why does a square wave appear "rounded" through an RC circuit? - Page 1
Forum section: electronics / beginners

author: ice595
author_rank: Regular Contributor
timestamp: October 04, 2014, 01:24:43 am
content: Title saids it all. Why would a square wave's sharp corners get rounded when passed through a RC circuit?is it A) its the classic RC circuit, the capacitor does not charge instantly........or B) the RC circuit is acting like a low pass filter and filters out all the high frequency components of a square waveor BOTH

author: rs20
author_rank: Super Contributor
timestamp: October 04, 2014, 01:30:02 am
content: Both. A) is the time-domain explanation, B) is the frequency-domain explanation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 04, 2014, 05:36:08 am
content: They are one in the same, the time and frequency domain aspects of the same characteristic.The deep equivalency between time and frequency, in fact, connects with the Heisenberg Uncertainty Principle. The same duality appears many times in science!Tim

author: ron
author_rank: Regular Contributor
timestamp: October 04, 2014, 06:27:53 am
content: https://en.wikipedia.org/wiki/Fourier_transform

author: hamster_nz
author_rank: Super Contributor
timestamp: October 04, 2014, 07:06:06 am
content: QuoteI somehow feel that the (b) answer is incomplete.If you put a step through an RC filter you don't get a smooth curve for the output. It has a sharp kink as the cap charges rapidly, then it flattens out as the voltage across the resistor drops.If it was as simple as answer B suggests the it would be symmetric and not have the sharp kink (which must have some high frequency components in it - how else can such a sharp transition occur)?Maybe the phase effects of the RC filter explains this when you have a periodic input.I might have to do some math tonight!

author: LvW
author_rank: Frequent Contributor
timestamp: October 04, 2014, 08:22:25 am
content: Quote from: T3sl4co1l on October 04, 2014, 05:36:08 amTim - may I ask you to explain a little the connection between the question under discussion and Heisenberg`s theorem? Thank you.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 04, 2014, 08:27:21 am
content: Quote from: hamster_nz on October 04, 2014, 07:06:06 amBecause it is indeed an oversimplified statement of the facts!QuoteOf course, it would be absurd to think a mere single pole RC filter would removeALLfrequencies above some arbitrary point (say, the cutoff frequency) -- the RC filter is the slowest of all known filters*, and passes a considerable amount beyond the (fairly arbitrarily defined) "cutoff frequency".*Synthesized with lumped constants (R, L, C), that is. There are a few "spooky physics" systems where diffusion (thermal, ionic, etc.) yield a bizarre sqrt(f) (i.e., a "half-th order filter") response -- sometimes over quite a wide frequency range! Which, relating back through the Fourier transform, connects to even heavier mathematical theory, such as fractional-order differential calculus, special functions and more!Tim

author: rs20
author_rank: Super Contributor
timestamp: October 04, 2014, 08:38:30 am
content: Quote from: hamster_nz on October 04, 2014, 07:06:06 amTo clarify, yes, once you take the phase effects into account, you will get the correct shape. Both with periodic and aperiodic signals.

