how to remove conj from equations? AND SOLVE THEM ?

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Rahul Jangid
Rahul Jangid 2022-8-4
评论： Walter Roberson 2022-8-4
hii i am getting this kind of equations can someone please help me to solve these equations
here "t" is a variable i need values of A1,A2,A3,A4,A5,A6,Q1,Q2,Q3,Q4,Q5,Q6.
conj((A4*sin(Q4 + (2^(1/2)*t)/5))/10 == 0)
conj((A5*sin(Q5 + (2^(1/2)*t)/5))/10 == 0)
conj((A6*sin(Q6 + (2^(1/2)*t)/5))/10 == 0)
conj((10^(1/2)*A1*sin(Q1 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 1)
conj((10^(1/2)*A2*sin(Q2 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 1)
conj((10^(1/2)*A3*sin(Q3 + (2*2^(1/2)*5^(1/2)*t)/25))/100 == 0)
conj((2^(1/2)*A4*cos(Q4 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*A5*cos(Q5 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*A6*cos(Q6 + (2^(1/2)*t)/5))/50 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A1*cos(Q1 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A2*cos(Q2 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
conj((2^(1/2)*5^(1/2)*10^(1/2)*A3*cos(Q3 + (2*2^(1/2)*5^(1/2)*t)/25))/1250 == 0)
THANKS IN ADVANCE

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采纳的回答

Walter Roberson
Walter Roberson 2022-8-4
At some point in your code, you construct a row vector of equations, and then you use the ' operator thinking that it will transpose the row vector into a column vector. However, ' is not the transpose operator: it is the conjugate transpose operator. You need to switch from using ' to using .' which is the transpose operator

更多回答（1 个）

Rahul Jangid
Rahul Jangid 2022-8-4
clc
clear all
close all
%% Intoroduction of variables
syms m k K1 K2 K3 K4 K5 K6 K7 K8 C1 C2 C3 C4 C5 C6 C7 C8 t
% sample data
a=4;
a1=2;
a2=(a-a1);
b=4;
b1=2;
b2=(b-b1);
c=4;
c1=2;
c2=(c-c1);
m=100;
J1=1000;
J2=1000;
J3=1000;
K=1;
C=1;
K1=K;K2=K;K3=K;K4=K;K5=K;K6=K;K7=K;K8=K;
C1=C;C2=C;C3=C;C4=C;C5=C;C6=C;C7=C;C8=C;
Xo = [0 0 0 1 1 0];
Vo = [0 0 0 0 0 0];
I = [Xo Vo].';
Mg=[m 0 0 0 0 0
0 m 0 0 0 0
0 0 m 0 0 0
0 0 0 J1 0 0
0 0 0 0 J2 0
0 0 0 0 0 J3];
%% STIFFNESS MATRIX 1
Kg = [K1+K2+K3+K4+K5+K6+K7+K8 0 0 0 -(K1+K2+K5+K6)*c1+(K3+K4+K7+K8)*c2 (K1+K2+K3+K4)*b1-(K5+K6+K7+K8)*b2
0 K1+K2+K3+K4+K5+K6+K7+K8 0 (K1+K2+K5+K6)*c1-(K3+K4+K7+K8)*c2 0 -(K1+K5+K4+K8)*a1+(K2+K3+K6+K8)*a2
0 0 K1+K2+K3+K4+K5+K6+K7+K8 -(K1+K2+K3+K4)*b1+(K5+K6+K7+K8)*b2 (K1+K5+K4+K8)*a1-(K2+K3+K6+K7)*a2 0
0 (K1+K2+K5+K6)*c1-(K3+K4+K7+K8)*c2 -(K1+K2+K3+K4)*b1+(K5+K6+K7+K8)*b2 (K1+K2+K5+K6)*c1^2+(K3+K4+K7+K8)*c2^2+(K1+K2+K3+K4)*b1^2+(K5+K6+K7+K8)*b2^2 -(K1+K4)*b1*a1+(K2+K3)*b1*a2+(K5+K8)*b2*a1-(K6+K7)*b2*a2 -(K1+K5)*c1*a1+(K2+K6)*c1*a2-(K3+K7)*c2*a2+(K4+K8)*c2*a1
-(K1+K2+K5+K6)*c1+(K3+K4+K7+K8)*c2 0 (K1+K4+K5+K8)*a1-(K2+K3+K6+K7)*a2 -(K1+K4)*b1*a1+(K2+K3)*b1*a2+(K5+K8)*b2*a1-(K6+K7)*b2*a2 (K1+K2+K5+K6)*c1^2+(K3+K4+K7+K8)*c2^2+(K1+K4+K5+K8)*a1^2+(K2+K3+K6+K7)*a2^2 -(K1+K2)*b1*c1+(K4+K3)*b1*c2+(K5+K6)*b2*c1-(K8+K7)*b2*c2
(K1+K2+K3+K4)*b1-(K5+K6+K7+K8)*b2 -(K1+K5+K4+K8)*a1+(K2+K6+K7+K3)*a2 0 -(K1+K5)*c1*a1+(K2+K6)*c1*a2-(K3+K7)*c2*a2+(K4+K8)*c2*a1 -(K1+K2)*c1*b1+(K3+K4)*b1*c2+(K5+K6)*c1*b2-(K7+K8)*c2*b2 (K1+K2+K3+K4)*b1^2+(K5+K6+K7+K8)*b2^2+(K1+K4+K5+K8)*a1^2+(K2+K3+K6+K7)*a2^2];
%% Equation of motion
format short
[U,w2] = eig(Kg,Mg);
%% Modal displacements
syms A1 A2 A3 A4 A5 A6 Q1 Q2 Q3 Q4 Q5 Q6 t real
for i = 1:6 % here [1 for x dir , 2for y dir, 3 for z dir, 4for rotation about x(i.e. alpha)]
x(i) = U(i,1)*A1*sin(sqrt(w2(1,1))*t+Q1) + U(i,2)*A2*sin(sqrt(w2(2,2))*t+Q2) + U(i,3)*A3*sin(sqrt(w2(3,3))*t+Q3) + U(i,4)*A4*sin(sqrt(w2(4,4))*t+Q4) + U(i,5)*A5*sin(sqrt(w2(5,5))*t+Q5) + U(i,6)*A6*sin(sqrt(w2(6,6))*t+Q6);
v= diff(x);
end
x;
v;
F = [x v].'
equ = F-I==0
S = solve(equ,[A1 A2 A3 A4 A5 A6 Q1 Q2 Q3 Q4 Q5 Q6])
can u please help me out .....
  1 个评论
Walter Roberson
Walter Roberson 2022-8-4
What difficulty are you encountering?
When I execute on my system, it thinks for a while, and eventually responds with 625 solutions.
The A* variables are all constant for the solutions, but there are 5 unique solutions for each of Q3, Q4, Q5, Q6, each of which appears in every combination, for a total of 5^4 = 625 solutions.
Some of the solutions can be simplified if you can assume that

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