通过傅里叶空间填充进行插值

Question

我最近尝试在matlab上实现一个在傅里叶域中使用zéro填充的插值方法的简单示例。 但是我无法正常工作，我总是有一个很小的频移，在傅立叶空间几乎看不到，但是它会在时空中产生巨大的误差。

由于在傅里叶空间中的zéro填充似乎是一种常见（且快速）的插值方法，我认为有些东西是我遗漏的：

这是matlab代码：

clc;
clear all;
close all;


Fe = 3250;
Te = 1/Fe;
Nech = 100;

F1 = 500;
F2 = 1000;
FMax = 1500;

time = [0:Te:(Nech-1)*Te];
timeDiscrete = [1:1:Nech];
frequency = (timeDiscrete/Nech)*Fe;

signal = cos(2*pi*F1*(time))+cos(2*pi*F2*(time))+cos(2*pi*FMax*(time));

%Compute the FFT
spectrum=zeros(1,Nech);
for k = timeDiscrete
    for l = timeDiscrete
        spectrum(k) = spectrum(k) + signal(l)*exp(-2*pi*j*l*k/Nech);
    end
end

%Compute de inverse FFT
reconstruction=zeros(1,Nech);
for k = timeDiscrete
    for l = timeDiscrete
        reconstruction(k) = reconstruction(k) + spectrum(l)*exp(2*pi*j*l*k/Nech);
    end
end
reconstruction=reconstruction/Nech;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%    Now interpolation will take place   %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Finterp = 6*Fe;
Tinterp = 1/Finterp;
TimeInterp = [0:Tinterp:(Nech-1)*Te];
[m,n] = size(TimeInterp);
NechInterp = n;
TimeInterpDiscrete = [1:1:NechInterp];

%Compute original signal value without any interpolation
signalResampled = cos(2*pi*F1*(TimeInterp))+cos(2*pi*F2*(TimeInterp))+cos(2*pi*FMax*(TimeInterp));

%Compute original signal interpolation by padding the fft and performing
%inverse fft on the result
semipaddedsize=floor(NechInterp/2);
padded_spectrum0 = zeros(1,semipaddedsize);
padded_spectrum0 = padarray(spectrum(1:Nech/2),[0 semipaddedsize-(Nech/2)],0,'post');
padded_spectrum = zeros(1,NechInterp);
padded_spectrum(1:semipaddedsize) = padded_spectrum0;
padded_spectrum(semipaddedsize+1:NechInterp-1) = conj(fliplr(padded_spectrum0));
% padded_spectrum = padarray(spectrum,[0 NechInterp-Nech],0,'post');
padded_timeDiscrete = [1:1:NechInterp];
padded_reconstruction = zeros(1,NechInterp);


for k = padded_timeDiscrete
    for l = padded_timeDiscrete
        padded_reconstruction(k) = padded_reconstruction(k) + padded_spectrum(l)*exp(2*pi*j*l*k/NechInterp);
    end
end
padded_reconstruction=padded_reconstruction/(1*Nech);


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%       Let's print out the result       %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

spectrumresampled=zeros(1,NechInterp);
for k = TimeInterpDiscrete
    for l = TimeInterpDiscrete
        spectrumresampled(k) = spectrumresampled(k) + signalResampled(l)*exp(-2*pi*j*l*k/NechInterp);
    end
end

figure(2);
plot(abs(spectrumresampled)/6,'g');
hold on;
plot(abs(padded_spectrum),'b');

figure(3);

% Ground truth : deterministic signal is recomputed
plot(TimeInterp,signalResampled,'g');
hold on;
% linear interpolation between subsampled points (matlab tracing tool)
plot(time,(reconstruction),'c');
hold on;
% Padding the spectrum and reconstructing it 
plot(TimeInterp,real(padded_reconstruction),'m');
hold on;

xlabel('Time in s','FontSize',16)
ylabel('Signal value (no unit)','FontSize',16)
title('\it{ Various signal reconstruction from fourier transform }','FontSize',16)
legend('True signal', 'Reconstruction with linear interpolation', 'Reconstruction with padded spectrum');

