Plotting error: Need to plot the evolution of symbolic matrix?

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];
N=norm(PHI);
ezplot(N,[0,1])

 Accepted Answer

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];
N = @(t) norm(PHI(t));
t = linspace(0, 1, 50);
for k = 1:numel(t)
Nt(k) = N(t(k));
end
figure
plot(t, Nt)
grid

4 Comments

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)
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
%% Not sure why is it not plotting determinant of the matrix in the same way.
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
That ran without error for me, and appears to do what you want.

Sign in to comment.

More Answers (0)

Categories

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!