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.

[help] A matrix problem??

Status
Not open for further replies.

v9260019

Member level 2
Member level 2
Joined
Jun 5, 2005
Messages
47
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,680
hello----all
i use three method to find Exp[At],where a is a matrix
A={{1,0,-1},{0,1,0},{0,0,1}}

Method 1: use computer (mathematica) find Exp[At]
-> Exp[At]={{Exp[t],1,Exp[-t]},{1,Exp[t],1},{1,1,Exp[t]}}

Method 2:use Cayley-Hamilton theorem
The answer is Exp[At]={{Exp[t],0,-Exp[t]t+Exp[t]t^2/2},{0,Exp[t],0},{0,0,Exp[t]}}

Method 3: use laplace transform
we all know the InverseLaplaceTransform[Inverse[sI-A]]=Exp[At]]
The answer is Exp[At]={{Exp[t],0,-Exp[t]t},{0,Exp[t],0},{0,0,Exp[t]}}

The three answers of Exp[At] are all different ,which one is right and what the relationship between the three methods ??????

Thanks a lot
 

The result by "method 3" is right and all others are incorrect.
This can be done manually. Note that
A^n={{1,0,-n},{0,1,0},{0,0,1}}
Therefore,
Exp[At]=I+At/1!+(At)^2/2! +...
The only thing special is the element at the upper right corner which is

-t/1!-2t^2/2! - 3t^3/3! - ...
=-t(1+t/1! +t^2/2! +...)=-tExp[t].
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top