1
$\begingroup$

There is a complex serie: $f(t_n)=\alpha_n+\beta_n i$, for $n = 1,...,N$,$t_n,\alpha_n$ and $\beta_n$ are known.When we have know that $f(t)$ has the following form: $$f(t)=Ae^{-iBt}$$ with unknown amplitude $A$ and unknown phase $B$, how to estimate the parameters $A$ and $B$ by using a numerical optimization method?

$\endgroup$
4

2 Answers 2

1
$\begingroup$

As written by Martín-Blas Pérez Pinilla, let us suppose that you want to find the optimum values of parameters $A$ and $B$ which minimize the objective function $$\Phi(A,B)=\sum _{n=1}^N (\alpha_n-A\cos (Bt_n))^2+(\beta_n+A\sin (Bt_n))^2=\sum _{n=1}^N r_n$$ Now, since you want the objective function to be minimum, write its derivatives with respect to $A$ and $B$ and set them equal to zero. This will then correspond to $$\sum _{n=1}^N \frac{dr_n}{dA}=0$$ $$\sum _{n=1}^N \frac{dr_n}{dB}=0$$ This corresponds respectively to $$\sum _{n=1}^N [A-\alpha _n \cos (B t_n)+\beta_n \sin (B t_n)]=0$$ $$\sum _{n=1}^N [A t_n (\alpha_n \sin (B t_n)+\beta_n \cos (B t_n))]=0$$ What is nice is that the first equation allows to explicit $A$ as a function of $B$; so, only the second equation is left and you can solve it using Newton method provided that you have a reasonable guess (notice than $A$ disappears from the second equation).

As written by Martín-Blas Pérez Pinilla, you could start your iterations computing the average value of the $$B_n= -\frac1{t_n}\arctan\frac{\alpha_n}{\beta_n}$$ over the entire data set.

$\endgroup$
9
  • $\begingroup$ Thanks for your inspiring answer, but I still did not get the right $B$. Here is my procedure: as you said, I should find the zero point of $\sum_{n=1}^N {dr_n\over dA}=0$, that is $\sum_{n=1}^N {[At_n(\alpha_nsin(Bt_n)+\beta_ncos(Bt_n))]}=0$. I should cancel $A$ from the left side, because it is nonzero, then I get $$\sum_{n=1}^N {t_n(\alpha_nsin(Bt_n)+\beta_ncos(Bt_n))}=0$$ Let $$f(B)=\sum_{n=1}^N {t_n(\alpha_nsin(Bt_n)+\beta_ncos(Bt_n))}$$, its derivative with respect to B is $f'(B)=\sum_{n=1}^N {t_n^2(-\alpha_ncos(Bt_n)+\beta_nsin(Bt_n))}$. I use $B_{k+1}=B_k-f'(B_k)$ to estimate B. $\endgroup$
    – Wang yan
    Apr 11, 2014 at 5:03
  • $\begingroup$ Your derivative is wrong (change the sign). And you iterative scheme must be $$B_{k+1}=B_k-f(B_k) / f'(B_k)$$ PLease let me know. $\endgroup$ Apr 11, 2014 at 7:12
  • $\begingroup$ Haha, it is a clerical mistake about the sign. On the other hand, $-f(B_k)/f'(B_k)$ and $-f'(B_k)$ have a same direction, but different step length. Actually, I use this $-f''^{-1}(B_k)f'(B_k)$ to be the compensation at every step. But still doesn't work. $\endgroup$
    – Wang yan
    Apr 11, 2014 at 7:33
  • $\begingroup$ Since you use Newton, you must use what I wrote. If I may suggest, plot $f(B)$ as a function of $B$ and see where it cancels. If you want, post the data or send them to me by e-mail. $\endgroup$ Apr 11, 2014 at 7:36
  • $\begingroup$ I use Newton to estimate both $A$ and $B$ simultaneously, so the direction $p$ is $$-[{{\partial \Phi}\over{\partial A}}\quad{{\partial \Phi}\over{\partial B}}]'$$ and step length is $$[{{\partial^2 \Phi}\over{\partial A^2}}\quad{{\partial^2 \Phi}\over{\partial A\partial B}};{{\partial^2 \Phi}\over{\partial A\partial B}}\quad{{\partial^2 \Phi}\over{\partial B^2}}]$$ $\endgroup$
    – Wang yan
    Apr 11, 2014 at 7:52
0
$\begingroup$

$$f(t)=Ae^{−iBt}=A(\cos(Bt)-i\sin(Bt))$$ $$\alpha_n+\beta_n i=f(t_n)=A(\cos(Bt_n)-i\sin(Bt_n))$$ $$A=|\alpha_n+\beta_n i|=\sqrt{\alpha_n^2+\beta_n^2}$$ $$B= -\frac1{t_n}\arctan\frac{\alpha_n}{\beta_n}$$ $$\cdots$$

$\endgroup$
3
  • $\begingroup$ Thanks for your answer, but I mean how to estimate them by numerical optimization method, like Newton's method. $\endgroup$
    – Wang yan
    Apr 10, 2014 at 7:51
  • $\begingroup$ Then, try a best fit, en.wikipedia.org/wiki/Curve_fitting, minimizing $\sum((\alpha_n-A\cos Bt_n)^2+(\beta_n+A\sin Bt_n)^2)$. $\endgroup$ Apr 10, 2014 at 8:35
  • $\begingroup$ This is what I am trying to do, but I found that it is difficult to estimate $B$ by using Newton's method. $\endgroup$
    – Wang yan
    Apr 10, 2014 at 9:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .