![Damped harmonic motion curve fit - 2019 10 09.png](https://www.mathworks.com/matlabcentral/answers/uploaded_files/241904/Damped%20harmonic%20motion%20curve%20fit%20-%202019%2010%2009.png)
Damped harmonic motion curve fit
39 次查看(过去 30 天)
显示 更早的评论
Hey,
I have a data set in matlab, when plotted it looks like this:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/241897/image.png)
My goal is to determen a damped sinusoidal equation that would fit this data set, I honestly dont even know how to start. I have included my code, but it isn't much.Any help is much appreciated. Thank you!
0 个评论
采纳的回答
Star Strider
2019-10-10
Try this:
D = load('Stashu Kozlowski DHM.mat'); % File Attached
x = D.x;
y = D.y;
y = detrend(y); % Remove Linear Trend
yu = max(y);
yl = min(y);
yr = (yu-yl); % Range of ‘y’
yz = y-yu+(yr/2);
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
zt = x(zci(y));
per = 2*mean(diff(zt)); % Estimate period
ym = mean(y); % Estimate offset
fit = @(b,x) b(1) .* exp(b(2).*x) .* (sin(2*pi*x./b(3) + 2*pi/b(4))) + b(5); % Objective Function to fit
fcn = @(b) norm(fit(b,x) - y); % Least-Squares cost function
[s,nmrs] = fminsearch(fcn, [yr; -10; per; -1; ym]) % Minimise Least-Squares
xp = linspace(min(x),max(x), 500);
figure
plot(x,y,'b', 'LineWidth',1.5)
hold on
plot(xp,fit(s,xp), '--r')
hold off
grid
xlabel('Time')
ylabel('Amplitude')
legend('Original Data', 'Fitted Curve')
text(0.3*max(xlim),0.7*min(ylim), sprintf('$y = %.3f\\cdot e^{%.0f\\cdot x}\\cdot sin(2\\pi\\cdot x\\cdot %.0f%.3f)$', [s(1:2); 1./s(3:4)]), 'Interpreter','latex')
The estimated parameters are:
s =
-1.398211481931498e+00
-6.142349926864338e+02
2.591368008158479e-04
-5.442228857001487e+00
-3.075267405594925e-15
and the fit is nearly perfect:
![Damped harmonic motion curve fit - 2019 10 09.png](https://www.mathworks.com/matlabcentral/answers/uploaded_files/241904/Damped%20harmonic%20motion%20curve%20fit%20-%202019%2010%2009.png)
更多回答(1 个)
Alex Sha
2019-10-10
The fellow results are little better.
Parameter Best Estimate
---------- -------------
b1 -1.36099782974822
b2 -599.110824641553
b3 0.000259106153388041
b4 1.22915310606227
b5 0.0138196119722517
0 个评论
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!