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(Adapted from this answer on dsp.SE) The reason that the impulse response (also called the unit pulse response for discrete-time systems) determines the output for arbitrary input $x$ to an LTI system is that The output of a linear time-invariant system in response to inout $x$ is the sum of scaled and time-delayed versions of the impulse response. ...


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I think about this in terms of an impulse containing content at all frequencies, so if you know how the system responds to an impulse, you have a complete description of its response. If you're comfortable with Laplace transforms, we can also say that the transfer function for a system can be computed by taking the Laplace transform of the impulse response -...


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It's not possible to recover the original signal $X$ based on $h$ perfectly !! $Filter$ or $conv$ are usually performed in time domain, so it's similar "But not exactly" to $y = conv(X,h)$ , So in that case after you know $h$ or in other words after you estimate $h$ , you need to use any equalizer like $ZF$ or $MMSE$ to get back $X$. Of course, ...


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In general, what you want is impossible! Not every operation is reversible. In frequency domain, that becomes obvious: a system's frequency response $H(f)$ might have actual zeros at some points, and thus, no inverse filter exists. Multiplication with zero can't be undone, since no matter what was at that frequency before filtering, it's 0 afterwards and we ...


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The scaling difference between your two output is exactly $32767= 2^{15}-1$ which is exactly the maximum amplitude of a 16-bit signed integer. I'm guessing, the difference is how you import the audio into the program: Matlab's $audioread()$ typically normalize the data to $[-1,1]$, i.e. divide by 32767. It would seem that whatever Python method you are using ...


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For an LTI system with impulse response $h(t)$ and transfer function $H(s)=\mathcal{L}\{h(t)\}$, an input signal $x(t)=e^{st}$ (note: no multiplication with the unit step function $u(t)$), results in an output $$\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&...


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Arbitrary shapes is A LOT more difficult than rectangular rooms. You either need to do image method with visibility checking or some sort of ray, particle or cone tracing algorithm. There are few commercial programs that do this (Ease, Odeon, Bose Modeller, etc. ) and I doubt you'll find a free one. In addition the much more complicated algorithms, these ...


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