# Problems in for loop with cell array

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Bruce Rogers 2021 年 7 月 1 日
コメント済み: Bruce Rogers 2021 年 7 月 2 日
Hello everyone,
I'm trying to get the phase shift of a normal sine wave with different sine waves that were produced by a robot. Every run of the experiment has a different period, so its y=sine(factor * x). I used Crossing_V7 function that was given to me a few days ago. When I used it with only one robot sine wave it worked fine, but now I want to use it for all five attempts with different parameters and I get the error message, that the length is different. Does anyone of you see the mistake? I'm pretty new to cell array calculations.
clear;
factor = [0.25, 0.5, 1.0, 2.0, 4.0];
z_irl = cell(1,5);
z_irl_temp = irl{:,4};
z_irl{:,1} = z_irl_temp;
z_irl_temp = irl{:,4};
z_irl{:,2} = z_irl_temp;
z_irl_temp = irl{:,4};
z_irl{:,3} = z_irl_temp;
z_irl_temp = irl{:,4};
z_irl{:,4} = z_irl_temp;
z_irl_temp = irl{:,4};
z_irl{:,5} = z_irl_temp;
%%
for j = 1:length(factor)
z_irl = z_irl{1,j};
num_irl = numel(z_irl);
t = (1:num_irl)*0.004;
A = 50;
b = 10; %Durchgänge
ti = 0.004;
%x = zeros(length(z_irl));
x(:,j) = 0:ti:(1/factor(j))*10*pi;
y(:,j) = -A*sin(factor(j)*x) + 389.387;
%y(j) = y(j)';
num_th = numel(x);
T = (1:length(x))*ti;
threshold = 389.387; % your value here
[t0_pos1(:,j),s0_pos1(:,j),t0_neg1(:,j),s0_neg1(:,j)]= crossing_V7(y(:,j),T(:,j),threshold,'linear'); % positive (pos) and negative (neg) slope crossing points %%Here I get the error message
[t0_pos2(:,j),s0_pos2(:,j),t0_neg2(:,j),s0_neg2(:,j)]= crossing_V7(z_irl(:,j),t(:,j),threshold,'linear'); % positive (pos) and negative (neg) slope crossing points
% ind => time index (samples)
% t0 => corresponding time (x) values
% s0 => corresponding function (y) values , obviously they must be equal to "threshold"
figure(7)
% plot(T,y,'g',t0_pos1,s0_pos1,'+g',t0_neg1,s0_neg1,'+g','linewidth',2,'markersize',12);grid on
% hold on
% plot(t,z_irl,'b',t0_pos2,s0_pos2,'+b',t0_neg2,s0_neg2,'+b','linewidth',2,'markersize',12);grid on
% hold off
% title('Nullstellen beider Wellen, Grün: Theorie, Blau: Realität')
% xlabel('time');
% ylabel('distance');
% time difference for positive slope crossing points
dt_pos_slope(j) = t0_pos1(j) - t0_pos2(j);
% time difference for negative slope crossing points
if numel(t0_neg1(j)) ~= 5
t0_neg1_temp(j) = 0.004;
t0_neg1(j) = [t0_neg1_temp(j),t0_neg1(j)];
end
if numel(t0_neg2(j)) ~= 5
t0_neg2_temp(j) = 0.004;
t0_neg2(j) = [t0_neg2_temp(j),t0_neg2(j)];
end
dt_neg_slope(j) = t0_neg1(j) - t0_neg2(j);
end
%%
function [t0_pos,s0_pos,t0_neg,s0_neg] = crossing_V7(S,t,level,imeth)
% [ind,t0,s0,t0close,s0close] = crossing_V6(S,t,level,imeth,slope_sign) % older format
% CROSSING find the crossings of a given level of a signal
% ind = CROSSING(S) returns an index vector ind, the signal
% S crosses zero at ind or at between ind and ind+1
% [ind,t0] = CROSSING(S,t) additionally returns a time
% vector t0 of the zero crossings of the signal S. The crossing
% times are linearly interpolated between the given times t
% [ind,t0] = CROSSING(S,t,level) returns the crossings of the
% given level instead of the zero crossings
