使用FFT和傅里叶级数展开FOR循环,我可以增加和减小信号的频率。但是,如果信号/数组太大,执行速度会非常慢(一个1x44100的数组需要约2分钟才能完成)。我确定这与for循环有关,但我不知道如何矢量化它以改善性能。请注意,这将用于长度为3到6分钟的音频信号。1x44100的数组仅为一秒钟,需要约2分钟才能完成。
请问有什么建议吗?
%create signal
clear all, clc,clf,tic
x= linspace(0,2*pi,44100)';
%Used in exporting to ycalc audio file make sure in sync with above
freq_orig=1;
freq_new=4
vertoff=0;
vertoffConj=0;
vertoffInv=0;
vertoffInvConj=0;
phaseshift=(0)*pi/180 ; %can use mod to limit to 180 degrees
y=sin(freq_orig*(x));
[size_r,size_c]=size(y);
N=size_r; %to test make 50
T=2*pi;
dt=T/N;
t=linspace(0,T-dt,N)';
phase = 0;
f0 = 1/T; % Exactly, one period
y=(y/max(abs(y))*.8)/2; %make the max amplitude here
C = fft(y)/N; % No semicolon to display output
A = real(C);
B = imag(C)*-1; %I needed to multiply by -1 to get the correct sign
% Single-Sided (f >= 0)
An = [A(1); 2*A(2:round(N/2)); A(round(N/2)+1)];
Bn = [B(1); 2*B(2:round(N/2)); B(round(N/2)+1)];
pmax=N/2;
ycalc=zeros(N,1); %preallocating space for ycalc
w=0;
for p=2:pmax
%
%%1 step) re-create signal using equation
ycalc=ycalc+An(p)*cos(freq_new*(p-1).*t-phaseshift)
+Bn(p)*sin(freq_new*(p-1).*t-phaseshift)+(vertoff/pmax);
w=w+(360/(pmax-1)); %used to create phaseshift
phaseshift=w;
end;
fprintf('\n- Completed in %4.4fsec or %4.4fmins\n',toc,toc/60);
subplot(2,1,1), plot(y),title('Orginal Signal');
subplot(2,1,2),plot(ycalc),title('FFT new signal');
如果有人想要查看输出结果,这里有一个图表,可以看到FOR循环确实非常慢。