Recent content by SYoung

Okay I think, after reading reading that link you gave me, I've got a better understanding of it now. Not 100% clear, but I was just curious about it and I'm really tired. Just had prelims/exams [SIZE="1"](what do you call it again? :smile:) the last 2 weeks. Anyways, Thanks a lot!

Well... nothing about these 4 terms, really. We didn't attend this part of the integrators to the matter at all but I'm just curious about it! ^^ For ω, I understand what 1/RC means. But not where it comes from, what's behind it? And j, 90°? Does this mean the real wavelength is 1/2π earlier...

Homework Statement The amplification of an integrator is: A = Vout/Vin = -Zf/Zin = +j * 1/(ωRC) The amplification of an differentiator is: A = Vout/Vin = -Zf/Zin = -jωRC Although, that's what my book says. Not that I doubt it.. But really, what do Zf, Zin, j...

Sure any example is fine! Ah so if you fill in ((6x+5)/3 - 14/3) in stead of (2x-3) you get: ((6x+5)/3 - 14/3)/(6x+5) = 1/3 - 14/3 So what you basically want is to have the same terms (or do you call it values?) in the denominator and numerator, right? Allowing you to easily divide it...

Thanks for the advise and help everybody. I'll review the polynomial long division in a few days. I've got exams this week so there's no time left for reviewing. Thanks again and see you around! :smile:

ah now I see.. how could I've been so blind <_< so you give the denominator and the numerator the same values so you can divide them by each other. Giving you the 1/2 - (3/2) * 1/(2x+1) in this case! Thanks a lot! [SIZE="1"]I was about to give up on this one since I'm running out of...

So as this tutorial says: (x-1)/(2x+1): x/2x = 1/2 then I carry the x and -1 below change the signs so they cancel out. so 2x+1 goes 1/2 times into x-1? That should be the first part of the integral.. looks like I'm dividing the diratives. But what about the -(3/2)?By the way, I'm really...

And that's what my problem is. I forgot how to do long division, it has been 5 years! That's why I asked if somebody could please show it step by step for this integral. I hope I can remember how it's done again once I see it. [SIZE="1"]I think you didn't read the whole first post

[SIZE="1"]This has been posted in another, but the wrong, section of the forum. That's why there are already quotes in it. Hello, Homework Statement I don't understand a step in the following integral: ∫(x-1)/(2x+1)dx = ∫(1/2)dx − (3/2)∫1/(2x+1)dx = (1/2)x − (3/4)ln|2x+1| + C The...

EDIT: I think I posted this in the wrong section of the forum. My apologies. So I have no need for further instructions in this thread. Thank you. Thanks! :smile: Wow that has been a looooong time! Could you please show it step by step how it's done in this case? I hope I can remember the...

Hello, I'm don't understand a step in the following integral: ∫(x-1)/(2x+1)dx = ∫(1/2)dx − (3/2)∫1/(2x+1)dx = (1/2)x − (3/4)ln|2x+1| + C The first step, where you get the 2 integrals ∫(1/2)dx and -(3/2)∫1/(2x+1)dx Where do (1/2)dx and -(3/2) come from? And where does (3/4) come...