I want to use IFFT to obtain the time signal from freqency data, being response amplitude operators (RAO's) and phase shifts of a vessel over frequency range 0.1-2.6 RAd/s, having 200 samples. I first used a very unrefined method to obtain the time domain response as follows:
for i=1:length(t)
surgedisp(:,i)=surge_amp(:,1).*cos(g(:,i)+surge_phase(:,1)+phase(:,1));
surge(1,i)=sum(surgedisp(:,i));
end
Here, g represents w*t, surge phase is the phase shift of the surge motion wrt the waves, phase is a random phase (multiple of 2pi) of the random (ocean)wave. (The values at each frequency)
For the use of IFFT, I needed to represent the surge_amplitudes and surge_phases into the complex form in order to use IFFT:
Z=surge_amp.*exp(j*surge_phase+j*phase), followed by IFFT(Z), and obtain the time signal.
I think I get totally wrong results and I don't know in which direction to look for. Can it have something to do with the j*wt part that I should incorporate into Z? And can someone tell me if the first method is equivalent to the IFFT method?
Related
I have a confusion about the way the LSTM networks work when forecasting with an horizon that is not finite but I'm rather searching for a prediction in whatever time in future. In physical terms I would call it the evolution of the system.
Suppose I have a time series $y(t)$ (output) I want to forecast, and some external inputs $u_1(t), u_2(t),\cdots u_N(t)$ on which the series $y(t)$ depends.
It's common to use the lagged value of the output $y(t)$ as input for the network, such that I schematically have something like (let's consider for simplicity just lag 1 for the output and no lag for the external input):
[y(t-1), u_1(t), u_2(t),\cdots u_N(t)] \to y(t)
In this way of thinking the network, when one wants to do recursive forecast it is forced to use the predicted value at the previous step as input for the next step. In this way we have an effect of propagation of error that makes the long term forecast badly behaving.
Now, my confusion is, I'm thinking as a RNN as a kind of an (simple version) implementation of a state space model where I have the inputs, my output and one or more state variable responsible for the memory of the system. These variables are hidden and not observed.
So now the question, if there is this kind of variable taking already into account previous states of the system why would I need to use the lagged output value as input of my network/model ?
Getting rid of this does my long term forecast would be better, since I'm not expecting anymore the propagation of the error of the forecasted output. (I guess there will be anyway an error in the internal state propagating)
Thanks !
Please see DeepAR - a LSTM forecaster more than one step into the future.
The main contributions of the paper are twofold: (1) we propose an RNN
architecture for probabilistic forecasting, incorporating a negative
Binomial likelihood for count data as well as special treatment for
the case when the magnitudes of the time series vary widely; (2) we
demonstrate empirically on several real-world data sets that this
model produces accurate probabilistic forecasts across a range of
input characteristics, thus showing that modern deep learning-based
approaches can effective address the probabilistic forecasting
problem, which is in contrast to common belief in the field and the
mixed results
In this paper, they forecast multiple steps into the future, to negate exactly what you state here which is the error propagation.
Skipping several steps allows to get more accurate predictions, further into the future.
One more thing done in this paper is predicting percentiles, and interpolating, rather than predicting the value directly. This adds stability, and an error assessment.
Disclaimer - I read an older version of this paper.
I'm a beginner in RL and want to know what is the advantage of having a state value function as well as an action-value function in RL algorithms, for example, Markov Design Process. What is the use of having both of them in prediction and control problems?
I think you mean state-value function and state-action-value function.
Quoting this answer by James MacGlashan:
To explain, lets first add a point of clarity. Value functions
(either V or Q) are always conditional on some policy π. To emphasize
this fact, we often write them as ππ(π ) and ππ(π ,π). In the
case when weβre talking about the value functions conditional on the
optimal policy πβ, we often use the shorthand πβ(π ) and πβ(π ,π).
Sometimes in literature we leave off the π or * and just refer to V
and Q, because itβs implicit in the context, but ultimately, every
value function is always with respect to some policy.
Bearing that in mind, the definition of these functions should clarify
the distinction for you.
ππ(π ) expresses the expected value of following policy π forever
when the agent starts following it from state π .
ππ(π ,π) expresses the expected value of first taking action π
from state π and then following policy π forever.
The main difference then, is the Q-value lets you play a hypothetical
of potentially taking a different action in the first time step than
what the policy might prescribe and then following the policy from the
state the agent winds up in.
