How to simplify a symbolic matrix

조회 수: 44 (최근 30일)
SOURAV KUMAR
SOURAV KUMAR 2021년 2월 15일
답변: Swatantra Mahato 2021년 2월 18일
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?

채택된 답변

Swatantra Mahato
Swatantra Mahato 2021년 2월 18일
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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