Info
Diese Frage ist geschlossen. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.
MATLAB error 'index exceeds matrix dimensions'
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
hello all
i want to run this function but i got index exceeds, i checked my loops but i could not saw the error.
where vn=vs=1 and M=100 and th1=1000 and Dr1=D1=0
function D=distortion(vn,vs,M,th1,x,y,Dr1,D1);
for i=1:M;
for j=1:M;
if i~=j
d1(i,j)=sqrt(((y(j)-y(i))^2+(x(j)-x(i))^2));
Dr1=exp(-d1(i,j)/th1)+Dr1;
end
end
end
for i=1:M
ds1(i)=sqrt(((y(j)-0)^2+(x(j)-0)^2));% point source at 0 0
cor2=exp(-ds1(i)/th1);
D1=cor2+D1;
end
D=(vs^2)-((vs^4)/(M*((vs^2)+(vn^2))))*(2*(D1-1))+((vs^6)/((M^2)*((vs^2)+(vn^2)))*Dr1);
end
0 Kommentare
Antworten (1)
Image Analyst
am 4 Feb. 2016
This worked for me.
function test1
vn=1
vs=1 ;
M=100 ;
th1=1000;
Dr1=0;
D1=0
y = rand(1,M);
x = rand(1,M);
D=distortion(vn,vs,M,th1,x,y,Dr1,D1)
end
function D=distortion(vn,vs,M,th1,x,y,Dr1,D1)
for i=1:M;
for j=1:M;
if i~=j
d1(i,j)=sqrt(((y(j)-y(i))^2+(x(j)-x(i))^2));
Dr1=exp(-d1(i,j)/th1)+Dr1;
end
end
end
for i=1:M
ds1(i)=sqrt(((y(j)-0)^2+(x(j)-0)^2));% point source at 0 0
cor2=exp(-ds1(i)/th1);
D1=cor2+D1;
end
D=(vs^2)-((vs^4)/(M*((vs^2)+(vn^2))))*(2*(D1-1))+((vs^6)/((M^2)*((vs^2)+(vn^2)))*Dr1);
end
What did you pass in for X and Y?
16 Kommentare
Walter Roberson
am 5 Feb. 2016
You see the importance of showing us the actual error instead of just saying that you get an error somewhere?
Diese Frage ist geschlossen.
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!