Fundamentals Friday: Wave impedance for engineers - Page 1
Forum section: electronics / general

author: Fensterstock
author_rank: Newbie
timestamp: December 04, 2015, 10:17:04 am
content: One of the great Mysteries no doubt: What does that "50 Ohm" blabla on coax cables mean? I'm sure we all have asked that question at some point. And investigating a bit further there comes the big unknown: wave impedance. Ok. Great. But what is it? What effect does it have? What is the difference between a 50Ohm and a 75Ohm cable?Explanations on that topic in forums, wikipedia, or even at universities normally start with a variation of "Ok. Let's keep it simple.....". Problem is: what follows is definitely not simple. They go on about phasors and complex numbers casually dropping names like Fourier or Laplace into their speech... naah!I'm sure this is not the wave impedance's fault. It just needs someone who is able to explain it in an understandable way.Basically just like in the "dezibels for engineers" episode.What do you think?

author: EEVblog
author_rank: Unknown
timestamp: December 04, 2015, 10:54:22 am
content: I actually had an idea for a physical build to explain waves, but I suspect it's not that easy to tweak to get right.Might have to put David2 back onto it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 04, 2015, 11:38:34 am
content: The ratio of inductance to capacitance is all it's about. Zo = sqrt(L/C). This works for lumped filters (where the product L*C defines the time constant, and therefore cutoff frequency), transmission lines, whatever.Just as distributed filters use different lengths and widths of transmission lines (relative to the system impedance Zo), lumped filters use different values of L and C to implement interesting frequency-dependent reflections.In a distributed line, the L and C are either imagined (infinitesimal) quantities, or an average, low frequency equivalent.When any engineer speaks of "inductance" or "capacitance", that's really what they mean (whether they realize it or not): the low frequency approximation, of an impedance which exhibits similar behavior (Z ~= j*2*pi*F*L or -j / (2*pi*F*C)), over some finite range of frequencies.Ratios are very powerful. Every time you see an impedance, that is derived from reactances rather than resistances (i.e., transmission through space, rather than absorption in a load, or emission from a source), then you can make some observation about its magnitude relative to the impedance of free space (377 ohms), or its length or delay, or both.Because of this, we can derive the LF equivalents, absolutely independent of geometry, just from knowing a few bulk properties of transmission line (and assuming that it's a simple geometry -- namely, a minimum phase network). Knowing that Zo is 377 ohms, and that a piece of coax is 50 ohms, and that its velocity factor is 0.67, and that it has no magnetic loading, we know that:- The velocity c_0 of an unloaded line, of any geometry, must be c. If c_0 < c, then it's lower by sqrt(mu_r * e_r).- The line isn't magnetically loaded, so mu_r = 1 and e_r = 1/0.67^2 = 2.23. (The published dielectric constant for most polyethylenes is 2.25. c_0 is higher, and e_r lower, for teflon or foamed dielectrics!)- Z_0 is lower than 377 ohms by the same factor, i.e., due to dielectric loading alone, the impedance of "unfree" space, in the cable, must be 252 ohms.- So the rest of the reduction in impedance (50 versus 250 ohms, a factor of 5) is due to increased capacitance and reduced inductance (split evenly, or a factor of 5 in each), due to geometry.- Finally, the permeability of free space is 1.257 uH/m, and the permittivity is 8.84 pF/m (note that sqrt(mu_0 / e_0) = 377 ohms). We have all the factors where the transmission line differs from this, so now we know:- Inductivity is 5 times lower, or 0.25 uH/m.- Capacitivity is 5 and 1/0.67^2 times higher, or 98 pF/m.If you look up the datasheet for a 50 ohm coax like RG-58, you'll find very close to these numbers (one reference for RG-58 gives c_0 = 65.9% of c, and 101.05 pF/m; but no inductance figure, but we know that must be L = Z_0^2 * C = 0.253 uH/m).You can repeat this process for anywhere there are two pieces of conductor nearby, and construct transmission line equivalents all over the place. Most of the time, you'll only be using this for the low frequency model anyway (e.g., transformer leakage inductance), but keeping in mind the deeper transmission line basis is very valuable when you need to consider frequencies above the low frequency limit.Tim

author: ataradov
author_rank: Super Contributor
timestamp: December 04, 2015, 05:20:42 pm
content: There is a wonderful video about it already