For example, suppose in state π Iβm one step away from a terminating
goal state and I get -1 reward for every transition until I reach the
goal. Suppose my policy is the optimal policy so that it always tells
to me walk toward the goal. In this case, ππ(π )=β1 because Iβm just
one step away. However, if I consider the Q-value for an action π
that walks 1 step away from the goal, then ππ(π ,π)=β3 because
first I walk 1 step away (-1), and then I follow the policy which will
now take me two steps to get to the goal: one step to get back to
where I was (-1), and one step to get to the goal (-1), for a total of
-3 reward.
All examples that I can find on the Internet just visualize the result array of the function computeSpectrum, but I am tasked with something else.
I generate a music note and I need by analyzing the result array to be able to say what note is playing. I figured out that I need to set the second parameter of the function call 'FFTMode' to true and then it returns sound frequencies. I thought that really it should return only one non-zero value which I could use to determine what note I generated using Math.sin function, but it is not the case.
Can somebody suggest a way how I can accomplish the task? Using the soundMixer.computeSpectrum is a requirement because I am going to analyze more complex sounds later.
FFT will transform your signal window into set of Nyquist sine waves so unless 440Hz is one of them you will obtain more than just one nonzero value! For a single sine wave you would obtain 2 frequencies due to aliasing. Here an example:
As you can see for exact Nyquist frequency the FFT response is single peak but for nearby frequencies there are more peaks.
Due to shape of the signal you can obtain continuous spectrum with peaks instead of discrete values.
Frequency of i-th sample is f(i)=i*samplerate/N where i={0,1,2,3,4,...(N/2)-1} is sample index (first one is DC offset so not frequency for 0) and N is the count of samples passed to FFT.
So in case you want to detect some harmonics (multiples of single fundamental frequency) then set the samplerate and N so samplerate/N is that fundamental frequency or divider of it. That way you would obtain just one peak for harmonics sinwaves. Easing up the computations.
The documentation for both of these methods are both very generic wherever I look. I would like to know what exactly I'm looking at with the returned arrays I'm getting from each method.
For getByteTimeDomainData, what time period is covered with each pass? I believe most oscopes cover a 32 millisecond span for each pass. Is that what is covered here as well? For the actual element values themselves, the range seems to be 0 - 255. Is this equivalent to -1 - +1 volts?
For getByteFrequencyData the frequencies covered is based on the sampling rate, so each index is an actual frequency, but what about the actual element values themselves? Is there a dB range that is equivalent to the values returned in the returned array?
getByteTimeDomainData (and the newer getFloatTimeDomainData) return an array of the size you requested - its frequencyBinCount, which is calculated as half of the requested fftSize. That array is, of course, at the current sampleRate exposed on the AudioContext, so if it's the default 2048 fftSize, frequencyBinCount will be 1024, and if your device is running at 44.1kHz, that will equate to around 23ms of data.
The byte values do range between 0-255, and yes, that maps to -1 to +1, so 128 is zero. (It's not volts, but full-range unitless values.)
If you use getFloatFrequencyData, the values returned are in dB; if you use the Byte version, the values are mapped based on minDecibels/maxDecibels (see the minDecibels/maxDecibels description).
Mozilla 's documentation describes the difference between getFloatTimeDomainData and getFloatFrequencyData, which I summarize below. Mozilla docs reference the Web Audio
experiment ; the voice-change-o-matic. The voice-change-o-matic illustrates the conceptual difference to me (it only works in my Firefox browser; it does not work in my Chrome browser).
TimeDomain/getFloatTimeDomainData
TimeDomain functions are over some span of time.
We often visualize TimeDomain data using oscilloscopes.
In other words:
we visualize TimeDomain data with a line chart,
where the x-axis (aka the "original domain") is time
and the y axis is a measure of a signal (aka the "amplitude").
Change the voice-change-o-matic "visualizer setting" to Sinewave to
see getFloatTimeDomainData(...)
Frequency/getFloatFrequencyData
Frequency functions (GetByteFrequencyData) are at a point in time; i.e. right now; "the current frequency data"
We sometimes see these in mp3 players/ "winamp bargraph style" music players (aka "equalizer" visualizations).
In other words:
we visualize Frequency data with a bar graph
where the x-axis (aka "domain") are frequencies or frequency bands
and the y-axis is the strength of each frequency band
Change the voice-change-o-matic "visualizer setting" to Frequency bars to see getFloatFrequencyData(...)
