleive that this is an important distinction. I beleive that, apart from some special cases, the zeros of a function in s domain or frequench domain does not have any special meaning in the time domain. The reason is simple. It is an integral transform. Every frequency component contributes to the response in the time domain.

 

[steve10]

You know, I love this kind of preface.

Steve10, you misunderstood at the beginning the question posed by Highlander-SP: you mistook what was “the function”, you confused whether the transfer function was f(t) or F

(s). This is not grave. Perhaps you are not familiar with some terms of circuit analysis employed in electrical engineering. This is not grave. The point is that you don’t

have reason for being so provocative and treat other people as they were ingorant. It’s funny for you? Not for me. It is pathetic.

Now, let’s go back to the question about zeroes.
....

What was that? You were wrong, but are trying to save your face? You got to find a better way to achieve that.


Actually, I still enjoy what is in your very first post along this line:

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’.

It reminds me what was written in the book about old schools. A wizened old man rushed into a class furiously, and shouted at a timid student:"Steve! Pay attentionto what I am

saying! You are the only one ...". "No, sir, I am just telling the truth ...", frightened Steve replied.

It sounds to me like you are used to the situation where everyone else keeps quiet while you are making speeches ... wow! did I use "pay attention .." to you? Did I use

"Analyse ..." without a "please"? Who do you think you are? You make me laugh and I don't see you are better than the person you are trying to help. Well, well, of course,

it's not "grave" (I like this word). The universe is still expanding ...


Oh, well, let's turn to something that might be more productive. So what we got here? You blamed me to misunderstand the "function" ...? Yes, I think I did and I explained why

I did. Here I explain it again if you have trouble to find it. To me, function F(s) = 12*exp(-2s) / [(s-4)(s-3)] is so clear that you can easily see everything about it. I

couldn't believe anyone has trouble about the zeros of this function. So, a meaningful question, I thought, would be to deduce some conclusions about the zeros of the function

in t - domain.


Now, let me ask you something. Here is Highlander-SP's original post:

The function

F(s) = 12*exp(-2s) / [(s-4)(s-3)]

has the poles: (4,0);(3,0)
and what is the zero of the function? How do i analize it?

Thanks

Highlander-SP might work in the eletrical enginieering or something similar, but where in this post he suggested this question comes from circuit analysis? Or do you think

that the circuit analysis is the only place that the Laplace Transform can be applied? Here are your arguments:

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.

...

If you are not sure if his question comes from the circuit analysis, how do you know that it's ok to mistake "singularity" as "zero"?

 

[claudiocamera]

Zorro,

In spite of promising myself that I coudnt carelless to this topic, I couldnt help to write again.

You are a smart guy and you have shown this in the message you posted, In my opinion regardless what somebody else wrote, you did a good job, you explained the topic in a great way. I can realize that you are very good at what you do. So, it is just an advice, dont care about steve10, he doesnt deserves our attention, this topic is closed and you closed it with your last message.

steve10 quote that it is not our business what he does, he is right, it is not our business that he is a freak, it is not our business that he has problem, Im sure this guy has no friends, probably he has problems with girls , probably he lives alone in front of a computer reading stupid estories like "wizened old man rushed into a class " ; if whales are mamals and other stupidits, probably he spend his spare time visiting porn sites, his behaviour point to these, anyway it is not our business . But it is our business dont give a rat about him.

Highlander is the other guy in this context, he keeps twindling with the same point and dont give a deffinition from what he wants, he is always quoted but dont give a clue, he keeps making question that are already answered, without giving a clear idea about what his purpose is, yet I dont know if he needs a circuit analysis approach or care about math definitions, which in the case of a practical engineering approach and in the way was presented by " steve-freak-10" is not suitable for the problem at hand.

 

[zorro]

I feel that there are two subjects in this discussion.
One of them is technical, and involves zeroes, singularities, circuit theory, complex analysis, etc. It is relatively simple to clarify, if people are well disposed.
The other subject is human, involves respect for people, honesty, superbia, and is much more difficult to deal with.


Steve10,
Regarding the first subject, in your last post you don’t mention nothing about my previous post. Can I assume that you agree with what I wrote?
Regarding the second subject, I feel that it is not worth to continue to respond to you.


It seems that there is a problem with some words I used (I omitted the word “please”; “grave” and “pay attention” are not well used). Sorry. My English is not so good (although I think it is clear enough for whoever disposed to understand me).
Steve10 suggests that I lacked good manners. Of course, I have to learn from him.

As I wrote, I feel that it is not worth to respond to Steve10’s unsustainable arguments. Instead, let me answer to other posts.


Highlander-SP,
Unfortunately, I don’t know if there is a way to include the delay term in a system description in Matlab Control System Tolbox. Maybe the System Identification Toolbox has that possibility.


Claudiocamera,
Thank you for your feedback.


Irfan1,
Let me write something about the relationship between zeroes in frequency domain and responses in time domain, starting from a more general point.

Let’s suppose we have a linear time-invariant (LTI) system with a single input a single output. The signals we will consider are not time-limited (they come from –Inf to Inf in time).
If the signal at the input has the form x(t)=exp(s*t) [and only in that case!], where s is any complex value, then the output y(t) has the same form: y(t)=H*exp(s*t), where H is a complex value. [This can be seen from the integral-differential equation that describes de LTI system].
The value of the constant H is dependent of s, thus the right way is to write y(t)=H(s)*exp(s*t) .
Here, H(s) is the transfer function of the system in Laplace domain. Thus,

y(t) = H(s)*x(t) when x(t)=exp(s*t)

This is an important concept of what a transfer function H(s) means.
This is a simple link between t-domain functions and s-domain functions. Other links are the initial-value and final-value theorems.

It can be noted from the above equation that if you put at the input x(t)=exp(s0*t), where s0 is a zero of H(s), then the ouput is 0. (I recall here the consideration of my previous post about the transient response).
It can be noted also that you could have a non-null ouput of the form y(t)=exp(sP*t) with null input, if sP is a pole of the system. From a mathematical point of wiew, this is a solution of the homogeneous integral-differential equation describing the system (when x(t)=0 for all t). Physically, each pole gives one of the ways in which the system evolves towards equilibrium losing its stored energy (if it is stable) when it has not forced response (i.e., with null input).

Regards

Z

 

[steve10]

That's another great word, "unsustainable". You called my arguments "unsustainable". Yes, those poor arguments claim "Infinity" is not a zero. Why do you change you position? "I ommitted ...". If you can omit, why don't you think others can do the same? Then, why do you think you have any reason to accuse others?

Now, let's get to your "sustainable" arguments. You complained that I didn't mention your previous post. I did. Here are you arguments:

F(s) has an essential singularity at Infinity because of the term exp(-2s). I agree with that.
But, for circuit analysis, F(s) is better specified saying that it is composed by the product (the cascade) of two transfer functions:
1) a function that has poles at s=4 and s=3, and unity gain at d.c.;
2) a function that, in time domain, represents a pure time delay of 2.
The first has the two poles (and in consequence two zeroes, that are at Inf).
The second does not contribute with poles nor zeroes to F(s), but with the essential singularity at Inf.

For circuit analysis purposes (response in frequency domain, transient response, stablity analysis, etc.) it is not useful to describe F(s) as a function with an essential singularity at Inf. Saying that, the only conclusion would be that standard methods of circuit analysis could not be applied.
Instead, describing F(s) in the other way, those methods are appropriate.

For instance, the doble zero at Inf says the behaviour of F(s) at “high frequencies” (in the imaginary axis): for frequencies high enough, the amplitude of the transfer function decays at 40 dB/decade, and the phase is +/-Pi. The additional term exp(-sT) doesn’t modify the amplitude response, and adds a linear phase term.

You mostly talked about circuit analysis, and I was asking you (right after your post) where in the Highlander-SP's very first post he suggested his function comes from circuit analysis? If there is no such suggestion, your arguments are irrelevant (maybe sustainable in your language?). Are you aware? If you want to take this chance to make a speech about "circuit analysis", you can. But why do I have to respond?

 

[claudiocamera]

Once again Zorro shows how to provide usefull information in a subject that was already over.
The Eingenfuntion and Envenvalue properties of Laplace transform are a relevant topic, It was shown in the context of Irfan1 question.

That is the difference between someone that has knoledge to share and someone who wants to find foes wherever he goes and whatever he does. ( even on Internet).

Well, since the aim here is answering question and since steve10 quoted that I study nothing, I will answer a question that I am sure what the answer is , Steve10's last question:

If you want to take this chance to make a speech about "circuit analysis", you can. But why do I have to respond?

Answer number 1:
You dont have to respond because you know nothing from circuit analysis.

Answer number 2:
You will respond because your profile is easily predictable, you want to have the last words, all over these messages you tried to have the last words. It is compulsive of yours.

Am I wrong ?

 

[muoinhohn]

HI
I do not why you discussed this problem in a manner that is not friendly. I think that we are on the forum to strike up a friendship, is not it ?

I found the definition in the dictionary of analysis :

"Zero of function For a function f : X → C, a point x0 ∈ X satisfying f (x0) = 0."

"Zero point of the kth order For a function f (z), analytic in an open set D ⊆ C
(or D ⊆ M, for M an analytic manifold), a point z0 ∈ D such that f (z0) = f (z0) =
· · · = f (k−1)(z0) = 0 and f (k)(z0) = 0. "

if we considere that infinity is a point ∈ C , then it will be zero point of the function.
else we conclude simply that the function has not zero point.

Regards

 

[steve10]

Zorro,
I apologize for whatever I said here that might have hurt you.

 

[hamidr_karami]

hi


f(t) = 12*Heaviside(t-2)*exp(4*t-8)-12*Heaviside(t-2)*exp(3*t-6)


bye

 

[jawadasheikh]

This function can also be written as F(s) = (exp(-2s) ) * {12/ [(s-4)(s-3)]}