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？？？

" e4 D; `% ?' C& ?4 i( j5 [/ e9 Z% O# [

function Faf = frft(f, a)
) |8 {7 f, D9 M* E) }% The fast Fractional Fourier Transform
% input: f = samples of the signal
% a = fractional power
! L3 l$ E' w6 G, O& }$ y% output: Faf = fast Fractional Fourier transform

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

f = f(;
N = length(f);
shft = 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;
* K' Q) R2 [4 gif (a==2), Faf = flipud(f); return; end;
if (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end
/ t0 e! R' J4 P/ i* m0 Yif (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
: E5 l1 E* v0 Q" e7 Dif (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
! b. M7 }. `1 B5 O" e5 wif (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
  K6 ?9 a+ G) ~alpha = a*pi/2;
$ b9 Z2 a( P7 \. |* ]3 \- Otana2 = tan(alpha/2);
sina = sin(alpha);
9 g5 U$ k8 @3 r0 @/ ?. Xf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
( E) b4 A# c* Mchrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);
f = chrp.*f;

% chirp convolution
! c1 n; v$ D* W+ \3 mc = 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)
0 ^' ?+ c/ N+ f" E" H% sinc interpolation

N = length(x);
& \8 w0 i6 J( e' fy = zeros(2*N-1,1);
y(1:2:2*N-1) = x;
7 O. m  S# @$ ?; f" i) p7 l' mxint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
xint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
: |5 j7 S5 u* ~' ^4 n* f6 lfunction z = fconv(x,y)
% convolution by fft

N = length([x(;y(])-1;
P = 2^nextpow2(N);
5 M. d5 U. }5 D' k: Uz = ifft( fft(x,P) .* fft(y,P));
z = z(1:N);

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