Control theory: Laplace versus state space representationby Mårten Tags: control theory, laplace transform, state space 

#1
Oct2108, 06:41 PM

P: 127

I'm taking a course in control theory, and have been wondering for a while what the benefits are when you describe a system based on the Laplace method with transfer functions, compared to when you use the state space representation method. In particular, when using the Laplace method you are limited to a system where the initial conditions all of them have to be equal to zero. The following questions then:
1) Is it correct that with the state space representation, the initial conditions could be whatever, i.e., you are not limitied to a system where they equal zero? 2) If so, why all this fuzz about using the Laplace method? Why not always use the state space representation? I'm most eager for an answer on question number 1 above... 



#2
Oct2108, 10:36 PM

P: 2,265

the main difference between the transfer function representation of a linear system and the statespace representation is that the former is concerned only with the inputoutput relationship while the latter is also concerned with what is going on inside the box. the statespace representation is more general and you can have all sorts of different statespace representations that appear to have the same inputoutput relationship. a consequence of this is what happens with polezero cancellation. you might think that you have a nice 2^{nd} order, stable linear system judging from the inputoutput relationship, but internally there might be an internal pole that is unstable but canceled by a zero sitting right on top of it. so before things inside blow up (and become nonlinear), everything looks nice on the outside while things are going to hell on the inside. you find out that things aren't so nice on the inside when some internal state (that is not observable at the outside) hits the rails. 



#3
Oct2208, 06:56 AM

P: 127

For instance, when using the Laplace transform on the following equation, we get: [tex]y'' + 4y' + 2y = u, y(0) = 0, y'(0) = 0,[/tex] [tex]s^2Y(s) + 4sY(s) + 2Y(s) = U(s),[/tex] [tex]Y(s) = \frac{1}{s^2+4s+2}U(s).[/tex] Now this simple result wouldn't be possible, if it wasn't for the initial conditions y(0) = 0 and y'(0) = 0. But when we have a state space representation, like the following [tex]X'(t) = AX(t) + BU(t),[/tex] [tex]Y(t) = CX(t) + DU(t),[/tex] [tex]X(t_0) = E,[/tex] then I thought it was possible to choose the initial conditions X(t_0) to what ever, e.g. [tex]X(t_0) = E = \{x_1(0),x_2(0),x_3(0)\}^T = \{2,7,6\}^T,[/tex] so I am not limited to initial conditions where all x_i(0) = 0 like in the Laplace transform case. Or am I thinking erroneously here? 



#4
Oct2208, 10:05 AM

P: 2,265

Control theory: Laplace versus state space representation[tex]s^2 Y(s) + 4s Y(s) + 2 Y(s) = U(s)[/tex] is not quite complete, in the general case. to go from [tex]y'' + 4y' + 2y = u[/tex] it should be [tex](s^2 Y(s)  s y(0))  y'(0)) + 4(s Y(s)  y(0)) + 2 Y(s) = U(s)[/tex] that's the "Laplace method". 



#5
Oct2208, 03:35 PM

P: 127

But what I haven't understood yet, is whether the state space representation method is limited to this as well or not (that all the initial values, each of them, are set to 0)? Are you instead allowed to use an initial value state vector X(t_0), as for instance, the one I used above in an earlier message? 



#6
Oct2208, 11:26 PM

P: 2,265

the main difference between this and the statespace model is that the statespace model is trying to model what is going on inside the box. it is more general than the simple inputoutput description of the system. 



#7
Oct2308, 12:05 PM

P: 95





#8
Oct2308, 02:35 PM

P: 2,265

[tex](s^2 Y(s)  s y(0))  y'(0)) + 4(s Y(s)  y(0)) + 2 Y(s) = U(s)[/tex] i meant what would happen if you solved for Y(s). it is not only a function of U(s), but also a function of the two initial conditions. [tex]Y(s) = \frac{1}{s^2 + 4s + 2}U(s) \ + \ \frac{s+4}{s^2 + 4s + 2} y(0) \ + \ \frac{1}{s^2 + 4s + 2} y'(0) [/tex] the simple one (assuming a completely relaxed system at t=0) would be [tex]Y(s) = \frac{1}{s^2 + 4s + 2}U(s) [/tex] or [tex]\frac{Y(s)}{U(s)} \equiv G(s) = \frac{1}{s^2 + 4s + 2}[/tex] 



#9
Oct2308, 07:19 PM

P: 127

But, still, I haven't got any answer on the question about the limitations on the initial values in the state space representations. There are no such limitations in the state space representation, are there? You can choose the initial values to whatever you like, can you? 



