# Tag Info

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No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response. For example the systems $$y(t) = g(t) x(t)$$ or $$y[n] = \sum_{k=0}^{k < n} x[n-k]$$ are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...

3

This happens frequently if your poles are reasonably close to the unit circle. Consider the following example %% TF2ZP is problematic fs = 44100; % 6th order lowpass, fc = 50Hz, sampled at 44.1kHz [z,p,k] = cheby2(6,80,50*2/fs); % to transfer function [b,a] = zp2tf(z,p,k); % back to zpk [z1,p1,k1] = tf2zp(b,a); display([p p1]); Displaying the poles side ...

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Any LTI system can be completely characterized (among other things) by it's transfer function or it's impulse response. If your filter represents an LTI system, that you can calculate it's output by either convolving the input with the impulse response or multiplying the transfer function with the spectrum of the input signal. In theory these things are ...

2

Neither. To me, filter classes using the notion of frequency bands (low-pass, high-pass, etc.) can be used safely in the linear case. And the bilateral filter is nonlinear. Edges are not really high-frequency: they often have sharp variations across the edge, but slow variation along it. I would consider the bilateral filter as an edge-preserving smoother, ...

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DO NOT PASS GO; DO NOT Collect 200; DO NOT Take Fourier transforms, or worse yet, FFTs You do the convolution exactly the way you would do any other convolution: start with the basic convolution integral (not what you wrote) and apply the properties of the signals that you are using to come up with an easier calculation. Begin with $$\int_{-\infty}^\infty ... 2 For the biquad section that is cascaded, the quantization issues regarding the pole locations are well understood. For a biquad transfer function:$$\begin{align} H(z) &= \frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}} \\ \\ &= \frac{b_0z^2+b_1z+b_2}{z^2+a_1z+a_2} \\ \\ &= b_0\frac{z^2+\frac{b_1}{b_0}z+\frac{b_2}{b_0}}{z^2+a_1z+a_2} \... 2 With infinite-precision arithmetic they will be stable. However, since you don't have infinite-precision arithmetic, you will have quantization issues even if you use 64-bit precision. These quantization issues can make your filter unstable. Even if your filter is stable, perhaps you will not get the frequency response wanted because of these quantization ... 2 You have a very narrow stop band which means that all the poles are crammed in a very small area of the complex plane, close to the unit circle. This can result in severe numerical problems, even for relatively small filter orders, even with floating point arithmetic. Another important point that you might not realize is that if you design a band pass or a ... 2 Your confusion is understandable. If you consider the definition of linear phase FIR filter and the associated symmetry conditions on their impulse responses, then you can arrive the conclusion that the first two cases $$h_1[n] = [0,0,0,1,0]$$ and $$h_2[n] = [0,0,0,0,1,0,0,0,0,0,1]$$ are non-symmetric. However, as you use zeros and ones in those ... 2 The convolution integral is a special case of the Fredholm equation of the first kind. https://en.wikipedia.org/wiki/Fredholm_integral_equation I believe that it covers linear time varying systems, as do linear time varying state space equations, so it’s a no but... kind of answer. 1 To unveil part of the mystery, let us recall how the convolution operation and the properties of linearity and time-invariance are related. In other words, if a discrete system\mathcal{S}$is linear and time-invariant, what would be the output for a discrete signal$x[n]$? To do that, let us rewrite the signal on the basis of Kronecker symbols$\delta_n$,... 1 I suppose that you obtain a transfer function of the desired discrete-time system in the form $$H(z)=\frac{b_0+b_1z^{-1}+\ldots +b_Nz^{-N}}{1+a_1z^{-1}+\ldots +a_Nz^{-N}}\tag{1}$$ From that transfer function you compute the coefficients of the second-order sections. Before doing that, you can make sure that the DC gain of$(1)$equals$1$(which of course ... 1 Bilateral Filter is indeed an Edge Preserving Filter. Moreover, due to being Spatially Variant Non Linear Filter it can be applied using Fourier Transform. Since it has no representation in Frequency Domain it is not well defined how to classify it into one of the categories: LPF, HPF, BPF or BSF. Nonetheless, let's try doing some analysis based on ... 1 Your diagram looks correct. Let's call the transfer function in the feedback loop$G(z)$. Consider the signal$w[n]$at the input to the delay line. Its$\mathcal{Z}$-transform satisfies $$W(z)=X(z)+G(z)Y(z)\tag{1}$$ where$X(z)$and$Y(z)$are the$\mathcal{Z}\$-transforms of the input and output sequences, respectively. The output is just a delayed ...

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I have since found this paper, which provides a nice framework for how to think about this problem.

1

I actually ran into this same paper and had the same problem. The paper linked below does a good job addressing the issue. Starting on page 13 the discuss the issue with this optical element. https://mstamenk.github.io/assets/files/OpticalRadon.pdf

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