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

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柊介 小山内
柊介 小山内 2022-12-8
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

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采纳的回答

Paul
Paul 2022-12-9
Ran in:
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)))
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柊介 小山内
柊介 小山内 2022-12-13
编辑：柊介 小山内 2022-12-13
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

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更多回答（1 个）

Walter Roberson
Walter Roberson 2022-12-8
you named your file fft.m which makes it impossible to call the Mathworks fft function. You need to rename your fft.m
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柊介 小山内
柊介 小山内 2022-12-9
sorry I forgot to add sample. If graph plot carried out correctly, X(f) is ploted as below.(green line)
You don't worry about amplitude. Shape of graph is most important. This graph shows how much larger is the GNLI(blue line) than X(f).
Walter Roberson
Walter Roberson 2022-12-9
编辑：Walter Roberson 2022-12-9
Ran in:
syms t f
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);
y(f)=fourier(x(t))
y(f) = 
char(x)
ans = '-(sin(2000000000*t*pi)*cos(600000000*t*pi))/(2000000000*t*pi*(1440000000000000000*t^2 - 1))'
Notice that the result has unevaluated calls to fourier(). That means that fourier() was unable to compute the fourier transform of that function.
I checked on Wolfram Alpha, which was able to come up with a transformation... but MATLAB is not able to do so.
https://www.wolframalpha.com/input?i=fourier+-%28sin%282000000000*t*pi%29*cos%28600000000*t*pi%29%29%2F%282000000000*t*pi*%281440000000000000000*t%5E2+-+1%29%29

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