New answers tagged

3

There's no need to use numerical methods here. The most straightforward way to compute the output is to see that the filter's impulse response is given by $$h(t)=\sum_{k=1}^Nr_ke^{s_kt}u(t)=\sum_{k=1}^Nh_k(t)\tag{1}$$ where $N$ is the filter order, $u(t)$ is the unit step function, and $r_k$ are the coefficients of the partial fraction expansion of $H(s)$: ...


1

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 explain it, feel free to ignore all the redundant info. Here you can download the archive, out of which you only need Filter.asy and filter.sub. Create a new ...


3

A plot of the normalized impulse responses, for the n = 2 through 10 Butterworth low pass filters, are given by H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976, p. 113. This is shown below. They do not give the h(t) expressions in the book, at least where I have looked thus ...


1

First you need to find (calculate or look up) the transfer function of that circuit. It has the form $$H(s)=\frac{a}{s^2+bs+c}\tag{1}$$ where the constants $a$, $b$, and $c$ depend on the values of the resistors and capacitors. The $3$ dB cut-off frequency $\omega_c$ can be found by solving $$\big|H(j\omega_c)\big|^2=\frac12\big|H(0)\big|^2=\frac12\left(\...


2

Converting the analog filter $H_a(s)$ into a digital filter $H_d(z)$ using the bilinear transform where T is the sampling period: $\Large H_d(z) = H_a(s)\bigg|_{s=\frac{2}{T}\frac{z-1}{z+1}}$ Example: Given a first order Butterworth filter $H_a(s) = \frac{1}{1+RCs}$ $H_d(z) = H_a\bigg( \frac{2}{T} \frac{z-1}{z+1} \bigg)$ $H_d(z) = \frac{1}{1+RC\Big(\...


Top 50 recent answers are included