# Tag Info

13

The FFT can only be performed over a limited chunk of data. The basic math is based on the assumption that the time domain signal is periodic, i.e. your chunk of data is repeated in time. That typically results in a major discontinuity at the edges of the chunk. Let's look at a quick example: FFT size = 1000 points, Sample Rate = 1000 Hz, Frequency ...

12

In reviewing Fred Harris Figures of Merit for various windows (Table 1 in this link) the Hamming is compared to the Hanning (Hann) at various values of $\alpha$ and from that it is clear that the Hann would provide greater stopband rejection (The classic Hann is with $\alpha =2$ and from the table the side-lobe fall-off is -18 dB per octave). I provided the ...

10

I'll use the shorthand window for "window function". With audio, any processing that creates something akin to pre-ringing or pre-echo will sound sloshy like a low bit-rate mp3. This happens when localized energy of a transient or an impulse is spread backwards in time, for example by modification of the spectral data in lapped transforms such as lapped ...

9

Aside from reduction of spectral leakage, there is a one major trade-off to be made when choosing a window function. Below you can see a figure with various parameters. Two of them are most important: Main Lobe Width Peak Side Lobe Level Width of the main lobe affects resolution of your analysis. I am sure you know that multiplication in time domain is ...

9

The cross pattern is typically a border effect, due to the periodicity induced by the standard implementation and hypotheses behind the Fast Fourier transform, when the image lacks periodicity from the right to the left, and the bottom to the top. In other words: if two opposite borders lacks continuity in values (when glued together), artifacts show. The ...

8

Windowing reduces spectral leakage. Say you start out with a $\sin(y) = \cos(\omega_0 t)$. The period is obviously $2 \pi/ \omega_0$. But if nobody told you that the period is $2 \pi/ \omega$ and you blindly choose the range $[0, 1.8 \pi/\omega_0]$ and take FFT of this truncated waveform, you will observe frequency components in other frequencies ...

8

Since the DFT is representable by multiplication with the Fourier matrix, your question is equivalent to asking what the eigenvectors of the Fourier matrix are. Actually, Wikipedia provides the answer (http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors). However, since the eigenvalues ($1, -1, i, -i$) are not simple, the ...

8

If done correctly, the order should not matter. You apply windowing to remove the jumps at the end of the signal segment by fading the values there to zero. Zero-padding then continues with zeros. Irregardless if padded or not, the windowing should only be applied to the original signal segment. The wrong way would be to zero-pad and then apply windowing to ...

8

We always want to apply some kind of a window function in order to minimize the effect of leakage. This makes rectangular window (lack of any windowing) case never used, this is why: Any tapering function used is almost always decreasing to zero at boundaries. This is why we are losing some data. In order to retrieve that somehow you will usually do 50% of ...

8

Is this meant to happen? Yes, that is absolutely a well known effect of using any window function. Taking a look through the Wikipedia article on window functions, we find that the rectangular window has the sharpest peak. This is quite simply due to the fact that the rectangular window acquires the most data, meaning that it can distinguish between ...

8

The concept of spectral leakage is not dependent on the context, and it's the same thing for the DTFT as it is for the DFT. It's probably helpful to remember that the DFT of a finite length sequence equals a sampled version of the DTFT of the same sequence: \begin{align}X(\omega)&=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\qquad\text{(DTFT)}\\ \tilde{X}[k]&... 8 Why is each window/frame overlapping? Windowing is a means to stationarize signals. Inside a small enough window, you can expect that the properties of the signal chunk do not vary too fast. And you now can use tools well-suited to stationary signals, like Fourier-based techniques. You can imagine non-overlapping rectangular windows, each defining a frame.... 8 Parseval's theorem will hold, but take into account that your signal in the time domain will no longer be x[n]. Namely, if you have that\sum_{n=0}^{N-1} \Big| x[n] \Big|^2 = \frac{1}{N} \sum_{k=0}^{N-1} \Big| X[k] \Big|^2$$then, if you window the signal x[n] with a window w[n], your signal will now be \hat{x}[n]=x[n]w[n], and the theorem will ... 7 Please refer to this paper on the optimum overlap percentage for the Blackman-Harris window, which is derived to be 66.1%. It has a lot of useful information on spectral analysis and windows. 7 Yes, it's possible to analyse sound the way ears do. For example, you could compute the DFT of a signal continuously using several Goertzel filters.$$ y_k[n] = e^{j2\pi k/N} y_k[n-1] + x[n] $$where k= 0,1,\ldots, N-1, so that y_0 is the DC or zero frequency term. Of course, this is an unstable filter, so some resetting or forgetting factor is ... 7 UPDATE: My previous response did not answer the OP's question. The following addresses the question directly: Bottom line: Prior to windowing in time, each sample in frequency is an IID Gaussian random variable since the Fourier Transform of an AWGN waveform in time results in an identically distributed waveform in frequency (Gaussian distributed and to be ... 6 zero-padding will not "improve resolution" in the sense that it would help you separate two peaks in the spectrum that are very close (too close) to each other. but zero-padding, as well as interpolation in the frequency domain, does improve resolution in the sense that it help you to more precisely estimate the location of an isolated peak in the spectrum. ... 6 More overlap means you end up with more windows (of a given length) per second of audio. More windows (of a given length) requires more FFTs which requires more MACs or FLOPs which generally requires more processing power. In return, more window overlap provides greater time locality of information (e.g. on average, random transient events are likely ... 6 just implement the Modified Bessel function. it's easy. i always like my window definitions centered about zero, since pretty much all of them are even symmetry. i'll do this in discrete-time, but it's essentially the same thing in continuous-time. Kaiser window:$$ w[n] \triangleq \begin{cases} \frac{1}{I_0(\beta)} I_0\left(\beta \sqrt{1 - \left(\frac{...

