FFT error 'not supported to carry out script fft as a function'

Question

柊介小山内 2022-12-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1873607-fft-error-not-supported-to-carry-out-script-fft-as-a-function

评论：柊介小山内 2022-12-13

I want to plot a graph as below. so I wrote a program using fft. but error message 'not supported to carry out script fft as a function' displayed. What should I do?

syms t f
T=5.0*10^(-10);
roll = 0.3;%roll-off β
A = pi*t/T;
x(t)= sin(A)/A*cos(roll*A)/(1-(2*roll*t/T)^2);
ht=matlabFunction(x(t));
y=fft(x(t));
X = f;
Y = y;
plot(X,Y);

formula of x(t),X(f) and graph I want to plot(green line) are shown as follows

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Paul 2022-12-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1873607-fft-error-not-supported-to-carry-out-script-fft-as-a-function#answer_1124402

在 MATLAB Online 中打开

Hi 柊介小山内,

fourier can return a closed form expression with a little help.

syms t w f real
T = sym(5.0)*10^(-10);
roll = sym(0.3);%roll-off β
A = sym(pi)*t/T;
x(t) = sin(A)/A*cos(roll*A)/(1-(2*roll*t/T)^2);

rewrite x(t) in terms of expoentials before taking the Fourier transform.

X(w) = simplify(fourier(rewrite(x(t),'exp'),t,w))

X(w) =

Convert to Thz

syms fThz

X(fThz) = X(2*sym(pi)*(fThz*1e12))

X(fThz) =

The plot doesn't look like yours, actually it looks like one cycle of yours. However, I also don't see how the the plots you've posted for X(f) match the equation you've posted for X(f)

xfunc = matlabFunction(X(fThz)/T);

figure

plot(-0.01:.00001:0.01,abs(xfunc(-0.01:.00001:0.01)))

9 个评论
显示 7更早的评论隐藏 7更早的评论

Paul 2022-12-10

编辑：Paul 2022-12-10

在 MATLAB Online 中打开

Define everything fully parameterically and use the actual notation in the problem statement.

syms t w f T beta real
%T = sym(5.0)*10^(-10);
%beta = sym(0.3);%roll-off β
%A = sym(pi)*t/T;
%x(t) = sin(A)/A*cos(roll*A)/(1-(2*roll*t/T)^2);
Pi = sym(pi);
x(t) = sin(Pi*t/T)/(Pi*t/T)*cos(Pi*beta*t/T)/(1 - (2*beta*t/T)^2);
X(w) = simplify(fourier(rewrite(x(t),'exp'),t,w));
X(f) = X(2*sym(pi)*f);

Verify that imag(X(f)) = 0

simplify(imag(rewrite(X(f),'sincos')))
ans = 0

So X(f) is just the real part:

X(f) = simplify(real(rewrite(X(f),'sincos')),100);

Compare the Matlab solution from fourier() with the given, closed-form solution.

Xclosed(f) = piecewise( ...

0 <= abs(f) < (1-beta)/2/T, T, ...

(1-beta)/2/T <= abs(f) < (1+beta)/2/T, T/2*(1 + cos(Pi*T/beta*(abs(f)-(1-beta)/2/T))), ...

0);

figure;

fplot(subs([X(f) Xclosed(f)],[T beta],[1 0.3]),[-1 1])

Looks good.

Convert to THz and sub in the actual values for T and beta

syms fTHz
Tval = 5e-10;
X(fTHz) = subs(X(fTHz*1e12),[T beta],[Tval 0.3]);

Sum up three shifted copies of X(fTHz)

xfunc = matlabFunction(X(fTHz)/Tval);

fval = -0.01:.00001:0.01;

figure

plot(fval,(xfunc(fval) + xfunc(fval-0.003) + xfunc(fval+0.003)))

Or, more compactly

figure

plot(fval,(sum(xfunc(fval+[0 0.003 -0.003].'))))

I have no idea if this approach is going to yield the correct curve in a mathematical sense, because I have no idea what that green curve really is. At a minimum, I do see that the y-axis is labeled PSD, which suggests some sort of squaring operation. Hopefully this Answer has provided a framework for you to get what you need, whatever that may be.

柊介小山内 2022-12-13

编辑：柊介小山内 2022-12-13

在 MATLAB Online 中打开

I don't know whether this is useful for you, my program I wrote part of it published here.

syms t f fTHz w
T=1/(5.0*10^10);
roll = 0.3;%roll-off β
a=(1-roll)/(2*T)
A = pi*t/T;
%x(t)= sin(A)/A*cos(roll*A)/(1-(2*roll*t/T)^2);
nch = 11;%number of mountain part
m =T/2*(1+cos(pi*T/roll))*(f-a);
A=zeros(size(nch));
for n =  -round(nch/2):round(nch/2)
    f= -a+50*10^9*n:1.0*10^9:a+50*10^9*n　
   x(f)=piecewise(-a+50*10^9*n<=f<=a+50*10^9*n,T,a+50*10^9*n<=f<=(1+roll)/(2*T)+50*10^9*n,m);
   A(n)=x(f);
end