>> t=[0.5 1 2 3 4 5 7 9];
v1=[6.36 6.48 7.26 8.22 8.66 8.99 9.43 9.63];
y=log(10- v1);
f=polyfit(t,y,1)
t0=-1/f(1)
v0=10-exp(f(2))
v2=10-(10-v0)*exp(-t/t0);
plot(t,v1,'bx',t,v2,'k:')
xlabel('时间t(s)'),
ylabel('充电电压(V)');
title('电容器充电电压与时间t的曲线');
f =-0.2835 1.4766
t0 = 3.5269
v0 = 5.6221
f =
-0.2835 1.4766
t0 =
3.5269
v0 =
5.6221