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？？？
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function Faf = frft(f, a)
( s1 q- X# S/ l  K8 I) i1 f* Q, t4 g% The fast Fractional Fourier Transform
% input: f = samples of the signal
3 ?# H3 [  d& J% a = fractional power
# f* A! T! n) E( B7 S: i  {" D) e5 X% output: Faf = fast Fractional Fourier transform

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

f = f(;
- \3 y. ~  }) t  J/ f3 }8 tN = length(f);
shft = rem((0:N-1)+fix(N/2),N)+1;
sN = sqrt(N);
2 h* u# Q6 Y/ m/ Z/ n, F: Q4 r6 Da = mod(a,4);

% do special cases
if (a==0), Faf = f; return; end;
/ C# o3 ]3 M5 N, y( wif (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
if (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
6 p6 B) B  X$ c2 r( Nif (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
- c9 g- j( b3 @5 e( E# @/ balpha = a*pi/2;
9 R+ w. m: P( d' v" W: T5 |tana2 = tan(alpha/2);
/ H* L2 X3 @8 K" M7 K7 i0 }6 l; ksina = sin(alpha);
f = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
4 U/ z5 l, W# ~0 U& |- Ichrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
3 B! v& w, }5 T$ X# p; S5 y' M) {f = chrp.*f;

% chirp convolution
c = pi/N/sina/4;
7 k: _, s/ }; D! i3 F0 b0 WFaf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
' x) Y0 L0 A1 q) R% d  WFaf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication
1 C( ]% I% b8 ~& p3 VFaf = chrp.*Faf;

% normalizing constant
, Q$ w3 _+ u  N  K8 EFaf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
  e! \+ z% F: P5 mfunction xint=interp(x)
# A+ `, t9 t. O8 C% sinc interpolation

N = length(x);
5 h, `- e7 T! Jy = zeros(2*N-1,1);
; g! v3 X$ p- [. Y4 j' K* ry(1:2:2*N-1) = x;
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
function z = fconv(x,y)
% convolution by fft

N = length([x(;y(])-1;
* I! |$ M9 u* Z1 H6 |" S! LP = 2^nextpow2(N);
& @& F, P3 q& h+ Oz = ifft( fft(x,P) .* fft(y,P));
) v$ D; V! K6 p& z7 B9 jz = z(1:N);

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