# Questions regarding normalized impedances with negative real part

Discussion in 'Wireless & RF Design' started by WBahn, Apr 1, 2012.

1. ### WBahn Thread Starter Moderator

Mar 31, 2012
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FYI: I am just getting into electromagnetics, so please keep explanations close to the basics and fundamentals, if possible. Thanks.

I have run across the claim, in numerous texts and online resources, that the real part of the normalized impedance (the ratio of load impedance to the characteristic impedance of the driving transmission line) can never be negative, except in the case of some active devices. The justification I have seen for this in several places is that, for passive circuits, negative resistance has no physical meaning.

Leaving all of that aside, my confusion stems from the fact that normalized impedance is not an impedance at all, it is the ratio of two impedances. Thus the real part of the ratio is not a "resistance" and the imaginary part is not a "reactance"; they are both dimensionless quantities.

More to the point, consider the following example:

Load: ZL = 50+j1000 ohms
Transmission line: Z0 = 50-j3 ohms

It would seem to me that the normalized impedance in this case would be:

Normalized impedance: Zn = ZL/Z0 = -0.2+j20

In fact, it would seem that as long as the transmission line has any reactive component at all, that a passive load could easily be constructed that would result in the real part of the normalized impedance being negative.

So what am I missing?

What would happen if I proceeded to build such a circuit? Is there something that makes such a circuit unrealizable?

Last edited: Apr 20, 2012
2. ### steveb Senior Member

Jul 3, 2008
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I've noticed that you've mentioned in some other threads that you would like an answer to this fist question you posted. I somehow missed this question when you originally posted it, but I looked at it a few days ago and have been giving it some thought.

I don't have a clear answer yet, but I expect to have one at some point. In the mean time, my post here will serve to bump your thread and get some other people to think about it.

3. ### WBahn Thread Starter Moderator

Mar 31, 2012
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Thanks. I didn't want to bump my own thread.

And I'm glad it has you scratching your head (misery loves company!).

Oh, for the record, I think this is only the second time I've mentioned it (but, then again, my short-term memory isn't what it used to be), but the first time was within the last couple of days. I wasn't planning on ever mentioning it again, but the opportunity that your post presented was just too good to pass up!

4. ### nsaspook AAC Fanatic!

Aug 27, 2009
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The real part of the equation is real energy. If you find a way to create negative energy (other than from my ex-wife) in this universe, let me know.

5. ### WBahn Thread Starter Moderator

Mar 31, 2012
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How can it be real energy? It is a dimensionless quantity.

And, if that is the case, then what about the example given? Where have I gone wrong? I have been told by a couple of people that work with them that transmission lines with significant reactance are the norm in many instances (PCB traces, for instance). I don't know if they tend to be capacitive or inductive because I didn't ask. But it doesn't matter.

I've also run across the explanation that a negative resistance means a greater than unity reflection coefficient. Intuitively, this seems reasonable that passive loads should not reflect a larger signal than impinges on them. Fine. But that still leaves the question of my example.

$
z_n = \frac{Z_L}{Z_0} = \frac{R_L+jX_L}{R_0+jX_0} = r_n + jx_n
$

Multiplying top and bottom by the complex conjugate of the denominator:

$
z_n = \frac{(R_L+jX_L)(R_0-jX_0)}{(R_0+jX_0)(R_0-jX_0)}
$

$
z_n = \frac{R_LR_0+X_LX_0}{R_0^2+X_0^2}+j \frac{R_0X_L-R_LX_0}{R_0^2+X_0^2}
$

The real part is therefore

$
r_n = \frac{R_LR_0+X_LX_0}{R_0^2+X_0^2}
$

Which is dimensionless.

The denominator is clearly always positive. Since we are talking passive compoments, we will stipulate that the resistive components of both are positive. But the numerator is negative if two conditions are met:

1) One of the reactive components is capacitive while the other is inductive.
2) The magnitude of the products of the reactive components is greater than the product of the resistive components.[/tex]

This condition will be satisfied if:

$
L > CR_LR_0
$

Where L is the inductance of whichever part is inductive and C is the capacitance of the other. If we are dealing with capacitances in the uF range, then we would only need inductances in the mH range to satisfy the constraints.

If I've made a mistake or am missing something, great. But I would very much like to know what it is.

6. ### Papabravo Expert

Feb 24, 2006
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You wrote Z0/ZL but the number you gave is the reciprocal ZL/Z0. So which is it?

7. ### WBahn Thread Starter Moderator

Mar 31, 2012
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Good catch. It is ZL/Z0. I will correct it.

8. ### nsaspook AAC Fanatic!

