blocks|key|5376155|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5376156|因此，我假设一个矩形窗口已经应用于我的数据。对吗？|blockquote|5376157|5376158|5376159|在某种程度上，窗口通常用于滤除由于信号的突然开/关状态引起的高频失真，或者减少或重新排序频谱泄漏(https://en.wikipedia.org/wiki/Spectral_leakage)|offset|length|5376160|如果要可视化fft，建议应用窗口，尤其是(非矩形)窗口。有关选项，请参见https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window。|5376161|请注意，您要在fft之前应用窗口。|5376162|5376163|还是我应该拆分它们？|5376164|5376165|5376166|嗯，这取决于你的要求。但一般来说，由于窗口的原因，采样时间越长，FFT在这段时间内的精度就越高，尽管这种技术并不少见，但这是最好的。|5376167|5376168|那么我应该得到哪种分辨率？|5376169|5376170|5376171|分辨率是采样率除以样本计数。|5376172|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Spectral_leakage|1|https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window^0|0|0|0|0|1D|1A|0|0|10|1Z|1|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1B|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1C|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1D|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|1E|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|1F|7|@]|8|@$H|1G|I|1H|1|1I]]|9|$]]|$1|J|3|K|4|5|6|1J|7|@]|8|@$H|1K|I|1L|1|1M]]|9|$]]|$1|L|3|M|4|5|6|1N|7|@]|8|@]|9|$]]|$1|N|3|-4|4|5|6|1O|7|@]|8|@]|9|$]]|$1|O|3|P|4|C|6|1P|7|@]|8|@]|9|$]]|$1|Q|3|-4|4|5|6|1Q|7|@]|8|@]|9|$]]|$1|R|3|-4|4|5|6|1R|7|@]|8|@]|9|$]]|$1|S|3|T|4|5|6|1S|7|@]|8|@]|9|$]]|$1|U|3|-4|4|5|6|1T|7|@]|8|@]|9|$]]|$1|V|3|W|4|C|6|1U|7|@]|8|@]|9|$]]|$1|X|3|-4|4|5|6|1V|7|@]|8|@]|9|$]]|$1|Y|3|-4|4|5|6|1W|7|@]|8|@]|9|$]]|$1|Z|3|10|4|5|6|1X|7|@]|8|@]|9|$]]|$1|11|3|-4|4|5|6|1Y|7|@]|8|@]|9|$]]]|12|$13|$4|14|15|16|9|$17|18]]|19|$4|14|15|16|9|$17|1A]]]]

<blockquote>
 Thus I am assuming that a rectangular window is already applied to my data. Is that correct?
</blockquote>

In a way, a window is commonly used to filter out high frequency distortion due to the sudden on/off state of the signal, or reduce or reorder spectral leakage (<a href="https://en.wikipedia.org/wiki/Spectral_leakage" rel="nofollow">https://en.wikipedia.org/wiki/Spectral_leakage</a>) 

It is recommended to apply a window, especially (non-rectangular) if you want to visualize the fft. See <a href="https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window" rel="nofollow">https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window</a> for options.

Be aware that you apply the window before the fft.

<blockquote>
 or should I split them up?
</blockquote>

Well, that depends on your requirements. But in general, its better not to, due to the windowing, the longer the sample, the more accurate the FFT for that period of time, although those kind of techniques are not uncommon to speed things up.

<blockquote>
 which resolution do I get then?
</blockquote>

The resolution is the sample rate divided by the sample count.

blocks|key|1218376|text|假设您的数据不是平稳的(换句话说，频谱内容是随时间变化的，例如语音或音乐就是这种情况)，那么您通常会希望选择一个窗口大小，在此期间数据可以被认为是某种平稳的。对于语音和音乐，典型的窗口大小可以是20ms的数量级。对于441FFT的采样率，这相当于882个样本，因此kHz大小为1024时可能是一个很好的起点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1218377|重叠连续的窗口也是很常见的，以获得更好的信号时变分量的时间分辨率。通常使用50%25的重叠，因此第一个样本块将是0..1023，第二个块将是512..1535，依此类推。|1218378|正如@Stefan的答案中已经建议的那样，在FFT之前，您应该对每个样本块应用合适的窗口函数。常用的窗口是Hamming和von+Hann+(也称为Hanning)。显然，窗函数需要与FFT的大小相同(例如，N=+1024)。|1218379|对于数据末尾的任何剩余大小小于N的样本块，只需用零填充即可。|1218380|上述操作的常用术语是生成spectrogram。它本质上是时间、频率、幅度/相位的3D数据结构，可以bd以各种不同的方式显示或用于进一步的频域处理。|offset|length|1218381|另请参阅这些密切相关的StackOverflow问题和答案：|1218382|1218383|Using+Apple's+Accelerate+framework,+FFT,+Hann+windowing+and+Overlapping|unordered-list-item|1218384|Giving+large+no.+of+samples+to+KissFFT.|1218385|Accelerate+framework+vDSP,+FFT+framing|1218386|1218387|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Spectrogram|1|https://stackoverflow.com/a/18932052/253056|2|https://stackoverflow.com/questions/30815295/giving-large-no-of-samples-to-kissfft/30821124#30821124|3|https://stackoverflow.com/a/19832881/253056^0|0|0|0|0|C|B|0|0|0|0|0|1Z|1|0|0|12|2|0|0|12|3|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1C|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1D|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@$J|1F|K|1G|1|1H]]|A|$]]|$1|L|3|M|5|6|7|1I|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1K|8|@]|9|@$J|1L|K|1M|1|1N]]|A|$]]|$1|R|3|S|5|Q|7|1O|8|@]|9|@$J|1P|K|1Q|1|1R]]|A|$]]|$1|T|3|U|5|Q|7|1S|8|@]|9|@$J|1T|K|1U|1|1V]]|A|$]]|$1|V|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]|16|$5|Z|10|11|A|$12|17]]|18|$5|Z|10|11|A|$12|19]]]]