Fourier Transform (aka Fast Fourier Transform/FFT)
Another way to think about "time domain vs frequency" is shown the diagram below, from Fast Fourier Transform wikipedia
getFloatTimeDomainData gives you the chart on on the top (x-axis is Time)
getFloatFrequencyData gives you the chart on the bottom (x-axis is Frequency)
a Fast Fourier Transform (FFT) converts the Time Domain data into Frequency data, in other words, FFT converts the first chart to the second chart.
cwilso has it backwards.
the time data array is the longer one (fftSize), and the frequency data array is the shorter one (half that, frequencyBinCount).
fftSize of 2048 at the usual sample rate of 44.1kHz means each sample has 1/44100 duration, you have 2048 samples at hand, and thus are covering a duration of 2048/44100 seconds, which 46 milliseconds, not 23 milliseconds. The frequencyBinCount is indeed 1024, but that refers to the frequency domain (as the name suggests), not the time domain, and it the computation 1024/44100, in this context, is about as meaningful as adding your birth date to the fftSize.
A little math illustrating what's happening: Fourier transform is a 'vector space isomorphism', that is, a mapping going bijectively (i.e., reversible) between 2 vector spaces of the same dimension; the 'time domain' and the 'frequency domain.' The vector space dimension we have here (in both cases) is fftSize.
So where does the 'half' come from? The frequency domain coefficients 'count double'. Either because they 'actually are' complex numbers, or because you have the 'sin' and the 'cos' flavor. Or, because you have a 'magnitude' and a 'phase', which you'll understand if you know how complex numbers work. (Those are 3 ways to say the same in a different jargon, so to speak.)
I don't know why the API only gives us half of the relevant numbers when it comes to frequency - I can only guess. And my guess is that those are the 'magnitude' numbers, and the 'phase' numbers are thrown out. The reason that this is my guess is that in applications, magnitude is far more important than phase. Still, I'm quite surprised that the API throws out information, and I'd be glad if some expert who actually knows (and isn't guessing) can confirm that it's indeed the magnitude. Or - even better (I love to learn) - correct me.
How should stereo (2 channel) audio data be represented for FFT? Do you
A. Take the average of the two channels and assign it to the real component of a number and leave the imaginary component 0.
B. Assign one channel to the real component and the other channel to the imag component.
Is there a reason to do one or the other? I searched the web but could not find any definite answers on this.
I'm doing some simple spectrum analysis and, not knowing any better, used option A). This gave me an unexpected result, whereas option B) went as expected. Here are some more details:
I have a WAV file of a piano "middle-C". By definition, middle-C is 260Hz, so I would expect the peak frequency to be at 260Hz and smaller peaks at harmonics. I confirmed this by viewing the spectrum via an audio editing software (Sound Forge). But when I took the FFT myself, with option A), the peak was at 520Hz. With option B), the peak was at 260Hz.
Am I missing something? The explanation that I came up with so far is that representing stereo data using a real and imag component implies that the two channels are independent, which, I suppose they're not, and hence the mess-up.
I don't think you're taking the average correctly. :-)
C. Process each channel separately, assigning the amplitude to the real component and leaving the imaginary component as 0.
Option B does not make sense. Option A, which amounts to convert the signal to mono, is OK (if you are interested in a global spectrum).
Your problem (double freq) is surely related to some misunderstanding in the use of your FFT routines.
Once you take the FFT you need to get the Magnitude of the complex frequency spectrum. To get the magnitude you take the absolute of the complex spectrum |X(w)|. If you want to look at the power spectrum you square the magnitude spectrum, |X(w)|^2.
In terms of your frequency shift I think it has to do with you setting the imaginary parts to zero.
If you imagine the complex Frequency spectrum as a series of complex vectors or position vectors in a cartesian space. If you took one discrete frequency bin X(w), there would be one real component representing its direction in the real axis (x -direction), and one imaginary component in the in the imaginary axis (y - direction). There are four important values about this discrete frequency, 1. real value, 2. imaginary value, 3. Magnitude and, 4. phase. If you just take the real value and set imaginary to 0, you are setting Magnitude = real and phase = 0deg or 90deg. You have hence forth modified the resulting spectrum, and applied a bias to every frequency bin. Take a look at the wiki on Magnitude of a vector, also called the Euclidean norm of a vector to brush up on your understanding. Leonbloy was correct, but I hope this was more informative.
Think of the FFT as a way to get information from a single signal. What you are asking is what is the best way to display data from two signals. My answer would be to treat each independently, and display an FFT for each.
If you want a really fast streaming FFT you can read about an algorithm I wrote here: www.depthcharged.us/?p=176