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

function Faf = frft(f, a)) v7 Z7 c: X+ t' p
% The fast Fractional Fourier Transform
& I6 {* s5 q0 L" b; p! i1 L% input: f = samples of the signal
5 o6 R/ [( l" F! P* R- R% a = fractional power9 p' L+ e0 C1 z1 V5 J
% output: Faf = fast Fractional Fourier transform

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

f = f(;/ w+ }6 s' t7 L, h0 _
N = length(f);; p5 S6 r6 |" H* c% O
shft = rem((0:N-1)+fix(N/2),N)+1;" l3 ^" f2 S) L/ A$ p& \! b
sN = sqrt(N);* |1 u) A/ @4 X( [* c1 P
a = mod(a,4);

% do special cases
* o; V0 v0 v: k8 b6 z1 h" qif (a==0), Faf = f; return; end;
+ }  U  [9 c. v/ [: iif (a==2), Faf = flipud(f); return; end;
7 l5 n$ K4 G+ I% Oif (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
. m' I  j5 z2 X" r, yif (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5, Z2 F% q0 t) u) z
if (a>2.0), a = a-2; f = flipud(f); end& W  M, p5 w* y$ l. R5 z
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
$ {; g/ g+ t' V% d& Kif (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
$ |6 ]2 s6 [0 e, N4 |alpha = a*pi/2;8 u7 B, {+ e8 r% a9 x
tana2 = tan(alpha/2);
# W  w  L- @0 D8 |sina = sin(alpha);- L9 I5 K( ?$ K' v% [6 l5 W; n
f = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication2 }0 ^* |, v' j. Y1 ~  r) T
chrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);' x% f  [, J  A" U" G+ S! j/ \
f = chrp.*f;

% chirp convolution/ E7 X3 @( C( D* x/ ~) E
c = pi/N/sina/4;
2 F) D7 m% S; K; OFaf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);3 ^6 j+ Y& g: O% k; `0 @, }& h: Q
Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication. y4 k+ C5 n3 v6 ?& D; e
Faf = chrp.*Faf;

% normalizing constant* I* |5 w* w. e
Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%
7 Q% Q* u( E( j# L7 ]+ C: ~function xint=interp(x)
3 t3 v  h& w, l$ i% sinc interpolation

N = length(x);
9 f3 c+ K5 _5 Xy = zeros(2*N-1,1);: v7 N, a; ^2 b% n
y(1:2:2*N-1) = x;
' S$ x+ v$ U8 v+ bxint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));; y& u/ s( y; p4 n: R
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
; E+ T* i) S: T5 J  N7 R$ hfunction z = fconv(x,y)
% Z& X5 ^" H9 N2 |/ E9 J3 b% convolution by fft

N = length([x(;y(])-1;2 x( U  n% f4 Y2 c0 k' C* d$ z) }/ Y
P = 2^nextpow2(N);2 |6 @5 f) R3 L+ u
z = ifft( fft(x,P) .* fft(y,P));
$ J/ Y! t5 x. x9 [' B: H8 A; w4 {/ ez = z(1:N);