Assuming your data is not stationary (in other words the spectral content is time-varying, as would be the case for e.g. speech or music), then you would typically want to pick a window size during which the data can be considered to be somewhat stationary. For speech and music a typical window size might be of the order of 20 ms. For a sample rate of 44.1 kHz this correspond to 882 samples, so an FFT size of 1024 might be a good starting point. 

It's also common to overlap successive windows, to get better time resolution for the time-varying components of your signal. A 50% overlap is commonly used, so your first block of samples would be 0..1023, the second block would be 512..1535, etc.

As has already been suggested in @Stefan's answer, you should apply a suitable window function to each block of samples, prior to the FFT. Commonly used windows are Hamming and von Hann (aka Hanning). Obviously the window function needs to be the same size as the FFT (e.g. N = 1024).

For any remaining block of samples of size &lt; N at the end of your data you can just pad with zeroes.

The commonly used term for the above operation is generating a <a href="https://en.wikipedia.org/wiki/Spectrogram" rel="nofollow noreferrer">spectrogram</a>. It's essentially a 3D data structure of time v frequency v magnitude/phase, which can bd displayed in various different ways or used for further frequency-domain processing.

See also these closely related StackOverflow questions and answers:

<ul>
<li><a href="https://stackoverflow.com/a/18932052/253056">Using Apple's Accelerate framework, FFT, Hann windowing and Overlapping</a></li>
<li><a href="https://stackoverflow.com/questions/30815295/giving-large-no-of-samples-to-kissfft/30821124#30821124">Giving large no. of samples to KissFFT</a>.</li>
<li><a href="https://stackoverflow.com/a/19832881/253056">Accelerate framework vDSP, FFT framing</a></li>
</ul>

I want to analyze a signal I get from my microphone port by using <code>portaudio</code> and <code>fftwpp</code>. For that I followed the explanation provided <a href="https://stackoverflow.com/a/3059084/2546099">here</a>. My questions concerning that are now: 
There it is stated that I should chunk a window out of the incoming data. My data is already chunked, after I am only recording for a short time, and afterwards process it. Thus I am assuming that a rectangular window is already applied to my data. Is that correct? 
Now I am getting 200k data points, should I directly put them into an array:

<pre><code> Array::array1&lt;Complex&gt; F(np,align);
 Array::array1&lt;double&gt; f(n,align); // For out-of-place transforms
 // array1&lt;double&gt; f(2*np,(double *) F()); // For in-place transforms

 fftwpp::rcfft1d Forward(n,f,F);
 fftwpp::crfft1d Backward(n,F,f);
 qDebug() &lt;&lt; "Putting " &lt;&lt; numSamples &lt;&lt; " into an array!";
 for(int i = 0; i &lt; numSamples; i++)
 f[i] = this-&gt;data.recordedSamples[i];
</code></pre>

or should I split them up? If I all put them in one array, which resolution do I get then? My sample rate is set to 44.1 kHz.

Further understanding of fftw processing of portaudio signals

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想使用portaudio和fftwpp分析从麦克风端口获得的信号。为此，我遵循了提供的解释。我现在的问题是：这里声明我应该从传入的数据中分出一个窗口。我的数据已经被分块了，在我只记录了很短的一段时间后，然后对其进行处理。因此，我假设一个矩形窗口已经应用于我的数据。对吗？现在我得到了200k个数据点，我是否应该直接将它...

问对端口音频信号fftw处理的再认识
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问对端口音频信号fftw处理的再认识EN