Zeros of the funcion in Laplace domain

[Highlander-SP]

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

 

[claudiocamera]

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

According to laplace transform property the term exp(-2s) corresponds to a shift of 2 in the time domain, so it does not affect the zeroes of the system. Thus we can conclud that the zeroes of the system are(is) the same of the system 12/ [(s-4)(s-3)]. This system has a zero in the infinity. That's the answer for you question

 

[Highlander-SP]

What's mean the zero of transfer function?
So, this t.f. hasn't zeroes or the zero is infinity?

 

[claudiocamera]

What's mean the zero of transfer function?
So, this t.f. hasn't zeroes or the zero is infinity?

As I wrote before , the zero is at infinity.

 

[steve10]

claudiocamera, I believe you are confused with the function itself and its laplace transform. Apparently, the function (s)he gives is the laplace transform, but not the function itself. We know that the function that has exp(-2s) as its laplace transform is δ(t-2), which is zero everywhere except t=2. I don't think you can deduce anything about the zeros of the function itself from the poles of the laplace transform. Here are a couple of examples.

L[(exp(bt)-exp(at))/(b-a)]=1/((s-a)(s-b)). Obviously, the function itself has zero at t=0.

L[(bexp(bt)-aexp(at))/(b-a)]=s/((s-a)(s-b)), where b>a, the laplace transform has the same two poles but the function itself doesn't have zero anymore.

 

[claudiocamera]

Dear steve10,

First in my explanation I've never deduce zeros from poles. So in a point I agree with you, we cant deduce anything about zero from poles.

Secondly the development of the question is right, it is using time shift property with doesnot affect the zeros in the "s" domain. When we talk about zeros and poles of a transfer function we talk in the s domain , right ? So I did not understand why you refered zero in the time domain t=0.

Finnaly, in the s domain the function has a zero at infinity.

Best regards.

 

[steve10]

You know? I am thinking if I should answer your call. Since no one picks it up, I'll take it.

My understanding about the question is that the original poster wants to conclude something about the zeros of the original function (in your language, in t-domain) based on the poles of its Laplace transform F(s). The reason I said so is that the function F(s) is so simple that I don't believe that he has any difficulty figuring out the zeros of F(s). In any case, if you need a confirmation, you should make your request to the original poster, instead of me. What can I possibly tell you?

By the way, have you counted how many times you claimed F(s) has zero at infinity? Since you are so sure about it, I suggest you pick up a simple book in Complex Analysis and proceed from the very beginning. You lack the basics. Anyway, good night.

 

[vale]

claudiocamera is right.
steve10, you should make clear what the question is before posting answer or giving advices to others.

 

[steve10]

claudiocamera is right.
steve10, you should make clear what the question is before posting answer or giving advices to others.

Are you suggesting a vote?
When someone posts a question, we all rush into taking a vote and then count how many people vote one answer against another. The one that takes most of the votes is labeled "CORRECT"? Oh, no, You are making laugh, buddy.
Where is this? A bulletin board easy for discussions, isn't it? The bottomline for the board is that it allows anyone make mistakes. We are all human and human makes mistakes.
For any questions, people may understand differently, some of them might be right and some might be wrong. This is NOT unusual. Actually, you can see wrong answers here and there from time to time. Now, when you say:" ... you should make clear what the question is before posting answer or giving advices to others ... ", what do you mean? I am NOT allowed to make mistakes? Thank you, I am so honored.
Lastly, when you say he is right, which part is he right about or everything is he right about? Care to elaborate on the specifics?

 

[claudiocamera]

Dear Vale,

Don't care about Steve10, I am convinced that he has problems. It is the second time that he does polemics in a topic with me.

Dear Steve,

You think I'm lacking the basics, thank you, the basics is easy to get, studying is enought. I think you lack cordiality, good manners, that is a little bit harder to fix.

I 'm sure you will answer, people like you always want to have the last words...

 

[Highlander-SP]

So..... the zero of the transfer function is infinity or not?...
What is the utility of the zero analysis from a transfer function?

 

[claudiocamera]

You gave the function F(s) = 12exp(-2s) / [(s-4)(s-3)] rearranging this function, it can be written as: exp(-2s) 12 / [(s-4)(s-3)] .

