Hoenig "Build and Use Electronic Devices" help with formulae p. 19/20 - Page 1
Forum section: electronics / beginners

author: MuseChaser
author_rank: Contributor
timestamp: February 28, 2022, 06:20:14 pm
content: Still struggling with trying to remember math skills I had many decades ago and haven't used since. At the suggestion of a mentor, I've paused working my way through the beginnings of the Horowitz/Hill AoE book and have moved into Stuart Hoenig's book, second edition. I can't get past one step between page 19 and 20, where he's discussing rudimentary determination of resistance and capacitance in terms of reactance for a filter. He describes the voltage across the capacitor, Vc, as =IXc, which makes sense, referring to Ohm's Law V=IR, with the reactance of the capacitor, Xc, subbing for R. So far, so good.The next step subs V/(Xr+Xc) for I in that formula, which I also understand, yielding Vc= VXc/(Xr+Xc). OK so far.He then writes that "Rewriting the above and recalling that Xc=1/(2πϝC), we obtain ---Vc/V = 1/(2πϝRC+1) "I know this is REALLY basic stuff, but I can't make the leap and figure out how he got there. I can get to ..Vc/V=Xc/(Xr+Xc)and then plug in for Xc ...Vc/V=(1/(2πϝC))/(Xr+1/(2πϝC))yielding Vc/V=Xr/(2πϝC) + (1/(2πϝC))^2 --- but then I'm done and I have a headache... and I'm not sure I'm even going down the right path. I also don't understand how R, the resistance of the resistor, reappears in his formula at the top of p.20, "Vc/V = 1/(2πϝRC+1)"As I said, I know this has to be incredibly basic to almost everyone here, but it's been a VERY long time since I did anything like this... can't believe I made it as far as Calc BC back in the late 70s, and can't remember this stuff. I'd be very grateful for any help.Here's picture of the excerpts from his book that has me baffled... first the circuit, then then the starting formula, and then the one I can't get to....

author: TimFox
author_rank: Super Contributor
timestamp: February 28, 2022, 07:16:30 pm
content: The correct equations are: (denotingw= 2 PIffor omega)VC/V= (1/jwC)/[R+(1/jwC)] = 1/(1 + jwRC) for the complex voltage ratio (which gives phase shift), after multiplying numerator and denominator by (jwC)The magnitude squared of the voltage ratio becomes|VC/V|2= 1/[1 + (wRC)2]R+ jXCdoes not equalR+XC

author: rstofer
author_rank: Super Contributor
timestamp: February 28, 2022, 07:26:35 pm
content: In the second equation, divide both sides by V to get Vc/V = Xc/(Xr+Xc) = (1/2πfC)/(Xr+1/2πfC)=(1/2πfC) / ( (Xr*2πfC+1)/(2πfC) ) = 1/(Xr(2πfc) + 1) but Xr = R so Vc/Vo = 1/(2πRC+1)The numerator fraction disappears when you invert and multiply by the denominator (after expanding to get a common denominator.I think...

author: TimFox
author_rank: Super Contributor
timestamp: February 28, 2022, 07:54:14 pm
content: No.The reactance "XC" is a real number, but the impedance of the capacitor is imaginary:ZC= -jXC.The impedance of the resistor is real:ZR=R

author: MuseChaser
author_rank: Contributor
timestamp: February 28, 2022, 11:20:09 pm
content: Thank you to BOTH of you. I've got it now. You're both right; in this section of the book, written to get people with very limited background knowledge and skills (like me) up and running quickly, Hoenig is the first to admit that some critical details are left 'til later. In this section, he writes, "In the first part of this section, we will cheat a little bit by leaving otu some complex and confusing material in order to provide you with theconceptof a filter..... (The real EEs who read this book will be screaming, "No!, The numbers won't come out right!" In the example I'm working through, he leaves out the phase variable and suggests for now that we just view Xr as R, at least if I understand the text correctly, when he says, "If we get a little sloppy,... we can define the resistance of a resistor as areactance, Xr."@rstofer provided me with the clue/refresher that I embarrassingly needed... how to deal with flipping the complex fractions and solving unlike denominators. I know... kid stuff... I just couldn't see it. I do now, and have successfully gone through all the steps myself.@TimFox - I'm not up to speed on what "j" represents yet, but I'm assuming it's the phase state that's being left out? In any case, I'll definitely keep this thread handy and refer to it once things get a little less "sloppy" and more specific in the book. Thank you very much!

