Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Laplace transform with boundary conditions

Status
Not open for further replies.
R

RelativelyGood

Newbie level 2
Joined
Mar 30, 2012
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,294
Here's the question:

Use laplace transforms to find X(t), Y(t) and Z(t) given that:

X'+Y'=Y+Z
Y'+Z'=X+Z
X'+Z'=X+Y

subject to the boundary conditions X(0)=2, Y(0)=-3,Z(0)=1.

Now I have learnt the basics of laplace transforms, but have not seen a question in this form before. How do I start the question, could someone for instance show me how to get X(t) and I'll try the rest knowing how to do it? I have other questions I need to do like this, but this looks like the easiest one.

Thanks
 

First some cleaning

(1) X'+Y'=Y+Z
(2) Y'+Z'=X+Z
(3) X'+Z'=X+Y

combining (1)-(2)+(3) , (1)+(2)-(3) y (2)+(3)-(1) is:

X' = Y
Y' = Z
Z' = X

Applying Laplace transform

s X(s) - x0 = Y(s)
s Y(s) - y0 = Z(s)
s Z(s) - z0 = X(s)

You could have applied LT from the beginning, only the system would have more terms, but obviously, the same final solution.

Now, you must solve this 3x3 system with unknowns X(s), Y(s), Z(s).
I don't know what methods for solving system you can use, but I will use matrix+software :smile:

[X(s);Y(s);Z(s)] = [s, -1, 0; 0, s, -1; -1, 0, s]^(-1)·[2; -3; 1]

That is:
X(s) = (2·s - 1)/(s^2 + s + 1)
Y(s) = - (3·s + 2)/(s^2 + s + 1)
Z(s) = (s + 3)/(s^2 + s + 1)

Now, you must calculate the inverse Laplace transform.
Again, I don't know what methods you can use, but I will use WolframAlpha :smile:

X(t) = InverseLaplaceTransform[(2·s - 1)/(s^2 + s + 1)]
Y(t) = InverseLaplaceTransform[- (3·s + 2)/(s^2 + s + 1)]
Z(t) = InverseLaplaceTransform[(s + 3)/(s^2 + s + 1)]

See the results on the links
 
_Eduardo_ said:
First some cleaning

(1) X'+Y'=Y+Z
(2) Y'+Z'=X+Z
(3) X'+Z'=X+Y

combining (1)-(2)+(3) , (1)+(2)-(3) y (2)+(3)-(1) is:

X' = Y
Y' = Z
Z' = X

Applying Laplace transform

s X(s) - x0 = Y(s)
s Y(s) - y0 = Z(s)
s Z(s) - z0 = X(s)

You could have applied LT from the beginning, only the system would have more terms, but obviously, the same final solution.

Now, you must solve this 3x3 system with unknowns X(s), Y(s), Z(s).
I don't know what methods for solving system you can use, but I will use matrix+software :smile:

[X(s);Y(s);Z(s)] = [s, -1, 0; 0, s, -1; -1, 0, s]^(-1)·[2; -3; 1]

That is:
X(s) = (2·s - 1)/(s^2 + s + 1)
Y(s) = - (3·s + 2)/(s^2 + s + 1)
Z(s) = (s + 3)/(s^2 + s + 1)

Now, you must calculate the inverse Laplace transform.
Again, I don't know what methods you can use, but I will use WolframAlpha :smile:

X(t) = InverseLaplaceTransform[(2·s - 1)/(s^2 + s + 1)]
Y(t) = InverseLaplaceTransform[- (3·s + 2)/(s^2 + s + 1)]
Z(t) = InverseLaplaceTransform[(s + 3)/(s^2 + s + 1)]

See the results on the links
Click to expand...


Thank you. I posted this in a hope to get an answer as thorough and clear, so that I can now complete harder questions, knowing the method. I have managed to do one by myself, by hand, so thanks!
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top