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

function Faf = frft(f, a)
- R' U0 T. O' I* O# J0 v8 O* T% The fast Fractional Fourier Transform2 A3 x, P7 a7 {2 `$ K9 R
% input: f = samples of the signal9 y- o( y9 i% S( h# n& ^
% a = fractional power
' H* J2 v6 ]* w+ S% output: Faf = fast Fractional Fourier transform

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

f = f(;! n2 C) U' p  o: z3 ~2 D
N = length(f);1 m- g* d7 B7 i4 a0 b0 B% _) d$ W
shft = rem((0:N-1)+fix(N/2),N)+1;# G- x8 R1 C6 M2 I. A* T6 a/ A; H
sN = sqrt(N);
/ |- f1 H5 }  r) Ya = mod(a,4);

% do special cases. d4 a; U- a5 p  {) C  X7 g) n6 h
if (a==0), Faf = f; return; end;
' x3 N, J+ X# A% N; w: b+ Pif (a==2), Faf = flipud(f); return; end;
3 C. Q/ G4 n' `3 i+ p; d! oif (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
% f& E* N, v; W$ |if (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
$ l* j5 {& P: K% u( F0 e/ u2 D# m: Hif (a>2.0), a = a-2; f = flipud(f); end, \/ g& D' Y+ X6 ]8 D2 @
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
( S* P$ l# }8 Q) sif (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5! w0 U- }  z, k$ b& m
alpha = a*pi/2;
' V# G! g7 q. ?0 Ftana2 = tan(alpha/2);
- A9 y7 l: R. Dsina = sin(alpha);
, [/ h! W% G$ |8 D# Cf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
8 Z! l& D2 X$ achrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
! B. p  R+ N9 b+ _% V" ?f = chrp.*f;

% chirp convolution
8 s6 s0 s+ s' t( D) L& ^c = pi/N/sina/4;
; o3 D8 a3 o0 B* AFaf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);
! E% k. V( V# y1 }& _Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication7 ^) k3 ?* O# J7 S% {
Faf = chrp.*Faf;

% normalizing constant
7 x" h: b$ e6 z3 X8 q$ k6 @Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

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4 I; w2 p: K& O7 Yfunction xint=interp(x)
; T: e* }% O- _3 a5 O. Q% sinc interpolation

N = length(x);1 L' A% p5 N+ U
y = zeros(2*N-1,1);
  t: v: t+ P6 T0 m4 Y7 B0 Y- Ly(1:2:2*N-1) = x;
' j  H* F8 O* a) e) axint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
+ \" b: l: l8 m, Pxint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
( f* y$ @$ S: k7 h: L) y: z) Bfunction z = fconv(x,y)
* ~+ U9 G* \+ Y% e$ D% convolution by fft

N = length([x(;y(])-1;1 ^; ]/ Z1 M. U( T1 y
P = 2^nextpow2(N);
2 w' p: X% G5 c: W$ H  d, E. ez = ifft( fft(x,P) .* fft(y,P));, C& y: f1 i" z% e
z = z(1:N);