bandwidth from 2D array

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fima v
fima v le 17 Fév 2017
Réponse apportée : KSSV le 17 Fév 2017
Hello,i have a 2D array of numbers, i need to take the peak value which always located at the same X=0 point and subtracts 3 and find the two points(positive X-axes and negative X axes) in this table which are the closest to these values.
for example if my value at x=0 is 7 then i need to find the two places int the Y axes which are closest to 4.
if if we have (-30,4.1) and (+32,4.3) then the bandwidth 30+32 is 62 i thought of running two parallel "for" one for positive axes one for negative axes and taking the closest point on each side by checking iteratively.
is there an easier way?
Thanks
  1 commentaire
KSSV
KSSV le 17 Fév 2017
Modifié(e) : KSSV le 17 Fév 2017
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KSSV
KSSV le 17 Fév 2017
clc; clear all ;
y = 2*sin(linspace(0,pi))' ;
x = [1:length(y)]' ;
p = [x y] ;
%%get maximum
[val,idx] = max(y) ;
% substract and add a value
dw = 0.2 ; % value to be substracted
y0 = val-dw ;
% get the nearest values
id = find(abs(y-y0)-dw <= 10^-3) ;
idx0 = id(1) ; idx1 = id(end) ;
% required points
p0 = [x(idx0) y(idx0)] ;
p1 = [x(idx1) y(idx1)] ;
% Bandwidth
BW = abs(x(idx0)-x(idx1)) ;
% plot
figure
hold on
plot(x,y,'r') ;
plot(x(idx), y(idx),'Or')
plot(p0(1),p0(2),'*g')
plot(p1(1),p1(2),'*b')

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