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laplace transform

Discussion in 'Math' started by shiva bharadwaj, Oct 8, 2008.

Sep 29, 2008
19
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Hi guys ,
in the expression s=σ+jω what does σ convey?
Fourier transform is a tool with which we can express any signal in the form of exponential signals,then what exactly is the purpose of a laplace transform? What is the physical significance of it?
can anybody suggest online resources for laplace transform?

Last edited: Oct 8, 2008
2. scubasteve_911 AAC Fanatic!

Dec 27, 2007
1,202
1
Dave said it best
http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx

Steve

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3. BenjaminSweet New Member

Jun 4, 2010
8
7
I am aware that this is an "old" post, but I thought that shiva asked in interesting question: "What is the physical significance of it?" I think that many students and perhaps professors alike might forget that all of those mathematical theories were derived to solve problems. In universities, it seems that "math" is often taught in the earlier years (presumably to build a foundation) in isolation of the application. It's "just math." While the professors, who have worked with these theories for many years, can appreciate the beauty and elegance of the theories (yes, I am serious! The theories ARE elegant and beautiful!), the students do not have this background and there is nothing for the theories to "stick to", so they just slide off...

Then, in the latter years of college the students take their "core" classes that use the theories from a year or two ago (or perhaps only a semester or two ago). By then, the concepts are distant memories but the teachers believe that the students "should already know that, you learned it in such-and-such class..."

If students were taught some "physical" or "practical" applications along with the theories, I think that they would be more likely to remember the concepts in the "core" classes.

So what "physically" does σ represent? (What "physically" does s = σ + jω represent for that matter?) In System "Modeling" (perhaps learned in Control Systems, Mechatronics, Dynamics, AC Circuits) a "physical" system is characterized by a "frequency domain" representation called the Transfer Function. It is the mathematical representation of the systems dynamic behavior (originally stated in differential equation form, derived from appropriate laws of physics) expressed in the Laplace transformed representation.

There are several benefits to the Laplace ("S-Domain") representation, not the least of which is that mathematical analysis and manipulation of the system is done with algebra rather than integro-differential equations. The Transfer Function represetnation also helps to identify the "characteristic" behavior of the system by looking at the roots of the Transfer Function denominator - the "Characteristic Equation". Note that these roots will ALSO determine the partial fractions in the Inverse Laplace Transform. The Characteristic Equation will determine whether the system will oscillate or not, whether the response will grow or decay over time, etc.

When systems have complex "poles" (roots of the Characteristic Equation), the system's response to an Impulse or Step input will oscillate with exponentially growing or decaying amplitude (depending on whether σ - the real part) is to the right or left of the imaginary axis.

For systems with complex poles s = σ ± jω, the imaginary part represent energy storage (in capacitance and inductance, potential and kinetic energy, etc.) and the real part, σ, represents energy dissipation. The energy dissipation can be due to electrical resistance, mechanical friction, etc.

If the real-part, σ, was 0, the poles would just be s = ±jω, which represents a system with constant (un-damped) oscillation.

I hope this is helpful information (if anybody is reading it!)

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4. endolith Member

Jun 21, 2010
27
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I've never understood why replying to old posts is discouraged/prohibited on some forums. It's certainly better than starting a new thread about the same topic and fragmenting the discussion over multiple threads in multiple different places.

Anyway, I'm replying to an old post, too.

That's what this diagram is showing, right? From Comparing Analog and Digital Complex Planes

So if the system has a pole at one of the dots, the impulse response of the system will be the curve shown at that dot? If you have a pole at the origin, that's an integrator, so the impulse response will be DC? If you have a complex pole, the impulse response will be a complex damped sinusoid. If you have a pair of complex poles, the impulse response will be a real damped sinusoid as the complex sinusoids rotating in opposite directions cancel out, etc.

This document helped me a lot in understanding the relationship between poles and zeros in the S plane and frequency response plots: TUTORIAL 733
A Filter Primer

I understand that the jω axis of the S plan is the normal frequency response of a transfer function, and so moving poles closer to this axis "pulls up" the frequency response into a peak, but I still don't really understand what σ is.

For instance, if you have a zero on the jω axis, then that's a null in the frequency response. If you input a signal at that frequency to the transfer function, you get exactly 0 output. If you move the zero away from the axis, to the left, it creates a dip in the frequency response, but it doesn't go all the way to 0. Now, with the zero at a point not on the frequency axis, is there still a signal you can input that will result in 0 output? Like, if you input a precisely-damped sinusoid of the right frequency, will you get zero output?

With normal filters that have complex poles (infinite amplitude at a certain σ+jω), is there a certain damped sinusoid you can input that results in infinite output?

5. JiaZheng New Member

Jan 28, 2013
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If you want to get frequency response, you force σ to be zero. Then system function becomes depending only on jω. If you want to get time domain representation of output signal, multiply input signal and system's Laplace transform and take the inverse transform.

σ in S is to match the root. Root is in the form of "σ+jω". So if you want to let (S - r) = 0 (you want to find the root). S must also in this form of "σ+jω". But as I said before, you can force σ to zero, then you just get frequency response.

Also we know Laplace transform is just a "transform". Once you finished manipulation in S domain, you may need to transform it back. When doing this, you will find these σ and jω are still contained in the inverse transform (time domain). So σ and jω will affect you time domain signal form directly. σ in is not created, it is brought by roots.

My understanding is superficial and doesn't explain the physical meaning. But I use this to understand a little about σ.

6. WBahn Moderator

Mar 31, 2012
24,560
7,700
For a two-bit explanation, the imaginary part captures the steady state response and the real part captures the transient response.

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