I want to analyze some audio and decompose it as best as I can into sine waves. I have never used FFT before and am just doing some initial reading and about the concepts and available libraries, like FFTW and KissFFT.
I'm confused on this point... it sounds like the DFT/FFT will give you the sine amplitudes only at certain frequencies, multiples of a base frequency. For example, if I have audio sampled at the usual 44100 Hz, and I pick a chunk of say 256 samples, then that chuck could fit one cycle of 44100/256=172Hz, and the DFT will give me the sine amplitudes at 172, 172*2, 172*3, etc. Is that correct? How do you then find the strength at other frequencies? I'd like to see a spectrum all the way from 20Hz to about 15Khz, at about 1Hz increments.
Fourier decomposition allows you to take any function of time and describe it as a sum of sine waves each with different amplitudes and frequencies. If however you want to approach this problem using the DFT, you need to make sure you have sufficient resolution in the frequency domain in order to distinguish between different frequencies. Once you have that you can determine which frequencies are dominant in the signal and create a signal consisting of multiples sinewaves corresponding to those frequencies. You are correct in saying that with a sampling frequency of 44.1 kHz, only looking at 256 samples, the lowest frequency you will be able to detect in those 256 samples is a frequency of 172 Hz.
OBTAIN SUFFICIENT RESOLUTION IN THE FREQUENCY DOMAIN:
Amplitude values for frequencies "only at certain frequencies, multiples of a base frequency", is true for Fourier decomposition, NOT the DFT, which will have a frequency resolution of a certain increment. The frequency resolution of the DFT is related to the sampling rate and number of samples of the time-domain signal used to calculate the DFT. Reducing the frequency spacing will give you a better ability to distinguish between two frequencies close together and this can be done in two ways;
Decreasing the sampling rate, but this would move the periodic repetitions in frequency closer together. (Remember NyQuist theorem here)
Increase the number of samples which you use to calculate the DFT. If only the 256 samples are available, one can perform "zero padding" where 0-valued samples are appended to the end of the data, but there are some effects to this which needs to be considered.
HOW TO COME TO A CONCLUSION:
If you depict the frequency content of different audio signals into individual graphs, you will find that the amplitudes differ abit. This is because the individual signals will not be identical in sound, and there is always noise inherent in any signal (from the surroundings and the hardware itself). Therefore, what you want to do is to take the average of two or more DFT signals to remove noise and get a more accurate represention of the frequency content. Depending on your application, this may not be possible if the sound you are capturing is noticably changing rapidly over time (for example speech, or music). Averaging is thus only useful if all the signals to be averaged are pretty much equal in sound (individual seperate recordings of "the same thing"). Just to clarify, from, for example, four time-domain signals, you want to create four frequency domain signals (using a DFT method), and then calculate the average of the four frequency-domain signals into a single averaged frequency-domain signal. This will remove noise and give you a better representation of which frequencies are inherent in your audio.
AN ALTERNATIVE SOLUTION:
If you know that your signal is supposed to contain a certain number of dominant frequencies (not too many) and these are the only ones your are interesting in, then I would recommend that you use Pisarenko's harmonic decomposition (PHD) or Multiple signal classification (MUSIC, nice abbreviation!) to find these frequencies (and their corresponding amplitude values). This is less intensive computationally than the DFT. For example. if you KNOW the signal contains 3 dominant frequencies, Pisarenko will return the frequency values for these three, but keep in mind that the DFT reveals much more information, allowing you come to more conclusions.
Your initial assumption is incorrect. An FFT/DFT will not give you amplitudes only at certain discrete frequencies. Those discrete frequencies are only the centers of bins, each bin constituting a narrow-band filter with a main lobe of non-zero bandwidth, roughly a width or two of the FFT bin separation, depending on the window (rectangular, von Hann, etc.) applied before the FFT. Thus the amplitude of spectral content between bin centers will show up, but spread across multiple FFT result bins.
If the separation of key signals is large enough and the noise level is low enough, then you can interpolate the FFT results to examine frequencies between bin centers. You may need to use a high quality interpolator, such as a Sinc kernel.
If your signal separation is smaller or the noise level is higher, then you may need a longer window of data to feed a longer FFT to gather sufficient resolution information. An FFT window of length 256 at 44.1k sample rate is almost certainly just too short to gather sufficient information regarding spectral content below a few 100 Hz, if those are among the frequencies you would like to see examined, as they can't be separated cleanly from a DC bias (bin 0).
Unfortunately, there's a degree of uncertainty in identifying the frequencies in a fixed sample of a signal. If you use a short FFT, then there's no way to tell the difference between frequencies over a fairly wide range. If you use a long FFT to get higher resolution in the frequency domain, then you can't detect frequency changes as quickly. This is inherent in the math.
Off the top of my head: If you want a 15kHz range at 1Hz increments, you need a 15000 point FFT, which at 44.1kHz means you'll get a frequency plot three times per second. (I may be missing a factor of 2 in there as I can't recall whether the Nyquist limit means you actually want a 30kHz bandwidth.)
