# What is the impedance of a matched transmission line

1. Oct 10, 2013

### Mr.Gronka

Hi,

I've been coming to physicsoforums for years, I love the community! This might be my first post though.

I am wondering what the impedance is of a transmission line with a matched stub on the end. The impedance of the line is 50 Ohms, and the impedance of the stub is 50 Ohms.

Is the resulting impedance 50 Ohms (simply the impedance of the line), or 100 Ohms (the impedance of the line summed in series with the impedance of the load)?

I want to know for the use of a voltage division equation which occurs at the start of the tx line.

Thanks!

2. Oct 10, 2013

### Averagesupernova

It's hard to tell what you mean. If you take a 50 ohm transmission line and attach a 50 ohm 'stub' to the end of it, it is no different than having a 50 ohm transmission line that is terminated in the same manner your stub is terminated of the combined length of the transmission line and the stub. You don't say how you are terminating your stub. Is it open? Shorted? Some finite impedance? 50, 75, 300 ohm? Tell us more. Generally a 'stub' is a specific length transmission line of a specific impedance to perform filtering or impedance matching.

3. Oct 10, 2013

### Mr.Gronka

Thanks!

To clarify, here is an image.

http://imgur.com/W6YgoE8

The green thing is a 50 Ohm stub, as I mentioned previously - in lab we call it the load. I am pretty sure it is just a resistor with a cap on it, meaning it would be a shorted stub. The line has a characteristic impedance of 50 Ohms.

So, does the line in series with the load have an equivalent impedance of 100 Ohms? Or does the fact that the load is matched to the line mean that the impedance of the load is negligible, such that the equivalent resistance is only 50 Ohms?

I can put an Ohm meter to the line (with one node touching the inner wire and one not touching the outer wire of the unconnected end of the line, which can be seen in the picture) and I get a resistance of around 62-70 Ohms. However, since impedance is related to frequency, I'm 99% sure this value is meaningless to me, aside from possibly being within a power of 10 of the actual impedance.

In fact, if the question is still confusing, think of it this way - I put a 50 Ohm load on a 50 Ohm line. If I connect the other end of that line to something, can I treat the load+line system as a 100 Ohm load? Or does something occur in the impedance domain of which I am being ignorant?

I feel like it's really a pretty simple question - there's nothing tricky here - I'm just not entirely sure.

4. Oct 10, 2013

### davenn

yes if you has said load instead of stub, then your question would have been more understandable earlier

no the transmission line is 50 Ohms, it has been matched correctly to a 50 Ohm load. This ensures maximum power transfer to the load with minimum reflection of energy back to the source

meaningless in that you are measuring DC resistance of the whole line ( including the load)
NOT the impedance. To state it simply, impedance is AC resistance. This is a result of the effects of inductive and capacitive reactance to an AC voltage

No you treat it as a system with a 50 Ohm load. The signal source output impedance must be matched to the transmission line and then the transmission line to the impedance of the load,
be that your resistor, an antenna, a speaker ... whatever

Dave

5. Oct 10, 2013

### Mr.Gronka

Thanks. That helps a lot.

I really had no idea what exactly a short or stub was, so I really appreciate that knowledge.

Bottom line though: I need to know the value of $Z_{L1.x}$. I'll explain:

I think I might need to go into more detail now, just to make sure. Here is a photo of the setup we are trying to analyze:
http://imgur.com/wa3zPHI

The picture that I took was of spool 1.x. What I want to find is the impedance of the load at the fault. I can calculate the reflection ratio occurring at the fault towards the generator $(\Gamma_F)$ from measurements taken. To determine the impedance of the fault load itself, I believe that I need to relate the reflection ratio $\Gamma_F$ to the impedance "seen" at the fault load $(Z_F')$:

$$\qquad \Gamma_F = \frac{Z_F' - Z_0}{Z_F' + Z_0}$$

Rearranging:

$$\qquad Z_F' = \frac{-Z_0 - Z_0 \Gamma_F}{\Gamma_F - 1}$$

Since we know the characteristic impedance is $50 \Omega$ and $Gamma_F$ was calculated to be roughly $-0.5$, we can solve for $Z_F'$:

$$\qquad Z_F' = \frac{-50 - 50 (-0.5)}{-0.5 - 1} = 16.667 \Omega = \frac{50}{3} \Omega$$

To solve for the impedance of the fault load itself, I believe that I want to use the equivalent impedance of the fault load in parallel with the system of the 1.x line+load $(Z_{L1.x})$. This gives me:

$$\qquad Z_F'^{-1} = Z_F^{-1} + Z_{L1.x}^{-1}$$

Rearranging:

$$\qquad Z_F' = \frac{Z_F Z_{L1.x}}{Z_F + Z_{L1.x}}$$

$$\qquad Z_F = \frac{-Z_F' Z_{L1.x}}{Z_F' - Z_{L1.x}}$$

To calculate this equation, I need to know the value of $Z_{L1.x}$ (or if I made an error in deriving my equations). I realize that it is a line with a characteristic impedance of $50 \Omega$ which is matched to a load of impedance $50 \Omega$. In fact, I doubt this calculation would be so simple if the line 1.x were not matched to its load.

