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1

I didn't know the convolution theorem for the DHT before, but it's pretty clear that if it exists, it must be about circular convolution, just like for the DFT. You're comparing that with acyclic convolution, so the results differing is no surprise.


2

You are on the right track, so here is a hint: Since these are all binary functions with time step of one (and I'm too lazy to type integrals), I'm just going to treat it like a sum with the index being the end-time. So you get $$y(4) = h(1)\cdot x(4) + h(2)\cdot x(3) + h(3)\cdot x(2) + h(4)\cdot x(1) $$ Given that both $|h| \le 1$ and $|x| \le 1$ it should ...


1

I´m not sure if this really applies to your problem since it may be another issue, but I can tell from my experience with acoustic impulse response measurements (only there you want to estimate the response from the input signal, but the deconvolution should be the same): In case your transfer function E is somehow bandlimited and has zero (or very low) ...


0

Cin = F^-1 [F(Cout)]/[F(E)] however FFT of the transfer function seems to be zero everywhere and the resulting recovered signal have a lot of noise. What you're building is effectively an equalizer, in this case, a zero-forcing equalizer. That comes with noise amplification: Where your F(E) is small, the division leads to large numbers – exactly where there ...


0

but is it possible to get only N point sequence for y[n] , Not without losing information. You can truncate the "tail" of the convolution but this does create an error. In most practical cases of convolution, the signal is much longer than the impulse response and it's often acceptable to discard the extra samples, but that really depends on your ...


1

I think you are using the wrong tool for the job. Semitones are spaced logarithmically but FIR filters have linear frequency resolution. If you want to reliably distinguish between the low E and the low F on the bass guitar you need a frequency resolution of better than 2 Hz which requires 10s of thousands of taps (at 48 kHz sample rate). That's why most ...


0

I stumbled across that wikipedia "ouput" image as well and no matter what I did with my kernels or values, my output never matched the example output on wiki. I took a close look at what kind of edges are shown and recognized, that the sobel filter is applied two times! So you take hy and hx and run your usual sobel on your input image. Then you ...


1

Assuming that the conditions are met to invoke Plancherel's (aka Parseval's) theorem, we can reformulate this in the frequency domain where the energy equality $E(x \star h) = E(x)$ can be expressed as $$\int_{-\infty}^{\infty}|X(f)H(f)|^2df = \int_{-\infty}^{\infty}|X(f)|^2df$$ I see no condition where energy is preserved unless $|H(f)| = 1$ (all-pass), the ...


1

I think any allpass filter should do that. Since they only change phase and not amplitude in the frequency domain, the time domain energy remains unchanged per Parseval's theorem. $$\sum x[n]^2 = \sum |X(k)|^2$$ An all pass only changes the phase of $X(k)$ so total energy is maintained. You could argue that an allpass is an IIR filter, so it's not suitable ...


1

Don't I need a two separate multipliers ? No. The moving average is an FIR filter hence you have $a_k = 0, k > 0$ DF1 and DF2 are for generic IIR filters. An FIR filter is a subset of an IIR filter, so technically you can implement it as DF1 or DF2, but there is really no reason to do that way since it's inefficient as compared to the alternatives. One ...


1

To start, notice that the diagrams are in the $\mathcal{Z}$ domain, multiplying by $z^{-1}$ is the delay of one sample. DF-I Writing the equation for $y_n$ of the DF-I $$y_n = \sum_{k=0}^{M} b_k x_{n-k} - \sum_{l=1}^{N} a_{l} y_{n-l}$$ Comparing with the equation of the moving average $$y_n = \sum_{k=0}^{M} \frac{1}{M+1} x_{n-k}$$ we see that we could ...


0

Most importantly, applying a shift in frequency, which the OP is proposing to do, will have no effect on the number of points needed to achieve a desired frequency resolution. The frequency resolution is given by the total number of samples alone and is finest when no further windowing is applied and results in a bandwidth (as the equivalent noise bandwidth) ...


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