You may also be interested in the Short-time Fourier transform. It doesn't solve the fundamental trade-off problem but in practice may get you what you want.
Related
The documentation of MALLET mentions following:
--num-iterations [NUMBER]
The number of sampling iterations should be a trade off between the time taken to complete sampling and the quality of the topic model.
MALLET provides furthermore an example:
// Run the model for 50 iterations and stop (this is for testing only,
// for real applications, use 1000 to 2000 iterations)
model.setNumIterations(50);
It is obvious that too few iterations lead to bad topic models.
However, does increasing the number of Gibbs sampling iterations necessarily benefit the quality of the topic model (measured by perplexity, topic coherence or on a downstream task)?
Or is it possible that the model quality decreases with the --num-iterations set to a too high value?
On a personal project, averaged over 10-fold cross-validation increasing the number of iterations from 100 to 1000 did not impact the average accuracy (measured as Mean Reciprocal Rank) for a downstream task. However, within the cross-validation splits the performance changed significantly, although the random seed was fixed and all other parameters kept the same. What part of background knowledge about Gibbs sampling am I missing to explain this behavior?
I am using a symmetric prior for alpha and beta without hyperparameter optimization and the parallelized LDA implementation provided by MALLET.
The 1000 iteration setting is designed to be a safe number for most collection sizes, and also to communicate "this is a large, round number, so don't think it's very precise". It's likely that smaller numbers will be fine. I once ran a model for 1000000 iterations, and fully half the token assignments never changed from the 1000 iteration model.
Could you be more specific about the cross validation results? Was it that different folds had different MRRs, which were individually stable over iteration counts? Or that individual fold MRRs varied by iteration count, but they balanced out in the overall mean? It's not unusual for different folds to have different "difficulty". Fixing the random seed also wouldn't make a difference if the data is different.
The following is how I understand the point of parameter sharing in RNNs:
In regular feed-forward neural networks, every input unit is assigned an individual parameter, which means that the number of input units (features) corresponds to the number of parameters to learn. In processing e.g. image data, the number of input units is the same over all training examples (usually constant pixel size * pixel size * rgb frames).
However, sequential input data like sentences can come in highly varying lengths, which means that the number of parameters will not be the same depending on which example sentence is processed. That is why parameter sharing is necessary for efficiently processing sequential data: it makes sure that the model always has the same input size regardless of the sequence length, as it is specified in terms of transition from one state to another. It is thus possible to use the same transition function with the same weights (input to hidden weights, hidden to output weights, hidden to hidden weights) at every time step. The big advantage is that it allows generalization to sequence lengths that did not appear in the training set.
My questions are:
Is my understanding of RNNs, as summarized above, correct?
In the actual code example in Keras I looked at for LSTMs, they padded the sentences to equal lengths before all. By doing so, doesn't this wash away the whole purpose of parameter sharing in RNNs?
Parameter Sharing
Being able to efficiently process sequences of varying length is not the only advantage of parameter sharing. As you said, you can achieve that with padding. The main purpose of parameter sharing is a reduction of the parameters that the model has to learn. This is the whole purpose of using a RNN.
If you would learn a different network for each time step and feed the output of the first model to the second etc. you would end up with a regular feed-forward network. For a number of 20 time steps, you would have 20 models to learn. In Convolutional Nets, parameters are shared by the Convolutional Filters because when we can assume that there are similar interesting patterns in different regions of the picture (for example a simple edge). This drastically reduces the number of parameters we have to learn. Analogously, in sequence learning we can often assume that there are similar patterns at different time steps. Compare 'Yesterday I ate an apple' and 'I ate an apple yesterday'. These two sentences mean the same, but the 'I ate an apple' part occurs on different time steps. By sharing parameters, you only have to learn what that part means once. Otherwise, you'd have to learn it for every time step, where it could occur in your model.
There is a drawback to sharing the parameters. Because our model applies the same transformation to the input at every time step, it now has to learn a transformation that makes sense for all time steps. So, it has to remember, what word came in which time step, i.e. 'chocolate milk' should not lead to the same hidden and memory state as 'milk chocolate'. But this drawback is small compared to using a large feed-forward network.
Padding
As for padding the sequences: the main purpose is not directly to let the model predict sequences of varying length. Like you said, this can be done by using parameter sharing. Padding is used for efficient training - specifically to keep the computational graph during training low. Without padding, we have two options for training:
We unroll the model for each training sample. So, when we have a sequence of length 7, we unroll the model to 7 time steps, feed the sequence, do back-propagation through the 7 time steps and update the parameters. This seems intuitive in theory. But in practice, this is inefficient, because TensorFlow's computational graphs don't allow recurrency, they are feedforward.
The other option is to create the computational graphs before starting training. We let them share the same weights and create one computational graph for every sequence length in our training data. But when our dataset has 30 different sequence lengths this means 30 different graphs during training, so for large models, this is not feasible.
