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？？？

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function Faf = frft(f, a)
% The fast Fractional Fourier Transform
0 B0 j; Y2 k$ ^4 l, ^/ k% input: f = samples of the signal
% a = fractional power
% output: Faf = fast Fractional Fourier transform

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

f = f(;
& j% _7 \1 o; c  I. r" J8 IN = length(f);
! m% m" J1 |8 gshft = 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;
if (a==2), Faf = flipud(f); return; end;
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
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
if (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;
tana2 = tan(alpha/2);
sina = sin(alpha);
9 J2 k9 h% w5 J& Vf = [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);
1 m0 d/ F' }, W0 c" ^7 f) Pf = chrp.*f;

% chirp convolution
c = pi/N/sina/4;
Faf = 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
Faf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

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

N = length(x);
6 N7 M% d7 h6 {, U8 wy = zeros(2*N-1,1);
7 [4 D; Z/ j; P% c4 H7 yy(1:2:2*N-1) = x;
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
function z = fconv(x,y)
% convolution by fft

N = length([x(;y(])-1;
& r) J; h0 b7 D# b3 H/ dP = 2^nextpow2(N);
* r- H1 ^/ _. @- N! dz = ifft( fft(x,P) .* fft(y,P));
z = z(1:N);

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好复杂