由于我的声誉，我无法发布结果图像，但是，图表很容易通过matlab生成。 我非常感谢对此代码或零填充fft的评论，以便进行插值。

提前谢谢

Answer 1

非常感谢你们这些建议，我决定回答我自己的问题，因为没有足够的空间可用于实验室：

@Try hard我确实定义了一个错误的离散时间向量，因为@Frederick也指出，我在填充向量时遇到了问题，谢谢你给我正确的“matlab方式”去做，我不应该有如此害怕fftshift / ifftshift，此外，如同@Frederick所提到的那样，使用带有'both'的padarray也可以完成这项工作。

折叠for循环也是正确执行matlab的重要步骤，我不会在训练目的中使用它来简化我的理解和绑定检查。

另外一个非常有趣的观点@Try在第一句中很难提及，而且我首先没有意识到，事实上，零填充只相当于通过sinc函数在时域中卷积我的数据

实际上我认为它相当于具有别名sinc函数的卷积，也称为dirichlet内核，当采样频率增加到无穷大时，它限制了经典的sinc函数（见http://www.dsprelated.com/dspbooks/sasp/Rectangular_Window_I.html#sec:rectwinintro）

我没有在这里发布整个代码，但我原来的程序的目的是比较dirichlet内核卷积公式，我在不同的框架中演示（使用傅里叶级数离散表达式的理论演示），sinc卷积{{3}和零填充，所以我应该得到一个非常相似的结果。

对于变迹问题，我认为真正的答案是，如果你的信号是有限的，你不需要其他变迹功能而不是矩形窗口。

如果你的信号没有带宽限制，或者有关采样率的别名，你需要减少光谱中混淆部分的贡献，这是通过频域滤波器=频域中的变迹窗口滤除它们来完成的。在时域中变成特定的插值核心。

Answer 2

您在时域中观察到的结果是由于sinc函数与原始数据的卷积而振铃。这是在频域中乘以矩形窗口的时域中的等价物，这实际上是零填充时的行为。别忘了apodize！

我在折叠循环之后重新发布代码（这会显着加速计算），重新定义时间和频率变量的范围（请参阅DFT的定义以了解原因），并删除其中一个填充操作你表演我坦率地不理解这一点。

clc;
clear all;
close all;

Fe = 3250;
Te = 1/Fe;
Nech = 100;
mlt = 10;
F1 = 50; 
F2 = 100; 
FMax = 150; 

time = [0:Te:(Nech-1)*Te];
%timeDiscrete = [1:1:Nech];
timeDiscrete = [0:1:Nech-1];
frequency = (timeDiscrete/Nech)*Fe;

signal = cos(2*pi*F1*(time))+cos(2*pi*F2*(time))+cos(2*pi*FMax*(time));
spectrum =  signal*exp(-2*pi*j*timeDiscrete'*timeDiscrete/Nech);
fspec = [0:Nech-1]*Fe/Nech;
reconstruction = spectrum*exp(2*pi*j*timeDiscrete'*timeDiscrete/Nech)/Nech;

figure
plot(time,signal)
hold on
plot(time,reconstruction,'g:')

% **** interpolation ****

Finterp = 6*Fe;
Tinterp = 1/Finterp;
TimeInterp = [0:Tinterp:(Nech-1)*Te];
NechInterp = length(TimeInterp);
%TimeInterpDiscrete = [1:NechInterp];
TimeInterpDiscrete = [0:NechInterp-1];

%Compute original signal value without any interpolation
signalResampled = cos(2*pi*F1*(TimeInterp))+cos(2*pi*F2*(TimeInterp))+cos(2*pi*FMax*(TimeInterp));

%Compute original signal interpolation by padding the fft and performing
%inverse fft on the result
padded_spectrum0 = spectrum;
padded_spectrum0(NechInterp) = 0;
fspecPadded = [0:NechInterp-1]*Finterp/NechInterp;

padded_reconstruction = padded_spectrum0*exp(2*pi*j*TimeInterpDiscrete'*TimeInterpDiscrete/NechInterp)/(1*Nech);
spectrumresampled = signalResampled*exp(-2*pi*j*TimeInterpDiscrete'*TimeInterpDiscrete/NechInterp);
fresampled = [0:NechInterp-1]*Fe/NechInterp;

