Programmatically find fzero of two RCL functions

Question

Doug Leaffer 2025-2-27

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2174608-programmatically-find-fzero-of-two-rcl-functions

评论： Star Strider 2025-2-27

RCL_vars.mat

How do I fix my code to programmatically find the fzero of two functions? I had to hand-code the functions funcVL and funcVR from model fit coefficients. I want to do that programmatically, then find fzero (intersection point) of the two function curves. Need some help here. Thank you

% find intersection (frequency) of two RCL functions
load RCL_vars.mat % creates 3 vectors, f, VL, VR from data
% function calls
createFitVL(f, VL)
createFitVR(f, VR)
% find intersection point, f
x = f;
funcVL = @ (x) -3.948*exp(-0.658*x)+3.898;
funcVR = @ (x) 3.856*exp(0.2436*x)+0.2436;
%use fzero with the function: @(x) funcVL(x) - funcVR(x).
soltn = fzero(@(x) funcVR(x)- funcVL(x), [0, 10])
%%
function [fitresult, gof] = createFitVL(f, VL)
%CREATEFIT(F,VL)
% Create a fit.
%
% Data for 'untitled fit 1' fit:
% X Input: f
% Y Output: VL
% Output:
% fitresult : a fit object representing the fit.
% gof : structure with goodness-of fit info.
%
% See also FIT, CFIT, SFIT.
% Auto-generated by MATLAB on 26-Feb-2025 15:57:28
%% Fit: 'untitled fit 1'.
[xData, yData] = prepareCurveData( f, VL );
% Set up fittype and options.
ft = fittype( 'a*exp(-b*x)+c', 'independent', 'x', 'dependent', 'y' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [0.317099480060861 0.950222048838355 0.0344460805029088];
% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
% Plot fit with data.
figure( 'Name', 'untitled fit 1' );
h = plot( fitresult, xData, yData );
legend( h, 'VL vs. f', 'untitled fit 1', 'Location', 'NorthEast', 'Interpreter', 'none' );
% Label axes
xlabel( 'f', 'Interpreter', 'none' );
ylabel( 'VL', 'Interpreter', 'none' );
grid on
end
%  
hold on
%% Fit: 'untitled fit 1'.
function [fitresult, gof] = createFitVR(f, VR)
[xData, yData] = prepareCurveData( f, VR );
% Set up fittype and options.
ft = fittype( 'a*exp(-b*x)+c', 'independent', 'x', 'dependent', 'y' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [0.257508254123736 0.840717255983663 0.254282178971531];
% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
% Plot fit with data.
figure( 'Name', 'untitled fit 1' );
h = plot( fitresult, xData, yData );
legend( h, 'VR vs. f', 'untitled fit 1', 'Location', 'NorthEast', 'Interpreter', 'none' );
% Label axes
xlabel( 'f', 'Interpreter', 'none' );
ylabel( 'VR', 'Interpreter', 'none' );
grid on
end

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Answer 1

Star Strider 2025-2-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2174608-programmatically-find-fzero-of-two-rcl-functions#answer_1560782

在 MATLAB Online 中打开

It seems that the curves don’t intersect.

The ‘funcVR’ function never crosses zero, although the ‘funcVL’ function does —

funcVL = @ (x) -3.948*exp(-0.658*x)+3.898;

funcVR = @ (x) 3.856*exp(0.2436*x)+0.2436;

xs = fzero(@(x) funcVL(x)-funcVR(x), randn)

Exiting fzero: aborting search for an interval containing a sign change because NaN or Inf function value encountered during search. (Function value at -1203.79 is -Inf.) Check function or try again with a different starting value.

xs = NaN

xv = linspace(-1E+2, 1E+2, 1E3);

yL = funcVL(xv);

yR = funcVR(xv);

ixx = find(diff(sign(yL-yR)))

ixx = 1x0 empty double row vector

Zpts = [fzero(funcVR, 0.5) fzero(funcVL, 0.5)]

Exiting fzero: aborting search for an interval containing a sign change because NaN or Inf function value encountered during search. (Function value at 3707.78 is Inf.) Check function or try again with a different starting value.

Zpts = 1×2

NaN 0.0194

figure

plot(xv, yR, DisplayName='funcVR')

hold on

plot(xv, yL, DisplayName='funcVL')

plot(Zpts(2), 0, 'mp', DisplayName=sprintf('funcVL(%.4f) = 0',Zpts(2)))

hold off

grid

legend(Location='best')

xlim([-100 100])

ylim([-5 5])

.

