Nitroxxx

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12 years, 337 days

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I will try to illustrate the problem some more. With laplace and in the time domain I get as solution: y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is illustrated enough above. When I construct the transferfunction for the differential equation I get: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by: Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) (Zero state response) This solution is different from the solution found with previous methods. I think that the fact that x(0)=10 is lost somewhere in the calculation but I can't find a way to add it. Normally when there are initial conditions voor y(t) and D(y)(t) that are different form zero I would add a zero input response to the zero state response calculated with the transferfunction. But in this case this is no solution because in the zero input response I can only plug conditions for y(t) because the input is zero. What is the problem here?
I will try to illustrate the problem some more. With laplace and in the time domain I get as solution: y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is illustrated enough above. When I construct the transferfunction for the differential equation I get: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by: Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) (Zero state response) This solution is different from the solution found with previous methods. I think that the fact that x(0)=10 is lost somewhere in the calculation but I can't find a way to add it. Normally when there are initial conditions voor y(t) and D(y)(t) that are different form zero I would add a zero input response to the zero state response calculated with the transferfunction. But in this case this is no solution because in the zero input response I can only plug conditions for y(t) because the input is zero. What is the problem here?
What you say is true but when you construct the transfertfunction for example, all initial condition are null. Then you get something like: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) When I now solve this in the time domain manual or with Maple I have as solution: restart: DV:=diff(y(t),t$2)+3*diff(y(t),t)+2*y(t)=diff(10*exp(-3*t),t); > Cond:=y(0)=0,D(y)(0)=0: > Opl:=dsolve({DV,Cond},y(t)): > Opl:=simplify(Opl); Opl := y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is totally different from what I get with Laplace. I do not find what I'm doing wrong. I followed these steps quite some times for other differential equations. Equations where x(t) was not derived. But for this equation I can't get two times the same solution.
What you say is true but when you construct the transfertfunction for example, all initial condition are null. Then you get something like: H(s) = s / (s^2+3s+2) The transform of x(t) is given by: X(s) = 10/(s+3) De output is given by Y(s) = X(s) H(s) y(t) = -15exp(-3t)+20exp(-2t)-5exp(-t) When I now solve this in the time domain manual or with Maple I have as solution: restart: DV:=diff(y(t),t$2)+3*diff(y(t),t)+2*y(t)=diff(10*exp(-3*t),t); > Cond:=y(0)=0,D(y)(0)=0: > Opl:=dsolve({DV,Cond},y(t)): > Opl:=simplify(Opl); Opl := y(t) = 15 (-exp(-2 t) + 2 exp(-t) - 1) exp(-t) This is totally different from what I get with Laplace. I do not find what I'm doing wrong. I followed these steps quite some times for other differential equations. Equations where x(t) was not derived. But for this equation I can't get two times the same solution.
The problem is clear now. I've found the right solutions thanks to method from JacquesC.
The problem is clear now. I've found the right solutions thanks to method from JacquesC.
I see you were able to find a general solution. For my problem it is sufficient enough to have an exact solution. Have some difficulties getting an exact solution out of this. By the way I've typed out a manually example from my handbook to show how It should about solved. It's kinda hard to translate it to maple. Download 2714_Example.mws
View file details
I see you were able to find a general solution. For my problem it is sufficient enough to have an exact solution. Have some difficulties getting an exact solution out of this. By the way I've typed out a manually example from my handbook to show how It should about solved. It's kinda hard to translate it to maple. Download 2714_Example.mws
View file details
It's indeed also working for opl now. There must have been some confusion here when using restart and preassigning a value to total which made it look like opl was the trouble maker.
It's indeed also working for opl now. There must have been some confusion here when using restart and preassigning a value to total which made it look like opl was the trouble maker.
My excuses. Previous time I've posted only the worksheet and few reacted to the topic because most would not open it. Download 2714_Curl.mws
View file details Edit: I just rememberd the command dsolve. Hmzz looks like it doesn't work for a vectorfunction.
My excuses. Previous time I've posted only the worksheet and few reacted to the topic because most would not open it. Download 2714_Curl.mws
View file details Edit: I just rememberd the command dsolve. Hmzz looks like it doesn't work for a vectorfunction.
It was indeed the name that caused the trouble. My problem is solved now. Thank you,
It was indeed the name that caused the trouble. My problem is solved now. Thank you,
I've been able to make following sum: "for i from 1 to 4 do total:=total+i end do;" The result is logic: 1,3,6,10. Now I want to make the sum of some terms I've already calculated and named by opl[i]. With I a posint. Now I write the command: "for i from 1 to 4 do total:=total+opl[i] end do;" This gives me the error of to many levels of recursion. I am not able to see the problem in what I've programmed. Thank you in advance,
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