assignment time again input impedance to a loaded cable - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: March 26, 2017, 07:46:32 pm
content: sso once again I have 1 of my dreaded assignment questions. this 1 is about the input impedance to a cable with a load.. The line impedance is simply given in ohms as a single number.. The load impedance is given as a complex number.. I am then given a formula to use.. Slight snag being that you have to use a value obtained from a previous part of the question which I have done.. Now if I plug all of the values into this formula as given I don't get the result I expect. The line impedance is 50 ohms the load impedance is 40 plus I 30 ohms which incidentally is 50 ohms.. Well although I've not Understood much of the underlying mat in this course what I do know is that if you load a line with an impedance equal to the impedance of the line you will see the line impedance no matter how long the line is. Now if I use the load in its complex form I get 60 ohms.. Yet if I calculate it's shall we say magnitude I'm not sure if that is the correct word which is 50 ohms and plug all of these 50 ohms into the formula unsurprisingly I come out with 50 ohms.. Is the problem here that the line impedance was not given as a complex number.. The only other clue is that the line is loss less but that still doesn't help me with the characteristics of line other than a fixed impedance of 50 ohms. The formula goes something like this:Zin = Zo x ((Zl cos B + iZo sin B) / (Zo cos B + iZl sin B))wwhere B is the phase change per length which I have been able to calculate.. Line impedance is given as 50 ohms on the load impedance is 40 + I 30 Ohms

author: Vtile
author_rank: Super Contributor
timestamp: March 26, 2017, 08:42:50 pm
content: So the load is inductive as there is +j30 ?Then the load is 50 ohms in angle of +36.87 ? Well I'm not getting the question down properly.. (these 50ohm nominal line impedance cases do go to a edge of my understanding as electrical and control person)You are after series resonance isn't it? (the 12pm is good time to try to blow the dust between ears .. always)

author: Andy Watson
author_rank: Super Contributor
timestamp: March 27, 2017, 12:02:22 am
content: Can you show your working? You have to calculate using complex numbers throughout.Transmission linescando transform impedances. It's been three decades since I last tried to get my head around this voodoo, however, assuming I've interpreted the Smith chart correctly - you could see the resistive component of Zin vary from 25 to 100 ohms - depending on wavelength of the line. Similarly, the reactive component could vary from -j38 to +j38.http://www.antenna-theory.com/tutorial/txline/transmission4.php

author: Simon
author_rank: Unknown
timestamp: March 27, 2017, 06:27:51 am
content: and this is the problem, you give some details, the question gives non, it says that it is a perfect transmission line of 50 ohms and then gives the load in complex number form.I attach a screenshot from microsoft mathematica that does the calculations, The top calcula;tion being the answer to the first calculation that required me to calculate the phase delay.

author: Andy Watson
author_rank: Super Contributor
timestamp: March 27, 2017, 07:45:12 am
content: I agree with your last calculation - and the Smith chart confirms the values.Code:octave:7> Z0 = 50;octave:8> ZL = 40 + 30j;octave:9> B=pi*4/5;octave:10> Zin = Z0 * (ZL*cos(B) + i*Z0*sin(B))/(Z0*cos(B) + i*ZL*sin(B))Zin = 25.4674 + 5.9024i

author: woodchips
author_rank: Frequent Contributor
timestamp: March 27, 2017, 05:58:04 pm
content: Presumably the cable is electrically long then the input impedance is simply 50 ohms. The tricky part comes at the end of the cable and the load, because the load impedance doesn't match the cable impedance then there will be some reflection. This will then travel back to the input where it will affect the input impedance seen, and where it will reflect again. Possibly easier to calculate the final result using this stepwise method.

author: Simon
author_rank: Unknown
timestamp: March 28, 2017, 06:51:38 am
content: I was under the impression that as both impedences are 50 ohms the total would be 50 or is it because one is atraight 50 and the other is a complex impedence. The line is supposed to be ideal

author: orolo
author_rank: Frequent Contributor
timestamp: March 28, 2017, 09:02:12 am
content: The phase difference in the load affects the behavior of the line + load. If your line has impedance \$ z\$ and your load has a phase shifted impedance \$ z\cdot e^{i\theta}\$, then the reflection coefficient at the load is:\$\Gamma \ = \ \displaystyle \frac{ze^{i\theta} - z}{ze^{i\theta} + z} \ = \ \frac{e^{i\theta} - 1}{e^{i\theta} + 1} \ = \ \frac{e^{i\theta/2} - e^{-i\theta/2}}{e^{i\theta/2} + e^{-i\theta/2}} \ = \ i \tan\theta/2 \$So, depending on the phase difference between line and load, you can have anything from total transmission to total reflection, with transmitted and reflected waves phase shifted by \$\pm\$ 90 degrees.Clearly, the phase difference between line and load matters a lot.