However, what is the impedance of $Z_{L1.x}$? My two guesses are $50 \Omega$ or $100 \Omega$. $100 \Omega$ makes a lot more sense to me, since that would be like adding the impedance of the line to the impedance of the load.

If $Z_{L1.x} = 50 \Omega$, then $Z_F = 25 \Omega$. If $Z_F = 100 \Omega$, then $Z_F = 20 \Omega$. The actual value of $Z_F$ is $\approx 22 \Omega$. Therefore, I think I am on the right track; however, I am not capable of determining which value is better.

Really though, the exact numbers are pretty irrelevant to the original question.

6. Oct 11, 2013

The whole point of matched impedance is wave propagation at the point of connection. So the Source's impedance should match the lines impedance to get maximum (wave) energy transferred to the line - at this point what is at the termination end does not matter it is characteristic, not part of a total circuit analysis.

Oh - found this general description : http://www.ece.uci.edu/docs/hspice/hspice_2001_2-269.html

Last edited: Oct 11, 2013
7. Oct 11, 2013

### sophiecentaur

A perfect transmission line doesn't 'have impedance' as such as it is always treated as having two ends and purely reactive elements in it (a two terminal device). It is more like a transformer which transforms the load impedance at one end to the source at the other end. The characteristic impedance of a line is the value of load impedance that will be unchanged by the presence of the line, and that is what we call 'matched'.
If you put a fault on the line then, at that point, you have the fault in parallel (or series, depending on the nature of the fault). Together, the fault and the downstream line (which is equivalent to 50Ω, when it has a matched load at the far end) constitute a modified load for the upstream line. To know how this will look at the source, you need to know the length of the line in λ, but it is straightforward to work this out with the transmission line equations or a Smith Chart.

8. Oct 17, 2013

### Mr.Gronka

Everyones' posts helped a little, so thanks! Although sophiecentaur hit the nail on the head!

9. Oct 18, 2013

### the_emi_guy

Have to respectfully disagree here.

Consider a really long, lossless, 50 ohm transmission line. The source is on the Earth and the 50 ohm matched termination at the far end is on the Moon.

Source is initially 0V (for a long time). Then we switch the source to 1V, a step function.

The source will supply a continuous 20mA of current into the transmission line, even while the wave is in flight and the 50 ohm termination at the other end still sees 0V. The power associated with this current (20mW) goes into depositing the E-field between conductors, and H-field around them as the wave propagates on.

From the source's perspective it looks just like a 50 ohm resistive load, even before the real resistive load at the far end comes into play. This is what we mean by "characteristic impedance". The source doesn't know that the power that it is pumping into the "port" is constructing fields around a pair of conductors vs. dissipating as heat from a resistor.

Because the termination is matched, when the wave reaches the termination all of the subsequent power supplied by the source goes into heat dissipation in the resistor as Ohms law takes over.

Last edited: Oct 19, 2013
10. Oct 19, 2013

### davenn

indeed

its not uncommon for us guys playing with UHF and microwave bands, just to use a longish length of lossy cable as a dummy load. Dont even need to terminate it with a load ( depending on freq/power level and coax losses)

if a transmission line, coax, openwire, whatever, doesn't 'have impedance' as such, then there would be no need for specifying that it has a characteristic impedance and any ol' pair of wires could be used between the TX and RX with no detrimental effects

Dave

Last edited: Oct 19, 2013
11. Oct 20, 2013

### Averagesupernova

the_emi_guy and davenn, I agree with you both. I had tried to wrap my head around sophie's last post and came to the same conclusion you both did. I have seen it in text books that a lossless transmission line that extends in one direction forever will measure whatever it's rated impedance is with an ohmmeter hooked to its one end. As far as not having an impedance goes, when the length of the line is a small fraction of a wavelength then we usually just ignore it.

12. Oct 20, 2013

### sophiecentaur

I have to agree with those various replies, as they are, strictly true. My original "doesn't 'have impedance' as such" was a bit over glib. But, in practical terms, I more or less hold to what I wrote initially. Of course, it's true that, if you apply a 1V step function to a 50Ω transmission line from a 50Ω source, the current step will, initially be 1/50A. Which is what would happen with a 50Ω load attached and that would be when the transmission line was presenting its characteristic impedance. But this current step is guaranteed to exist only until the step reaches the end of the line and any reflection appear back at the drive point. The final, steady current will depend upon the impedance or the load on the end. (A time domain reflectometer gives a great picture of what's happening along a line for the first few hundred ns, with the levels of the signal representing the various line impedances on a feeder run.) When a load is connected to a source via a transmission line, the impedance that the source will see is the load impedance, transformed by the line. That's just a practical way of looking at it; the line just modifies what the source sees of the load - hence the use of lengths of line as matching networks.
A real length of feeder has series and parallel resistance and will serve well enough as a good load but that's an added complication and more of a nuisance under most circs. (And you need to pay through the nose for high quality transmitter feeders!).