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# Tag Info

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The frequency resolution of your FFT is determined by its length and the sampling rate. In your second plot you show Fs = 10e6 and 3200 samples. Assuming you don't zero-pad these samples when you pass them to the FFT this gives you a frequency resolution of 3125Hz: $$3215 = \frac{10 \times 10^6}{3200}$$ So at lower frequencies you have frequency bins as ...

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Most of the time, a non-parametric PSD estimation based on FFT alone (called a Periodogram) will provide a random looking spectrum of the numerical data. To reduce those random variations, or to get a smoother looking estimate, you can do one of the two things: Use Periodogram averaging such as Welch's method. Use a model based parametric PSD estimate ...

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There are some libraries which calculate the non-uniform Fourier transform (non-uniform may refer to time or frequency domain sampling or both) like this one. But you must know both accuracy and spectral resolution of estimated spectrum doesn't increase by increasing the number of frequency bins, and these are more related to the windowing and the averaging ...

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Dr. Manuel Kuehner, You are close. You need to take the square root of the linear values squared. $$P_{\mbox{total_linear}}=\sqrt{p_1^2+p_2^2+...}$$ $$P_{\mbox{total_dB}}=20 log_{10}\left( P_{\mbox{total_linear/20E-6}} \right)$$ FYI: I wrote a MATLAB function to do exactly as you request. It is here Looking to read? See page 16 of this book: https://...

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don't need the solution. A hint would be good to start. This is bog-standard cyclic-prefix OFDM, and that's really well-covered in literature.

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Different methods exist for estimating PSDs. They can be broadly classified as parametric or non-parametric methods. Indeed you should learn which method they used to obtain the associated PSD. An FFT based PSD estimation is a non-parametric method such as Periodogram and its variants. This method will produce random looking non-smooth output result which ...

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Since your signal isn't sampled uniformly some strange things might happen when you apply FFT and look at the results. What you should do is estimate the Uniform DFT of the Non Uniform Time Series. One easy way to do it is use the reference code and analysis I posted on the question - Frequency Analysis of a Signal Without a Constant Sampling Frequency (...

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My guess would be that the DC peak is part of the transient response, thus it decreases to zero over time T as the oscillating parts of your signal continue. But I don't know what your signal is so that's a guess

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You might be referring to windowing artifacts, which are caused by a the finite length default rectangular window or other window applied to segment your signal for a finite length FFT. Such artificial windowing artifacts can be removed by using an infinite length FT (for integrable functions), or the full length of time of the existence of the transmitter ...

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As the comments say: it really depends on your signals. If the energy in the "don't care" region is significantly larger than the energy at the frequency of interest, it's helpful to pre-filter it first. Otherwise you may see "spectral leakage" of the high energy components into your target area. On the downside: the pre-filter will affect the group delay ...

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As Hilmar pointed out, loudness, is a frequency dependent psychoacoustic metric (like most if not all of them). Taken directly from Pulkki's book: Sounds between the threshold of hearing and threshold of pain are perceived with increasing ‘strength’ or ‘volume’. This subjective feature of sound is called loudness. The loudness level has been ...

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No The sensitivity of human auditory is very dependent on frequency and without any type of frequency weighting, a raw RMS estimate will be pretty much useless. There are some standard frequency weighting curves, most popular being the "A" weighting. See https://en.wikipedia.org/wiki/A-weighting. The sensitivity is also dependent on the absolute level ...

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True that a chirp signal helps to get the FRF, but every time we change the frequency we can't reach the steady state, so this will cause bias in the estimation. As an advice try to use the multisine excitation, they are more suitable for such cases.

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First off, a small thing: The sampling rate usually has to be one of the standard sampling rates. Closest to your 1024*40 would be 44100. The hardware has to support the sampling rate. Some drivers will resample to the given rate, but that will be somewhat slow as it is in software. Try using a proper sampling rate. The real bottle neck is the plotting. ...

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You call a lot of methods and functions that are not included so it's hard to read. Here is how I debug this step by step. Verify your audio framework. Do NOTHING in the process() function other than copying the input to the output Verify simple processing. Now add multiplication with 0.5 or something simple like this. Verify the FFT based processing. Do ...

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This is not a knowledgeable answer, but more of a comment including one possible "how-to" tip with a screen capture of what was done. The FFT was saved as a TIFF; reopened; and ultimately an inverse FFT returned the original image (although not quite - it looks smaller, and it contains two images that I don't know how to interpret - the second image is not ...

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Before I address your question, I'll suggest that using a frequency domain technique (FFTs) to demodulate inherently time domain data (a time series of sequential data symbols) won't be very practical. Symbol timing recovery is inherently a time domain process. ISI encountered in common modulations like GFSK/GMSK means you really have more than 2 ...

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I would like to post here the link to the accepted answer that I obtained to the question in StackOverflow, as I think it is relevant to this forum. The original answer by francis is here: post in StackOverflow

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The length of the diagonal line will be, in a sense, a measurement of the sharpness of the sharpest edge in that direction. If you apply the proper coordinates to your axes, you could quantify this.

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If you rotate an image (2D function), its 2D Fourier transform also rotates. Those lines mean you have something like a barcode oriented in that direction in your image. In one direction it has lots of variations and behaves like a white noise with flat spectrum and in the other direction it has very slow variations. I'm not sure what do you mean by the ...

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Is the output below reasonable for a 440HZ sine wave? Yes. Why is the peak at 430 and not 440? Because you have chosen your FFT grid this way. It appears are you using a 1024 point FFT with a 44.1 kHz sample rate. 440Hz is not part of that grid If I want to get the amplitudes of all frequencies in a range, e.g. 20Hz-20kHZ (20,21...20k) how would ...

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