Index exceeds matrix dimensions
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
abdulaziz alofui
le 17 Avr 2014
Commenté : Image Analyst
le 18 Avr 2014
this is my code:
pathloss2 = zeros(50,8);
f = 2500;
height = 5; % height of the base station(b/w 10 t0 80 meters)
do = 100; % constant
% d=500; %d>do
% s=8; % typical value of standard deviation (b/w 8 and 10 dB)
wavelength = 3.0e+8/f;
atype = 4.6; % model parameters
btype = 4;
ctype = 3.6;
gamma = (atype-btype*height+ctype/height);
A = 20*log10(4*3.14*do/wavelength);
deltaPLf = 6*log10(f/2000);
deltaPLh = -10.8*log10(height/1.5);
UserLocationX = randi(50, 1, 50);
UserLocationY = randi(50, 1, 50);
AccessPointX = randi(50, 1, 8);
AccessPointY = randi(50, 1, 8);
dis = sqrt((UserLocationX(:,1)-AccessPointX).^2 + (UserLocationY(:,2)-AccessPointY).^2);
for k=1:50
for l=1:8
PL = A+10*gamma*log10(dis(k,l)/do*1000);
pathloss2(k,l)= PL + deltaPLf + deltaPLh + 30;
end
end
0 commentaires
Réponse acceptée
Image Analyst
le 18 Avr 2014
Bring dis inside the loop. Try this:
UserLocationX = randi(50, 1, 50);
UserLocationY = randi(50, 1, 50);
AccessPointX = randi(50, 1, 8);
AccessPointY = randi(50, 1, 8);
for k=1:50
for l=1:8
distance = sqrt((UserLocationX(k)-AccessPointX(l)).^2 + (UserLocationY(k)-AccessPointY(l)).^2);
PL = A+10*gamma*log10(distance/do*1000);
pathloss2(k,l)= PL + deltaPLf + deltaPLh + 30;
end
end
4 commentaires
Plus de réponses (2)
Walter Roberson
le 17 Avr 2014
UserLocationX = randi(50, 1, 50) is going to build UserLocationX as a row vector. UserLocationX(:,1) then asks for a particular column out of that row vector, and since there is only one row in the row vector the result is going to be a scalar.
Watch out for row versus column access.
0 commentaires
Voir également
Catégories
En savoir plus sur Discrete Math dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!