Output doesn't display a value, just an empty space.

Question

Anna_P 2025-7-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2178516-output-doesn-t-display-a-value-just-an-empty-space

评论： Anna_P 2025-7-17

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));

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Answer 1

Stephen23 2025-7-16

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2178516-output-doesn-t-display-a-value-just-an-empty-space#answer_1568023

编辑：Stephen23 2025-7-16

在 MATLAB Online 中打开

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...

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Stephen23 2025-7-16

编辑：Stephen23 2025-7-16

在 MATLAB Online 中打开

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]);
x_S1 = 5.500000 , y_S1 = +0.022215 .
% 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]);
x_pvS1 = 5.500000 , y_pvS1 = +0.019493 .
% 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]);
x_d2S1 = 0.980000 , y_d2S1 = -0.039983 .
x_d2S1 = 5.480000 , y_d2S1 = +0.001776 .
x_d2S1 = 9.980000 , y_d2S1 = -0.039983 .
% 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]);
x_dp2S1 = 0.920000 , y_dp2S1 = -0.065128 .
x_dp2S1 = 4.380000 , y_dp2S1 = +0.002983 .
x_dp2S1 = 6.540000 , y_dp2S1 = +0.003272 .

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?

Stephen23 2025-7-16

编辑：Stephen23 2025-7-17

在 MATLAB Online 中打开

"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)
ans = 1×4
    0.1000    1.8800    9.1000   10.8800
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Y(idx)
ans = 1×4
    0.3489    0.3542    0.3492    0.3539
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Method two: GRADIENT

d2y = gradient(gradient(Y, X), X);
idx = find(diff(sign(d2y)) ~= 0);
X(idx)
ans = 1×4
    0.1000    1.8800    9.1000   10.8800
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Y(idx)
ans = 1×4
    0.3489    0.3542    0.3492    0.3539
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

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
Xi = 1×4
    0.1184    1.8811    9.1189   10.8816
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Yi = interp1(X,Y,Xi)
Yi = 1×4
    0.3536    0.3539    0.3539    0.3536
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

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
Xj = 1×4
    0.1188    1.8807    9.1193   10.8812
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Yj = interp1(X,Y,Xj)
Yj = 1×4
    0.3536    0.3540    0.3540    0.3536
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Lets take a look at the inflection points from method 4:

plot(X,Y)

hold on

plot(Xj,Yj,'rx')

●

Stephen23 2025-7-17

编辑：Stephen23 2025-7-17

在 MATLAB Online 中打开

"... 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)
x_min = 5.5017
y_min = interp1(X,Y,x_min)
y_min = 0.0222

Anna_P 2025-7-17

@Stephen23 THANK YOU!)

I'm not desperate anymore! Going to explore and study HELP section with more clarity now!

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Output doesn't display a value, just an empty space.

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Output doesn't display a value, just an empty space.

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