# Tag Info

4

Assuming you meant to produce something similar to the green line: What about $$\text{output}[n] = \max\{\text{input}[n-k], \text{input}[n-k+1], \ldots ,\text{input}[n]\}$$ i.e. you just find the maximum along a sliding window over the last $k$ input values?

3

The way I understand the problem is each sample of the output is a linear combination of the samples of the input. Hence it is modeled by: $$\boldsymbol{y} = H \boldsymbol{x}$$ Where the $i$ -th row of $H$ is basically the instantaneous kernel of the $i$ -th sample of $\boldsymbol{y}$. The problem above is highly ill poised. In the classic ...

1

In other words, $h[n]$ is ideal but its frequency response $H(e^{jω})$ is practical and realistic. This is not true. To compute $H(e^{j\omega})$, you must evaluate the summation $\sum_{n=-\infty}^{\infty}h[n]e^{-j\omega n}$. A real system can't do this because $h[n]$ is not finite length. Say you choose to use a rectangular window $w[n]$ (with length $2M+1$)...

1

TL;DR: Time shift in doable with LTI systems. Subtracting $1$ is non linear. In discrete systems, integer shifts can be implemented with filters that are zero everywhere, except at the shift location. This can be implemented via circulant matrices, akin to convolution. This can be made compatible with Operation 1 (taking border effects into accounts). The ...

1

As an answer probably require to have more details on the look-up table (smoothed and regularity of the kernels), here is a couple of recent papers, including a review: Satellite image restoration in the context of a spatially varying point spread function, 2010 Efficient shift-variant image restoration using deformable filtering, 2012 Fast Approximations ...

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Only action needed is to change the value of M and study the plot .

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