ERROR Subscript indices must either be real positive integers or logicals.

3 visualizzazioni (ultimi 30 giorni)
clear all; close all; clc
fs=200; %sampling freq.
dt =1/fs;
N0=fs/50; %number of samples/cycle
m=3; %no. of cycles
t = dt*(0:200); %data window
fi=50; %Frequency test
ww=wgn(201,1,-40);
size(transpose(ww))
X=sin(2*pi*fi*t + 0.3)
v = bsxfun( @plus, X , ww );
tmax=1;
% v : as function of time
% fs : sampling frequency (Hz)
% tmax : time of final estimation
% to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)),50*512,1)
% to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)+randn(1)*.1),50*512,1)
n=N0-1:-1:0;
f0=50;
f=50.88;
Hc=2/N0*cos(2*pi*n/N0+pi/N0);
Hs=-2/N0*sin(2*pi*n/N0+pi/N0);
t_est=[];
f_est=[];
j_max=tmax*fs;
for j=1:j_max+1
x=v((j-1:j+N0-2)*dt);
c(j)=x*Hc';
s(j)=x*Hs';
if(j>N0)
Ac(j-N0)=sqrt(sum(c(end-N0+1:end).^2)/N0);
As(j-N0)=sqrt(sum(s(end-N0+1:end).^2)/N0);
cc(j-N0)=c(end-N0+1:end)*Hc';
ss(j-N0)=c(end-N0+1:end)*Hs';
if(j>2*N0)
Acc(j-2*N0)=sqrt(sum(cc(end-N0+1:end).^2)/N0);
Ass(j-2*N0)=sqrt(sum(ss(end-N0+1:end).^2)/N0);
ccc(j-2*N0)=cc(end-N0+1:end)*Hc';
ccs(j-2*N0)=cc(end-N0+1:end)*Hs';
ssc(j-2*N0)=ss(end-N0+1:end)*Hc';
sss(j-2*N0)=ss(end-N0+1:end)*Hs';
ff=f0*N0/pi*atan(tan(pi/N0)*((ccc(j-2*N0).^2+ccs(j-2*N0).^2)./(ssc(j-2*N0).^2+sss(j-2*N0).^2)).^.25);
t_est=[t_est;(j-1)*dt];
f_est=[f_est;ff];
end
end
end
t_est
f_est
plot(t_est,f_est,'red')
o=rms(fi);
c=rms(f_est)
RMSE = sqrt(mean(c - o).^2)
t_est;
f_est
plot(t_est, f_est,'red')
hold on
RMSE = sqrt(mean((f_est-fi).^2))
xlabel('time')
ylabel('frequency')
title('three LDFT white noise')
plot (t_est,fi*ones(size(t_est)))
hold off
Subscript indices must either be real positive integers or logicals.
Error in threeDFT (line 35)
x=v((j:j+N0-2)*dt);

Risposta accettata

VBBV
VBBV il 2 Dic 2020
for j=2:j_max+1
x=v((j-1:j+N0-2)*dt);
Use a positive non zero integer subscripts. In your case it is zero
  1 Commento
common fernando
common fernando il 3 Dic 2020
GIVE ME THE SAME ERROR
Subscript indices must either be real positive integers or logicals.
Error in threeDFT (line 34)
X=v((j-1:j+N0-2)*dt);
SORRY, i'M NOT GOOD IN MATLAB

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Più risposte (1)

Walter Roberson
Walter Roberson il 2 Dic 2020
for j=1:j_max+1
x=v((j-1:j+N0-2)*dt);
j starts at 1. j-1 starts at 0. 0 multiplied by anything is 0. Therefore your first index into v would be 0, which is not a permitted index.
If you add one to the indices to avoid this problem, such as starting with j = 2, then j-1:j+N0-2 would be 1:2+4-2 ==> 1:4 which is plausible. Then you multiply that by dt which is 1/fs so dt = 1/200. This would give you an index of (1:4)/200 which would not be integer indices.
Therefore your problem is not just an "off-by-one" problem: you are confusing indexing (relative positions into an array) with the value that the location is intended to denote. You cannot index at 1/fs: you can only index at a value that is associated with 1/fs .
In most cases, your first index into an array should work out to be either 1 or 2 (unless you are reading from a static data array based upon user input or device readings or date)
  2 Commenti
common fernando
common fernando il 3 Dic 2020
YES reading from a static data array based upon user input or device readings or date)
RELATED FS
Walter Roberson
Walter Roberson il 4 Dic 2020
When I say reading from a static data array based upon user input or device readings or date, I mean things like:
temp = input('Enter temperature ');
property_idx = floor(temp*10) + 1;
property_value = material_density(property_idx);
or
basetime = datetime('now');
hidx = hms(basetime) + 1;
proj_elec = electrical_projections_by_hour(hidx);
-- cases where the data is not being stepped through in sequence.
Your code, though, is trying to start at the beginning of the data and step through all of the data eventually. And in such a case, your first array index should almost always work out as either 1 or 2.
You need it to be the case that the index of the "next" item in the array is one more (or a constant integer more) than the current index. The index of the "next" item in the array should never be 1/120 more.
Do not confuse the index of an item and the time associated with the index.
timestamp = (0:149) / 120;
y = sin(timestamp * 2 * pi * 5);
Now y(2) is the data value associated with the time given in timestamp(2), which is 1/120. You would not index the array y with y(1/120) to get the y value associated with the time 1/120: you would index y(2) to get the y value associated with the time given in timestamp(2) where timestamp(2) = 1/120 .
Do not confuse array indexing with formulas. You might have a formula
Y = @(t) sin(t * 2 * pi * 5)
such that Y(1/120) would compute the result associated with the value 1/120, but a formula is not an array, where the numbers in () tell you about relative location of the information.

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