when running lsqcurvefit to fit the complex function: F = a1*exp(-1i*x*a2) + a3*exp(-1i*x*a4), i get a2 and a4 as complex estimates, however these values should be real. how to solve this?

7 次查看（过去 30 天）
joseph chahin
joseph chahin 2017-12-15
评论： Matt J 2017-12-22
when runing lsqcurvefit to fit the complex function: F = a1*exp(-1i*x*a2) + a3*exp(-1i*x*a4), i get a2 and a4 as complex estimates, however these values should be real. how to solve this?

请先登录，再回答此问题。

回答（4 个）

Walter Roberson
Walter Roberson 2017-12-15
lsqcurvefit defines that the xdata and ydata must be real valued. In your model
F = a1*exp(-1i*x*a2) + a3*exp(-1i*x*a4)
with x and a2 and a4 real-valued, -1i*x*a2 and -1i*x*a4 are guaranteed to be complex valued (or 0), and exp() of a complex value is complex.
If a1 and a2 are real-valued then the two summands would be complex, and it is quite unlikely that their imaginary components will exactly balance out to give a real value -- but ydata must be real.
If a1 and a2 are complex valued then although it is possible for the a1*exp() terms and a3*exp() terms to happen to be real-valued, it is not likely.
If you are counting on complex components happening to cancel out or happening to multiply to give real values, you should probably be using a different formulation of your model.
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2017-12-15
Please describe more clearly what you want to do.
Are you trying to find real-valued a1, a2, a3, a4, with real-valued x and y, such that
a1*exp(-1i*x*a2) + a3*exp(-1i*x*a4) - y
is real-valued ? So, in other words,
imag(a1*exp(-1i*x*a2)) == -imag(a3*exp(-1i*x*a4))
so that the imaginary part exactly cancels to 0?
joseph chahin
joseph chahin 2017-12-16
编辑：Walter Roberson 2017-12-16
I have a complex data y which should exhibit:
y = a1*exp(-1i*xdata*a2) + a3*exp(-1i*xdata*a4)
where a2 and a4 should be real positive values (a1 and a3 can be complex). Now my function is defined as:
F = myfun(a,xdata)
F = a1*exp(-1i*xdata*a2) + a3*exp(-1i*xdata*a4)
when performing:
[p,resnorm] = lsqcurvefit(@myfun,a0,xdata,(ydata), [],[],opts);
i am getting a2 and a4 as complex values whereas these should be real and positive. please let me know if is not clear? thx.

请先登录，再进行评论。

Matt J
Matt J 2017-12-15
编辑：Matt J 2017-12-15
Convert F to an equivalent real-valued function (but with higher dimensional output) as follows,
F_modified=@(a,xdata) [real(F(a,xdata(:))) ; imag(F(a,xdata(:)))];
Similarly, convert your ydata to a real equivalent,
ydata_modified = [real(ydata(:)); imag(ydata(:))];
Pass the modified inputs to lsqcurvefit in place of the original.
  11 个评论
显示 9更早的评论
joseph chahin
joseph chahin 2017-12-22
Dear Matt, the real value should always be possitive. shall I set this in LB,UB parameters or is there any option to include this condition in the function F. Thx. Br.

请先登录，再进行评论。

YT
YT 2017-12-15
编辑：YT 2017-12-15
I haven't used lsqcurvefit myself, but if you want the real value of a complex number, you can use
a2 = real(a2);
a4 = real(a4);
Torsten
Torsten 2017-12-18
Write your function F as
y = (a11+1i*a12)*(cos(a2*xdata)-1i*sin(a2*xdata))+(a31+1i*a32)*(cos(a4*xdata)-1i*sin(a4*xdata)) =
(a11*cos(a2*xdata)+a12*sin(a2*xdata)+a31*cos(a4*xdata)+a32*sin(a4*xdata))+1i*(-a11*sin(a2*xdata)+a12*cos(a2*xdata)-a31*sin(a4*xdata)+a32*cos(a4*xdata))
and fit simultaneously real(ydata) against
a11*cos(a2*xdata)+a12*sin(a2*xdata)+a31*cos(a4*xdata)+a32*sin(a4*xdata)
and imag(ydata) against
-a11*sin(a2*xdata)+a12*cos(a2*xdata)-a31*sin(a4*xdata)+a32*cos(a4*xdata)
Best wishes
Torsten.
  4 个评论
显示 2更早的评论
joseph chahin
joseph chahin 2017-12-20
Thanks. I will give it a try. I see that you suggested lsqnonlin, however i am using lsqcurvefit. could you please tell why? does it matter?
Torsten
Torsten 2017-12-20
编辑：Torsten 2017-12-20
I used lsqnonlin because the handling of xdata and ydata is easier.
To use "lsqcurvefit", you will have to double "xdata" as
Xdata = [xdata,xdata];
,supply the corresponding ydata as
Ydata = [real(ydata),imag(ydata)];
and call "lsqcurvefit" as
x = lsqcurvefit(@fun,x0,Xdata,Ydata);
Further, you will have to change res because the subtraction of rydata and iydata is no longer necessary.
From the numerical point of view, both solvers are equivalent ; only the Interface is a bit different.
Best wishes
Torsten.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by