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

function Faf = frft(f, a)
; M/ @$ {, |+ }3 ]5 a$ `% The fast Fractional Fourier Transform
3 o, L* E4 v& n, ?  t9 b6 c% input: f = samples of the signal
3 n' @# Q( f- B: ^; L; X/ w3 K% a = fractional power
6 Z% H5 U' I; a# n; K* l% output: Faf = fast Fractional Fourier transform

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

f = f(;0 P! j* N  S  u! q" W! w
N = length(f);
. I1 K; ?9 ]  v% v+ @shft = rem((0:N-1)+fix(N/2),N)+1;& ^) V8 Q  r  U( F3 k8 B! ]
sN = sqrt(N);; a; g& J* A( T$ _+ Y
a = mod(a,4);

% do special cases
* j3 C* i( s. b6 ^# Z1 }" Gif (a==0), Faf = f; return; end;) J! U2 ~# ^$ C, [2 x  `" `0 ~
if (a==2), Faf = flipud(f); return; end;" t$ H2 c) {4 J2 W
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
7 x. B! B& k3 |, `7 `0 ]: ~if (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5  P9 e7 R  ~- D! r
if (a>2.0), a = a-2; f = flipud(f); end( L5 |  {. I! h$ m* h2 M3 z% M
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end' |$ D4 {1 f1 S: ~! t2 Z
if (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5# }" a; n* x8 O) ^; W. h$ F
alpha = a*pi/2;% U$ N7 N0 x8 H3 f% n2 Z
tana2 = tan(alpha/2);
8 d. `# o/ S( s$ f9 B: m5 lsina = sin(alpha);
8 K9 j$ ~* D0 Y2 t2 L5 k7 {f = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
& n( [) |% Z. `7 @chrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);- g* a) K: _- ~# z+ \3 B% v
f = chrp.*f;

% chirp convolution
1 Y1 t$ V- a) \; f( Ec = pi/N/sina/4;
1 @  B( I% C2 ?9 E7 _Faf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
& g. @: J& E' c; d4 n9 t7 t, yFaf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication- X. `, M" M# L6 F6 s$ T$ Z0 v
Faf = chrp.*Faf;

% normalizing constant
) q, @) D+ X" C) n. W4 \Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%8 T& t: J0 }/ G
function xint=interp(x)4 L- |$ F( w+ R1 N$ [* o! o
% sinc interpolation

N = length(x);/ S6 _$ ?  l! ^. B+ E
y = zeros(2*N-1,1);
. M5 t' d# T: m8 ~( _6 E3 ly(1:2:2*N-1) = x;) E8 R9 {. T4 v/ f- |7 h
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));# }$ M' g' Q& ]' p4 D" t
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%. |- ]. A$ a$ F
function z = fconv(x,y)
6 y: D3 ~$ ?( k% convolution by fft

N = length([x(;y(])-1;7 n* `' D6 q! k
P = 2^nextpow2(N);, h  }; @: \2 J4 k7 A; D
z = ifft( fft(x,P) .* fft(y,P));
6 W& q+ Y) o: W0 Z8 _* Tz = z(1:N);