# Tag Info

5

In general there is no straightforward analytical solution. As you know, you need to solve $$\left|H(e^{j\omega_c})\right|=\frac{1}{\sqrt{2}}\tag{1}$$ for $\omega_c$, where it is assumed that the maximum filter gain equals $1$. For Butterworth filters, the specified cut-off frequency always equals the $3\textrm{ dB}$ frequency. This is not the case for other ...

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? ...

4

The best book I know that exclusively treats digital filters is Digital Filter Design by Parks and Burrus. But any good textbook on signal processing has one or more chapters about properties and design of digital filters. And since the most common IIR filter designs are based on analog filters, these books usually also treat the classic analog filter ...

3

For continuous-time systems, you obtain the frequency response by evaluating the transfer function $H(s)$ on the imaginary axis $s=j\omega$ (assuming stability). In discrete-time, you get the $\mathcal{Z}$-transform instead of the Laplace transform, and the imaginary axis is replaced by the unit circle. So the frequency response is obtained by evaluating the ...

3

Online, I learned a lot from the sections of Introduction to digital filters with audio applications by Julius O. Smith III. In addition to the series of books by Matt L., I suggest another from Manolakis, and two from Maurice Bellanger (one of the references is in French, there might be a translation), who cared for precision in filter coefficients ...

3

If you are "observing" the source, this implies there is some sort of information you are looking to get out of it, whether it be the total background noise, interference levels etc. Do you find "value and use" in the RF signal's magnitude versus time? What about the RF signal's phase versus time? The IQ representation gives us both of ...

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 ...

2

FIR filters that have coefficients symmetric about their center coefficient(s) are linear phase. Digital pulse filters commonly in use, including a Root-Raised-Cosine filter, are FIR and have this property, so they will have will have a linear phase response. For all digital modulations, I would avoid non-linear phase response filters.

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 ...

2

Unless mentioned otherwise withing the context the classic interpretation of Second Derivative Gaussian Filter is indeed (a) in your question: $$L \left( x, y, \theta \right) = \cos \left( \theta \right) {g}_{xx} \left( x, y \right) + \sin \left( \theta \right) {g}_{yy} \left( x, y \right)$$

2

This sounds counter intuitive to many, but as long as the difference operator and the smoothing kernel are linear and space-invariant, they can be applied in any order, and thus are often combined in a single convolution operator (for more computational efficiency), for the same. For some intuition, consider that, either for the linear smoothing and the ...

2

I've used this method in Octave: function [c] = Hz_at_3dB(b, a, fs, samples) [H,W] = freqz(b,a,samples); magresp = 20*log10(abs(H)); maxresp = max(magresp); [I,~] = find(magresp < maxresp-3.0103,3,'first'); c=(fs/2 * W(I(1)))/pi; EDIT: SECTION 8.4: STANDARD RESPONSES - https://www.analog.com/media/en/training-seminars/design-...

2

Assuming that you understand the left-hand side of Eq. $(8.47)$, for understanding the right-hand side you need to know that $-1=e^{j\pi}$, and that $e^{j2k\pi}=1$. So in order to obtain all $2N$ roots of $(-1)^{\frac{1}{2N}}$ you rewrite $-1$ as $$-1=e^{j\pi}e^{j2\pi k}\tag{1}$$ from which you get $$(-1)^{\frac{1}{2N}}=e^{j\frac{\pi}{2N}}e^{j\frac{2\pi k}{... 1 Let me slightly change notation and denote the two modulation frequencies by \omega_1 and \omega_2. You need |\omega_1-\omega_2|>2\omega_{max} to be satisfied, otherwise the shifted spectral of x(t) and y(t) will overlap. If this condition is satisfied, you can retrieve x(t) by demodulation (multiplication with \cos(\omega_1t)) and (scaled) ... 1 Assuming that we're really talking about an analytic complex-valued signal with no negative frequency components, then a sampling frequency \omega_s>\omega_h-\omega_l will guarantee that the shifted spectra don't overlap, i.e., there will be no aliasing. However, it's not guaranteed that you'll have an image of the spectrum centered at DC. This is only ... 1 This is from the mathworks documentation on cheby2 For digital filters, the stopband edge frequencies must lie between 0 and 1, where 1 corresponds to the Nyquist rate—half the sample rate or \pi rad/sample. For discrete-time signals, we use the normalized frequency \omega in radians (per sample), defined as$$\omega=\frac{2\pi f}{f_s}\tag{1}$$where ... 1 You're using the function cheb2ord in the wrong way. Assuming that your lower stopband edge is 800 Hz and your upper stopband edge is 2000 Hz (note that a bandpass filter has two stopband edges, as well as two passband edges), and if the sampling frequency is 6000 Hz, then the following command should do what you intended to do: [b, a] = cheby2(10, 60, ... 1 Not sure what your application requirements are but there seem to be some issues with your implementation. Your pass band ripple of 5 dB is way too big. Something like 0.1dB would be much more typical Elliptic filters are poor choice for cross overs. The create significant phase distortions in the cross over region that you can easily see if you sum your ... 1 Stopband attenuation is just how much the filter attenuates components of the input signal that lie in the filter's stopband. Stopband attenuation is actually a function of frequency, but very often the term is used to refer to the minimum stop band attenuation, which is usually achieved at the stop band edge(s). The term stopband ripple is used for filters ... 1 Hint: Check for time-invariance of the system by working out the output for a delayed input x[n-n_0]. Is it equal to the delayed output y[n-n_0]? 1 It's important to specify at which frequency you want unity gain. But assuming you mean DC (\omega=0), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$It's also common to normalize the denominator coefficients such that a=1. In your example that would give a = [... 1 I think that with a 250 MHz sampling clock, you should be able to generate a signal with a 125 MHz max frequency. Also, what type of DDS do you use? Is it based on a table of sine/cosine or is it a CORDIC algorithm? It has an impact on precision. Is the signal path from DDS to DAC ok? Do they work at the same sampling frequency? If not, a resampling module ... 1 As suggested by Matt L., total variation denoising is probably the tool of choice. If the staircase morphology can be trusted, I would suggest looking at the notion of piecewise constant approximation. It is often meant to approximate generic functions along the idea if Riemann sums. However, this is exactly your model here, so you can introduce jump ... 1 The Laplace transform of the original differential equation is$$Cs V_o(s) = V_i(s) \left( \frac{1}{R_1} \right) - V_o(s) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)R_1 R_2 C s V_o(s) = R_2 V_i(s) - (R_1 + R_2) V_o(s).$$In the Laplace domain, the ratio V_o(s)/V_i(s) is the transfer function for the circuit,$$H(s) = \frac{ V_o(s) }{ V_i(s) } = \frac{...

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