error message: integral calculation

Question

柊介小山内 2022-11-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1863598-error-message-integral-calculation

评论：柊介小山内 2022-12-2

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

I want to calculate a formula as followed.

So I wrote a program as below. But in this program, error message" integralCalc/finalInputChecks (line 522). input function must be type of single or double. function hundle found." was displayed.

I tried to use fun1*fun2 in spite of fun, but error messages"function mtimes(input parameter of type 'function_handle' not defined.)" and integralCalc/finalInputChecks (line 314) displayed.

What should I do?

a = 0.2;      %field loss coefficient α
y = 1.3;       % fiber non-linearity coefficient　γ
Ns = 20;        %number of span
Ls = 100;       %span length         
b2 = 20.7;     %dispersion coefficient
%% formula of　ρ
PI = pi;
fun1 = @(z)exp(2*z-2*a*z);
Le = (abs((integral(fun1,0,100))))^2 ;%square of span effective length
fun = @(z,f,f1,f2)exp(1j*4*PI^2*(f1-f)*(f2-f)*b2*z)*exp(2*z-2*a*z);%fun1*fun2
%fun2 = @(f,f1,f2)exp(1j*4*PI^2*(f1-f)*(f2-f)*b2*z); %second element of integration
p = (abs(integral(@(z) fun,0,Ls)))^2/Le%ρ

Thank you.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Torsten 2022-11-26

编辑：Torsten 2022-11-26

So you set g = 1 and beta3 = 0 ?

You didn't give values to f, f1 and f2. If you use "integral", you must give numerical values to all the constants involved except z .

Or do you aim at a symbolic solution ? Then use "int".

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2022-11-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1863598-error-message-integral-calculation#answer_1112613

在 MATLAB Online 中打开

p = (abs(integral(@(z) fun,0,Ls)))^2/Le%ρ

@(z)fun is an anonymous function that ignores its input argument and returns the anonymous function handle fun. It does not execute fun on the value of z. Just pass fun at that point instead of @(z)fun

18 个评论
显示 16更早的评论隐藏 16更早的评论

Walter Roberson 2022-11-27

The fun that you pass to integral() must be a handle to a function that accepts one parameter.

If the "arrayvalued" option is not used, then the function must return single precision or double precision results exactly the same size as its input (which will not be a fixed size), and for each input value, the result must be the same as-if you had invoked the function on that scalar (so be sure to use element-by-element operations rather than matrix operations.)

If the "arrayvalued" option is used, then the function will be passed a scalar value each time, and must return a single precision or double precision array that is the same size each time.

The function must be deterministic -- no random numbers.

All of this does not mean that the calculation cannot involve symbolic variables -- but the calculation must return single or double precision. For example it would be legal for the calculation to do double(vpasolve(Expression,x))

f1,f2,f being symbolic parameters is a problem if the evaluation of the function cannot get rid of f1, f2, f by the time of return.

If you are trying to do a symbolic integration that returns in terms of f, f1, f2, then you cannot use integral and must instead use int

柊介小山内 2022-11-28

在 MATLAB Online 中打开

thank you for teaching. I can calclate ρ, but I can't integrate ρ respect to f1,f2.

alfa = 0.2;      %field loss coefficient α
gamma = 1.3;       % fiber non-linearity coefficient　γ
Ns = 20;        %number of span
Ls = 100;       %span length         
beta2 = 20.7;     %dispersion coefficient
%% formula of　ρ
PI = pi;
syms z f f1 f2
assume(z>0);Le = (abs((int(exp(-4*alfa*z^2),z,0,Ls))))^2;
assume(f,'real');   
assume(f1,'real');  
assume(f2,'real'); 
Le = (abs((int(exp(-4*alfa*z^2),z,0,Ls))))^2;
p = (abs(int(exp(1j*4*pi^2*(f1-f)*(f2-f)*beta2*z)*exp(2*z*(1-alfa)),z,0,Ls)))^2/Le
%% formula of GNLI 
pint1 = int(p,f1,-0.5,0.5);%integration about f1
pint2 = int(pint1,f2,-0.5,0.5);%integration about f2

error message didn't displayed, but calclation didn't finish.

calclation still continue time goes by 5 minutes.

Is this a problem of my PC(windows10) spec? or something wrong with my code?

notice:

ρ=

(16*abs(a*i/(b*f*f1 + b*f*f2 - b*f1*f2 - b*f^2 + ci) - (exp((f^2*d*i)/g)*exp(-(f*f1*d*i)/g)*exp(-(f*f2*d*i)/g)*exp((f1*f2*d*i)/g)*exp(160)*h*i)/(a*f*f1 + a*f*f2 -a*f1*f2 - a*f^2 + k*i))^2)/(5*pi*erf(40*5^(1/2))^2)

alphabet a～h and k indicate very large constant integer.

I want to calclate formula as below,so I have to integrate ρ respect to f1,f2.( assume all Gwdm=1)

thank you.

