Find a parameter that satisfy the equation

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Francesco Paolo De Simone
评论： Francesco Paolo De Simone 2020-11-15
I have in a .txt file 101 samples of acceleration taken for t=[0:0.1:10].
I also have that the expression of the mathematical through which i can estimate the acceleration :
acc_samp = load('samples.txt') ;
syms y(t) b
mat_model = diff(y(t), t, t) == (20*cos(5*t)*exp(-10*t))/3 - (b*diff(y(t), t))/3 - (100*y(t))/3 ;
interval = (0:0.1:10) ; % vector containing the time instants in which the solution will be evaluated
Dy = diff(y,t) ;
dis = dsolve(mat_model, [y(0)==0, Dy(0)==0]) ; % symbolic expression of the displacement y(t,b)
acc = diff(dis,t,2) ; % symbolic expression of the acceleration yddot(t,b)
How it is possible to find a b such that acc_samp and acc are equal for t=[0:0.1:10]?
Thank you

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回答（1 个）

Raunak Gupta
Raunak Gupta 2020-11-13
Hi,
It is my assumption since the variable b is not declared numerically the acc variable will be in terms of b only. If that is the case, then you may want to equate the two vectors to find the out the value of b. But note that there are 101 samples in acc_samp so it will create 101 equations when equating the variables acc and acc_samp. This can give multiple solutions for b. So, a better approach will be to minimize the difference between two vectors given a single value of b. This answer can help set up the optimization problem.

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