I have a hard time understanding how one would come up with \$H(j\omega )\$ given a circuit for an unknown filter.

The circuit looks like this: enter image description here

I know that: \$H(j\omega ) = \frac{Y(j\omega )}{X(j\omega )}\$ , \$\delta = \frac{R}{2L}\$ , \$\omega _{0} = \frac{1}{\sqrt{LC}} \rightarrow \omega _{0}^{2} = \frac{1}{LC}\$ also \$\underline{P} = j\Omega = \frac{j\omega }{\omega _{0}}\$

I'm not expecting spoonfeed since i want to learn, but general direction at what i should look at or hints would be appreciated.

  • \$\begingroup\$ when you mention Digital filter design in your title, this is a whole different fishpond from this entirely. then you are looking at discrete time signals, stuff such as FIR/IIR filters and all the fun and games that go along with that en.wikipedia.org/wiki/Finite_impulse_response \$\endgroup\$ – Thefoilist Nov 27 '18 at 22:36
  • \$\begingroup\$ how I would start is by expressing everything in phasor domain, such that you have all your impedances, and you can use the same reduction techniques you learnt in your first year electronics class. then you can get to a transfer function fairly quickly, and you can make a bode plot from that and see all the characteristics \$\endgroup\$ – Thefoilist Nov 27 '18 at 22:38
  • \$\begingroup\$ @Thefoilist Well goal is to model the analog filter with an IIR-Filter. I know dsp basics but im lacking knowledge regarding analog signals. Also didn't have any electronics class since im studying computer science and am kinda doing an excursion into signal processing. Anyways, i would need H(jw) to be able to do z-transform on it and eventually calculate the filter coefficients. \$\endgroup\$ – liquid Nov 27 '18 at 22:55
  • \$\begingroup\$ yes, now I understand :) I think this (electronics-tutorials.ws/accircuits/ac-waveform.html) will be a useful start for you in Phasor notation, in particular chapters 4-8. Here is a tool to help you check your equivalent series impedance calculations as well to build your confidence :) keisan.casio.com/has10/… \$\endgroup\$ – Thefoilist Nov 27 '18 at 23:07
  • \$\begingroup\$ also, what is the V0 block representing? I havent seen this notation before \$\endgroup\$ – Thefoilist Nov 27 '18 at 23:08

I have a hard time understanding how one would come up with H(jω) given a circuit for an unknown filter.

Ignoring the gain stage because it's trivial, you need to calculate Vo/Vi where Vo is the voltage across the resistor: -

\$\dfrac{V_O}{V_I} = \dfrac{R}{sL + \frac{1}{sC}+R} = \dfrac{sCR}{s^2LC + sCR + 1} = \dfrac{s\frac{R}{L}}{s^2 + s\frac{R}{L}+\frac{1}{LC}}\$

The denominator is of the standard form: -

$$s^2 + s2\zeta\omega_n + \omega_n^2$$

Where \$\omega_n\$ is the natural resonant frequency = \$\dfrac{1}{\sqrt{LC}}\$

\$\zeta\$ (or zeta) is the damping ratio.

  • \$\begingroup\$ Thanks! So many questions tho, since i'm really just getting me feet wet with all of this. Is it correct that you could chose \$s = j\omega\$ ? Also would you happen to have a source where i can read about this? I think i'm lacking some basics to fully understand whats going on. \$\endgroup\$ – liquid Nov 28 '18 at 12:31
  • \$\begingroup\$ There's an interactive tool here that covers low pass and high pass 2nd order RLC with the math but the band-pass is still under construction. Yes, s = jw and the site mentioned uses derivations with jw. \$\endgroup\$ – Andy aka Nov 28 '18 at 12:45
  • \$\begingroup\$ Cheers. Well so it's a RLC Bandpass 2nd order:)! Found a transfer function, that looks abit different compared to the one you gave me? \$\frac{U_{a}}{U_{e}} = \frac{1}{\frac{LC}{s^{2}+s\frac{1}{RC}+\frac{1}{LC}}}\$ with cutoff frequency \$f_{0} = \frac{1}{2\pi\sqrt{LC}} \$ \$\endgroup\$ – liquid Nov 28 '18 at 13:08
  • \$\begingroup\$ That isn't a bandpass filter; it's a low pass filter. Where did you get the formula? \$\endgroup\$ – Andy aka Nov 28 '18 at 13:13
  • \$\begingroup\$ Strange, got it from here bottom of the page \$\endgroup\$ – liquid Nov 28 '18 at 13:19

You could express the impedances of the components in terms of s, i.e. sL, 1/sC, R. This should allow you to determine the transfer function H(s). This, I reckon, will give you a zero at the origin and two poles on the left half of the s-plane. Once you have a transfer function in terms of s, you can transform this to the z-plane. One method is the bilinear transform. Substituting for s will lead to an H(z). H(z) can be realised as an IIR filter.

  • \$\begingroup\$ one can also remain in the frequency domain for the initial calculations, mapping the frequency domain to the Z-domain is done by wrapping the linear frequency axis around the unit circle \$\endgroup\$ – Thefoilist Nov 27 '18 at 23:48

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