Replace data inside a loop.

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Kugen Raj
Kugen Raj 2012-4-5
for N=10:10
count2=count2+1;
for ii=1:last
a=abs(rand(1,N))
for n =1:N
count=0;
for angle=0:360
count=count+1;
Xd=(D*cos(angle*pi/180));
Yd=(D*sin(angle*pi/180));
d1(n,count)= sqrt((Yd-y(n))^2+(Xd-x(n))^2);%distance between destination and sensor for 0 to 360 degree.
k(n,count)=((2*pi)/lambda)*(d1(n,count)); %phase lag for each sensor.
yy(n,count)=(exp(-i*k(n,count))*c(n)*a(n)); %weight multiplication to correct the phase lag.
end
Y=sum(yy);
YY=abs(Y);
p=20*log10(YY/max(YY));%normalized power in dB.
end
end
end
I want to replace the (a) value for each loop of (for ii=1:last) loop. i want to replace the new (a) value with the old (a) value. how can I do that?
  1 个评论
Jan
Jan 2012-4-5
RAND replies non-negative values only. Therefore ABS() is not required.

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回答（1 个）

Jan
Jan 2012-4-5
Please explain how the first a value is defined, if you replace each a by the former value.
Usually such problems are solved like this:
aOld = 0;
for i = 1:100
a = rand;
% Your computations here, e.g.:
disp(a - aOld);
...
aOld = a;
end
  3 个评论
显示 1更早的评论
Jan
Jan 2012-4-5
Again: The ABS() is not required here.
Does this mean, that your problem is solved?
Kugen Raj
Kugen Raj 2012-4-5
well, that is because i want the a value to be a positive value. If i don't use abs(), i would get a negative value too.

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