Output doesn't display a value, just an empty space.
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So, I've got 2 pulses (imp and S0) and their sum (Si). I've got to find minimum of sum (Si). And I approximate the sum with a polynome. I took derivatives (1-st and 2-nd order) of the sum and polynome.
I analyze derivatives, and decided to look for minimum of sum (Si), polynome and 2-nd derivatives.
But, in some cases the result is just empty, but size of variable is 1x999 double.
lb = -5;
ub = 15;
x_imp = linspace(lb, ub, 1001);
x = linspace(lb, ub, 1001);
imp = sech(x_imp);
S0 = sech(x-10);
imp1 = sech(x - 1);
S1 = 0.5 * (imp1 + S0);
xmin = -1;
xmax = 10;
% find inflection point Si
range_x = x(x>=xmin & x<=xmax);
range_S1 = S1(x>=xmin & x<=xmax);
MinPtsS1 = islocalmin(range_S1);
fprintf ('x_S1 = %f , y_S1 = %f .\n', range_x(MinPtsS1), range_S1(MinPtsS1));
% polynom approximation pv
[p_S1, ~, mu] = polyfit(x, S1, 26);
pv_S1 = polyval (p_S1, x, [], mu);
% find inflection point of polynom pv
range_x = x(x>=xmin & x<=xmax);
range_pvS1 = pv_S1(x>=xmin & x<=xmax);
MinPts_pvS1 = islocalmin(range_pvS1);
fprintf('x_pvS1 = %f , y_pvS1 = %f .\n', range_x(MinPts_pvS1), range_pvS1(MinPts_pvS1));
% DIFF find inflection of diff2 Si, corresponding to inflection point of Si
dS1 = 20*diff (S1);
d2S1 = 10*diff(dS1);
range_x = x(x>=xmin & x<=xmin);
range_d2S1 = d2S1(x>=xmin & x<=xmin);
MinPts_d2S1 = islocalmin(range_d2S1);
fprintf('x_d2S1 = %f , y_d2S1 = %f .\n', range_x(MinPts_d2S1), range_d2S1(MinPts_d2S1));
% DIFF find inflection of diff2 polynom Si, corresponding to inflection point of Si
dpS1 = 20*diff(pv_S1);
dp2S1 = 20*diff(dpS1);
range_x = x(x>=xmin & x<=xmax);
range_dp2S1 = dp2S1(x>=xmin & x<=xmax);
MinPts_dp2S1 = islocalmin(range_dp2S1);
fprintf('x_dp2S1 = %f , y_dp2S1 = %f .\n', range_x(MinPts_dp2S1), range_dp2S1(MinPts_dp2S1));
Accepted Answer
How many values fulfill your logical comparisons? Exactly one.
range_x = x(x>=xmin & x<=xmin);
range_d2S1 = d2S1(x>=xmin & x<=xmin);
should be
range_x = x(x>=xmin & x<=xmax);
range_d2S1 = d2S1(x>=xmin & x<=xmax);
Due to FPRINTF processing all input arguments in linear order you will also need to modify the FPRINTF input arguments (e.g. by defining a matrix), or use e.g. COMPOSE with column vectors.
There might be other bugs too...
10 Comments
I modified the FPRINTF calls:
lb = -5;
ub = 15;
x_imp = linspace(lb, ub, 1001);
x = linspace(lb, ub, 1001);
imp = sech(x_imp);
S0 = sech(x-10);
imp1 = sech(x - 1);
S1 = 0.5 * (imp1 + S0);
xmin = -1;
xmax = 10;
% find inflection point Si
range_x = x(x>=xmin & x<=xmax);
range_S1 = S1(x>=xmin & x<=xmax);
MinPtsS1 = islocalmin(range_S1);
x_S1 = range_x(MinPtsS1);
y_S1 = range_S1(MinPtsS1);
fprintf ('x_S1 = %f , y_S1 = %+f .\n', [x_S1;y_S1]);
% polynom approximation pv
[p_S1, ~, mu] = polyfit(x, S1, 26);
pv_S1 = polyval (p_S1, x, [], mu);
% find inflection point of polynom pv
range_x = x(x>=xmin & x<=xmax);
range_pvS1 = pv_S1(x>=xmin & x<=xmax);
MinPts_pvS1 = islocalmin(range_pvS1);
x_pvS1 = range_x(MinPts_pvS1);
y_pvS1 = range_pvS1(MinPts_pvS1);
fprintf('x_pvS1 = %f , y_pvS1 = %+f .\n', [x_pvS1;y_pvS1]);
% DIFF find inflection of diff2 Si, corresponding to inflection point of Si
dS1 = 20*diff (S1);
d2S1 = 10*diff(dS1);
range_x = x(x>=xmin & x<=xmax);
range_d2S1 = d2S1(x>=xmin & x<=xmax);
MinPts_d2S1 = islocalmin(range_d2S1);
x_d2S1 = range_x(MinPts_d2S1);
y_d2S1 = range_d2S1(MinPts_d2S1);
fprintf('x_d2S1 = %f , y_d2S1 = %+f .\n', [x_d2S1;y_d2S1]);