This is why we need padding. We pad all sequences to the same length and then only need to construct one computational graph before starting training. When you have both very short and very long sequence lengths (5 and 100 for example), you can use bucketing and padding. This means, you pad the sequences to different bucket lengths, for example [5, 20, 50, 100]. Then, you create a computational graph for each bucket. The advantage of this is, that you don't have to pad a sequence of length 5 to 100, as you would waste a lot of time on "learning" the 95 padding tokens in there.
my question is if a periodic signal is fourier transformed, will the bandwidth be equal to its frequency in time domain?
For example, if a sine wave has the frequency wc, then in frequency domain of fourier transform, it will have two impulses at -wc and +wc, right ? So the bandwidth or the highest frequency in frequency domain is supposed to be wc, same as the frequency of the original signal in time domain.
But is this true for any periodic signal ?
And another thing, how can I generate fourier transform of a periodic signal? Doesn't it violate the first condition of fourier transform ?
I need it because Nyquist theorem states that, If a continuous time signal contains no frequency components higher than B Hz, then the sampling frequency should be FS>2B. In exercise, I am given the sum of a bunch of sine sine functions and I need to calculate the minimum sampling frequency. So I need to know the highest frequency component?
First of all, the bandwidth of a periodic signal is usually defined as the difference between it's highest and lowest frequency component. It is the frequency range in which the signal's spectral density is nonzero. If you have a periodic signal with frequency components down to zero, you have a low-pass or baseband signal where the bandwidth is equal to it's highest frequency component.
The Nyquist theorem states that perfect reconstruction of a continuous-time signal (analog signal) with finite bandwidth is possible with a discrete sequence of samples (digital signal), if certain constrains are satisfied.
In the case of a baseband signal (frequency components down to zero), the sampling rate has to be bigger than two times the highest frequency component of the signal.
In the case of a bandpass signal (all frequency components higher than zero), the sampling rate has to be higher than two times the bandwidth of the signal e.g . This leads to a range of different possible sampling frequencies in the range:
fc is the center frequency of the bandpass signal, B is the bandwidth and
m is an arbitrary, positive integer ensuring that fs is at least two times the bandwidth of the periodic signal.
Generally, as I can recall from a prior career that involved radar image processing, the "bandwidth" is the highest 3db portion of the power spectrum. The power spectrum isn't the same as the FFT output; power = square of the amplitude (FFT outputs complex numbers, so this would be the square of the magnitude of each complex vector).
If you're doing an FFT, remember to swap the halves of the frequency domain output before figuring out the bandwidth, because the output of an FFT starts with the middle frequency and goes to the highest, then continues with the minimum frequency and ends before the middle freq.
I am trying to come up with an efficient way to characterize two narrowband tones separated by about 900kHz (one at around 100kHZ and one at around 1MHz once translated to baseband). They don't move much in freq over time but may have amplitude variations we want to monitor.
Each tone is roughly about 100Hz wide and we are required to characterize these two beasts over long periods of time down to a resolution of about 0.1 Hz. The samples are coming in at over 2M Samples/sec (TBD) to adequately acquire the highest tone.
I'm trying to avoid (if possible) doing brute force >2MSample FFTs on the data once a second to extract frequency domain data. Is there an efficient approach? Something akin to performing two (much) smaller FFTs around the bands of interest? Ive looked at Goertzel and chirp z methods but I am not certain it helps save processing.
Something akin to performing two (much) smaller FFTs around the bands of interest
There is, it's called Goertzel, and is kind of the FFT for single bins, and you already have looked at it. It will save you CPU time.
Anyway, there's no reason to do a 2M-point FFT; first of all, you only want a resolution of about 1/20 the sampling rate, hence, a 20-point FFT would totally do, and should be pretty doable for your CPU at these low rates; since you don't seem to care about phase of your tones, FFT->complex_to_mag.
However, there's one thing that you should always do: look at your signal of interest, and decimate down to the rate that fits exactly that. Since GNU Radio's filters are implemented cleverly, the filter itself will only run at the decimated rate, and you can spend the CPU cycles saved on a better filter.
Because a direct decimation from 2MHz to 100Hz (decimation: 20000) will really have an ugly filter length, you should do this multi-rated:
I'd try first decimating by 100, and then in a second step by 100, leaving you with 200Hz observable spectrum. The xlating fir filter blocks will let you use a simple low-pass filter (use the "Low-Pass Filter Taps" block to define a variable that contains such taps) as a band-selector.
I have read the following sentence in a documents talk about the drawback of short time Fourier transform, and he is said :
the drawback is that once you choose a particular size for the time
window, that window is the same for all frequencies
So what is the relation between frequencies and the size of window. If we have a high frequency component in a part of a signal how will not be able to detect this frequency if the size of the window is not smaller/bigger enough?
Furthermore, he is said about wavelet transform :
Wavelet analysis allows the use of long time intervals where we want
more precise low-frequency information, and shorter regions where we
want high-frequency information
I feel that the answer has a relation with nyquest rate somehow
For sampled data, the number of orthogonal sinusoidal FT basis vectors below half the sample rate increases with the length of the STFT window, and the bandwidth of each DFT/FFT result bin for each basis vector decreases. If the window is too short, then each DFT result might detect not only your high frequency component of interest, but a greater bandwidth of adjacent frequencies.