According to the time shifting property:

lets call ---> L--->-- the laplace transform.

x(t) ---> L--->--X(s)
x(t-to) ---> L--->--exp(-tos) X(s)

As someone quote: exp(-2s) ---> L--->--δ(t-2) . Considering that multiplication in the s domain is convolution in time domain, you have x(t)*d(t-2) =x(t-2)---> L--->--exp(-2s) X(s) which confirms the shifiting property and what I have explained in the first mail and will explain again:

Considering explained above, multiply a function for exp(-tos) has no effect in pole and zero analysis. Thus we can analyse F'(s)=12 / [(s-4)(s-3)] and have the same conclusion.

Zeros are the values of "s" that make numerator null and Pole are the values of "s" that make denominator null , it is easy to see that s=4 and s=3 are the poles.

Now we will see why zero is at infinity:

Another way to write an s function is normaling it, for example, lets the function : H(s) = (s-a) it is easy to see tha s = a is a zero. Normalizing the funcion is the same that: H(s) = a (s/a -1) when you do that, it is clear that the value that goes under the s term is the zero of the function. Lets now analyse the case:

F(s) = 12 / [(s-4)(s-3)] can be written as : 12 [ s/infinity +1] / 12 [( s/4 -1)(s/3 -1)] since s/infinity = 0, continuing => [ s/infinity -1]/[( s/4 -1)(s/3 -1)] , so the values under the s in the denominator are 4 and 3 that are the poles of the function, and the value under the s in numerator is the zero of the function, this value in infinity. Thats why we can afirm that the zero is at infinity.

Understood ?
About the second question:

The importance of zeros are related with the frequency response, it is also important related with the inverse function, once the zero of a function will be the poles of its inverse, so it is important to analysis of minimum phase systems.

One of the well know application of inverse system are equalizers.

Now I would like your oppinon, as the poster of the original message, do you think the explanation was satisfatory ? did you notice some lack of basis in what was explained ?

 

[steve10]

To claudiocamera,
You know what? If I were you, I would make a simple post to admit the error. That would make you better off. Nobody is perfect and everyone makes errors. Your parents, family members would be proud of you.
Here is what you have been doing. Look at the second post, you wrote:

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

After I pointed out you lack the basics of Complex Analysis, you wrote above:

F(s) = 12 / [(s-4)(s-3)] can be written as :

Apparently, you are trying to cheat on the original poster. I am so surprised that you are so dishonest.

You are the only one that really needs help, the behavior, the Complex Analysis, ... If I were you, I would hide myself in a corner.



To Highlander-SP,

I am still not sure what you mean by "transfer function". Assume L(f(t))=F(s). Is the transfer function f(t) or F(s)?
If you are looking for the zeros of F(s), F(s) does NOT have zeros in the finite complex plane, and INFINITY IS NOT THE ZERO OF THE F(s) either, but it's F(s)'s essential singular point. Yes, INFINITY is the zero of 12 / [(s-4)(s-3)]. That's why I said claudiocamera is dishonest. He is trying to confuse others with the two functions.

 

[Highlander-SP]

Claudio, thanks for your answer!

. If the zero of the function is infinity, what it's effect in the frequency response?
. Why we have to normalize the function? What's it's meaning?
. Where can i found good e-books or examples about zeros and normalizing functions?
. How can i do the poles and zeroes diagram in matlab:

num = [? ? ?];
den = [1 -7 12];
transfer_function = tf(num, den)
rlocus(transfer_function)

 

[claudiocamera]

Steve10,

you should pay attention in the tópics. I wrote F'(s)=12 / [(s-4)(s-3)] . Did you pay attention in the " ' " after F ? Did you notice that before I wrote F(s) = 12exp(-2s) / [(s-4)(s-3)] the second was just to say that exp(-2s) has no effect on zero analysis, if you aren't so dumb you would have realized that.


So where is the dishonesty ?

Once are you so Dumb, I let you know that I also took off the original expression the sign "*" because I used it after as a convolution. This is to avoid your dumbness of sayind that I modify this too.

It is amazing that it is the second time that you are trying to confuse things. The question was already answered since the first time, give up of findind error in it, look in another post of mine, shearch other messages of mine in the board , maybe you find an error in some other message, not in this one.

You just confirmed what I wrote early , you lack good manners.

You should be the one to hide in a corner, because you are LOOSER!

