exceeded number of array elements

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Maurilio Matracia
Maurilio Matracia 2020-10-22
评论： Stephen23 2020-10-22
Hello everyone,
could you please help me figure out how to fix this error? It is showing up already for B=1 and C=1
Here's the code:
clear,
% [ THIS IS test_I_analysis ]
%% PARAMETERS
H = [100 ]'; h=1 ; var = 2e5 ;
lambdaA = 1e-6 ; lambdaT = 0.15 ;
lambda = [lambdaA, lambdaA, lambdaT] ;
Sa = [4.88 9.6117 12.0810 27.304]' ; Sb = [0.429 0.1581 0.1139 0.0797]' ;
alpha = [2, 3, 3] ;
alpha= [2.37 3 3]
tau = 0.31623 ; sigma_n2 = 10^-13 ;
p = [1.585, 1.585, 10] ;
eta = [0.9972 0.9943 1]
xi = eta .* p ;
m = [2 1 1] ; % Shape parameters for Nakagami fadings
Ru=600
R= 500 ;
for A=1:2
dd{A,3} = @(d) (xi(3) / xi(A))^(1/alpha(3)) * d.^(alpha(A) / alpha(3)) ;
dd{3,A} = @(d) ( (xi(A) / xi(3))^(1/alpha(A)) * d.^(alpha(3) / alpha(A)) ) .* ( d>=dd{A,3}(H) ) ...
+ H .* ( d<dd{A,3}(H) ) ;
end
dd{2,1} = @(d) (xi(1) / xi(2))^(1/alpha(1)) * d.^(alpha(2) / alpha(1)) ;
dd{1,2} = @(d) ( (xi(2) / xi(1))^(1/alpha(2)) * d.^(alpha(1) / alpha(2)) ) .* ( d>=dd{2,1}(H) ) ...
+ H .* (d<dd{2,1}(H)) ;
d = @(z) sqrt(z.^2 + H^2) ;
for C = 1:3
dd{C,C} = @(z) z ;
for B = 1:3
D{B,C} = @(z) sqrt( dd{B,C}(d(z)).^2 - H^2 ) ;
D{B,3} = @(z) dd{B,3}(d(z)) ;
end
D{3,C} = @(z) sqrt(dd{3,C}(z).^2 - H^2) ;
D{C,C} = @(z) z ;
end
xm = @(Beta) Ru(1)*cos(Beta) - sqrt(R^2 - Ru(1)^2*sin(Beta).^2) ;
BetaX = asin(R/Ru(1)) ;
for u = 1:2
xX{u} = @(Beta) Ru(u)*cos(Beta) + sqrt( R^2 - (Ru(u)*sin(Beta)).^2 ) ;
Betai{u} = @(z) real( (z>=0).*acos( (z.^2 + Ru(u)^2 - R^2) ./ ( (z>0)*2*z*Ru(u) + (z==0)) ) ) ;
end
PLoS = @(z) 1 ./ (1+Sa(1)*exp( -Sb(1)*(180/pi*atan(H./z)-Sa(1)) ) ) ;
integrandK{1} = @(z) z ./ (1+Sa(1)*exp( -Sb(1)*(180/pi*atan(H./z)-Sa(1)) ) ) ;
integrandK{2} = @(z) z .* (1 - 1/(1+Sa(1)*exp( -Sb(1)*(180/pi*atan(H./z)-Sa(1)) ) ) ) ;
for A=1:2
I_{A} = @(s,z) (1 - (m(A)/(m(A) + s .* xi(A) .* (d(z)).^-alpha(A)) ).^m(A) ) .* integrandK{A}(z) ;
for B = 1:3
LapI{B,A} = @(s,x) exp(-2*pi*lambda(A) * integral(@(z) I_{A}(s, z), D{B,A}(x),Ru(u)-R,'ArrayValued',1) ) ...
.* exp(lambda(A)*integral(@(Beta) integral(@(z) I_{A}(s,z), xm(Beta),xX{1}(Beta),'ArrayValued',1), -BetaX,BetaX,'ArrayValued',1 ) ) ;
LapBC{B,A,1} = @(s,x) LapI{B,A}(s, x) .*(Ru(1)-R>D{B,A}(x)) ;
end
end
for u = 1:length(Ru)
r2{u} = @(Beta,z) Ru(u)^2 + z.^2 - 2*Ru(u)*z.*cos(Beta) ;
I_T{u} = @(Beta,s,z) exp(-r2{u}(Beta,z)./(2*var)) .* (1-(m(3)/(m(3)+s.*xi(3).*z.^-alpha(3))).^m(3)) .* z ;
for B=1:3
LapBC{B,3,u} = @(s,x) exp(-lambda(3)/sqrt(2*pi*var) * integral(@(z) integral(@(Beta) I_T{u}(Beta,s,z), 0,2*pi,'ArrayValued',1), D{B,3}(x), Ru(u)+R,'ArrayValued',1) ) ;
LapB{B,u} = @(s,x) LapBC{B,1,u}(s,x) .* LapBC{B,2,u}(s,x) .* LapBC{B,3,u}(s,x) ;
end
end
% [THIS IS test_I]
u=1;
DsI = [0 0 0] ;
xLapI = zeros(3);
LapBC_ = {1 1 1; 1 1 1; 1 1 1}
for B=1:2
for C = 1:2
LapBC_{B,C} = LapBC{B,C,u}(tau/xi(B)*DsI(B)^alpha(B), xLapI(B,C)) ;
end
end
  1 个评论
Maurilio Matracia
Maurilio Matracia 2020-10-22
The weird thing is that I am actually able to pone s = 0 and x=0 to compute LapI{1,1}(s,x), but I then get the error when I implement it in the for loop

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回答（1 个）

Stephen23
Stephen23 2020-10-22
Ru is scalar, but in multiple locations you try to access Ru(2), e.g.:
for u = 1:2
xX{u} = @(Beta) Ru(u)*cos(Beta) + sqrt( R^2 - (Ru(u)*sin(Beta)).^2 ) ;
% ^^^^^ ^^^^^
and
LapI{B,A} = @(s,x) exp(-2*pi*lambda(A) * integral(@(z) I_{A}(s, z), D{B,A}(x),Ru(u)-R,'ArrayValued',1) ) ...
% ^^^^^
  2 个评论
Maurilio Matracia
Maurilio Matracia 2020-10-22
Thank you very much! What did you exactly do to find out where the mistake was?
Stephen23
Stephen23 2020-10-22
Funnily enough I used Octave, because it gives an error message which names the exact variable related to the error:
̀error: Ru(2): out of bound 1
error: called from
temp0>@<anonymous> at line 56 column 23
temp0>@<anonymous> at line 58 column 39
temp0 at line 78 column 17
The MATLAB Editor is great for most debugging, but for cases like this just the line number is not enough.

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