# Tag Info

## Hot answers tagged minimum-phase

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More taps. You don't have anywhere near enough taps for a filter that steep. Start large with 8192 or so cut to desired accuracy, if needed Due to the low number of tabs you are seeing the effect of "circular" hilbert transform. See for example: http://andrewduncan.net/air/ How do you know you have zeros outside the unit circle? Calculating the roots of a ...

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You made a minimum-phase filter but with a different magnitude response than the original linear phase filter. What you have to do to keep the magnitude the same is to reflect the zeros outside the unit circle into the circle. Since for linear phase filters the zeros are symmetrical with respect to the unit circle (or on the circle), you simply have to ...

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The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with $$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$ can be calculated in the following way. First, note that $$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$ From this table we know that $$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$ We also ...

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as a related aside question i posted this question about minimum-phase filters and the phase-magnitude relationship. let $N$ be the FFT size you will use. (often $N$ is a power of two, but it doesn't have to be.) the target magnitude response is $$G[k] \qquad \text{for } 0 \le k \le \tfrac{N}{2}$$ $G$ is the magnitude at DC. $G[\tfrac{N}2]$ (if $N$...

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I commend you for using an intuitive algorithm. However, there are already established algorithms with far better performance. Phase recovery algorithms work by filtering the error signal down to zero. An example is the Costas loop structure shown in the figure below for QPSK. Let's start with phase error detectors (Highlighted in yellow). The arctan ...

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If I see it correctly, your transfer function is $$H(z)=\frac{1-\frac{1}{2}z^{-1}}{z^{-1}}=z-\frac{1}{2}$$ This system is not a minimum-phase system because its pole is at infinity (as you've already pointed out). You could use a trivial all-pass filter with impulse response $h[n]=\delta[n-1]$, i.e. a simple delay by one sample to change your original (non-...

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one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ...

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It is generally impossible to transform a given minimum-phase FIR system into a linear phase FIR system with the same magnitude response. There is one special case for which this is possible, and that is if the zeros of the minimum phase system inside the unit circle haven even multiplicity. Because in that case you can, for each zero location, mirror half ...

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If $H(\omega)=e^{\alpha(\omega)+j\phi(\omega)}$ is a minimum phase frequency response, then the attenuation $\alpha(\omega)$ and the phase $\phi(\omega)$ are related by the following Hilbert transform relationship: $$\phi(\omega)=-\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\alpha(\Omega)}{\omega-\Omega}d\Omega$$ So $\phi(\omega)$ is uniquely determined by $\... 2 To your second definition it should be added that you only consider causal transfer functions, because it is not difficult to find a smaller phase lag with a non-causal system: A minimum-phase system is a causal and stable system with a phase lag that is smaller than the phase lag of any other causal and stable system with the same magnitude response. ... 2 Just looking at it I can see it's related to control theory. If you make a contrived open-loop gain $$H(z) = \frac{A(z)D(z)}{B(z)C(z)}$$ and then wrap it with unity-gain feedback, you get $$G(z) = \frac{A(z)B(z)C(z)D(z)}{A(z)D(z) + B(z)C(z)}$$. So any stability test that takes$H(z)$as an open-loop gain will equally be a test to see if your sum is a ... 2 An LTI system is said to be minimum-phase if the system and its inverse system are causal and stable. That's implying that all poles and zeros must be strictly inside the unit circle. An all-pass filter's zero-pole pairs are reflections of each other across the unit circle so that an all-pass filter absorbs the poles or zeros. For a stable all-pass filter, ... 1 Why is it that reflecting any poles or zeros of a rational function across the unit circle gives a minimum phase system? It doesn't. You are starting with a wrong assumption. Here's an example, it seems reflecting any poles or zeros would result in the zeros being outside the circle? Correct. Hence it's nonsense and you shouldn't do it. Assuming that the '... 1 two transfer functions HL and HR can each be represented as a minimum-phase filter (MPF) plus a pure delay. That is generally not true and it's easy enough to disprove it by counter example. Let's simply look at a first order all pass $$H(z) = \frac{p-1/z}{1-p/z} = H_{MP}(z) \cdot H_{AP}(z)$$ which has a pole at$p$and a zero at$1/p$. The "minimum ... 1 What they mean here is that the real-valued amplitude function of the linear phase FIR filter that you provide to the function must be non-negative, because it is interpreted as the desired squared magnitude of the minimum-phase filter. This amplitude function is referred to as zero-phase response, which is the frequency response of the filter when the ... 1 I don't think you will have much luck there. Minimum phase means that all the roots of all polynomials$A,B,C,D$are inside the unit circle. That means that the product of two polynomials will also have their roots inside the unit circle so$A \cdot D$,$B \cdot C$, and$B \cdot D$are all good. However it's NOT good for$A \cdot D + B \cdot C$. The roots of ... 1 Here is one answer, if someone can improve on this I will select it as the "right" answer (also comments very welcome on obvious flaws with this approach): Given Cauchy's argument principle, the number of zeros outside the unit circle is given by the number of encirclements of the origin for the frequency response of the filter as plotted on a complex plane.... 1 As you have pointed out, the nyquist stability criterion is more general, moreover, is the only stability criterion. Nevertheless, the Bode plot give us the exact same information that the nyquist plot does, so the stability of a system can be determined with either the bode or the nyquist plot, some times its easier with the former and other times with the ... 1 HINT: (because it's a homework problem) Apply the$\mathcal{Z}$-transform to both equations. Use the first equation to express$Q(z)$in terms of$X(z)$, and plug that into the second equation to obtain$Y(z)$in terms of$X(z)$. From this you can easily get the transfer function$H(z)$. Compute the poles and zeros of$H(z)$and determine the constant$k$... 1 The transfer function of a discrete-time minimum phase system described by a linear difference equation with constant coefficients can be written as $$H(z)=h\cdot\frac{\prod_{k=1}^M(1-c_kz^{-1})}{\prod_{k=1}^N(1-d_kz^{-1})},\qquad |c_k|\le 1,\quad |d_k|<1\tag{1}$$ where$h$is the first sample of the corresponding impulse response, and$c_k$and$...

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