author: Benta
author_rank: Super Contributor
timestamp: February 28, 2022, 11:49:14 pm
content: Quote from: MuseChaser on February 28, 2022, 11:20:09 pmNo, it's much more basic.Natural numbers can be laid out on a horizontal line. Like: -2, -0.33, 0, 0.5, PI, 10 etc.They're one-dimensional on an X-axis.Complex numbers are two-dimensional and have both an X- and a Y-axis component. The "j" denotes the Y-part.So for the same numbers as above, the complex notation would be:-2+j0, -0.33+j0, 0+j0, 0.5+j0, PI+j0, 10+j0But what if the "j" part is not zero? Example: 1+j1. This gives the X/Y coordinates 1,1 and is a vector.The length of the vector is sqrt(2) (Pythagoras) and the angle is PI/4 or 45 degrees.Now you have your magnitude and angle (which may be interpreted as phase, depending on context).

author: Kleinstein
author_rank: Super Contributor
timestamp: March 01, 2022, 12:23:21 pm
content: In the EE context j is used for the complex unit or square root of -1. In normal math the letter i is used, but this could be confused with some current and thus the letter j.The complex number reprentation is commonly used for AC signals. So an AC voltage is decribed as (A+jB) * exp( j*w*t) instead of a C*cos( w*t + phase).

author: rstofer
author_rank: Super Contributor
timestamp: March 01, 2022, 04:10:02 pm
content: OP:As you wander through electronics, you may want to visit KhanAcademy for refreshing in math. They now want to specialize in K-14 which is a change I haven't thought about.https://www.khanacademy.orgThey also have an EE program which is clearly above K-14https://www.khanacademy.org/science/electrical-engineeringFor more advanced things,https://www.3blue1brown.com/is very good. The animations are excellent!If you find you have to plot graphs of functions (like 1-et/RCplot as y=1-e-x), the capacitor charge equation, thenhttps://www.desmos.com/calculatoris the place to go. You can see how fast the capacitor gets to nearly 100% at 6 * Tau where Tau is the time constant R (Ohms) * C (Farads). Pay attention to the fact it is Farads, not microfarads or nanofarads. If you use a calculator and x=6, you will find that at Tau = 6, the capacitor voltage is 99.75+% of the applied charging voltage. See attached graph from Desmos.If you want to find solutions to things,https://www.symbolab.com/is truly useful.If you don't mind spending a few bucks,https://calcworkshop.com/is an excellent site where instruction is more like a classroom/lecture type of thing. Jenn, the site owner, does an outstanding job. I subscribed for 4 years while my grandson got his degree in Applied Mathematics. Worth every penny! Especially when compared to books and tuition.

author: TimFox
author_rank: Super Contributor
timestamp: March 01, 2022, 05:30:31 pm
content: I have posted this anecdote before on this site, but it may bear repeating.The late Professor Ugo Fano at the University of Chicago was giving a lecture on how to compute macroscopic quantities with quantum mechanics.His example was electrical polarization in a dielectric as a function of frequency.He set up the equations for a "perturbation" calculation, which involved the Hamiltonian (energy) of the E-field interacting with bound electrons.He then expressed the external E-field as a Fourier expansion, an integral over frequencywof termsE(w) exp(iwt) .A theoretically-minded student in the front row objected, "Dr Fano, that Hamiltonian is not Hermitian!", by which he meant that energy is real-valued, but the individual terms in the integral were complex.Dr Fano replied by erasing "i" and replacing it with "j", proclaiming that now it was Hermitian.