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

function Faf = fRFt(f, a)) O" K* G( n, A1 A2 {7 Y. F
% The fast Fractional Fourier Transform
, M( ]2 _+ j9 s* z, N9 [% input: f = samples of the signal
# J  X: q+ h4 \& C2 ^( a4 v9 v% a = fractional power
0 h  `- O* o0 f$ c0 G" y6 E% output: Faf = fast Fractional Fourier transform

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

f = f(;! x& Y' E  J  _; y+ j) q
N = length(f);
& u5 `8 k1 ~6 p- c: W$ L3 v0 X# Kshft = rem((0:N-1)+fix(N/2),N)+1;
1 a$ E2 u0 y" L# B: v6 csN = sqrt(N);* @/ z* U/ H) S" R, M6 j
a = mod(a,4);

% do special cases+ o' @$ E& U8 I8 g& L  @
if (a==0), Faf = f; return; end;
# ^3 g5 N( q, aif (a==2), Faf = flipud(f); return; end;" J# z8 L% D8 ?* ?. M" @1 E$ ^
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
' p" t/ W. L0 m  Z" K1 S8 e/ Lif (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
, Q( q) W; n3 f; v. H. {if (a>2.0), a = a-2; f = flipud(f); end
% f* ~2 c8 r+ H9 h2 B# |if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end; T/ i4 t+ F, |  e- q1 w4 b
if (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
8 @+ F7 F3 P. j* f  jalpha = a*pi/2;; W2 r/ p2 b& w: }& _
tana2 = tan(alpha/2);8 w& \4 }  G1 |8 x3 t. O
sina = sin(alpha);2 t8 @+ f. T. n0 K
f = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
% `9 q% c6 g6 w2 _$ echrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);* ?1 f" a- u; {" ]1 E
f = chrp.*f;

% chirp convolution5 l, X, E! @/ m8 _' ]
c = pi/N/sina/4;
, Y% }: r* c5 l% \Faf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);4 E' I! o% n* Z: J0 `. C* [
Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication) Z7 J. E% S; l1 `+ {
Faf = chrp.*Faf;

% normalizing constant
4 r( L$ D8 a( Z) L9 d7 c. dFaf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%( b8 k& F0 r/ e* D2 B8 L- k( y8 v
function xint=interp(x)7 p+ h$ D: J! }) d, u, `7 {/ E
% sinc interpolation

N = length(x);
: Z- |1 |- G) v1 T+ |6 Yy = zeros(2*N-1,1);
5 u1 V! ?  D4 E  \! n1 P5 }6 Ty(1:2:2*N-1) = x;
. d8 f- l, s" v6 C# Y% @) N" Fxint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));( X8 I  ?8 P1 G$ ~
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
  `  |6 T0 x0 r0 T) f& a8 Ofunction z = fconv(x,y)
7 D; Q- f& G0 \+ ?2 t0 G  O1 |" L% convolution by fft

N = length([x(;y(])-1;
: }; _1 j% k1 `) e8 d/ P- J! yP = 2^nextpow2(N);8 @& x' L# d' q, ]4 k
z = ifft( fft(x,P) .* fft(y,P));
3 c- Z9 p& V" r% Q# o2 }z = z(1:N);