author: tggzzz
author_rank: Super Contributor
timestamp: October 04, 2014, 08:44:26 am
content: Quote from: hamster_nz on October 04, 2014, 07:06:06 amSince you are prepared to do some work in order to learn, you stand a better than average chance of succeedingHave fun, safely.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 04, 2014, 09:05:06 am
content: Quote from: LvW on October 04, 2014, 08:22:25 amThe "ten thousand mile" overview is simply:E&M deals with the solutions to differential equations (namely, Maxwell's equations), defined by boundary conditions (e.g., the RF fields inside a metal box).Quantum mechanics deals with the solutions to differential equations (namely, Schrodinger's equation), defined by boundary conditions (e.g., the energy potentials around an atom).In E&M, the tradeoff arises between length or frequency and time: a high frequency has a short wavelength or period, and vice versa. (Ok, so we weren't talking about E&M in depth, just a very simple discrete circuit case. What really matters is, basically, that you can draw up a differential equation for the system, which you can here.)In QM, the tradeoff arises between momentum and position: a high certainty in momentum has a small certainty in position, and vice versa. (Or various equivalent quantities, such as energy and time, etc.)It's still probably hard to see, at a glance, what the significance is, or whether this has provided any illumination with respect to QM. But the fact is simple: both are dealing with waves (one of energy, the other of matter), and both have the same X-vs-Y tradeoff, a necessity of their shared common mathematical basis.A very different, but perfectly equivalent way to think of it, is this:Suppose you have a signal that has a very well defined frequency. How long does it persist for?Well, the most ideal frequency is a pure sine wave.It goes on from t = -infty to +infty.So, it persists forever. (And, it's worth noting, is non-causative: it has existed for all time, and will continue to exist for all time. No real waveform humanity has ever played with, or that can even exist in the real world for these last 13.8 or so billion years, can ever beexactlya sine wave, as such. Unless you believe in... God Waves, or Intelligent Waveform Design, or something like that...)Or...Suppose you have a signal that has a very well defined time. What frequency is it?Well, the most ideal impulse is a Dirac delta (a spike of infinite height and infinitesimal width at t = 0, and zero elsewhere; such that the integral over time equals 1, so you can perform calculus on it -- admittedly, it's not a "real" function).The FT of delta(t) is 1 (give or take a constant). That is to say... it containseveryfrequency. Um... so, since there's no peak... it's alsonofrequency... great.Then, how about...Since an infinite wave won't help us, nor will an indefinite frequency, suppose we compromise. Maybe a wave that goes to zero at both ends will also have a reasonably defined frequency.So, you might ask: is there a function, which when I take the FT of it, returns itself -- give or take some constants (maybe it's scaled or offset, but it generallylooksthe same)?It turns out, there is an infinite number of these -- eigenfunctions, they're called (why they're called this, again, goes back to ugly theoretical roots, which don't matter so much for present purposes). A great example is the function exp(-x^2): the Gaussian function. If you've ever heard of an ideal Gaussian step response, or frequency response -- wonder no more, because they are, in fact, the same form! (This function shares the unique property that it is maximally flat (no peaking) in both frequency and time -- the Bessel filter is often touted for its step response, but it does ring slightly, particularly the higher order formulations: Bessel functions are not eigenfunctions of the Fourier transform. Likewise, the Butterworth filter is "maximally flat" in frequency, and definitely -- and necessarily -- rings in the time domain. An example of a maximally-flat-step-response filter might be a repeated-real-pole filter, like a chain of R+C + op-amp followers -- which has a decidedly slow frequency cutoff, almost certainly not optimal.)Now, if we really want to get down to brass tacks, we have to know that "down to a constant" part, too. It turns out, if you make the time-domain Gaussian narrower, the frequency gets wider, and vice versa. In the extreme example, you could make it infinitely sharper, so that it has approximately zero duration -- thus approximating the Dirac delta impulse, which has such a broad frequency range that it apparently contains all frequencies.As it happens, the ratio between time width and frequency width is a constant.This constancy,is the Heisenberg uncertainty principle.Plain and simple.Not for QM necessarily, but in a much more general sense, as a direct result of the Fourier transform, or any systems which exhibit analogous qualities (which are a lot, so it's very useful).Tim

author: LvW
author_rank: Frequent Contributor
timestamp: October 04, 2014, 11:18:39 am
content: Hi Tim - thank you for this long elaboration. At a first glance, I didn´t get everything but, of course, I agree that every signal still contains transients (steady state cannot be reached in finite time).However, I must admit that there was a misunderstanding between us: I thought you were claiming that the mathematical connection beween time and frequency DOMAIN has some relation to Heisenberg.Thank youLvW

author: max_torque
author_rank: Super Contributor
timestamp: October 04, 2014, 12:36:52 pm
content: After reading that, i'm uncertain how Heisenberg was ever so certain about his uncertainty...... ;-)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 04, 2014, 07:08:39 pm
content: Quote from: LvW on October 04, 2014, 11:18:39 amMmmm, as far as direct theoretical or historical connection, I don't know. I'm sure the mathematical formalism was there, in a broad sense, though. That theory being about generally solving differential equations, especially equations of motion (E&M kicked off in the 19th century, but it wasn't of much relevance until the 20th; mechanics and fluid dynamics were the stars of the mid-to-late 19th century!).Tim

author: Zeta
author_rank: Contributor
timestamp: October 05, 2014, 06:56:57 pm
content: Quote from: T3sl4co1l on October 04, 2014, 05:36:08 amlol what?

author: macboy
author_rank: Super Contributor
timestamp: October 05, 2014, 07:24:17 pm
content: Quote from: hamster_nz on October 04, 2014, 07:06:06 amThe RC is merely a first order low pass, and it is a perfect explanation if you run through the math.

author: c4757p
author_rank: Super Contributor
timestamp: October 05, 2014, 07:48:38 pm
content: Quote from: hamster_nz on October 04, 2014, 07:06:06 amWell yeah, it's not a brick-wall filter with 0dB gain in the passband and -inf gain in the stopband, it has a smooth rolloff.