# why range of x axis of time is not showing

2 visualizaciones (últimos 30 días)
shiv gaur el 9 de Mzo. de 2022
Comentada: Walter Roberson el 15 de Mzo. de 2022
Tstart = 0.0;
Tend = 100.0;
Nt = 2000;
dT = (Tend-Tstart)/Nt;
X0 = 1;
Y0 = 2;
Z0 = 3;
N = 20;
k1=1
k2=3
k3=1
k4=5
I=0.5
s=4;
xr=-1.6;
r=0.0001;
% Initialize coefficient arrays
T = zeros(Nt+1,1);
X = zeros(Nt+1,1);
Y = zeros(Nt+1,1);
Z = zeros(Nt+1,1);
a = zeros(N+1,1);
b = zeros(N+1,1);
c = zeros(N+1,1);
T(1) = 0.0;
X(1) = X0;
Y(1) = Y0;
Z(1) = Z0;
for j = 2:Nt+1
a(1) = X(j-1);
b(1) = Y(j-1);
c(1) = Z(j-1);
for k = 1:N
a(k+1)=(b(k)-k1*a(k).^3+k2*a(k).^2-c(k)+I)/(k+1);
b(k+1)=(k3-k4*a(k).^2-b(k))/(k+1);
c(k+1)=r*(s*(a(k)-xr)-c(k))/(k+1);
end
x = a(1);
y = b(1);
z = c(1);
for k = 2:N+1
x = x + a(k)*dT^(k-1);
y = y + b(k)*dT^(k-1);
z = z + c(k)*dT^(k-1);
end
T(j) = T(j-1) + dT;
X(j) = x;
Y(j) = y;
Z(j) = z;
end
figure(1)
plot(T,X,'Color','red')
hold on
figure(2)
plot(T,Y,'Color','red')
hold on
figure(3)
plot(T,Z,'Color','red')
hold on
figure(4)
plot3(X,Y,Z,'Color','red')
hold on
the time range is coming 1.5 in
place of 100
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Rik el 9 de Mzo. de 2022
Serious question: why do you refuse to use formatting?

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### Respuestas (2)

Walter Roberson el 9 de Mzo. de 2022
Because after that your values go to infinity or nan. Automatic selection of plot axes range is based only on the finite coordinates of what is drawn.
##### 2 comentariosMostrar NingunoOcultar Ninguno
shiv gaur el 9 de Mzo. de 2022
Editada: Walter Roberson el 9 de Mzo. de 2022
why other type is showing time value on x axis
Tstart = 0.0;
Tend = 1000.0;
Nt = 20000;
dT = (Tend-Tstart)/Nt;
X0 = 0.0;
Y0 = 1.0;
Z0 = 0.0;
N = 20;
SIGMA = 10;
R = 28;
B = 8/3;
%
% Initialize coefficient arrays
T = zeros(Nt+1,1);
X = zeros(Nt+1,1);
Y = zeros(Nt+1,1);
Z = zeros(Nt+1,1);
a = zeros(N+1,1);
b = zeros(N+1,1);
c = zeros(N+1,1);
T(1) = 0.0;
X(1) = X0;
Y(1) = Y0;
Z(1) = Z0;
for j = 2:Nt+1
% Compute Taylor coefficients
a(1) = X(j-1);
b(1) = Y(j-1);
c(1) = Z(j-1);
for k = 1:N
SB = 0.0;
SC = 0.0;
for i= 0:k-1
SB = SB + a(i+1)*c(k-i);
SC = SC + a(i+1)*b(k-i);
end
a(k+1) = (SIGMA*(b(k) - a(k)))/k;
b(k+1) = ((R*a(k) - b(k)) - SB)/k ;
c(k+1) = (-B*c(k) + SC)/k ;
end
x = a(1);
y = b(1);
z = c(1);
for k = 2:N+1
x = x + a(k)*dT^(k-1);
y = y + b(k)*dT^(k-1);
z = z + c(k)*dT^(k-1);
end
% Prepare for T = T + dT
T(j) = T(j-1) + dT;
X(j) = x;
Y(j) = y;
Z(j) = z;
end
figure(1)
plot(T,X,'Color','red')
hold on
figure(2)
plot(T,Y,'Color','red')
hold on
figure(3)
plot(T,Z,'Color','red')
hold on
shiv gaur el 9 de Mzo. de 2022
pl resolve this problem for time rannge

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Davide Masiello el 9 de Mzo. de 2022
That's happening because all the values of X,Y, and Z after T = 1.5 are NaN, which will not be plotted by MatLab.
I guess there might be a mistake in your equations that causes them to predict your variables to be NaN.
##### 12 comentariosMostrar 10 comentarios más antiguosOcultar 10 comentarios más antiguos
Walter Roberson el 15 de Mzo. de 2022
I had my program continue calculatign with symbolic Y0.
Before the end of the 7th j loop, some of the expressions involve Y0 to the power of over 320000, and coefficients over 10 to the million.
The arithmetic round-off involved in the calculations is horrible.
If your Y0 is anything with absolute value greater than 1 then for certain your values would explode beyond the range of double precision; if your Y0 is anything with absolute value less than 1, then most of the terms are going to underflow.
Walter Roberson el 15 de Mzo. de 2022
I substituted in 10^-10 for Y0 in c(6) . The result was less than -2 times 10 to the one million.
There is just no possibility of getting useful numbers out of the equations you have given.

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