Aug 27, 2009
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Take a look at the Smith Chart, it's an easy way see the effects of lumped elements in a transmission line. I think you need to DE-normalise your result to get the true result.

Page 21:
Smith Charts Page 1 of 27 The Smith Chart , invented by Phillip H ...

Last edited: Apr 20, 2012
9. ### WBahn Thread Starter Moderator

Mar 31, 2012
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The Smith Chart is nothing more than a plot of reflectance coefficient as a function of normalized impedance. It doesn't tell you anything more than what calculating the normalized impedance tells you. For the example I have given, the reflection coefficient is greater than one.

10. ### steveb Senior Member

Jul 3, 2008
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I think you are asking a good and interesting question. I have a preliminary answer, which is best described as an opinion right now. I'd need to give it more thought to be sure, but you've been waiting on this one, so I'll give my best attempt.

First, I think you are basically correct from a mathematical perspective, and I don't see any flaw in your example. However, I'm willing to bet that your numbers are not representative of practical cases (note, I haven't checked them in detail yet).

I think that the characteristic impedance will not generally have such a large imaginary component in practical cases. I base this on the basic formulas for lossy transmission lines. Take a transmission line with characteristic capacitance and inductance per unit length C and L. Losses are typically considered to be an inline resistance per unit length R (wire resistance) and a parallel conductance per unit length G (dielectric loss). The lossy components R and G manifest themselves in the attenuation constant for the wave α and the characteristic impedance Zo. Now, assuming the loss is low (otherwise the transmission line is not useful) and taking first order terms only (as an approximation) we can see the following.

$\alpha = {{R}\over{2\sqrt{L/C}}}+{{G\sqrt{L/C}}\over{2}}$

$Z_o=\sqrt{L/C}(1+j(G/(2\omega C)-R/(2\omega L))$

Notice how the two lossy components R and G tend to cancel in the Zo formula, but they add in the attenuation coefficient. Also, the imaginary part of Zo is divided by the frequency, which tends to make is small, and even smaller as frequency goes up.

Hence, the effect of the wave attenuating may be noticeable in a practical case, and the imaginary part of the Zo might be quite small, and small enough to keep the real part of the normalized impedance greater than zero.

EDIT: looking through another EM field book seems to confirm my opinion above. The complex part of Zo is generally very small for practical transmission lines at frequencies of interest.

Last edited: Apr 21, 2012
11. ### WBahn Thread Starter Moderator

Mar 31, 2012
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Thanks, this is much along the lines of what I was thinking and hoping for.

It's not completely satisfying because it should still be possible to make a lousy transmission line that has a noticeable reactance and combine it with a load that makes the normalized resistance negative and, therefore, still raise the basic question of what are the implications and what is the reflectance coefficient.

Also, note that in the equation for Z0 you have, they tend to cancel each other only if they are of very nearly equal size. In my example the difference is only -6%, which isn't all that large.

I've tried going to cable manufacturer's websites and getting numbers for {L,C,G,R} per foot and haven't had much luck. Of course, I have nomincal L and C numbers from the geometry and can probably estimate G and R values from knowing the materials and dimensions.

I don't see any intrinsic reason to believe that (G/C - R/L) is always going to be close to zero. While I can certainly believe that cable manufacturers design their cables to make this as true as possible, homemade transmission lines or lines laid on on PCBs or in ASICs would seem to be far more likely to have a significant mismatch.

With a little more digging, I can probably be convinced that a good argument can be made that, in practice, the reactive part of the transmission line characteristic impedance is sufficiently small, especially in light of the inverse dependence of frequency, so as to make creating a normalized impedance, in a passive system, with a normalized impedance that has a negative real part impractical.

But that will still leave me with the following:

1) Is it impossible? and
2) If not, what does it mean if you construct such a thing.

It will also leave me unable to just spout the absolutist claims in so many texts about why it has to be positive (unless the answer to question #1 above is shown to be YES).

Last edited: Apr 21, 2012
12. ### steveb Senior Member

Jul 3, 2008
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You're missing a piece here. Don't forget the frequency part. Also, your R/L is upside down. The term is (G/C-R/L)/(2w), where w is angular frequency, 2pi times the cyclic frequency f in Hertz.

You're right that there is not going to be an exact cancelation. I only mean that there will be a lessening of the imaginary part. Compare this with the attenuation which sees both loss mechanisms adding together. Notice that at high frequency, the imaginary part gets very small. There is no frequency in the formula for attenuation. Now imagine a practical case where G and R are high enough to make the imaginary part of Zo significant. What will happen to attenuation? It will get so large that the idea of propagating waves will be moot because the signal will attenuate before traversing a distance of one wavelength.

