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FIR filter with impulse response $h[n] = {-\frac{1}{2}, \frac{1}{2}}$ means that : $$y[n] = \frac{1}{2}x[n-1] - \frac{1}{2}x[n]$$ This is in some sense a mirror operation of Moving Average of two samples. This is Moving Difference (samples reversed). A High Pass Filter. So, each output sample is the difference of current input sample with previous input ...


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I'd try a very "tinkery" approach here: Erode the image, so that the black area is shrunk by a fixed radius of pixels from its border (say, 5px). Dilate the resulting image by the same amount measure the amount of difference between original and processed image. The idea is that something that is a locally convex border doesn't suffer through erosion (it's ...


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Sample rate conversation is easy in theory but tricky in practice. Assuming you want to convert to the standard rate of 44.1 kHz (not 44 kHz), you have an awkward conversion ratio. $3800 =2^3 \cdot 5^2 \cdot 19$ and $441 = 3^2 \cdot 7^2$ are mutually prime that means that rational sample rate conversion is impractical,so you need irrational sample rate ...


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Your digital signal is originally sampled at $380KHz$ and you want to downsample it to sampling rate $44KHz$. Therefore, you will require fractional sampling rate change. You cannot just downsample to $44KHz$ because $\frac{380}{44}$ is not an integer. First upsample by a factor of $11$ and then downsample by a factor of $95$ to reach your goal. Since you ...


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If you don't want to use the PyPI package for bm3d, you can use ffmpeg and run the bm3d filter as an OS command- command="ffmpeg -i "+input_image_path+" -filter_complex bm3d=sigma=30/255:block=4:bstep=2:group=1:hdthr=10000:estim=basic /path/to/output/directory/output.png" os.system(command) This takes lesser computation time.


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As @hotpaw2 says, the FFT assumes periodic boundary conditions, that is the next point on the right should be equal to the first point on the left. It's not the case in your code because the time $t=10$ is included in your time array. The next point on the right is thus $t=10+dt$ and you get the discontinuity because $y(10+dt) \neq y(0)$. You can fix this by ...


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THe DWT does not really know about the actual sampling. It do cares about the relative frequency span. And it has not knowledge about pre-processing, like linear filtering. So in your case, the detail coefficients of the first level of wavelet coefficients correspond, more or less, to the [128/2-128]/2 frequency range. Because of the initial [0-30] low-pass ...


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If you are not interested in picking out the frame that bubbles start forming and since the transition is (clearly) from order (clearly formed shapes) to disorder (mostly noise), you can use a feature based approach. That is, characterise the whole image with a few numbers. An obvious feature here is image entropy. The typical (but not the only) way to ...


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First you will need to build the frequency tables for each note scale ex. [B, A, G, F, E, D, C], you will need say what scale you will use to compare with each window pitch, of course you also have to worry that your frequency table includes major or minor chords. But the essence of one basic autotunes works by adjusting to the chromatic scale, In other ...


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Are you using scipy by any chance ? If yes , this might help https://scikit-dsp-comm.readthedocs.io/en/latest/_modules/sk_dsp_comm/digitalcom.html check for functions "QPSK_tx" and "QPSK_bb"


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