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标题: FRFT程序运行时说Not enough input arguments. 是怎么回事？？？ [打印本页]

作者: cpupygu 时间: 2019-8-2 14:14
标题: FRFT程序运行时说Not enough input arguments. 是怎么回事？？？
如题：FRFT程序运行时说Not enough input arguments. 是怎么回事？？？

function Faf = frft(f, a)8 u' V Q0 F$ m) |
% The fast Fractional Fourier Transform4 c2 Z) N- }% S& U
% input: f = samples of the signal6 |$ O' W" [1 q' @ a3 S
% a = fractional power
! I1 ?2 ]: @' P6 i$ ^$ q, P7 g% output: Faf = fast Fractional Fourier transform

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

f = f(;
( O* p. ^1 j2 V2 A, N; ]6 EN = length(f);" N$ w( S) M4 T
shft = rem((0:N-1)+fix(N/2),N)+1;
# e. X" ` B4 B+ z$ IsN = sqrt(N);
* D3 g; X1 t: _1 Z8 [4 u! j3 ja = mod(a,4);

% do special cases3 ]- ]+ F6 d4 S6 Z. j* L
if (a==0), Faf = f; return; end;% R1 s3 T2 u3 ~9 m- E/ f
if (a==2), Faf = flipud(f); return; end;
. B- V) l' T0 Jif (a==1), Faf(shft,1) = fft(f(shft))/sN; return; end & y! N. b" }6 O. _: t, R0 k8 [/ h
if (a==3), Faf(shft,1) = ifft(f(shft))*sN; return; end

% reduce to interval 0.5 < a < 1.5
+ L3 C8 `7 X8 y Q9 m: Y5 {* C- pif (a>2.0), a = a-2; f = flipud(f); end0 S' u- R1 k8 r
if (a>1.5), a = a-1; f(shft,1) = fft(f(shft))/sN; end
9 L8 l. I( q$ v; f, Eif (a<0.5), a = a+1; f(shft,1) = ifft(f(shft))*sN; end

% the general case for 0.5 < a < 1.5
" [% {- w1 _# M% t. ralpha = a*pi/2;
, L, d; S7 y6 v( W$ T" k% \1 i6 b8 c0 xtana2 = tan(alpha/2);$ x9 G, I. S$ E+ c s4 b& T. @) Y, V
sina = sin(alpha);
# j8 T1 H. ^" z: Z8 ?, n( x" tf = [zeros(N-1,1) ; interp(f) ; zeros(N-1,1)];

% chirp premultiplication
) z, E) \, {! x. v r: kchrp = exp(-i*pi/N*tana2/4*(-2*N+2:2*N-2)'.^2);' T7 [0 W& R' H! A5 _7 g5 _
f = chrp.*f;

% chirp convolution) m0 Z I9 R; x$ i5 w- p
c = pi/N/sina/4;
4 e+ t1 D G6 b; G: GFaf = fconv(exp(i*c*(-(4*N-4):4*N-4)'.^2),f);6 j: v6 I' ?1 t* f
Faf = Faf(4*N-3:8*N-7)*sqrt(c/pi);

% chirp post multiplication- |0 L1 q/ W$ {
Faf = chrp.*Faf;

% normalizing constant
5 P% e! A/ y H/ yFaf = exp(-i*(1-a)*pi/4)*Faf(N:2:end-N+1);

%%%%%%%%%%%%%%%%%%%%%%%%%; H1 w' L: Q, {( P, O
function xint=interp(x): W: n7 d% n" s' a3 R/ U$ |
% sinc interpolation

N = length(x);# G$ z% M ^9 h" Z3 W8 S
y = zeros(2*N-1,1);+ x9 Q' W& a, Z
y(1:2:2*N-1) = x;( D( G3 z/ @5 p% H1 t9 q, w
xint = fconv(y(1:2*N-1), sinc([-(2*N-3)2*N-3)]'/2));
9 H u3 }/ X( a' Txint = xint(2*N-2:end-2*N+3);

%%%%%%%%%%%%%%%%%%%%%%%%%
5 _ W. W4 C5 L @7 wfunction z = fconv(x,y)- e4 C3 s0 k2 z" g- e& [* n
% convolution by fft

N = length([x(;y(])-1;
, z3 f. @6 X3 `% d/ k4 y" DP = 2^nextpow2(N);, j. L, u% M* u5 X9 F3 ~# f+ b
z = ifft( fft(x,P) .* fft(y,P));% L4 `8 d4 q4 ?- o* q# V
z = z(1:N);

作者: xixihahaheihei 时间: 2019-8-2 16:49
好复杂

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