bandwidth from 2D array

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fima v
fima v el 17 de Feb. de 2017
Respondida: KSSV el 17 de Feb. de 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 comentario
KSSV
KSSV el 17 de Feb. de 2017
Editada: KSSV el 17 de Feb. de 2017
Giving the data would help us to help you..

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Respuestas (1)

KSSV
KSSV el 17 de Feb. de 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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