Timeline for Find output given state space representation
Current License: CC BY-SA 3.0
22 events
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| Aug 31, 2017 at 11:28 | comment | added | John Katsantas | That is true in many cases unfortunately. There are many times I think they are trying to cover too much stuff in too little time.Anyway,I'll try to do as you said and be able to face the problem this way. Thanks for your time and help pasaba. | |
| Aug 31, 2017 at 11:26 | comment | added | pasaba por aqui | Off-topic: Why do schools teach so many things? They do not leave time to learn anything. | |
| Aug 31, 2017 at 11:20 | comment | added | pasaba por aqui | (numbering rows and columns 1,2,..): \$ A_{2,1} = 0 \$ thus \$ x_2(t) \$ is unrelated of \$ x_1(t) \$; \$ B_1=0 \$ thus \$ x_2(t) \$ is unrelated to \$ u_1(t) \$, \$ x_2'(t)=2u_2(t) \$; finally, as C=(0,1), output y(t) is only related to \$ x_2(t) \$, \$ y(t)=x_2(t)\$. As result, the system is simply \$ y'(t)=2u(t) \$, an integrator. | |
| Aug 31, 2017 at 11:09 | comment | added | pasaba por aqui | No, do not start by reading books nor taken notes (or going to magisterial class): relax, take time, and thing about what you have in front. Try to visualize and give sense to each part of the problem. Books are for reference, or to stablish an ordenation in the matter. | |
| Aug 31, 2017 at 11:05 | comment | added | pasaba por aqui | Be careful that in this latest example you are assuming u(t) is the Heaviside function, when you say U(s)=1/s. | |
| Aug 31, 2017 at 11:04 | comment | added | John Katsantas | Man, I want to understand everything the way you do. Any textbooks you can suggest? | |
| Aug 31, 2017 at 10:58 | comment | added | John Katsantas | You are right but it doesn't affect the result I think. The x(0) still vanishes. | |
| Aug 31, 2017 at 10:56 | comment | added | pasaba por aqui | I think there are an error in sign of \$ T_{1,2} \$ | |
| Aug 31, 2017 at 10:46 | comment | added | John Katsantas | Can you have a look ? i.sstatic.net/GCw3l.jpg | |
| Aug 31, 2017 at 10:31 | comment | added | pasaba por aqui | In the case of the original question, the only C that mades zero y0(t) is (0,0) | |
| Aug 31, 2017 at 10:20 | comment | added | John Katsantas | I solved a similar problem and the final formula y(t)=C(x(t) +x0(t)) gives Cx0(t)=0 and I'm only left with Cx(t) | |
| Aug 31, 2017 at 8:45 | comment | added | pasaba por aqui | Fixed some matrix permutations | |
| Aug 31, 2017 at 8:44 | history | edited | pasaba por aqui | CC BY-SA 3.0 |
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| Aug 31, 2017 at 8:35 | history | edited | pasaba por aqui | CC BY-SA 3.0 |
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| Aug 31, 2017 at 8:29 | comment | added | pasaba por aqui | No, initial conditions can not be "cancelled" (if you are moving at 100 km/h, you can accelerate or decelerate, but not ignore you are moving). If C is not a constant but c(t) it means that the system is not time invariant. In this case, usually is easier calculation in time space y(t)=c(t)x(t) than calculation in Laplace space L-1{convolution(L{c(t)},L{x(t)})} | |
| Aug 30, 2017 at 19:53 | vote | accept | John Katsantas | ||
| Aug 30, 2017 at 19:53 | comment | added | John Katsantas | Is there a chance the initial conditions are not in y(t)? I mean C could cancel them out right? | |
| Aug 30, 2017 at 19:45 | history | edited | pasaba por aqui | CC BY-SA 3.0 |
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| Aug 30, 2017 at 19:42 | comment | added | pasaba por aqui | Taken into account that C is a constant, y(t)=Cx(t) and Y(s)=CX(s), you can do the multiplication in time or Laplace space. | |
| Aug 30, 2017 at 19:39 | comment | added | John Katsantas | Very helpful. I assume I can also find X(s), replace in Y(s)=CX(s) and then taking the inverse Laplace transform find y(t). Is that correct? | |
| Aug 30, 2017 at 14:42 | history | edited | pasaba por aqui | CC BY-SA 3.0 |
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| Aug 30, 2017 at 14:36 | history | answered | pasaba por aqui | CC BY-SA 3.0 |
