Maybe a picture is worth a thousand words? Here's how a Gaussian pulse of variable width and its derivatives look like: As others have said, Dirac is a distribution, hence the Gaussian pulse, and its ...

Throughout the answer I will use the mathematical notations, that is, the mathematica equivalent of expressing the magnitude response of a filter in frequency domain. For this, $x$ will be used ...

You could try the exponential window: $$w_n=\frac{ \exp \left[\alpha \sqrt{1-\left(\frac{n-M}{M}\right)^2}\right]}{\exp(\alpha)}$$ $$\alpha=-427.5*10^{-6} A_s^2+0.1808*A_s-3.516$$ or the hyperbolic ...

You can use the Hilbert transform, then multiply the real part with the sine of the angle you wish to transform, and the imaginary part wth the cosine. A quick code in Octave: t=[0:0.01:4]; s=hilbert(...

Since your frequencies f2 and f3 are a multiple of f1=1k, and since your time vector is until 5 ms, that means your signal fits for a whole 5 periods, harmonics and all, so it's just ripe for a FFT. ...

Well, as far as intuitive thinking goes, I was thinking something along these terms: if you have two numbers, $a\neq b$, and you do an average between them, the average will always come between those ...

If the filter has a fixed order then increasing the sampling rate will make its response scale with frequency, thus the transition width increases relative to the sampling frequency. For example: if $... View answer 2 votes If you look at the formula and plot it, you get what you see in the other link on dsp.ee. But if you add the inflexion points and plot the consituent functions, it should become more clear: The ... View answer 2 votes I might be a tad late to this, but I'll only reply to the part about the "similarities" between the (analog) Bessel and Gaussian. They are not the same. The Bessel filter is meant to ... View answer 2 votes It looks like you're using LTspice. If so, in the LTspice group you'll find a little free utility, ltsputil, which does exactly what you want, maybe a bit more. There are also questions in the group ... View answer 2 votes I don't know if this is what you really want but, inspired by your 2nd attempt, I thought about the triangular window, which doesn't have to have the ends null (like Bartlett), but which has variable ... View answer Accepted answer 1 votes This is what it looks to me what you're doing: you're preparing a signal with mixed frequencies, then you're making a sin/cos matrix of multiple sines which are to be multiplied with the test signal, ... View answer 1 votes Looking at this page, the part where they explain "Signal Dynamics and Demodulation Bandwidth" (search for suppress), they give an example of AM demodulation by comparing simple filtering ... View answer 1 votes Well, this is one way you could do it: with LTspice, use the white() function. There are also rand() ([0,1] V, pulses) and random() ([0,1] V, smooth pulses), but white(2*fs*time) will give you what ... View answer Accepted answer 1 votes What was previously posted as an answer should have been an extended comment -- sorry for that. There simply was not enough space for what I thought I'd write about how the Octave code seemed to ... View answer 1 votes It looks like artifacts due to the derivative. I used this code in Octave: fs=300; t=[0:1/fs:2]; c=chirp(t, 20, 2, 100); s=c.*(1+0.5*cos(2*pi*10*t)); h=hilbert(s); m=abs(h); a=diff(unwrap(arg(h)))/2/... View answer 1 votes If you want to have a "numerical grasp" and you're not afraid of getting a little bit dirty, you can check the numbers with LTspice. I don't know how well you know to work with it, so I'll just ... View answer 1 votes The filter has one kernel, and adding L filters (in series) makes the input of the first filter appear convolved, L times, with the same kernel, and repeated convolution leads to a Gaussian response. ... View answer 1 votes One way to make a non-linear out of a FIR means solving for the zeroes, keeping only the inside/outside ones, then rebuilding the impulse response, and all this with equiripple filters due to the ... View answer Accepted answer 1 votes I may have fond the answer. The$Q\$ matrix, made from the sum of the Hankel and Toeplitz forms, stems from: $$cos(k\omega)cos(n\omega) = \frac{1}{2}\left[cos((k-n)\omega)+cos((k+n)\omega)\right]$$ ...

A quick test shows that you are imposing the order to be 33, but that's not enough to guarantee that the resulting transition width will cover your needs. The low order can't cover everything so the ...

Just because eigenvalues apply in linear algebra doesn't mean that the same meaning applies everywhere. Think of the Fourier transform: does it only apply to the study of heat transfer? Same here. The ...