% DIFF find inflection of diff2 polynom Si, corresponding to inflection point of Si
dpS1 = 20*diff(pv_S1);
dp2S1 = 20*diff(dpS1);
range_x = x(x>=xmin & x<=xmax);
range_dp2S1 = dp2S1(x>=xmin & x<=xmax);
MinPts_dp2S1 = islocalmin(range_dp2S1);
x_dp2S1 = range_x(MinPts_dp2S1);
y_dp2S1 = range_dp2S1(MinPts_dp2S1);
fprintf('x_dp2S1 = %f , y_dp2S1 = %+f .\n', [x_dp2S1;y_dp2S1]);
Taking a look at that data:
plot(range_x,range_S1, range_x,range_pvS1, range_x,range_d2S1, range_x,range_dp2S1)
hold on
plot(x_S1,y_S1,'ok')
plot(x_pvS1,y_pvS1,'r+')
plot(x_d2S1,y_d2S1,'gd')
plot(x_dp2S1,y_dp2S1,'bx')
Are those the inflection points that you would expect to see?
Anna_P
on 16 Jul 2025
Paul
on 16 Jul 2025
The inflection points marked on the graph don't look like inflection points of the function that's plotted. Is the logic for determining inflection points correct?
Anna_P
on 16 Jul 2025
Anna_P
on 16 Jul 2025
The red "inflection S1" points are based on the original data, not the polynomial, and appear to be incorrect. So the fundamental issue is not with polynomial fit, but with the criteria for finding an inflection point.
Going to a polynomial order higher than 26 will likely cause even more problems in finding inflection points as compared to the original curve (at least in this example), and there likely is already problems with such a high order polynomial (can't say for sure because I did not plot pv_S1) in finding its inflection points as compared to the underlying data (at least for this example).
"maybe you could show an "easy" way to find zeros of 1-st derivative? Or there is a "100%" working function with gradient?"
Here are some approaches to find the inflection points of a set of data, which are zeros of the 2nd derivative. We assume that the data represents some reasonably well-behaved curve, e.g.:
X = linspace(-5, 15, 1001);
Y = 0.5 * (sech(X-1) + sech(X-10));
Method one: DIFF
The most straightforward approach since inflection points occur where the second derivative changes sign.
d2y = diff(Y,2);
idx = 1+find(diff(sign(d2y)) ~= 0);
X(idx)
Y(idx)
Method two: GRADIENT
d2y = gradient(gradient(Y, X), X);
idx = find(diff(sign(d2y)) ~= 0);
X(idx)
Y(idx)
Method 3: FZERO and INTERP1
Returns inflection points interpolated from the original X and Y data:
d2y = gradient(gradient(Y, X), X);
f_d2y = @(x) interp1(X, d2y, x, 'spline');
sc = find(diff(sign(d2y)) ~= 0);
Xi = nan(size(sc));
for k = 1:numel(sc)
idx = sc(k);
x1 = X(idx);
x2 = X(idx + 1);
Xi(k) = fzero(f_d2y, [x1, x2]);
end
Xi
Yi = interp1(X,Y,Xi)
Method 4: FZERO & SPLINE
Returns inflection points interpolated from the original X and Y data:
pp = spline(X, Y);
pp2 = fnder(pp,2); % Requires Curve Fitting Toolbox
breaks = pp.breaks;
Xj = [];
for k = 1:numel(breaks)-1
segment = @(x) ppval(pp2, x);
try
zpt = fzero(segment, breaks([k,k+1]));
catch
continue
end
Xj(end+1) = zpt;
end
Xj
Yj = interp1(X,Y,Xj)
Lets take a look at the inflection points from method 4:
plot(X,Y)
hold on
plot(Xj,Yj,'rx')
Anna_P
on 17 Jul 2025
"... I need the min point of polynomial, around x = 5, which is not presented in all 4 methods.. "
X = linspace(-5, 15, 1001);
Y = 0.5 * (sech(X-1) + sech(X-10));
[p_S1, ~, mu] = polyfit(X, Y, 26);
pv_S1 = polyval(p_S1, X, [], mu);
dp_S1 = polyder(p_S1);
dpoly = @(xq) polyval(dp_S1, (xq-mu(1))./mu(2));
x_min = fzero(dpoly, 5)
y_min = interp1(X,Y,x_min)
Anna_P
on 17 Jul 2025
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