Laplace plane meaning in control engineering ?

[Femme_physics]

We use this plane called the "Laplace plane" to solve problem in control-engineering, second order systems.

Can anyone help explain the Laplace plane in simple words? I find the wiki article too fancy for me... it says that it "is a mean or reference plane about whose axis the instantaneous orbital plane of a satellite precesses"..

But I'm doing control engineering. I'm confused how these two are related if I solve problems that looks like an electronics circuit and I have to use the laplace plane


For instance,

A capacitor in electronics is "C", in the Laplace plane, it's 1/CS

An inductor in electronics is L, in the Laplace plane, it's LS

Resistance, Battery and Voltage seem to be the same in both planes...

I'm just trying to understand what it means-- this plane. I do know what second order system means.

 

[AlephZero]

The "Laplace plane" described in the Wiki article is something different. The usual name for what you want is the "s-plane" (there is also a similar thing called the z-plane which is used in digital signal processing).

The "s" in a Laplace transform is a complex number, which represents a frequency and an amount of damping. The imaginary part is the frequency, and the real part is the amount of damping (negative represents a response that decays to nothing, positive represents a response that grows exponentially).

It turns out that the behaviour of many linear systems is represented in the s plane by ratio of two polynomials like
$$\frac{a_0 + a_1s + a_2s^2 + \cdots}{b_0 + b_1s + b_2s^2 + \cdots}$$.
This can be converted into partial fractions like
$$\frac{A_1}{s-B_1} + \frac{A_2}{s-B_2} + \cdots$$
A plot showing the position of the B's on the s-plane contains most of the important information about the response of the system, and (with practice!) you can interpret what it tells you about the steady state response, impulse response, and step response of the system, without having to grind through the math.

 

[DragonPetter]

I think it helps to understand that, as mentioned above, the s-plane is a graphical plot of a transfer function's pole and zero values, which is a function of the variable s = sigma + jw. The transfer function is the function you get when you take the laplace transform of a differential equation. It is mapping the time domain information (variable t) into frequency domain information (variable s), and the s-plane is a graphical representation of this mapping. I think Laplace transforms were originally invented to solve differential equations, but now they also provide a lot of insight into the behavior of systems described by differential equations, and we can quickly look at this information graphically by plotting the poles and zeroes of the function.

The laplace plane is simply plotting the real and imaginary poles/zeroes of a transfer function. The horizontal axis is the real part, which is called sigma, and the vertical axis is the imaginary part, represented by jw. We use complex numbers as a consequence of euler's identity that let's us describe single frequency sine waves in a more convenient way, rather than as a trig function. For the signal to be real, any complex number pole or zero has to have a complex conjugate, and so you will always see zeros or poles as pairs if they are not plotted on the horizontal real axis. The distance that the points are from the origin as well as their distance from each other and the angles between them is related to the magnitude and phase information you see in bode plots, which are another graphical represntation of the transfer function.

I wish I could give you more insightful and intuitive information, but I mostly understand the math mechanically, and I'm still trying to get to the bottom of it myself and that would take a concentrated effort of studying. Studying Fourier series and transforms, and knowing how they are different from laplace transforms is also very beneficial for intuitively understanding what it all means.

 

[I like Serena]

It's the plane of the Laplace Transform, or s-plane, which you can find here in wiki.

(In circuit analysis it is the Laplace Transform of the voltage across the component divided by the one of the current through the component.)In circuit analysis such a transform is called the impedance, with which you can calculate as you would with Ohm's law.
Usually the symbol Z is used to represent the impedance.You already know that Ohm's law says: $V = I \cdot R$.

The same applies for impedances.
Meaning:

$V = I \cdot R$ for resistors

$V = I \cdot \frac{1}{C \cdot s}$ for capacitors

$V = I \cdot (L \cdot s)$ for inductorsWe say for instance that the impedance Z of an inductor is:

$Z = L \cdot s$This also means that you can apply KVL and KCL with inductors and capacitors.
You would treat an inductor as a resistor with resistance $L \cdot s$.

 

[DragonPetter]

It's the plane of the Laplace Transform, or s-plane, which you can find here in wiki.

(In circuit analysis it is the Laplace Transform of the voltage across the component divided by the current through the component.)


In circuit analysis such a transform is called the impedance, with which you can calculate as you would with Ohm's law.
Usually the symbol Z is used to represent the impedance.


You already know that Ohm's law says: $V = I \cdot R$.

The same applies for impedances.
Meaning:

$V = I \cdot R$ for resistors

$V = I \cdot \frac{1}{C \cdot s}$ for capacitors

$V = I \cdot (L \cdot s)$ for inductors


We say for instance that the impedance Z of an inductor is:

$Z = L \cdot s$


This also means that you can apply KVL and KCL with inductors and capacitors.
You would treat an inductor as a resistor with resistance $L \cdot s$.

Good point. The complex impedance of energy elements being represented in the s domain is most obvious when you consider that s can be considered an operator that means to differentiate when you mulitply with it, or integrate when you divide by it.

If you consider the time domain relations for C and L:

capacitor:
$i(t) = C\frac{dV(t)}{dt}$

inductor:
$V(t) = L\frac{di(t)}{dt}$

and then, if you exchange the derivative with the s as a differentiation operator you get:
capacitor:
$i(s) = CV(s)s$, so arrange for the definition of impedance $\frac{V}{i}$, you get $\frac{V(s)}{i(s)} = \frac{1}{Cs}$

inductor:
$V(s) = Li(s)s$, so arrange for the definition of impedance $\frac{V}{i}$, you get $\frac{V(s)}{i(s)} = Ls$

 

[Femme_physics]

Thanks, I even made sure to ask my teacher about it and that confirms it :) Reading all the replies I get a bigger picture. He told me what DragonPetter said in his last reply. I appreciate the replies.

 

[Ouabache]

I understand you are taking a practical engineering program which may have omitted this math. Typically in EE (elec engr), & CE (comp engr) programs, they require 'signals & systems theory' as a prerequisite to control systems. In signals, they teach everything you ever wanted to know about s-planes, z-planes, $j \omega$-planes and then some. So when it is presented again in control systems, it is already clear. As you have seen, using the LaPlace operator makes the math; working with complex impedances (Z) of capacitance & inductance, a whole lot easier. (The alternative is solving KVL equations with integrals and differential operators in them). :yuck: