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)
; @) S# l9 R7 E# t2 W" c4 F% The fast Fractional Fourier Transform
) _7 V* N" p5 c: o7 i. L% input: f = samples of the signal
8 e1 |* D) w: _8 i! j: s2 i, A% a = fractional power
% output: Faf = fast Fractional Fourier transform

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

f = f(;
4 l; W2 R7 D* l! [N = length(f);
shft = rem((0:N-1)+fix(N/2),N)+1;
sN = sqrt(N);
a = mod(a,4);

% do special cases
if (a==0), Faf = f; return; end;
; C5 q) Z; `5 {9 j0 o5 Dif (a==2), Faf = flipud(f); return; end;
8 @9 F  @( J; ?+ f! sif (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
; p8 w# v0 g* P3 l9 t% ~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
! T9 b8 o3 o) s! v8 d( uif (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;
( l/ O# f1 B& S" o. J4 u. Jtana2 = tan(alpha/2);
sina = sin(alpha);
f = [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
c = pi/N/sina/4;
. ]3 a# E  L* L2 L+ |2 W: pFaf = 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
3 w' b) {, U. d8 C: x/ Z5 ]$ YFaf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
function xint=interp(x)
% sinc interpolation

N = length(x);
y = zeros(2*N-1,1);
y(1:2:2*N-1) = x;
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
  f% S' |  a/ F4 p9 Gxint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
function z = fconv(x,y)
+ J8 Y  C, A+ r% convolution by fft

N = length([x(;y(])-1;
P = 2^nextpow2(N);
z = ifft( fft(x,P) .* fft(y,P));
" l8 J5 b- m( Xz = z(1:N);

5 y; }0 ]# n2 F$ T, f3 @3 k, Z, y

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好复杂