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？？？
: P3 P: o% B8 ~5 I) r! }4 q

function Faf = frft(f, a)
% The fast Fractional Fourier Transform
% input: f = samples of the signal
% a = fractional power
. B% L5 x4 U2 M2 s% output: Faf = fast Fractional Fourier transform

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

f = f(;
; F0 W) [! G6 i  aN = length(f);
shft = rem((0:N-1)+fix(N/2),N)+1;
sN = sqrt(N);
a = mod(a,4);

% do special cases
5 O1 i3 p0 i/ e# ]9 K: m0 a7 |if (a==0), Faf = f; return; end;
if (a==2), Faf = flipud(f); return; end;
  _1 O! m; D! H; Gif (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
% W, u$ }, A3 t1 {$ b6 w  B8 Hif (a>2.0), a = a-2; f = flipud(f); end
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
8 |9 M$ Z, l& ?& q: V" y) Tif (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);
2 I, I- B+ L( y9 T/ g: K! Nf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
chrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
f = chrp.*f;

% chirp convolution
8 _  ^$ j; I4 uc = pi/N/sina/4;
Faf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
$ @, ?4 h' h8 N6 e; d5 p5 VFaf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication
Faf = chrp.*Faf;

% normalizing constant
8 U/ h3 V; M+ O! u2 K& BFaf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
( t, p2 h5 v' `3 k7 {" Sfunction xint=interp(x)
% sinc interpolation

N = length(x);
* q& Y: b+ Q, ~, }1 n1 {" Z% H8 qy = zeros(2*N-1,1);
% U: |- [0 ~4 s& J6 G$ {; ~+ Iy(1:2:2*N-1) = x;
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
# Q( ~" T9 o- O4 B8 oxint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
function z = fconv(x,y)
7 }1 K& f  Z! U* B% |% convolution by fft

N = length([x(;y(])-1;
! K% w/ d% @! x: A  J1 bP = 2^nextpow2(N);
6 E& w# y' q5 {) f2 y  Cz = ifft( fft(x,P) .* fft(y,P));
6 z' E3 j) L: _! q& e4 U1 kz = z(1:N);

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好复杂