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Star Strider 2025-2-27

在 MATLAB Online 中打开

I realise that. The functions don’t intersect, however subtracting one from the other and getting the minimum of that difference is straightforwarrd —

funcVL = @ (x) -3.948*exp(-0.658*x)+3.898;

funcVR = @ (x) 3.856*exp(0.2436*x)+0.2436;

Exiting fzero: aborting search for an interval containing a sign change because NaN or Inf function value encountered during search. (Function value at -1311.57 is -Inf.) Check function or try again with a different starting value.

xs = NaN

xv = linspace(-1E+2, 1E+2, 1E3);

yL = funcVL(xv);

yR = funcVR(xv);

ydif = yR - yL;

funcD = @(x) funcVR(x)-funcVL(x);

[xmin,ymin] = fminsearch(funcD, randn)

xmin = 1.1283

ymin = 3.3005

figure

plot(xv, yR, DisplayName='funcVR')

hold on

plot(xv, yL, DisplayName='funcVL')

plot(xv, ydif, DisplayName='funcVR-funcVL', LineWidth=3)

plot(xmin, ymin, 'rs', DisplayName='Minimum')

hold off

grid

legend(Location='best')

xlim([-1 1]*15)

ylim([-5 20])

text(xmin,ymin,sprintf(' \\leftarrow (%.4f, %.4f)',xmin,ymin), Horiz='left')

●

Change or negate ‘funcD’ if you want to subtract ‘funcVR’ frrom ‘funcVL’.

.

Steven Lord 2025-2-27

在 MATLAB Online 中打开

Yes, and from the presence of this line in your createFitVL local function it seems you've already used it.

% Auto-generated by MATLAB on 26-Feb-2025 15:57:28

See this page in the documentation for Curve Fitting Toolbox.

Or if you don't want to generate code, rather than manually creating the anonymous function yourself you can use the fitresult output argument (created by the fit function) from the local functions directly in the anonymous function you pass to the zero finder. Let's make two fits to the census data included in MATLAB:

load census
fitobj1 = fit(cdate, pop, 'poly1')
fitobj1 = 
     Linear model Poly1:
     fitobj1(x) = p1*x + p2
     Coefficients (with 95% confidence bounds):
       p1 =       1.216  (1.045, 1.387)
       p2 =       -2212  (-2535, -1889)
fitobj2 = fit(cdate, pop, 'poly2')
fitobj2 = 
     Linear model Poly2:
     fitobj2(x) = p1*x^2 + p2*x + p3
     Coefficients (with 95% confidence bounds):
       p1 =    0.006541  (0.006124, 0.006958)
       p2 =      -23.51  (-25.09, -21.93)
       p3 =   2.113e+04  (1.964e+04, 2.262e+04)

We can see from a plot of the two fits that they do cross.

plot(fitobj1)

hold on

plot(fitobj2)

●

Let's computing one of the crossing points using a starting guess of 1900. Here I'm directly evaluating the two fits inside crossingFcn.

crossingFcn = @(x) fitobj1(x)-fitobj2(x);
format longg
crossingValue = fzero(crossingFcn, 1900)
crossingValue = 
          1950.55300708195

Let's show the crossing point fzero found for this starting point. I made the circle a bit bigger and gave it a descriptive name for the legend.

figure

plot(fitobj1);

hold on

plot(fitobj2);

plot(crossingValue, fitobj1(crossingValue), 'ko', ...

MarkerSize = 10, DisplayName = "Crossing point")

A different starting point could give the other crossing.

crossingValue = fzero(crossingFcn, 1800)
crossingValue = 
          1829.44699291805

Star Strider 2025-2-27

在 MATLAB Online 中打开

If I understand correcttly what you want to do, it is relatively straightforward to get the function intersections.

This gives the single intersection at (1.8292,2.7132) —

funcVL = @ (x) -3.948*exp(-0.658*x)+3.898;

funcVR = @ (x) 3.856*exp(-0.2436*x)+0.2436;

xv = linspace(-2E+1, 1.5E+1, 1E3);

yL = funcVL(xv);

yR = funcVR(xv);

ixx = find(diff(sign(yL-yR)));

idxrng = max(1,ixx-1) : min(numel(xv),ixx+1);

xisx = interp1(yR(idxrng)-yL(idxrng), xv(idxrng), 0)

xisx = 1.8292

yisx = interp1(xv(idxrng), yR(idxrng), xisx)

yisx = 2.7132

ydif = yR - yL;

funcD = @(x) funcVR(x)-funcVL(x);

[xmin,fv] = fminsearch(@(x)norm(funcD(x)), randn)

xmin = 1.8292

fv = 4.5517e-05

[xmin,fv] = fzero(@(x)funcD(x), randn)

xmin = 1.8292

fv = 4.4409e-16

figure

plot(xv, yR, DisplayName='funcVR')

hold on

plot(xv, yL, DisplayName='funcVL')

% plot(xv, ydif, DisplayName='funcVR-funcVL', LineWidth=3)

plot(xisx, yisx, 'rs', DisplayName='Intersection')

hold off

grid

legend(Location='best')

xlim([-5 10])

ylim([-1 5])

text(xisx,yisx,sprintf(' \\leftarrow (%.6f, %.6f)',xisx,yisx), Horiz='left', Vert='middle')

●

This appears to provide a reasonable answer for the intersection.

.

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Programmatically find fzero of two RCL functions

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Programmatically find fzero of two RCL functions

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