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Just think of the value of $t$ where $x(0)$ appears. For $x(t+t_0)$ it is at $t=-t_0$, which corresponds to a left shift if $t_0>0$. However, for $x(-t+t_0)$ the value $x(0)$ occurs at $t=t_0$, which is to the right of its original position if $t_0>0$. You can think of deriving $x(-t+t_0)$ from $x(t)$ in two different ways: invert the time axis: $x(-t)... 0 the impulse-response and frequency-response of a signal Only systems have impulse responses and frequency responses, signals don't. I assume that what you mean here. A System describes the relation ship between its input signal and its output signal. For an LTI system, that relationship can be captured through either the transfer function or the impulse ... 0 I'm struggling a little with this notation because I don't usually do any significant image processing. Many LTI filters have an inverse and it seems like this is where your discussion is leading. I'm assuming your image filter with three values could be expressed something like this:$h[n] = h_0\delta[n+1] + h_1\delta[n] + h_2\delta[n-1]$If so, getting the ... 2 The green curve in your plot corresponds to the correct result. The red curve is wrong, because the weights in the vector cnvweights are in the wrong order (left and right halves are interchanged). The correct way to compute those weights is cnvweights <- convolve(rep(1/n,n), rev(rep(1/n,n)), conj = TRUE, type = "open") According to the R ... 0 Here’s are pretty good source on the subject: http://www.dspguide.com/ch15/4.htm A single moving average is a boxcar. Two is a triangle. Beyond that, it will start to approach a Gaussian. 2 Since these are two different integral equations, during substitution you cannot use the same variables for both the integrals. On substituting$r_1(t-\tau)$to$r(t)$we get, $$r(t) = \int \int h_2(\tau,\nu)e^{j 2\pi\nu(t-\tau)} \int \int h_1(\tau_1,\nu_1)e^{j 2\pi\nu(t-\tau-\tau_1)}s(t-\tau-\tau_1)d\tau_1 d\nu_1 d\tau d\nu$$ Now substitute$\tau = \tau', \...

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As for the why, it's always good to come up with alternatives, even if one doesn't immediately see their benefit. In this case it's about efficiency, especially if the FFT length is a prime. The resulting algorithm can also be used to solve the more general problem of computing (samples of) the $\mathcal{Z}$-transform on circles or spirals in the complex ...

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The fact that your source is 30m away means that your ratio of direct to reverberant sound is likely quite low. I think this makes your task much more difficult if not impossible. Do you detect any directionality in the original recording, or just a sense of ambience ? Bob

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It would be better to show that property using pencil and paper, because you're fooling yourself with that Matlab code. Of course we have $$\delta[n-n_0]\star g[n]=g[n-n_0]\tag{1}$$ Now define $h[n]=g[n-n_1]$. According to $(1)$ we have $$\delta[n-n_0]\star h[n]=h[n-n_0]\tag{2}$$ And, consequently, with $h[n]=g[n-n_1]$, $$\delta[n-n_0]\star g[n-n_1]=g[n-n_0-... 1 As most of the answers already provided state, this is quite tricky and rather difficult to achieve faithful decomposition of the sound field. Since you are considering a pair of microphones you could consider two different methods to decompose the impinging sound field into idealized plane waves. Coincidence microphones: Here you have to use the magnitude ... 1 What you are describing is an FIR filter, such that all the denominator coefficients are zero, save the basis, a=1. So you could do something like: data_smoothed = scipy.signal.filtfilt(window, 1, data_noisy) There is a notable point. The DC gain of the filter is equal to the sum of the coefficients for FIR filters. Your window is likely normalized to ... 0 Can we assume this is a dry two-channel recording, i.e. there is no "production" after effect to add more reverb, or tweak the phases etc...? Do you have access to the original set-up? The theoretical approach would be to determine the impulse response for the first recording, for each channel (L & R). If you don't have it, try to estimate it ... 0 That's tricky Basically you need to first recover the original left/right content and then re-render through the new microphone geometry. In this case you could do a time-frequency analysis and look for content that's correlated (similar phase) but has significant inter-channel level differences. You would re-render it by reducing the level difference (based ... 1 Which algorithms can be used to "re-spatialize" this recording, i.e. try to virtually "move the microphones", and recreate a new stereo signal, for example with an AB mic positioning? In general, this is a beamforming "problem", but it cannot be done in exactly the way it is described here. With a setup like this, you can ... 0 The original U-Net paper used nearest neighbor interpolation as far as I know. This is also the default upsampling method in TensorFlow. My own anecdotal advice is to not use transposed convolutions in U-Net. It will only make your CNN slower and won't really increase your F1 score (or whatever metric you're using). I would recommend nearest neighbor ... 0 1D nearest neighbor interpolation is K = \begin{bmatrix}1 & 1 & 0\end{bmatrix}^T or K = \begin{bmatrix}0 & 1 & 1\end{bmatrix}^T. This depends on whether your library flips the kernel (i.e. cross correlation vs convolution). To increase the sampling frequency just increase the kernel size e.g. \begin{bmatrix}0 & 1 & 1 & 1 &... 2 Systematically, you could just solve the problem by finding a solution A that satisfies the following two equations:$$(A-1)^2=A-1\\A^2=A

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I have no access to your audio files so I've downloaded: IR from here (mono/r1_omni.wav) - it's a really long one Anechoic recording from here (operatic-voice/mono/singing.wav) Resampled voice signals: Final convolved signal: As for your questions: 1. As you did the plot of IR in logarithmic scale it's clearly visible that towards its end there is ...

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