FRFT程序运行时说Not enough input arguments. 是怎么回事？？？

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如题：FRFT程序运行时说Not enough input arguments. 是怎么回事？？？

function Faf = frft(f, a)
/ Y' {. ?. v& J1 ^3 K1 ^3 R& V% The fast Fractional Fourier Transform
% input: f = samples of the signal
% a = fractional power
" E& ?% [& j/ }8 s% output: Faf = fast Fractional Fourier transform

error(nargchk(2, 2, nargin));

f = f(;
7 R2 z* J# K  s' a) R: R  hN = length(f);
* R9 ]% x. }+ ^- @; bshft = rem((0:N-1)+fix(N/2),N)+1;
sN = sqrt(N);
a = mod(a,4);

% do special cases
& Y: X/ L0 f& }if (a==0), Faf = f; return; end;
: d9 J0 [% R* O' y* Y& p/ t, Bif (a==2), Faf = flipud(f); return; end;
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
if (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
. _; V9 k1 K8 G1 Z5 P" u* {if (a>2.0), a = a-2; f = flipud(f); end
4 x! z/ l1 i8 i, x! L+ Z  L! p& J+ Sif (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
if (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
alpha = a*pi/2;
tana2 = tan(alpha/2);
sina = sin(alpha);
' L* z' H; ~( |$ w3 A$ Kf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
4 p; z8 ?2 g+ u5 O0 D0 r# U0 Kchrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
f = chrp.*f;

% chirp convolution
c = pi/N/sina/4;
Faf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication
Faf = chrp.*Faf;

% normalizing constant
0 w4 F3 e8 _! ]% {$ w- _Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
- y0 N+ Q& [" W% Z( n. m, D0 ~5 [function xint=interp(x)
$ Q* C, d3 h; r- v" f7 F+ \% sinc interpolation

N = length(x);
: o$ N: y9 F1 S- O& `; yy = zeros(2*N-1,1);
9 B$ g8 c6 F# ]& b" \, Q' Z8 ny(1:2:2*N-1) = x;
- r  g! n: ^0 g6 }% Y+ j. Zxint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
- }) p9 ^9 I' `% }xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
5 G0 }0 m- v" y$ H6 {" r8 A  @( S# W3 Afunction z = fconv(x,y)
% convolution by fft

N = length([x(;y(])-1;
P = 2^nextpow2(N);
z = ifft( fft(x,P) .* fft(y,P));
$ \7 @1 i5 T. U; n# cz = z(1:N);

8 B: G6 x6 o" V+ W

xixihahaheihei

好复杂