Need to plot the determinant of the matrix for t 0 to 1?
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Shauvik Das
le 7 Déc 2019
Commenté : Star Strider
le 7 Déc 2019
syms t
PHI=[ 1, -t, -t/3 - (2*exp(-3*(-t)))/9 + 2/9, (2*-t)/3 + (2*exp(-3*(-t)))/9 - 2/9;
0, 1, (5*exp(-3*(-t)))/12 - (3*exp(-t))/4 + 1/3, 2/3 - exp(-t)/4 - (5*exp(-3*(-t)))/12;
0, 0, exp(-3*(-t))/4 + (3*exp(-t))/4, exp(-t)/4 - exp(-3*(-t))/4;
0, 0, (3*exp(-t))/4 - (3*exp(-3*(-t)))/4, (3*exp(-3*(-t)))/4 + exp(-t)/4];
PHIT=transpose (PHI);
B=[0;1;2;1];
BT=transpose (B);
GRAMi = PHI*B*BT*PHIT
GRAMfinal=int(GRAMi,t, 0, t)
A= det(GRAMfinal) %% Absolute determinant of the matrix
N = @(t) (A(t));
t = linspace(0, 1, 50);
for k = 1:numel(t)
Nt(k) = N(t(k));
end
figure
plot(t, Nt)
grid
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Réponse acceptée
Star Strider
le 7 Déc 2019
Try this:
PHI = @(t) [ 1, t, t/3-(2*exp(-3*t))/9+2/9, (2*t)/3+(2*exp(-3*t))/9-2/9;
0, 1, (5*exp(-3*t))/12-(3*exp(t))/4+1/3, 2/3-exp(t)/4-(5*exp(-3*t))/12;
0, 0, exp(-3*t)/4+(3*exp(t))/4, exp(t)/4-exp(-3*t)/4;
0, 0, (3*exp(t))/4-(3*exp(-3*t))/4, (3*exp(-3*t))/4+exp(t)/4];
PHIT = @(t) transpose(PHI(t));
B=[0;1;2;1];
BT=transpose(B);
GRAMi = @(t) PHI(t)*B*BT*PHIT(t);
GRAMfinal = @(t) integral(GRAMi, 0, t, 'ArrayValued',1)
A = @(t) det(GRAMfinal(t));
N = @(t) (A(t));
t = linspace(0, 1, 50);
for k = 1:numel(t)
Nt(k) = N(t(k));
end
figure
plot(t, Nt)
grid
xlabel('t')
ylabel('N(t)')
producing:
5 commentaires
Star Strider
le 7 Déc 2019
My pleasure.
If my Answer helped you solve your problem, please Accept it!
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