How to fit a log curve to a scatterplot?

Could anyone advise me on how to a fit a curve to this plot?

The y-axis is decibels and so logarithmic.

I have tried the help page of lsqcurvefit but clearly using the wrong inputs as the curve doesn't align with the points.

x=slopeDist; %18 distances in km
y=PSD; %18 dB values from 50-90
scatter(x,y)

p = polyfit(x,y,1); 
f = polyval(p,x); 
plot(x,y,'o',x,f,'-') 
legend('data','linear fit') 
%fits a line but we want a curve!

%Log curve? Doesn't work...
fun = @(x,xdata)x(1)*exp(x(2)*xdata);
x0 = [100,-1];
x = lsqcurvefit(fun,x0,xdata,ydata)

times = linspace(xdata(1),xdata(end));
plot(xdata,ydata,'ko',times,fun(x,times),'b-')
legend('Data','Fitted exponential')
title('Data and Fitted Curve')

[Walter Roberson]

I tried your data using cftool. All of the models produced fairly bad fits, except for the piecewise interpolations (such a cubic spline), unless you go for something like an 8 term fourier (each term has two coefficients, so 2*8 = 16 coefficients, and you only have 17 datapoints, so it is not surprising you can get something pretty close. But it is also useless, taking wild swings.)

Answer 1

[Star Strider]

If you have the Statistics and Machine Learning Toolbox, use the lsline funciton.

Otherwise use polyfit and polyval.

x = 1:10;
y = 10.^(-x/20) + randn(size(x))/10;

figure
scatter(x, mag2db(y), 'p')
grid
lsline

EDIT — Added plot.

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[Louise Wilson]

Thanks for your help! I am not sure what the purpose of mag2db() is as the values I have are already in dB? This fits a line but I am looking to fit a curve?

[Star Strider]

The mag2db function transforms its argument from linear to decibels.

I have no idea what you are starting with, so I have no idea what sort of fit you want.

Since your data are already in dB, changing the code slightly:

x = linspace(0,6E5,18);                                         % Create Vector
y = 10.^(-x*5E-6) * 10^3.5 + rand(size(x))*100;                 % Original Data
dBy = 20*log10(y);                                              % Convert To dB

figure
plot(x, dBy, 'p')
grid
refline
xlabel('Distance')
ylabel('dB')

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[Louise Wilson]

Thanks, I will run this code again to produce the data and show you the raw data that I am plotting with. For the plot you show, I am wondering if it is possible to fit a curver rather than a straight line. So, the y intercept would be around 70 rather than ~64.

[Star Strider]

The plot image you posted showed a straight line linear regression. The regression fit would have to be changed if the data changed. I cannot anticipate what that would be at this point, however a likely choice would use polyfit and polyval.

I will let you experiment with those.

[Louise Wilson]

Thanks. The reason I have a straight line regression is because that's all I know how to do. The data hasn't changed. I have tried with polyfit too but it produces the same output, a straight line.

p = polyfit(xdata,ydata,1); 
f = polyval(p,xdata); 
plot(xdata,ydata,'o',xdata,f,'-') 
legend('data','linear fit') 

[Star Strider]

This is the best I can do —

LD = load('data.mat');
outsort = sortrows(LD.out);                     % Not Sorting The Rows Creates Problems For The Regression
x = outsort(:,1);
y = outsort(:,2);

fcn = @(b,x) b(1).*exp(b(2).*x) + b(3);
B = fminsearch(@(b)norm(fcn(b,x)-y), [90; 0.001; 60] );
yfit = fcn(B,x);
expstr = @(x) [x(:).*10.^ceil(-log10(abs(x(:)+(x==0)))) floor(log10(abs(x(:)+(x==0))))];

figure
scatter(x, y)
hold on
plot(x, yfit, '-r')
hold off
grid
text(2.5E+5, 77.5, sprintf('$y = %.1f\\ e^{%.1f\\times 10^{%d} \\ x} +%.1f$',B(1),expstr(B(2)),B(3)), 'Interpreter','latex', 'FontSize',15);

I added the equation for the regression. Remove that if it is not necessary. (The ‘expstr’ function generates the matissa and exponent from ‘B(2)’ here. It looks better than displaying all the leading zeros in the regression equation. If you do not want to display the regression equation, the function is not otherwise necessary for the code.)

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