Index exceeds matrix dimensions

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abdulaziz alofui
abdulaziz alofui 2014-4-17
评论： Image Analyst 2014-4-18
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

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采纳的回答

Image Analyst
Image Analyst 2014-4-18
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 个评论
显示 2更早的评论
abdulaziz alofui
abdulaziz alofui 2014-4-18
Index exceeds matrix dimensions.
this line : PL = A+10*gamma*log10(distance(k,l)/do*1000);
Image Analyst
Image Analyst 2014-4-18
That's not the code I gave you!!!!

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更多回答（2 个）

Sara
Sara 2014-4-17
The variable dis is 1 by 8 while you ask the code to access dis(k,l) with k from 1 to 50
  2 个评论
abdulaziz alofui
abdulaziz alofui 2014-4-17
no the variable dis its mean the distance b/w 50 random userlocation and 8 random access point
Sara
Sara 2014-4-17
I ran the code as is and that's what acme out. You'll need to change that variable expression if it is not what you expected.

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Walter Roberson
Walter Roberson 2014-4-17
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.

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