% ind = CROSSING(S,[],level) as above but without time interpolation
% [ind,t0] = CROSSING(S,t,level,par) allows additional parameters
% par = {'none'|'linear'}.
% With interpolation turned off (par = 'none') this function always
% returns the value left of the zero (the data point thats nearest
% to the zero AND smaller than the zero crossing).
%
% [ind,t0,s0] = ... also returns the data vector corresponding to
% the t0 values.
%
% [ind,t0,s0,t0close,s0close] additionally returns the data points
% closest to a zero crossing in the arrays t0close and s0close.
%
% This version has been revised incorporating the good and valuable
% bugfixes given by users on Matlabcentral. Special thanks to
% Howard Fishman, Christian Rothleitner, Jonathan Kellogg, and
% Zach Lewis for their input.
% Steffen Brueckner, 2002-09-25
% Steffen Brueckner, 2007-08-27 revised version
% Copyright (c) Steffen Brueckner, 2002-2007
% brueckner@sbrs.net
% M Noe
% added positive or negative slope condition
% check the number of input arguments
error(nargchk(1,4,nargin));
% check the time vector input for consistency
if nargin < 2 | isempty(t)
% if no time vector is given, use the index vector as time
t = 1:length(S);
elseif length(t) ~= length(S)
% if S and t are not of the same length, throw an error
error('t and S must be of identical length!'); %This is the error message I get
end
% check the level input
if nargin < 3
% set standard value 0, if level is not given
level = 0;
end
% check interpolation method input
if nargin < 4
imeth = 'linear';
end
% make row vectors
t = t(:)';
S = S(:)';
% always search for zeros. So if we want the crossing of
% any other threshold value "level", we subtract it from
% the values and search for zeros.
S = S - level;
% first look for exact zeros
ind0 = find( S == 0 );
% then look for zero crossings between data points
S1 = S(1:end-1) .* S(2:end);
ind1 = find( S1 < 0 );
% bring exact zeros and "in-between" zeros together
ind = sort([ind0 ind1]);
% and pick the associated time values
t0 = t(ind);
s0 = S(ind);
if strcmp(imeth,'linear')
% linear interpolation of crossing
for ii=1:length(t0)
%if abs(S(ind(ii))) > eps(S(ind(ii))) % MATLAB V7 et +
if abs(S(ind(ii))) > eps*abs(S(ind(ii))) % MATLAB V6 et - EPS * ABS(X)
% interpolate only when data point is not already zero
NUM = (t(ind(ii)+1) - t(ind(ii)));
DEN = (S(ind(ii)+1) - S(ind(ii)));
slope = NUM / DEN;
slope_sign(ii) = sign(slope);
t0(ii) = t0(ii) - S(ind(ii)) * slope;
s0(ii) = level;
end
end
end
% extract the positive slope crossing points
ind_pos = find(sign(slope_sign)>0);
t0_pos = t0(ind_pos);
s0_pos = s0(ind_pos);
% extract the negative slope crossing points
ind_neg = find(sign(slope_sign)<0);
t0_neg = t0(ind_neg);
s0_neg = s0(ind_neg);
end

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### 採用された回答

David Hill 2021 年 7 月 1 日
Why not just use fit() function
x=0:.01:2*pi;
y=(5+.05*randn(size(x))).*sin((3+.05*randn(size(x))).*x);
f=fit(x',y','sin1');
period=f.b1;
##### 3 件のコメント表示非表示 2 件の古いコメント
Bruce Rogers 2021 年 7 月 2 日
Now I get what you mean. Thanks David, it helped, but I used another method, because otherwise I would had to change too much on the program

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