5

This is a rather general answer : I would say that, although each window function has specific characteristics that make it more or less suitable for specific situations, in this context you're not going to see a huge difference in the STFT of a windowed speech signal. In particular, the two windows that you're comparing are quite similar, so I'd expect ...

5

You don't HAVE to zero pad, ever, unless there are constraints on the length of data required by the FFT algorithm - often it's a case of speed. Many FFT's (like the Numerical Recipes code) require a power-of-two length series, some (like Kiss-FFT) only require it to be an even number, but being a multiple of 2,3 and 5 makes it run much faster. In Matlab'...

5

Filters and windows complement each other, in a way. This is due to the convolutional lemma of the fourier transform. Let me expand on that. A filter would be a convolution of a signal with another. In a technical sense, a window has a fixed (supposedly short) size and is seen as "cutting out" a short part of the signal. Often, a DFT would follow. In the ...

5

If you follow the reference link no. 43 from Wikipedia, then you will end up on this website of Stanford University. They are providing all necessary theory behind DPSS window, together with this MATLAB function (not to mention, that MATLAB already has the dpss function) : function [w,A,V] = dpssw(M,Wc); % DPSSW - Compute Digital Prolate Spheroidal ...

5

There are only heuristic formulas for estimating the filter order. For a Kaiser window (which is probably the most frequently used window for filter design) the required filter order can be estimated from [1] $$M=\frac{A-8}{2.285\,\Delta\omega}\tag{1}$$ where $A=-20\log\delta$ ($\delta$ is the maximum deviation from the desired response), and $\Delta\omega$...

5

Knowing the energies of $x_1$ and $x_2$ is not sufficient for determining the energy of $x_3=x_1x_2$. What you can do is determine an upper bound for the energy of $x_3$ given the energies of $x_1$ and $x_2$ and their maximum values: $$E_3=\sum_{k}\big|x_1[k]x_2[k]\big|^2\le\begin{cases}\max_k\big|x_1[k]\big|^2\sum_k\big|x_2[k]\big|^2=\max_k\big|x_1[k]\big|^... 5 You could try the exponential window:$$w_n=\frac{ \exp \left[\alpha \sqrt{1-\left(\frac{n-M}{M}\right)^2}\right]}{\exp(\alpha)}\alpha=-427.5*10^{-6} A_s^2+0.1808*A_s-3.516$$or the hyperbolic cosine window:$$w_n=\frac{\cosh \left[ \alpha \sqrt{1-\left(\frac{n-M}{M}\right)^2} \right]}{\cosh(\alpha)}\alpha=-325.1*10^{-6}*A_s^2+0.1677*A_s-3.149$$... 5 I'd like to apply zero padding to it, for better frequency bin resolution. First of all, let's state it one more time that zero padding does not improve frequency resolution of DFT. It'll only interpolate the existing spectrum on a finer frequency grid, but will not add any new information to it, otherwise. In order to improve the true frequency resolution ... 5 While the Fourier transform, discrete or continuous, can be regarded as unitary transform i.e a naturally norm preserving change between orthonormal bases in a normed complex vector space, the windowed FT does not in general possess this quality. And non-unitary operators cannot be turned into unitary ones by re-scaling. Here is why: Unitary operators can ... 5 In this document you can find the coefficients of a seven-term Blackman Harris window. Ignoring the bizarre notation, it seems like the window is defined by$$w[n]=\sum_{k=0}^6a_k\cos\left(\frac{2\pi kn}{N}\right),\quad n=0,1,\ldots,N-1\tag{1}$$where N is the window length and the coefficients a_k are given by$$\begin{align} a_0&=0....

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