## 261 Reputation

16 years, 113 days

## Illustration...

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?

## Illustration...

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?

## Strange...

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.

## Strange...

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....

The problem is clear now. I've found the right solutions thanks to method from JacquesC.

## The problem is clear now....

The problem is clear now. I've found the right solutions thanks to method from JacquesC.

## I see you were able to find...

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...

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...

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...

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.

## Sorry...

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.

## Sorry...

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...

It was indeed the name that caused the trouble. My problem is solved now. Thank you,

## It was indeed the name that...

It was indeed the name that caused the trouble. My problem is solved now. Thank you,

## I've been able to make...

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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