for loop jump an element of an array

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Davide
Davide 2024-3-27
评论： Davide 2024-3-30
Hello there,
it's my first post here, I'm writing a code to perform linear interpolation of a sinusoid and then a resampling with an higeher sampling frequency but i'm facing some trouble since the foor loop seems to jump one value of the sampling time vector and that's a problem when i'm walking through the array to find the overlapping times of the two sampling time vectors.
f0 = 10; % Hz
T0 = 1/f0; % period of sin [s]
fc = 100; %Hz
T = 1/fc; % first sampling period [s]
N = floor(10*(fc/f0)); % number of samples in 10 periods
samples_time = 0:T:(N-1)*T;
samples_value = zeros(1,N);
% first sampling
for i=1:N
samples_value(i) = sin(2*pi*f0*samples_time(i));
end
fc2 = 200; % Hz
T2 = 1/fc2; % second sampling period [s]
N2 = floor((N-1)*(T/T2)+1); % number of samples
samples_time2 = 0:T2:(N2-1)*T2;
for j=1:N2
if samples_time2(j) == 0.64
disp("found");
end
%disp(samples_time2(j));
end
The foor loop seems not to find the value 0.64 in the array, it works with other values. I cannot find a solution, it's very strange. (I'm running R2023b)
Hope someone can help me, thanks

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回答（2 个）

Voss
Voss 2024-3-27
Ran in:
From the documentation for the colon operator:
"x = j:i:k creates a regularly-spaced vector x using i as the increment between elements. The vector elements are roughly equal to [j,j+i,j+2*i,...,j+m*i] where m = fix((k-j)/i). However, if i is not an integer, then floating point arithmetic plays a role in determining whether colon includes the endpoint k in the vector, since k might not be exactly equal to j+m*i."
I'd expand that to add, if i is not an integer, then floating point arithmetic plays a role in determining whether colon includes any intended value beyond the first.
To avoid this problem, construct your time vectors using a spacing of 1 in colon, i.e.:
samples_time = (0:N-1)*T;
samples_time2 = (0:N2-1)*T2
instead of:
samples_time = 0:T:(N-1)*T;
samples_time2 = 0:T2:(N2-1)*T2;
(Yes, the same problem of missing values happens in samples_time as well.)
Some illustrations:
f0 = 10; % Hz
T0 = 1/f0; % period of sin [s]
fc = 100; %Hz
T = 1/fc; % first sampling period [s]
N = floor(10*(fc/f0)); % number of samples in 10 periods
Constructing samples_time both ways:
samples_time = 0:T:(N-1)*T
samples_time = 1×100
0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500 0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300 0.2400 0.2500 0.2600 0.2700 0.2800 0.2900
<mw-icon class=""></mw-icon>
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samples_time_test = (0:N-1)*T
samples_time_test = 1×100
0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500 0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300 0.2400 0.2500 0.2600 0.2700 0.2800 0.2900
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They look the same, but they are not:
isequal(samples_time,samples_time_test)
ans = logical
0
In particular, 0.64 is in the one constructed using a colon spacing of 1 but not the one using T spacing:
find(samples_time == 0.64)
ans = 1×0 empty double row vector
find(samples_time_test == 0.64)
ans = 65
There is an element in samples_time that is very close to 0.64 (about 1e-16 away), but it's not exactly 0.64:
[d,idx] = min(abs(samples_time-0.64))
d = 1.1102e-16
idx = 65
whereas in samples_time_test that element is exactly 0.64:
[d,idx] = min(abs(samples_time_test-0.64))
d = 0
idx = 65
The same is true for the second set of times.
fc2 = 200; % Hz
T2 = 1/fc2; % second sampling period [s]
N2 = floor((N-1)*(T/T2)+1); % number of samples
samples_time2 = 0:T2:(N2-1)*T2
samples_time2 = 1×199
0 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450 0.0500 0.0550 0.0600 0.0650 0.0700 0.0750 0.0800 0.0850 0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200 0.1250 0.1300 0.1350 0.1400 0.1450
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samples_time2_test = (0:N2-1)*T2
samples_time2_test = 1×199
0 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450 0.0500 0.0550 0.0600 0.0650 0.0700 0.0750 0.0800 0.0850 0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200 0.1250 0.1300 0.1350 0.1400 0.1450
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isequal(samples_time2,samples_time2_test)
ans = logical
0
find(samples_time2 == 0.64)
ans = 1×0 empty double row vector
find(samples_time2_test == 0.64)
ans = 129
And 0.64 isn't the only missing value either, in either vector:
% elements in samples_time_test that aren't in samples_time:
samples_time_test(~ismember(samples_time_test,samples_time))
ans = 1×13
0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.8200 0.8300 0.9200 0.9300 0.9400 0.9500
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% elements in samples_time2_test that aren't in samples_time2:
samples_time2_test(~ismember(samples_time2_test,samples_time2))
ans = 1×28
0.5150 0.5250 0.5350 0.5450 0.5550 0.5650 0.5750 0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.7850 0.7950 0.8050 0.8150 0.8200 0.8250 0.8300 0.9200 0.9300 0.9350 0.9400 0.9450 0.9500 0.9550
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  4 个评论
显示 2更早的评论
Stephen23
Stephen23 2024-3-29
"...can I have the same problem?"
Yes. Using two different methods can result in two different values. That is nature of binary floating point numbers.
Davide
Davide 2024-3-30
Sure, of course. As soon as I get home I will provide you the complete code

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the cyclist
the cyclist 2024-3-27
You should not use strict equality to compare floating point numbers.
Instead, check equality with a small tolerance, e.g.
if abs(samples_time2(j) - 0.64) < 1.e-6

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