代码搜索：FFt 有哪些应用？
找到约 10,000 项符合「FFt 有哪些应用？」的源代码
代码结果 10,000
www.eeworm.com/read/427917/6843426
m qiuzhi.m
function [A,P,A1,P1,A0,P0,y]=qiuzhi(S,N,num,den)
y=filter(num,den,S);
%y=y+normrnd(0,0.06,1,N+1);
ham=hamming(N+1);%加窗函数
%S=S.*ham';
%y=y.*ham';
Y = fft(S,N); %做FFT
Y0=fft(y,N);
A1 =abs(Y(1:N/
www.eeworm.com/read/395847/8150402
txt readme.txt
;*************************************************************************************
;*********************** SECTION 1: FFT LIBRARY ***********************************
;*************************
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asm c2cx0064.asm
.file "c2cx0064.asm"
.title "0064 point DIT Radix-2, Complex FFT"
.width 120
N .set 64 ; NUMBER OF POINTS FOR FFT
***************************
www.eeworm.com/read/394773/8207279
asm c2cx0016.asm
.file "c2cx0016.asm"
.title "0016 point DIT Radix-2, Complex FFT"
.width 120
N .set 16 ; NUMBER OF POINTS FOR FFT
***************************
www.eeworm.com/read/394773/8207303
asm c2cx0032.asm
.file "c2cx0032.asm"
.title "0032 point DIT Radix-2, Complex FFT"
.width 120
N .set 32 ; NUMBER OF POINTS FOR FFT
***************************
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asm c2cx0128.asm
.file "c2cx0128.asm"
.title "0128 point DIT Radix-2, Complex FFT"
.width 120
N .set 128 ; NUMBER OF POINTS FOR FFT
***************************
www.eeworm.com/read/294636/8214634
m convfft.m
function [out] = convfft(z1,z2)
%CONVFFT FFT-based convolution and polynomial multiplication.
% C = CONVFFT(A, B) convolves vectors A and B. The resulting
% vector is length LENGTH(A)+LENGTH
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m exa090300_2.m
%-----------------------------------------------------------------------------
% exa090300_2.m, for fig. 9.3.2,
%-----------------------------------------------------------------------------
clea
www.eeworm.com/read/392767/8327269
c intfft_brin.c
//-----------------------------------------------------------------------------
// IntFFT_BRIN.c
//-----------------------------------------------------------------------------
// Copyright 2003 Cy
www.eeworm.com/read/292487/8352356
m answer1.m
%正弦干扰陷波LMS算法
clear all
%正弦信号产生 噪声s 正弦信号v
%输入信号x=v+s
[s,pv]=rands(8,pi/3);
v=randv(pv);
x=v+s;
%步长u，期望d
N=10;
d(1)=0;
for i=1:99
d(i+1)=x(i);
end %d要对x进行移位，第一个值补0
u=0.00028;%u应再