author: Vtile
author_rank: Super Contributor
timestamp: March 28, 2017, 10:58:29 am
content: Quote from: Simon on March 28, 2017, 06:51:38 amThere is no such thing as non-complex impedance.Both are complex, but line as given (I always wonder where this 50 ohms comes from) seems to have only 50 ohms resistive part (50 ohms in angle of zero), while the load do have 40 ohms resistive part and a 30 ohms reactance part (which might be a combination of inductance and capacitance, but with more inductance in this case ending up to 30 ohm worth of inductive reactance). Complex calculus is vector calculus in the end, so you need always take account both parts of the impedance (or admittance). Complex numbers are superset of real numbers. I think you have some hick-up on the complex calculus, you might want to revisit them and the unit circle representations of it. If you ie. add them you get Z0+Zload=(50+j0)+(40+j30)=((50+40)+j(0+30))=90+j30. Just as polynomials.\$ i = \sqrt{-1} \$Freethinking part, take with grain of salt (as this is just a prototype level of thinking from me).it could actually in circuit theory be thought (while not anywhere noted as far as I know)\$ \pm i = \pm \sqrt{-1}\$ where \$ -i = -\sqrt{-1}= \frac{1}{i}\$ which is capacitance (or electric field). The positive is as described anywhere.The smiths chart stuff that this seemed to be is out of my direct curriculum.@orolo should \$ z\cdot e^{i\theta}\$ be \$ \left | z \right |\cdot e^{i\theta} = \left | z \right |\angle\theta \$, where \$ \left | z \right |= \left | a+i*b \right |=\sqrt{ a^{2}+b^{2} } \$ or am I now mixing something.

author: orolo
author_rank: Frequent Contributor
timestamp: March 29, 2017, 05:06:33 am
content: Quote from: Vtile on March 28, 2017, 10:58:29 amIt doesn't matter, the cable impedance z can be any impedance (except 0), and the load is a phasor with the same module and forming an angle theta with z. If the cable impedance is real, as in Simon's problem, things can be written as above, but for calculation purposes it works the same with arbitrary z. Thinking of Smith charts, the unit circle is being mapped into the imaginary axis; only the angle between impedances matters.@SimonTo compute the impedance of the line + load, two things matter:1) The length of the cable causes a delay from source to load, which appears as factor B in your equation.2) Part of the signal may be reflected back to the source. Since the reflection coefficent is (ZL - Z0)/(ZL + Z0), it will only be zero (no reflection) if ZL = Z0. So there will be reflections if ZL and Z0 are differentas complex numbers. So it is not enough that Z0 and ZL have the same module, they must also have the same argument.Imagine you connect a sinusoidal function generator to the cable, giving a voltage Vi. To compute the impedance, you must obtain Vi/Ii, the ratio of voltage to current into the cable. If you leave the generator working long enough to reach a steady state, you get:a. The generator injects a signal Vi that propagates from source to load along the cable. The current at the source generated by this signal is Vi/Z0.b. When the injected signal reaches the load, it will have been delayed by an angle B, since it travelled along the line: let's call it Vi(B). If Z0 and ZL are mismatched, a reflected wave will arise, with value G*Vi(B), where G is the reflection coefficient. That reflected wave will travel back to the source, getting an additional delay B when it reaches the source.c. So back at the source, you have a reflected signal G*Vi(2B), which causes a current G*Vi(2B)/Z0 that must be added to the injected current Vi/Z0.d. The total current at the source is: Ii = Vi/Z0 + G*Vi(2B)/Z0. Injected current plus reflected current, doubly delayed.From there, using Zi = Vi/Ii, G = (ZL - Z0)/(ZL + Z0) and Vi(2B) = Vi*exp(-2*i*B), and some algebra, you arrive to the sin/cos formula you used in the original post. Between the effects of delay B and reflection coefficient G, the equivalent impedance at the source can be a very different value from both Z0 and ZL.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 29, 2017, 12:17:41 pm
content: FYI, line matching occurs when the load resistance equals the line's characteristic impedance.The characteristic impedance is 50 ohmsreal, so 40 + j30 is not a match.When you have a complex source/load, the best match is given by thecomplex conjugate, i.e., negate the complex part: 40 - j30 ohms.The line is not 40 - j30 ohms, therefore it is not matched, and you can't take the easy way out of this problem.It is suggestive, that the load |Z| happens to be 50 ohms, but this doesn't appear to be relevant. Red herring, perhaps?Tim