# electromagnetism homework

Discussion in 'Homework Help' started by liam1234, Jan 5, 2015.

1. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
Q1(a) A solution of the second order partial differential equation (d^2f)/(dx^2) = (1/c)(d^2f)/(dt^2) is of the form (x+-ct). Explain the relationship between this requirement, and standing wave solution found by "separation of the variables" which are of the form h(x)g(t).

Q1(b) If the length of the short-circuited transmission line is between lambda/4 and lambda/2 for a given operating frequency, what can we tell about the impedance at the end of this transmission line?

help would be much appreciated!

2. ### WBahn Moderator

Mar 31, 2012
20,213
5,743
This is not the Homework Done For You forum. YOU are expected to show YOUR best attempt to answer YOUR homework -- WE can then try to give you feedback, hints, and suggestions to HELP you get from where you are to where you need to end up.

3. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
Q1(a) in the notes it shows getting the second order of phase and then its just a matter of putting two answers together. I don't understand it.

Q1 (b) i get that lambda/4 and lambda/2 are opposite sides of the smith chart but I'm not sure if this is relevant to the question or not. I'm not sure how to relate just what I'm given to impedance.

The answers I'm looking for are only a sentence are two long. I half know what I'm looking for but would like to be gently pushed in the right direction.
cheers

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
For Q1 (b):
You plot the line load impedance as 0+j0 Ohms at the center left hand side of the chart. So that's where the Tx line has zero length. Looking at the Smith Chart, as you increase the line length from zero to 0.25 lambda (half revolution) up to 0.5 lambda (full revolution) what corresponding impedance do you read off the chart on the same radius circle?

5. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
Is it just that it has 0 impedence? it's on the very outside circle and you could also say it's real because it's on the horizontal?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The impedance is zero at the load. It has neither real nor imaginary non-zero magnitude. Something akin to the origin of a complex impedance 'plane'. All possible complex impedances must be encompassed within & on the Smith Chart outer boundary circle - including infinite values.
The point is that as one begins moving along the (lossless) line towards the source (generator) the impedance looking back along the line towards the load changes to something non-zero. Look at the Smith Chart and you will hopefully see this. You will note that on the outermost circle the real value remains zero until exactly a quarter wavelength is traversed - while the imaginary value continually changes over that length. What value does the real value approach as one approaches the quarter wavelength position? Look at the circles of constant resistance as one approaches this position. They are decreasing in radius as they increase in magnitude - approaching zero radius at the quarter wavelength point. What 'value' is the constant resistance circle of infinitely small radius? How about the imaginary component?
You may get a better grasp of how this goes if you consider the case of a small non zero load resistance with zero imaginary part. Watching how the impedance varies along a more inwards concentric lossless transmission line circle may be conceptually easier to interpret - thereby better informing you regarding impedance variation along the outermost circle.

Last edited: Jan 6, 2015
7. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
ok I realise I have been using lambda/2 and lambda instead of what's given.

when you say move a quarter wavelength from 0+j0 you end up at 0+j1. there is no circles near it? I don't understand you here. I get that the imaginary values change but the real value will stay 0. at 0+j1 I know this is a special cause but cant remember why.

the circles of resistance are just the real values?

at lambda/2 that is the point each circle is touching so then you can say it is on the circle 1+j1 which is what you want?

thanks for your help. sorry I'm taking so long to cop this, I haven't done it in quite a while now. It's revision for an exam

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The impedance changes to 0+j1 when you move lambda/8 - not a quarter wavelength.
As I later edited in my previous post - it may be helpful to observe impedance changes along a loss-less line for a small non-zero purely resistive load value - say 0.01 ohms on the normalized Smith Chart or 0.5 ohms on a 50 ohm based case.
For the normalized case of 0.01 [ohms] at the line load point I have approximately from a Smith Chart App....
1. @ 0.05 lambda from load Z=0.01+j0.33
2. @ 0.10 lambda from load Z=0.02+j0.73
3. @ 0.15 lambda from load Z=0.04+j1.38
4. @ 0.20 lambda from load Z=0.12+j3.07
5. @ 0.25 lambda from load Z=89.5+j0.00
Notice what happens at the quarter wavelength position - the impedance rises very sharply and becomes purely real. What happens if the load is actually zero? The loss-less line impedance approaches infinite resistance at the quarter wave point.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
By way of a further concrete example to illustrate my earlier comments, I've attached the case of a normalized load impedance of 0.03+j0 [ohms]....

What you may notice is that as the distance from the load reaches 0.25 lambda, the locus of the line impedance intersects a normalized resistance circle of approx 30 ohms. Extending the concept to the case of a zero load resistance, the line impedance locus would intersect a circle of infinitesimal radius at 0.25 lambda, which would indicate an infinite impedance "looking back" towards the load.

• ###### Smith Chart 1.pdf
File size:
389.7 KB
Views:
36
Last edited: Jan 6, 2015
10. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
so simply the answer is, " it's purely real and has no reflection". is that the simple answer or am i completely wrong here?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I'm convinced there is something hypnotic about the Smith Chart that leaves the newcomer bamboozled.
Keep in mind the important point, that the Smith Chart is (at its heart) a plot of complex refection coefficient vs complex impedance. The reflection coefficient magnitude lies between 0 & 1. As you progress outwards from the geometric center of the chart, the reflection coefficient varies from 0 at the origin to 1 at the outermost circle. So circles concentric with the geometric center represent loci of constant magnitude reflection coefficient. An ideal Tx line with any given load end impedance will present as a circle of constant reflection coefficient concentric with the geometric center.

In short, you are wrong. Having a purely resistive driving point impedance at some position along an unmatched line doesn't imply the absence of reflection at that position.

The reflection coefficient on an ideal Tx line with a shorted load end will have a relection coefficient magnitude of 1 at any point along the line between load and generator.

Last edited: Jan 7, 2015
12. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
so could i say that when i reach lambda/8 (very top of the smith chart) the termination is purely inductive, at lambda/4 it is open and real, 3lambda/8 (the very bottom) its purely capacitive. at lambda/2 it's back where you started so it will be the same impedance as when you started?

please please just tell me the answer if thats not it. i'm getting so frustrated with this and seem to be going around in circles!!!

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Well that is quite humorous, intentional or otherwise.

Very good, you are getting close. Frustrating as this is, you seem to have come to a place of better understanding, rather than one of confusion.

So the impedance is (everywhere) either purely inductive or capacitive reactive except at one single point along the line - up to but not including the first half wavelength from the load. On the perimeter of the chart you never actually "intersect" a non-zero, finite real component of line impedance. It's either a zero value real component of the complex impedance or infinity plus a zero reactive term. The latter occurring only at the point exactly one quarter wavelength from the load. The only point of deviation is at multiples of half wavelength where the impedance is zero. The process repeats over half wavelength intervals as one moves further towards the generator along whatever line length exists between generator and load.

My only final comment is that your use of the word "termination" is confusing. One usually reserves that term for the load impedance connected at the end of the line, rather than at any arbitrary point between the generator and load. I used the term "driving point impedance" in an early post, but even that isn't really satisfactory, as one normally reserves that for the impedance "seen" by the generator looking into the line towards the load. The best I can offer at the moment is "the impedance seen looking towards the load" at the point of interest on the line.

You may think this is all purely artificial pointless speculation.....

Some years ago I remember visiting a facility with a very high power microwave heating chamber. To prevent microwave leakage from the chamber along the sealing periphery the designer had created a continuous shorted quarter wave path at the working microwave frequency - which provided a high impedance blocking of microwave radiation traveling towards and potentially escaping the chamber at the seal.

Last edited: Jan 8, 2015
14. ### liam1234 Thread Starter New Member

Jan 5, 2015
7
0
unfortunately it was unintentional

so I was talking to my lecturer and I asked him about the question and he said all he was looking for was that "it is purely capacitive". That is the full answer! easy peesy!

thanks for the help though!