Passive RC filter calculator - Page 1
Forum section: electronics / projects

author: Max12345
author_rank: Newbie
timestamp: October 28, 2013, 04:38:51 pm
content: HiI have built a circuit using passive RC filters. With the little bit of knowledge I have since gained, I should have gone with an active design using an opamp. Oh well...I have attached some information with 3 questions please.Thank youMax

author: minime72706
author_rank: Frequent Contributor
timestamp: October 28, 2013, 08:15:25 pm
content: From what I've seen, my friend, people here are typically against doing others' homework. You obviously just made an account today with that in mind.I'll help you with one part of your question:j? is complex frequency, though actually technically that's just the imaginary part. The whole term is (?+j?) where ? (sigma) is the real part and ? is the imaginary part. When you learn Laplace transforms in the context of electronics, which I doubt you're doing right now, you find out that s, the complex variable used in the transform, is ?+j?. When you just pay attention to frequency, such as for the Fourier transform, you only analyze using j?. This is a big one for students: When some teachers may write a transfer function, they may write G(j?) or just G(?) and they may even use them interchangeably in the same lecture. They mean the same thing and it's just preference, really. G(j?) is technically the most correct, but engineers don't seem to go out of their way to make mathematicians happy. G(s) is something different entirely and is basically the same as saying G(?+j?). The ? in (?+j?) allows you to capture amplitude effects as well as frequency. You can model both decay and growth of the amplitude, which is important in real-life situations. Working with frequency only assumes that the signals go on forever unchanged, which, given that the universe will eventually experience heat death eventually, is impossible.EDIT: WTF? What kind of EE-related forum doesn't use proper unicode? The lowercase omega is just '?'...

author: Max12345
author_rank: Newbie
timestamp: October 28, 2013, 09:47:47 pm
content: Lol, I am far from being a student, in the formal sense of the word anyway...Actually I registered on 9 October.Thanks for taking the trouble to explain the lower case omega. I appreciate it.I discovered that it doesn't really matter which way round the 2 halves are put together, although the practice is to place the Lowpass section first.Do you know of a calculator by any chance?ThanksMax

author: minime72706
author_rank: Frequent Contributor
timestamp: October 29, 2013, 02:10:54 am
content: Yes, it shouldn't matter, though I wonder if there's anything to be said about input and output impedance. Glad that you followed my explanation despite the Greek characters not coming through.sigma plus j (or i) omegaI would solve the thing using mesh analysis along with basic Laplace transforms, but there's probably a less-advanced way because I believe I was given problems like this prior to learning Laplace. (i.e. a capacitor becomes 1/Cs, an inductor is Ls, and a resistor is just R). Just pretend the sigma in the s term doesn't exist.It's been a long time since I've used this stuff.

author: Max12345
author_rank: Newbie
timestamp: October 29, 2013, 06:04:26 am
content: Thanks minime72706, I'll look into it.Max

author: xchip
author_rank: Regular Contributor
timestamp: October 29, 2013, 10:02:02 am
content: I thought that representing the input signal as jw would mean that you are making the amplitude of the input signal as 1 to make computations simpler...A*exp(jw) -> 1 * exp(jw) -and then you write this as jw to simplify

author: minime72706
author_rank: Frequent Contributor
timestamp: October 29, 2013, 08:53:36 pm
content: You can still have a magnitude with no real part - remember it's the length of the hypotenuse of the right triangle that is formed by the magnitude of the real and imaginary part that determines the signal magnitude. A e^st where s is purely imaginary describes an oscillating sinusoidal signal and when s has both a real and imaginary part, you break it apart and end up with e^at * e^jwt.The real exponent behaves as you'd expect - it grows or decays. I used 'a' in lieu of sigma and 'w' in lieu of omega.

author: xchip
author_rank: Regular Contributor
timestamp: October 30, 2013, 01:03:51 am
content: I never said otherwise

author: minime72706
author_rank: Frequent Contributor
timestamp: October 30, 2013, 02:09:06 am
content: That's not the way I understood your post. Doesn't really matter, though.