Torsten 2022-11-30

编辑：Torsten 2022-11-30

在 MATLAB Online 中打开

Are you sure about exp(2*z*(1-alfa)) in the function p ? Shouldn't it be exp(-2*z*(1-alfa)) ? Or exp(-2*alfa*z) as in your graphics ?

alfa = 0.2;      %field loss coefficient α
gamma = 1.3;       % fiber non-linearity coefficient　γ
Ns = 20;        %number of span
Ls = 100;       %span length         
beta2 = 20.7;     %dispersion coefficient
%% formula of　ρ
PI = pi;
syms z 
syms f f1 f2 real
assume(z>0);
Le = int(exp(-4*alfa*z^2),z,0,Ls);
p = int(exp(1j*4*pi^2*(f1-f)*(f2-f)*beta2*z)*exp(2*z*(1-alfa)),z,0,Ls);
p = real(p)^2+imag(p)^2;
p = p/Le^2;
ht = matlabFunction(p)
ht = function_handle with value:
    @(f,f1,f2)(1.0./erf(sqrt(5.0).*4.0e+1).^2.*((imag((exp(f.^2.*8.172032444101988e+4i).*exp(f.*f1.*-8.172032444101988e+4i).*exp(f.*f2.*-8.172032444101988e+4i).*exp(f1.*f2.*8.172032444101988e+4i).*exp(1.6e+2).*7.0368744177664e+14i)./(f.*f1.*5.750556604705831e+17+f.*f2.*5.750556604705831e+17-f1.*f2.*5.750556604705831e+17-f.^2.*5.750556604705831e+17+1.125899906842624e+15i))-((f.*f1.*5.750556604705831e+17+f.*f2.*5.750556604705831e+17-f1.*f2.*5.750556604705831e+17-f.^2.*5.750556604705831e+17).*7.0368744177664e+14)./((f.*f1.*5.750556604705831e+17+f.*f2.*5.750556604705831e+17-f1.*f2.*5.750556604705831e+17-f.^2.*5.750556604705831e+17).^2+1.267650600228229e+30)).^2+(real((exp(f.^2.*8.172032444101988e+4i).*exp(f.*f1.*-8.172032444101988e+4i).*exp(f.*f2.*-8.172032444101988e+4i).*exp(f1.*f2.*8.172032444101988e+4i).*exp(1.6e+2).*7.0368744177664e+14i)./(f.*f1.*5.750556604705831e+17+f.*f2.*5.750556604705831e+17-f1.*f2.*5.750556604705831e+17-f.^2.*5.750556604705831e+17+1.125899906842624e+15i))-7.922816251426434e+29./((f.*f1.*5.750556604705831e+17+f.*f2.*5.750556604705831e+17-f1.*f2.*5.750556604705831e+17-f.^2.*5.750556604705831e+17).^2+1.267650600228229e+30)).^2).*(1.6e+1./5.0))./pi
pint = @(f)integral2(@(f1,f2)ht(f,f1,f2),-0.5,0.5,-0.5,0.5);
NLI = @(f)(16/27)*gamma^2*double(Le)*pint(f) % NLI of 1span
NLI = function_handle with value:
    @(f)(16/27)*gamma^2*double(Le)*pint(f)
NLI(0.1)
ans = 2.2036e+137

Walter Roberson 2022-11-30

在 MATLAB Online 中打开

The below code takes several hours to run, mostly the last step. The function handle it produces starts with

@(f)exp(-3.96e+2./5.0).*integral(@(f2)integral(@(f1)exp(3.96e+2./5.0).*1.804851387845415e-35.*(

with the nested calls to integral(). It contains quite a number of very large coefficients beyond 1e250.

The portion I looked at did not have any calls to real() or imag() -- but I can only see a small portion because it is such a long expression.

If you do take this approach I recommend that you use the 'file' option for matlabFunction, and that you specifically ask for optimization to be false when you use the 'file' option (optimization has been producing incorrect results in recent MATLAB releases; I do not know if it is fixed in R2022b.)

This all gives you a NLI as a symbolic expression in f, and gives you NLIF as function handle to process values of f numerically. It is not a closed-form formula: you are unlikely to get a closed form formula unless you do something like taylor series (keeping in mind that taylr series of an exponential is pretty inaccurate unless you restrict the range of inputs a fair bit.)

alfa = 0.2;      %field loss coefficient α
gamma = 1.3;       % fiber non-linearity coefficient　γ
Ns = 20;        %number of span
Ls = 100;       %span length         
beta2 = 20.7;     %dispersion coefficient
PI = sym(pi);
syms z f f1 f2
assume(z>0);
Le = (abs((int(exp(2*alfa-2*alfa*z),z,0,Ls))))^2;
assume(f,'real');   
assume(f1,'real');  
assume(f2,'real'); 
Le = ((int(exp(2*alfa-2*alfa*z),z,0,Ls)))^2;
p = (int(exp(1j*4*pi^2*(f1-f)*(f2-f)*beta2*z)*exp(2*z*(1-alfa)),z,0,Ls))^2;
p = real(p)^2*imag(p)^2;
p = expand(p)/Le;
%% formula of GNLI
%%p = matlabFunction(p);
pint1 = int(p,f1,-0.5,0.5); %integration about f1
pint2 = int(pint1,f2,-0.5,0.5); %integration about f2
NLI = (16/27)*gamma^2*Le*pint2; % NLI of 1span
NLIF = matlabFunction(NLI);