 

[steve10]

The orignal poster is giving you another chance to clean up yourself. Quibbling makes you look even uglier. You only need one line to tell about that function the original poster posted, but wasted too much space only paint the whole thing even darker.

The exponential function is written on the first a few pages if not the very first page of any books on complex analysis. You even have no idea how it behaves at infinity, but you keep teling people how to learn complex analysis. What a shame! Jesus Christ, I hope your family members are not reading your posts.

Now, I just noticed that the original poster is REALLY taking INFINITY as the zero of the function he posted ...... now what do you think?

Added after 9 minutes:

To claudiocamera,

Do you have the courage to say to the original poster:"Your function doesn't have a zero anywhere. I am sorry for the confusion"?

No, you can't. You are just a liar, a coward, an idiot,....

 

[me2please]

Padé Approximation

(1,1) approximation:
\[e^{T_ds} \approx \frac{1- \frac{T_ds}{2}}{1+ \frac{T_ds}{2}}\]

(2,2) approximation:
\[e^{T_ds} \approx \frac{1- \frac{T_ds}{2}+ \frac{(T_ds)^2}{12}}{1+ \frac{T_ds}{2} + \frac{(T_ds)^2}{12}}\]


small delay:
\[e^{T_ds} \approx \frac{1}{1+ \frac{T_ds}{2}}\]

 

[claudiocamera]

1-The zero at the infinity has no efect in the frequency response . Its is a concept that is good to you know, since you will get across it in some text . for instance you will read that butterworth , chebyshev filters have zeros at infinity. Take a look in any classical filters book and you will see the application of this concept.

2- You dont have to normalise the function, I just did it to show why it is said that the function has zero at infinity .

3-There are a lot of books where you can find good signals and systems theory: Simon Haykin, Oppenheim, Lathi. I have already seen Haykin in Oppenheim in this forum. For root locus , stability and others related, you can look up discrete time control books like Ogata.
4- the set of commands in Matlab are:
>> num=[0,0,12];
>> den=[1,-7,12];
>> sys1=tf(num,den)

Transfer function:
12
--------------
s^2 - 7 s + 12

>> rlocus(sys1).

You can use others to help your analysis, such as:

>> pzmap(sys1)
>> bode(sys1)

The results can lead you to intepret that function has no zero, Matlab interpret zero at infinity as empty set, but if you know the concept you will interpret correctly.

See for example a way to find the function with the zpk comand:
>> z=[];
>> p=[3,4];
>> k=12;
>> sys2=zpk(z,p,k)

Zero/pole/gain:
12
-----------
(s-3) (s-4)

Take care, after all these you should know what z=[] means.

I would like to state clearly that I stand the point : The zero of the given function is at Infinity.

 

[steve10]

It's nice that Pisoiu has unlocked the topic, but (s)he said that, if it's gonna happen again, (s)he is gonna delete the topic.
Relax, I don't see anyone needs this topic anymore ...

 

[claudiocamera]

I trully believe that it is not up to me decide if someone need or not of something, thats why I have a different view of the remark above, for instance in this case, the original poster had questions, and again I tried to answer. That is the aim of forums, people try to answer or discuss each other questions.

By the way, I was discussing with some palls about zero at infinite and one of them had a simple explanaition: " A zero in a transfer function is gonna null the numerator and consequently it´s gonna null the function, if the numerator has a order less than the denominator, the function will gonna have zeros at infinite, since when 's' goes to infinity the functions goes to null"
It is realy true and it remind me the Z transform, that is usually expressed as z^(-1) form. So if you write :

F(s) = 12 / [(s-4)(s-3)] as

F(s) = 12. (s ^-2) / [(1-4s^-1)(1 -3s^-1)] It is clear that the poles is gonna be in 4 and 3 and we gonna have 2 zeros at infinity.

Another friend of mine said " Remember the root locus analysis, one of the rules is that you have to have the same number of poles and zeros".

So in my oppinion this topic is still useful , because it helps clarifying a concept. For instance I believe it was helpful to the original poster, and it was helpfull for me, because I've learned different ways to show the concept that has been questioned.

I believe it was helpful for those who disagreed, in spite of the fact that I think the adivice of the pall Vale should have been considered more seriously.

Finally I am oppen minded to discuss, if some doubts is still hovering. But only discuss in decent terms.

Best Regards.