curve fitting for the optimal value of a constant
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Hey here, I am trying to find the value of k(unknown constant) and for that I have collected data points from a curve and then fit into other(as attached). I had found the result i.e., 0.1300 but it isn't matching with the result which is 1.
So, can anybody tell me if there is any mistake.
Thankyou
r=0:1*10^-9:50*10^-9;
l=50*10^-9;
e=1.6*10^-19;
V=((1.44).*exp(3.257*((r./l).^2)).*besselk(0,(3.257*((r./l).^2))))./(e.^2);
R=r/l;
plot(R,V)
h=findobj(gca, 'Type', 'line');
R=h.XData
V=h.YData
k0 = rand()*10;
R=[0.0200,0.0400,0.0600,0.0800,0.1000,0.1200,0.1400,0.1600,0.1800,0.2000,0.2200,0.2400,0.2600,0.2800,0.3000,0.3200,0.3400,0.3600,0.3800,0.4000,0.4200,0.4400,0.4600,0.4800,0.5000,0.5200,0.5400,0.5600,0.5800,0.6000,0.6200,0.6400,0.6600,0.6800,0.7000,0.7200,0.7400,0.7600,0.7800,0.8000,0.8200,0.8400,0.8600,0.8800,0.9000,0.9200,0.9400,0.9600,0.9800,1.0000];
V=1.0e+38*[3.8070,3.0381,2.5965,2.2900,2.0580,1.8734,1.7216,1.5937,1.4841,1.3888,1.3051,1.2309,1.1646,1.1050,1.0510,1.0020,0.9572,0.9161,0.8783,0.8433,0.8110,0.7810,0.7530,0.7269,0.7025,0.6796,0.6581,0.6379,0.6189,0.6009,0.5839,0.5678,0.5525,0.5380,0.5243,0.5111,0.4987,0.4868,0.4754,0.4645,0.4541,0.4442,0.4346,0.4255,0.4167,0.4083,0.4002,0.3924,0.3849,0.3777];
bestk = lsqcurvefit( @(k,R)(1./((sqrt((k.^2)+(R.^2))).*(e.^2))), k0, R, V)
0 Kommentare
Antworten (0)
Siehe auch
Kategorien
Mehr zu Nonlinear Optimization finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!