How to simplify a symbolic matrix
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Hello everyone,
I was trying the following code:
clc
clear all
close all
syms a k1 k2;
A=[exp(-i*k1*a) 0; 0 exp(i*k1*a)];
B=[k2+k1 k1-k2; k1-k2 k2+k1];
C=[exp(i*k2*a) 0; 0 exp(-i*k2*a)];
D=[exp(i*k2*a) 0; 0 exp(-i*k2*a)];
E=[k2+k1 k2-k1; k2-k1 k2+k1];
F=[exp(-i*k1*a) 0; 0 exp(i*k1*a)];
T=((((A*B)*C)*D)*E)*F;
T=T/(4*k1*k2);
T=simplify(T);
fprintf('T11=\n%s \n',char(T(1,1)));
fprintf('T12=\n%s \n',char(T(1,2)));
fprintf('T21= \n%s \n',char(T(2,1)));
fprintf('T22= \n%s \n',char(T(2,2)));
In this code, i am trying to evaluate T matrix;
I want to simplify the individual components of T matrix {i.e., T(1,1) , T(1,2) , T(2,1) & T(2,2) }
I searched it on internet and found "simplify" will perform the above task.
But , for the above code, the result is yet unsimplified,
i.e., i am getting T(1,1) output as
(exp(a*k1*(-i))*(exp(a*k1*(-i))*exp(a*k2*(2*i))*(k1 + k2)^2 - exp(a*k1*(-i))*exp(a*k2*(-2*i))*(k1 - k2)^2))/(4*k1*k2)
i.e.,
but we can see that from the inner bracket can be further taken out to simplify the result ;
hence how to simplify the results of the above code?
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採用された回答
Swatantra Mahato
2021 年 2 月 18 日
Hi Sourav,
As mentioned in the documentation for "simplify" there is no universal idea to the simplest form of an expression. You may want to try out different Name-Value Pair arguments mentioned in the documentation to get the desired form suitable for your use case
As an example,
executing
T=simplify(T,'Steps',20);
instead gives
T12=
-(exp(-a*k2*2i)*(k1^2 - k2^2)*(exp(a*k2*4i) - 1))/(4*k1*k2)
while
T=simplify(T,'Steps',30);
gives the result
T12=
-(sin(2*a*k2)*(k1^2 - k2^2)*1i)/(2*k1*k2)
Hope this helps
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