Extracting coefficient Matrix and vector from available polynomial Expressions with exponential terms

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Hello All,
I have stuck in one problem, where I have variable matrix vector available and I want to extract the coefficient matrix and variable vector from that original polynomial vector matrix. For e.g. if A is matrix as:
and I want to extract its coefficients and variable vector as given below:
So my question is how to compute or Extract Matrix Y and Matrix Z from A which comprises of expressions with exponential terms of eigenvalues of system matrix.
Is there any direct command? I tried commands like - coeffs, sysvar, equationsToMatrix but in all those cases the variables are fixed.
So, please help me in solving this above problem.
Thank You.

采纳的回答

Star Strider
Star Strider 2019-5-12
You have not shown what you already tried or what the results were.
In R2019a, this is possible with a for loop (since coeffs apparently doesn’t like matrices):
syms lambda1 lambda2 t
A = [exp(-lambda1*t)+3*exp(-lambda2*t); -2*exp(-lambda1*t)+4*exp(-lambda2*t)];
for k = 1:size(A,1)
[Ac{k},trm{k}] = coeffs(A(k,:));
end
Cfs = [Ac{1}; Ac{2}]
Trm = [trm{1}; trm{2}]
producing:
Cfs =
[ 1, 3]
[ -2, 4]
Trm =
[ exp(-lambda1*t), exp(-lambda2*t)]
[ exp(-lambda1*t), exp(-lambda2*t)]
That may be the best you can hope for.
  13 个评论
Rajani Metri
Rajani Metri 2021-3-25
😃😉 Ok, Thank You STAR
I am entering that Cfs matrix manually now, may be that is what I have to continue.
Thank you again 😊.

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更多回答(1 个)

Rajani Metri
Rajani Metri 2019-5-15
If the coefficients of lyamda in A matrix are either 0 (ZERO) or if the eigenvalues are repeated then there will be t term along with exponential term (For e.g. e^(-2t) and t.e^(-2t). So in these two cases Z Matrix is not correctly computed.
Apart from these two cases the above program given by Star Strider work.
Thank You.

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