% **** print out ****

figure(2);
hold on;
plot(fspec,abs(spectrum),'c');
plot(fresampled,abs(spectrumresampled)/6,'g--');
plot(fspecPadded,abs(padded_spectrum0),'m:');
xlabel('Frequency in Hz','FontSize',16)
ylabel('Signal value (no unit)','FontSize',16)
legend('True signal', 'Reconstruction with resampled spectrum', 'Padded spectrum');

figure(3);
% Ground truth : deterministic signal is recomputed
plot(TimeInterp,signalResampled,'g');
hold on;
% Padding the spectrum and reconstructing it 
plot(TimeInterp,real(padded_reconstruction),'m:');
xlabel('Time in s','FontSize',16)
ylabel('Signal value (no unit)','FontSize',16)
title('\it{ Various signal reconstruction from fourier transform }','FontSize',16)
legend('True signal', 'Reconstruction with padded spectrum');

这是由于频域填充导致的可怕失真信号的图像：

通过首先应用fftshift使光谱居中并在中心光谱的交替边上填充，然后反转fftshift操作，可以实现一些改进：

Nz = NechInterp-Nech;
padded_spectrum0 = ifftshift([ zeros(1,floor(Nz/2)) fftshift(spectrum) zeros(1,floor(Nz/2)+rem(Nz,2)) ]); % replaces (NechInterp) = 0;
fspecPadded = [0:NechInterp-1]*Finterp/NechInterp;

然后你得到这个更好的内插时域信号，因为填充操作不会导致频谱中的这种突然下降（但是进一步修改可能仍然可以进行一些改进）：

Answer 3

行。一个问题是你为padded_reconstruction做IDFT的方式。你定义TimeInterp和NechInterp的方式使得复指数的元素不正确。这说明了不正确的频率。

然后有一个问题，即在傅立叶域（pt 50）中包括两次中点。它几乎为零，所以它没有出现一个非常明显的问题，但它应该只被包含一次。我重写了这一部分，因为我很难按照你的方式完成它。我保持非常接近。如果我这样做，我会使用fftshift然后使用padarray（......，'both'），这将节省必须将零置于中间的工作。如果你这样做是为了获得学习经验并且试图不使用matlab工具（例如fftshift），那就永远不要了。

我也重新定义了你定义时间的方式，但为了公平起见，我认为它可以按你自己的方式工作。

我已用％＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;

指示更改

Fe = 3250;
Te = 1/Fe;
Nech = 100;

F1 = 500;
F2 = 1000;
FMax = 1500;

time = [Te:Te:(Nech)*Te];  %<<<<<<<<<<
timeDiscrete = [1:1:Nech];
frequency = (timeDiscrete/Nech)*Fe;

signal = cos(2*pi*F1*(time))+cos(2*pi*F2*(time))+cos(2*pi*FMax*(time));

%Compute the FFT
spectrum=zeros(1,Nech);
for k = timeDiscrete
    for l = timeDiscrete
        spectrum(k) = spectrum(k) + signal(l)*exp(-2*pi*j*l*k/Nech);
    end
end

%Compute de inverse FFT
reconstruction=zeros(1,Nech);
for k = timeDiscrete
    for l = timeDiscrete
        reconstruction(k) = reconstruction(k) + spectrum(l)*exp(2*pi*j*l*k/Nech);
    end
end
reconstruction=reconstruction/Nech;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%    Now interpolation will take place   %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Finterp = 6*Fe;
Tinterp = 1/Finterp;
TimeInterp = [Tinterp:Tinterp:(Nech*6)*Tinterp];  %<<<<<<<<<<
[m,n] = size(TimeInterp);
NechInterp = n;
TimeInterpDiscrete = [1:1:NechInterp];

%Compute original signal value without any interpolation
signalResampled = cos(2*pi*F1*(TimeInterp))+cos(2*pi*F2*(TimeInterp))+cos(2*pi*FMax*(TimeInterp));

%Compute original signal interpolation by padding the fft and performing
%inverse fft on the result
padded_spectrum = zeros(1,NechInterp); %<<<<<<<<<<
padded_spectrum(1:floor(Nech/2-1)) = spectrum(1:floor(Nech/2-1)); %<<<<<<<<<<
padded_spectrum(end-floor(Nech/2)+1:end) = spectrum(floor(Nech/2)+1:end); %<<<<<<<<<<

padded_reconstruction = zeros(1,NechInterp);
for k = TimeInterpDiscrete %<<<<<<<<<<(no reason for new variable)
    for l = TimeInterpDiscrete %<<<<<<<<<<(no reason for new variable)
        padded_reconstruction(k) = padded_reconstruction(k) + padded_spectrum(l)*exp(2*pi*j*l*k/NechInterp);
    end
end
padded_reconstruction=padded_reconstruction/(1*Nech);