4

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 ...


4

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 $$ G[k] \qquad \text{for } 0 \le k \le \tfrac{N}{2} $$ $G[0]$ is the magnitude at DC. $G[\tfrac{N}2]$ (if $N$...


3

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-...


3

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 ...


3

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 ...


3

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 ...


2

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 $\...


1

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. ...


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 $$H(z)=h[0]\cdot\frac{\prod_{k=1}^M(1-c_kz^{-1})}{\prod_{k=1}^N(1-d_kz^{-1})},\qquad |c_k|\le 1,\quad |d_k|<1\tag{1}$$ where $h[0]$ is the first sample of the corresponding impulse response, and $c_k$ and $...


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 ...


Only top voted, non community-wiki answers of a minimum length are eligible