Need to plot the determinant of the matrix for t 0 to 1?

Question

Shauvik Das 2019-12-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/495437-need-to-plot-the-determinant-of-the-matrix-for-t-0-to-1

评论： Star Strider 2019-12-7

采纳的回答： Star Strider

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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Answer 1

Star Strider 2019-12-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/495437-need-to-plot-the-determinant-of-the-matrix-for-t-0-to-1#answer_405324

在 MATLAB Online 中打开

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:

1Need to plot the determinant of the matrix for t 0 to 1 - 2019 12 07.png

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Star Strider 2019-12-7

在 MATLAB Online 中打开

First —

The easiest way is just to negate ‘t’ whenever you call ‘PHI’:

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)')

Replaces it elsewhere as necessary.

Second —

syms s t 
SI(s) = [s,0,0,0;0,s,0,0;0,0,s,0;0,0,0,s];
A = [0,1,1,0; 0,0,-2,1;0,0,0,1;0,0,3,-2];
SIMA(s) = SI(s)-A;
SIMAINV(s) = inv(SIMA(s));
PHI = ilaplace(SIMAINV)

proiduces:

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]

You can easily transform that to a numeric matrix as:

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];

You will have to manually remove the spaces between the binary operations, specifically the ‘+’ and ‘-’, because MATLAB otherwise considers spaces as delimiters, and the matrix will throw a vertcat error when you use it without first eliminating them.

Shauvik Das 2019-12-7