In s-domain, is this unit Amps, or Coulombs ? - Page 1
Forum section: electronics / beginners

author: MathWizard
author_rank: Super Contributor
timestamp: May 22, 2023, 02:51:27 am
content: When you take the Laplace transform of the current of a capacitor, and so you take the LT of(C times (dv/dt)), and I get the right answer, like the pic, but how are the units for the steady state/initial condition term, (-Cv(0)) , not just Coulombs ? When I do the integration by parts, I don't see it either, but the 1st term is admittance times volts = amps. And then when they draw it they label the current source equivalent value as just Cv(0).What am I missing here, I'm looking at a few pdfs, but they are just notes. Unless i(t) goes to I(s) with units [sec*C/s]and v(t) goes to V(s) with units [sec*V] from integration, (and I don't mean convolution *)https://guitar.ucsd.edu/mauricio/courses/mae140/lectures/6sdomain.pdf

author: Benta
author_rank: Super Contributor
timestamp: May 22, 2023, 10:24:43 am
content: Amperes.The equations say it all:I: currentV or U: voltagecoulombs would be Q.

author: jwet
author_rank: Frequent Contributor
timestamp: May 22, 2023, 11:35:06 am
content: I haven't thought about Laplace in 50 years- they're worth learning but the fact that I haven't thought about them since school should tell you something...I think that cheat sheet that you posted is incorrect and the units are inconsistent as you've noted.It seems like the initial condition for current should be the current at t(0) and those for voltage would we a voltage. Cv is indeed charge on the cap- current is its derivative. Similarly Li is the flux in the inductor- its derivative is voltage.The units should be consistent I think whether s domain or time domain. Talk to the author of that slide, your prof or TA? Please post the outcome if I'm wrong.Interesting morning distraction.

author: Benta
author_rank: Super Contributor
timestamp: May 22, 2023, 06:50:51 pm
content: Quote from: jwet on May 22, 2023, 11:35:06 amThere's no Cv. What you have is C x VC(s)All the equations are consistent and deal only with voltage, current, admittance and impedance.The fonts chosen are extremely unfortunate and make reading the equations very difficult, also the lack of multiplication dots obfuscate everything.Write it out on a piece of paper in a readable fashion and you'll see the equations are OK.EDIT.As an example, this is what the IC(s) equation should be read as:

author: MathWizard
author_rank: Super Contributor
timestamp: May 24, 2023, 12:04:27 am
content: Well again what am I missing, the last term is capacitance times volts, so [C^2/J][J/C]= as usual, and in the integral that term comes from dv/dt getting integrated, and then multiplied by C. So isn't the unit of I(s) actually coulombs, or does Cv(t=0) get multiplied by a hidden (1/) to make it amps? Otherwise it still looks to me like adding apples and oranges. With the s=jw on the 1st term, yes it should be I(s) is still amps, but yeah where's the thing to make the last term match.I should try a few examples, I was reviewing some other stuff yesterday.

author: jasonRF
author_rank: Regular Contributor
timestamp: May 24, 2023, 09:19:08 am
content: Quote from: MathWizard on May 22, 2023, 02:51:27 amYes, this is correct. You can tell from the definition $$F(s)=\int_0^\infty \, f(t) \, e^{-st} dt $$The \$ dt \$ has units of time, so the units of \$ F(s) \$ will be [time * units of \$ f(t)\$].jason

author: jwet
author_rank: Frequent Contributor
timestamp: May 24, 2023, 01:59:21 pm
content: Mathwizard- I agree with you, I don't get it it. Please post if you get it to make sense. Nothing said so far looks correct to me but all this theory is getting a bit foggy after all these years.

author: switchabl
author_rank: Frequent Contributor
timestamp: May 24, 2023, 03:35:06 pm
content: Indeed, [X(s)] = [x(t)] * time because of the integration (or [X(s)] = [x(t)] / frequency if you prefer).Left side:[Ic(s)] = current * timeRight side:[s * C * Vc(s)] = frequency * capacitance * (voltage * time)= 1/time* (current * time /voltage) * (voltage*time)= current * timeand[C * vc(0)] = capacitance * voltage= (current * time /voltage) *voltage= current * time

author: jwet
author_rank: Frequent Contributor
timestamp: May 24, 2023, 03:47:38 pm
content: Switchabl- this is very good and I have to go have some quiet time in a Starbucks with some old books. Q=CV and Q=It. I must have known this once...?

author: MathWizard
author_rank: Super Contributor
timestamp: May 26, 2023, 11:03:21 am
content: OK thanks, of any of the video's I watched on it, they never mentioned that, and I guess I'm rusty on physics , so I wasn't sure or remembering it for sure.I still never made it back to that step of the examples, I'm trying to nail down some more stuff about convolution integrals, so that don't seem so hand-wavy.