5

$$\left . H_d(z) = H_c(s) \right |_{s = \frac{2}{T\,s}\frac{z-1}{z+1}}$$ describes a transfer function in the $z$ domain that you can easily translate into a difference equation and realize in software. $$\left . H_d(\omega) = H_c(\Omega) \right |_{\Omega = \frac{\omega}{T\,s}}$$ describes an idealized frequency response that you would like $H_d$ to have ...


4

Short Answer: Basic RLS (no forgetting, no weird weighting, etc.) is ALWAYS Lyapunov stable. If the regressor sequence for the LS problem is persistently exciting--which is data and problem dependent, not algorithm dependent--then RLS is exponentially stable. So I don't know what you mean by "LMS is more stable than RLS"--more stable in what sense? ...


3

A discrete-time first-order high pass filter with unity gain at Nyquist and a zero at DC is described by the following difference equation: $$y[n]=\frac{1+\alpha}{2}\big(x[n]-x[n-1]\big)+\alpha y[n-1],\qquad -1<\alpha<1\tag{1}$$ Its transfer function is given by $$H(z)=\frac{1+\alpha}{2}\frac{1-z^{-1}}{1-\alpha z^{-1}}\tag{2}$$ Evaluating the squared ...


3

I checked your code in my PC, you need just to delete the delay added before the filter. For example, you can use: U_aft_fil = U_aft_fil(fil_delay+1:end); Then when filtering it again at the receiving side, you delete it again : U_r_fil = U_r_fil(fil_delay+1:end); Good luck


2

HINT: The frequency response has the form $$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$ which can be rewritten as $$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$ Now note that the term in brackets is purely real-valued. I trust that you can take it from here.


2

It is perfectly possible to build a notch filter with either FIR or IIR techniques that has little delay for frequencies away from the notch. However frequencies in and close to the notch must have rather large group delay, and therefore if there is a requirement for linear phase or constant group delay across all frequencies, then group delay inevitably is ...


2

Here are couple examples: % R is the resistance value (in ohms) % C is the capacitance value (in farrads) % fs is the digital sample rate (in Hz) % Constants RC = R * C; T = 1 / fs; % Analog Cutoff Fc w = 1 / (RC); % Prewarped coefficient for Bilinear transform A = 1 / (tan((w*T) / 2)); % using Bilinear transform of % % 1 ( 1 - z^-1 ...


1

This is from the mathworks documentation on cheby2 For digital filters, the stopband edge frequencies must lie between 0 and 1, where 1 corresponds to the Nyquist rateā€”half the sample rate or $\pi$ rad/sample. For discrete-time signals, we use the normalized frequency $\omega$ in radians (per sample), defined as $$\omega=\frac{2\pi f}{f_s}\tag{1}$$ where $...


1

You're using the function cheb2ord in the wrong way. Assuming that your lower stopband edge is $800$ Hz and your upper stopband edge is $2000$ Hz (note that a bandpass filter has two stopband edges, as well as two passband edges), and if the sampling frequency is $6000$ Hz, then the following command should do what you intended to do: [b, a] = cheby2(10, 60, ...


1

Not sure what your application requirements are but there seem to be some issues with your implementation. Your pass band ripple of 5 dB is way too big. Something like 0.1dB would be much more typical Elliptic filters are poor choice for cross overs. The create significant phase distortions in the cross over region that you can easily see if you sum your ...


1

Stopband attenuation is just how much the filter attenuates components of the input signal that lie in the filter's stopband. Stopband attenuation is actually a function of frequency, but very often the term is used to refer to the minimum stop band attenuation, which is usually achieved at the stop band edge(s). The term stopband ripple is used for filters ...


1

This is just as it turns out when you do the math. The discrete-time Fourier transform (DTFT) of the sampled continuous-time impulse response $h(t)$ is $$H_d(e^{j\omega T})=\sum_nh(nT)e^{-jn\omega T}\tag{1}$$ With $$h(nT)e^{-jn\omega T}=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}\delta(t-nT)dt\tag{2}$$ this can be written as $$\begin{align}H_d(e^{j\omega T})&...


1

Your probably your best option here is "frequency warping". You can calculate the poles and zeros from the coefficients, warp the poles and zeros and then recalculate the coefficients again. Warping is a procedure the applies a conformal mapping to the poles/zeros. Specifically a conformal mapping that maps the unit circle onto itself will maintain ...


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