#10
Oct2308, 11:40 PM

P: 2,265

with either model, you can put in whatever initial values you want. but the inputoutput model only allows you to put in initial values for the output(s) and the various derivatives of the output (up to the order of the system). since the inputoutput model doesn't even think about the internal states, then i guess you can't choose initial values of internal states. 



#11
Oct2408, 09:31 PM

P: 127

Next thing then is: With the state space model, you have initial values for the internal states, that is X(t_0). But what about for the output, Y(t)? There's no need for initial values there? Sorry, if this is obvious... 



#12
Oct2408, 10:22 PM

P: 2,265

those initial conditions (for y(t)) are fully determined by the initial conditions for the states in X(t). they cannot be independently specified. and with the statespace model, with all of those states (an equal number of states to the order of the system), you need not and will not have initial conditions for higher derivatives (like you did for y'(0)). just having initial conditions for every element in the X(0) vector is good enough. 



#13
Oct2608, 09:54 PM

P: 127

Before we go on  thank you very much for this personal class I get here in control theory! It helps me a lot!
Then comes the ultimate question: Why don't always use the statespace model in preference to the Laplace transform model? What benefits does the Laplace transform model have, which the statespace model doesn't have? 



#14
Oct2608, 10:54 PM

P: 2,265





#15
Nov1008, 02:01 PM

P: 127

Sorry for late reply, been away for a while...
1) With the state space representation the system is described by N^{2} numbers in the Amatrix. But most of the numbers in the Amatrix are just zeros and ones. So it seems that the number of significant numbers in the Amatrix are just 2N+1. So then the state space representation is not so much more complicated? 2) If a system is described by this ODE above, how could there be any internal information that is being revealed if putting up this system on a state space representation, compared to when using the transfer function to describe it? 



#16
Nov1008, 11:46 PM

P: 2,265

this "controllable" and "observable" stuff is in sorta advanced control theory (maybe grad level, even though i first heard about it in undergrad when i learned about the state space representation). i dunno if i can give it justice in this forum. i would certainly have to dig out my textbooks and relearn some of this again so i don't lead you astray with bu11sh1t. so far, i'm pretty sure i'm accurate about what i said, but i don't remember everything else about the topic. 



#17
Nov1108, 04:59 PM

P: 127

Okey, I think I understand now. Correct me then if something of the following is wrong.
The expression of G(s) as I was talking about above, could actually be the result of a lot of boxes so to say. That is, G(s) is a black box, and inside that black box there could be several other boxes with signals going in and out, and even feedback loops. Actually, the number of configurations inside the black box, are infinite. That corresponds to the infinite number of state space representations for G(s). But, let's now say that we know exactly what happens inside the black box. E.g., the black box is the ODE describing a bathtub with water coming in, and water going out. Let's say for simplicity, even though this is probably not physically correct, that this bathtub is described by [itex]G(s) = \frac{1}{s^2+4s+2}[/itex]. Then, this bathtub system could just have one state space representation, since we know the system perfectly inside. Is that correct? (I here disregard the fact that you could manipulate the rows in the Amatrix, and make different matrixoperations, but it's still really the same equivalent matrix.) So in this case, I could as well solve the matrix equation, as solving the Laplace equation, it is as simple. Except that if I do it the matrix way (i.e. the state space representation way), I could have whatever initial values, it won't do it more complicated, as it would do in the Laplace way. Is that correct? 



#18
Nov1408, 01:03 PM

P: 127

Okey, since I got no reply to my previous post above, I guess that means everything there was correct...?
Then to the following question: One of the benefits with Laplace transforms, is that you can just mulitply boxes in a row to get the transfer function for all the boxes together. In general, it's pretty simple to handle different configurations with boxes in series and in parallel, when doing transfer functions. Is it possible as well to put up a state space representation when you have several boxes in series and so on, and how is that done then? Maybe that's a drawback to state space representations, that it's not done so easily as with transfer functions? 


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