# poles and zeros of Laplace transform

Discussion in 'Mathematics and Physics' started by PG1995, Mar 23, 2013.

1. ### PG1995Active Member

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Thank you, Steve.

I will focus on the quoted parts below from your previous post because so far I have not started studying bode plots therefore I will talk about the remaining part once I'm into bode plots. I'm doing self-study. Please note that I'm basing queries on example 14.2 and two of the queries are being repeated from my previous post because as far as I can see your reply didn't address them. Perhaps, the reason for this being that they are too silly. If that's so then please guide me on the right path of understanding. Thanks.

Q1:
Are you implying that there are also other kinds of Bode plots which contain other information in addition to magnitude and phase?

Regards
PG

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2. ### steveBWell-Known MemberMost Helpful Member

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Q1: No, i wasn't trying to imply that, but still it might be true. I have always seen Bode plots show magnitude and phase, but Wiki says this is "usually" the case. Perhaps there are other plots that are also called Bode plots, but generally most people imply mag/phase when they mention Bode plots.

Q2: You are interpreting correctly. The σ=0 implies Fourier transform and here the signals are pure sine waves. That's why we do Bode plots with s=jω. Doing so tells us how the system responds to sine waves of various frequencies.

Q3: No, you have it wrong. s=-2 implies that σ+jω=-2, which means σ=-2 and ω=0. We are talking about complex numbers here. You are not allowed to turn a real number into an imaginary number willy-nilly. So, in my previous post I was trying to point out that poles and zeros are typically not on the jω axis. They can be sometimes, but typically they are not.

Q4: Is not relevant since you had it wrong in Q3.

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3. ### PG1995Active Member

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Thanks a lot, Steve.

Regards
PG

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5. ### steveBWell-Known MemberMost Helpful Member

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I'm confused at to what you have plotted in Fig. 1. The transfer function is a complex value as a function of the variable s, which is also a complex value. How can you plot a complex number versus a complex number on a 2D plot? Bode plots use only the real value ω for the input, and then use two separate plots to show magnitude (real number) and phase (real number). So, what you plotted makes no sense really. So the complex valued output is resolved into two separate real 2D plots.

On the issue of Fourier versus Laplace. The transfer function is a Laplace Transform relation. Fourier transforms are a subset of Laplace transforms. So, why are you worried about the poles and zeros which relate to the Laplace transform, when you consider Bode plots which are related to Fourier Analysis and sine waves only. Poles and zeros can be anywhere in the complex plane, but the Bode plot information comes only from the jω axis portion of the complex plane. They don't directly relate. It's not clear to my why you are trying to force them together.

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6. ### PG1995Active Member

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Thank you, Steve.

But I'm sorry because I'm still not clear. So, please bear with me. Kindly have a look here and see if you can guide me. Thanks for the patience.

Regards
PG

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7. ### misterTWell-Known MemberMost Helpful Member

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s is always defined as: s = σ + jω
The zeros z = 0 and z = -2 are correct.

Location of the zeros/poles has little to do with finding out the fourier transform. The poles and zeros pretty much do define what the fourier transform looks like. Poles create "peaks" to the frequency response and if the pole happens to be on the imaginary axis, then the "peak" goes to (positive) infinity. And zeros go to negative infinity. If the pole or zero is not "spot-on" on the imaginary axis, then the peaks are "milde, or smoother".. They do not go to infinity on the Fourier plot (imaginary axis values).

Laplace transform is just an expansion of Fourier transform.

Fourier is integral of:
f(t)*e^(-jωt)

That can be expanded to Laplace; Integral of:
[f(t)*e^(-jωt)] * e^(-σt)

Combine the two exponentials and you get
f(t)*e^-(σt+jωt)

And for convenience, define s = σ+jω and you get integral of:
f(t)*e^(-st)

Which is the Laplace transform "as we know it". So, When you have Laplace transform of something, you can always find out the Fourier transform by setting σ = 0.

http://www.electro-tech-online.com/custompdfs/2013/07/CH32PDF-1.pdf
(Take a look at the picture on page 597. Or read the whole chapter. It is very good.)

Some links that you may find very interesting and useful:
http://www.dspguide.com/pdfbook.htm

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8. ### steveBWell-Known MemberMost Helpful Member

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PG,

You said all of a sudden s=σ+jω and this confuses you. The problem is that s is always σ+jω. Always and forever. That's how it is defined. The only reason jω is substituted for s (obviously with σ=0) is to generate the Bode plot, or think about the Fourier transform. Poles and zeros always relate to s, not jω. That's why we keep saying, the Bode plot is not relevant for poles and zeros.

So, don't worry about 1* and 2* and 3* etc. All you have is the Laplace transform of h(t) to get H(s) which is the transfer function. The function H(s) defines the poles and zeros. Then, if you want the Bode plot look at H(jw), with all s values replaced by jw.

I'm probably still missing what you are looking for, but I don't know how to make it any clearer than that.

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9. ### steveBWell-Known MemberMost Helpful Member

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PG,

You said all of a sudden s=σ+jω and this confuses you. The problem is that s is always σ+jω. Always and forever. That's how it is defined. The only reason jω is substituted for s (obviously with σ=0) is to generate the Bode plot, or think about the Fourier transform. Poles and zeros always relate to s, not jω. That's why we keep saying, the Bode plot is not relevant for poles and zeros.

So, don't worry about 1* and 2* and 3* etc. All you have is the Laplace transform of h(t) to get H(s) which is the transfer function. The function H(s) defines the poles and zeros. Then, if you want the Bode plot look at H(jw), with all s values replaced by jw.

I'm probably still missing what you are looking for, but I don't know how to make it any clearer than that.

10. ### PG1995Active Member

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Thank you, misterT, Steve.

@misterT: Thanks for the PDF. The text was good but unfortunately I didn't have the time to go through the complete chapter. Someday, I will read it all.

I was being quite silly. This text might be helpful here.

I understand how one can plot FT of a function. But in case of LT, as you say, we have complex input and complex output, so how do we plot it? Like how the phase, magnitude etc. relationships are established. This is how it's plotted in PDF provided my misterT in his post above (page #587). Please help me. Thanks very much.

Regards
PG

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11. ### steveBWell-Known MemberMost Helpful Member

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So what help are you looking for? The only question you asked above is "so how do we plot it?", but you noted misterT's reference, which shows how it is plotted.

12. ### misterTWell-Known MemberMost Helpful Member

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There is the real valued FT and the "more sophisticated" Complex Fourier Transform. Best I can do is to point you again to a chapter from the same book:
Chapter 31: The Complex Fourier Transform

EDIT: The chapters are not in html format yet. Download the pdf at upper left corner to read the chapter. Also the chapter 30 about complex numbers is excellent. Very practical and gives a great insight to complex numbers.

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13. ### PG1995Active Member

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Thank you, Steve, MrT.

steve: I will update my query later.
mrt: I have already downloaded that ebook. I believe you also referred me to this book in the past. That's a good book.

Regards
PG

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14. ### PG1995Active Member

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Hi

The transfer function H(w)=Y(w)/X(w) is found using Fourier transform and the transfer function H(s)=Y(s)/X(s) using Laplace transform. If you don't mind, what's the fundamental difference the two in terms of usability? When one of the two is preferred over the other? Personally, I have always found the H(w) easier to understand. Thanks.

Regards
PG

Last edited: Nov 10, 2013
15. ### steveBWell-Known MemberMost Helpful Member

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The fundamental difference is that the Laplace transform is more general. This implies greater ability to find regions of convergence and usable solutions. This is again a case where one way is easier at first, but the other way is easier long term. The only way to cross the barrier and appreciate the other side is to use the methods in practical cases. I can't really think of too many advanced texts that would use H(w) over H(s)

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16. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Sometimes you'll find a newer theory that require one or the other so you'd have to have both theories in order to understand the newer theory. So it's good to be at least a little familiar with both. In other words, you might run into a new theory that you want to understand and because that writeup uses Fourier in part of the explanation you'd have to know Fourier, and if it uses Laplace then you'd have to know Laplace.

If i remember right, they are similar except that Fourier just works with the 'AC' part of the fully complex variable 's', while Laplace uses the complete complex variable 's'. So the variable in Fourier is just imaginary (jw), while the variable in Laplace is full complex s=a+jw. The limits thus must be different too and there are specific rules for the real part.

I dont use Fourier that much anymore except in harmonic analysis. Once i learned Laplace and all the wonderful theories that follow it i was done with Fourier for circuit analysis except in some specific cases. For example (again from memory) the "one ohm energy" calculation uses Fourier.

Could we live without Fourier? I dont think so because it simplifies some things. Like who would want to go to the store to buy apples and have to use complex math to count the number of apples they are going to purchase.

Maybe we could look at some examples if you care to and have the time.

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17. ### steveBWell-Known MemberMost Helpful Member

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Yes.

And probably where you see H(w) most is after using Laplace to find s-domain transforms, you may then substitute s=jw into H(s) to get the Bode plots, and you would typically display this as magnitude and phase of H(jw). So, the Fourier answer falls directly out of the Laplace transform work you do.

18. ### PG1995Active Member

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Thank you for the offer, MrAl. We will do it but not now because I'm preparing for the exams so I won't be able to concentrate on what you say fully and this way your help won't be utilized fully. Thanks.

Regards
PG

19. ### PG1995Active Member

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Hi

I'm still struggling with the interpretation of poles and zeros. How mathematical poles and zeros affect real circuits. I thought that to start with a series RC circuit would be a good idea.

The following transfer functions are for a series RC circuit. The first one is for the voltage across capacitor and the second one is for voltage across resistor.

Both transfer functions have a pole at s=-1/RC where s=σ+jw. For transfer function Hr(s) zero exists at the origin.

What do pole and zero represent in this case? The cutoff frequency for a series RC circuit is

Thank you.

Regards
PG

Note to self: The content from post #2 is relevant here.

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20. ### steveBWell-Known MemberMost Helpful Member

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Poles are places where the transfer function blows up. Zeros are places where the transfer function is zero.

This then leads to the question of what the words "places" and "where" mean. The places in the s-domain represent complex exponential signals.

If the poles and zeros are on the jw axis, then you will see the effect clearly and directly in a Bode plot. The magnitude plot will either go to infinity for a pole, or go to zero for a zero.

If the poles and zeros are not on the jw axis, then they effect the Bode plot indirectly, but sometimes still in a simple way, as you pointed out with the first order system.

There are also issues of stability (poles in the right half place) and regions of convergence (locations where the Laplace transform does not converge) to consider, which complicates the details, but that's basically it.

You will still feel uncomfortable, but as I've pointed out in the past, you just have to work with systems, poles and zeros to get experience, and then the usefulness and insight will come to you. There are some questions that you should not expect a clear answer to, and this is one of them. The answer is to get to work and dig the answers out of the dirt.

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