author: nfmax
author_rank: Super Contributor
timestamp: December 04, 2015, 05:32:20 pm
content: Consider two parallel straight wires, of infinite conductivity. If you apply a voltage between them (somehow) so that one is positive with respect to the other, they will attract each other by electrostatic attraction. Now if you pass a current along each - the same current but in opposite directions - the magnetic interaction between them will try to push them apart. Given a specified current, adjust the voltage until the two forces exactly cancel out, so there is no net attraction or repulsion. The voltage divided by the current is the characteristic impedance of the transmission line formed by the two wires. Since the wires are lossless (and this is DC) the impedance is purely resistive.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 04, 2015, 09:39:35 pm
content: ^ A good way of putting it, at DC.It may seem very different and arbitrary that such a construction should produce such a result, but indeed, they are equivalent.The physical explanation is that, both the E and B fields (which are both vector quantities, i.e., having direction -- namely, that the magnetic field goes "around" the wires (remember the right-hand rule), and the electric field spans between them) produce physical force, and the cross product of them (which is also a vector, having a direction) is known as the Poynting vector, and indicates the direction of power flow. If the Poynting vector is exactly along the axis of the transmission line, then there is no loss or radiation, and the power simply flows from one end to the other.(Normally, the Poynting vector is very slightly canted into one or the other conductor, a result of finite conductivity. Indeed, the amount by which it's pointed inwards gives the loss coefficient of the system!)Tim

author: retrolefty
author_rank: Super Contributor
timestamp: December 04, 2015, 09:54:40 pm
content: Quote from: EEVblog on December 04, 2015, 10:54:22 amLecher Wires! I seem to recall some construction project in an ancient ARRL manual showing a movable incandescent lamp sliding between two parallel wires along with printed scale. Would be a cool retro project.

author: Fensterstock
author_rank: Newbie
timestamp: December 07, 2015, 12:39:08 am
content: Thanks a lot for your explanations! Especially Tim's calculation example was very helpful. Found some datasheets and tried to do my homework...I'll watch the video tomorrow. Looks a bit like the neat'n'tidy version of "Why is it so?"nfmax said to consider two wires.... this reads so incredibly simple, but honestly I'm still in the process of trying to put this all together in my head.Lecher lines... that sounds a bit more familiar again (Austria is not so far from where i live...). I'll have a look into that as well.

author: Fensterstock
author_rank: Newbie
timestamp: December 07, 2015, 01:01:41 am
content: Oh my God!! I think I almost get it! This is an amazing feeling when you finally understand something you never thought you ever would....You see, I'm just a hobbyist repairing old radios and I learned all that electronics stuff by trial, error and ancient books from the local library. I never thought about Poynting vectors... yetBut of course I always want to know what holds the world together at its core.So thank you so much for taking some of your time to explain it to me!

author: Maxlor
author_rank: Frequent Contributor
timestamp: December 07, 2015, 05:59:29 pm
content: So, a short at the end of a transmission line reflects the signal without inverting it, and the original and reflected signals are then added. If we make the line very short, say only three of bars in the video, would the middle bar swing at twice the amplitude of the first? I think not, but I can't say why exactly. Can someone explain it to me?

author: ataradov
author_rank: Super Contributor
timestamp: December 07, 2015, 06:07:55 pm
content: Can you point to the time for the experiment you are describing?But in general, no, it is impossible to double the frequency by using only additions. If I understand your question correctly, the middle bar will just stay put due to destructive interference.

author: rfeecs
author_rank: Frequent Contributor
timestamp: December 07, 2015, 06:40:32 pm
content: Quote from: Maxlor on December 07, 2015, 05:59:29 pmA shortdoesinvert the voltage wave. (Because at the short you have zero voltage)An open reflects without inverting.So if you had a zero length open, that would be just the generator open, and with no termination you get twice the voltage.If you had a zero length short, you short out the generator and get zero voltage.

author: rfeecs
author_rank: Frequent Contributor
timestamp: December 07, 2015, 10:06:28 pm
content: Quote from: nfmax on December 04, 2015, 05:32:20 pmThis is really a bizarre explanation for impedance. Never heard it before.I don't know a simple physical explanation for why it happens to be true.I don't see that it has anything to do with Poynting vectors.Instead of an infinite line, consider a finite terminated line. Then you have the same situation as the attached picture from the Wikipedia article on Poynting vector.It doesn't matter what termination resistor you use. Impedance matched or not, the direction of the Poynting vectors don't change.

author: Fensterstock
author_rank: Newbie
timestamp: December 09, 2015, 10:08:30 am
content: Quote from: ataradov on December 04, 2015, 05:20:42 pmThat was exactly what i was looking for. Covered all questions i had about that topic and is really easy to understand (no Poynting vectors, etc...).