Anyway, that's the basic crux of the argument. Try some real numbers. The values of C and L are available for coaxial RF cables. If you can't find the G and R numbers, then assume G=0 and calculate R from the copper loss of the central conductor. If you know the wire gauge, you can look up the resistance per meter (or calculate it from copper conductivity and wire cross sectional area).

13. ### steveb Senior Member

Jul 3, 2008
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By the way, can you post the references that make the claim you are mentioning. I have not seen such a claim in any of my books.

14. ### WBahn Thread Starter Moderator

Mar 31, 2012
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Thanks for pointing out the error. I think I am so used to righting L/R (the time constant) that my fingers just took over. Had I sanity checked the units I would have caught it and I would certainly have caught it had I tried to use that in an equation because it totally screws up the units.

I didn't forget the frequency dependence (in fact, mentioned it explicity a bit further along). I was focusing on the factor that would need to vanish if the desired result were to be identically true.

Which claim are you referring to? That the real part of the imaginary impedance must be positive because negative resistance has no physical meaning (in passive circuits) or that it must be positive because the reflection coefficient must be less than unity? I will look through and post as many as I can, but remember that my original post referred to both texts and online resources. I didn't keep track of the online stuff I found.

Here's the one from the book I have handy right now:

"To start with, the normalized reactance may assume both positive and negative values, whereas the normalized resistance cannot be negative (negative resistances are physically meaningless)."

Fawwaz T. Ulaby, Fundamentals of Applied Electromagnetics, 5th Ed, Pearson Prentice Hall, 2007, p74.

BTW, my understanding is that this (in 6th Ed) is presently one of the standard textbooks in engineering electromagnetics.

15. ### Wendy Moderator

Mar 24, 2008
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I'm following this with interest, and don't really have anything to offer.

Seems that it would also overlap with transformers to some degree. Maybe they have literature to offer?

16. ### WBahn Thread Starter Moderator

Mar 31, 2012
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Perhaps.

NSAspook pointed me to a discussion on another forum that I just got done reading. I followed much, but not all of it. It touched not only on this topic but another question that has been bothering me. I need to look at it closer in order to see if I can follow that one.

But the assertion (and I can't give it any stronger credence than that right now) by at least one of the posters was the on one thing that transmission line theory is only an approximation that is very good for transmission lines with negligible reactive components but that it departs substantially if this is not the case. I can believe that this is the case and that a more rigorous analysis using field theory might resolve the problem. It's at least something to hang my hat on for the moment.

17. ### steveb Senior Member

Jul 3, 2008
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469
The claim I was referring to was this one that you made reference to.

"I have run across the claim, in numerous texts and online resources, that the real part of the normalized impedance (the ratio of load impedance to the characteristic impedance of the driving transmission line) can never be negative, except in the case of some active devices.

This is consistent with the quote you gave from the book reference above. So far I don't see the evidence for author's claims. I don't see an error in your math, so I don't think there is a problem in principle with negative real components of the normalized impedance. I don't see any physical principle which is violated yet. The main one to worry about is conservation of energy, but it is clear that as long as R and G are positive, the wave attenuation is positive and energy is dissipated during propagation. As far as reflection or transmission of greater power than is incident, I don't see how that will happen in your example. (although, I still have not run the numbers myself, and plan to do that later this weekend)

I don't agree with the idea that transmission line theory fails with reactive components. If we are careful to not make simplifying assumptions, the theory holds up, and it is field theory that we are using here. Of course, we apply the field theory to very perfect geometries and assumptions about materials, but those are reasonably good assumptions and approximations for most practical transmission lines.

As a point of interest, I found a note in one of my books that the condition of G/C=R/L is called Heaviside's condition for a distortionless line. So, apparently it is a design criterion sometimes, which indicates that complex Zo can happen in some practical cases, which was one of your points. Although, I still think the imaginary component will be much smaller than in your example. Still, you are talking about principle here.

My present conclusion is that your math is correct and that there is no fundamental issue with normalized real part being negative, although I think in many (possibly even most) practical cases it should be true. I think the burden falls on anyone making a claim to the contrary, to show a physical principle violated by your example, and not just say it is physically meaningless.

The note from your reference (negative resistances are physically meaningless) is dubious to begin with, as you've noted. First of all, normalized resistance is not the same as resistance, and second of all, it's not true. Negative resistance is a power generator, rather than a power dissipator. It may not be physically realizable (typically), but that is a different statement.

One possible interpretation of "physically meaningless" is that the condition may create so much loss, that wave propagation is not possible. In other words, what is the "physical meaning" of a waveguide that kills your wave over a distance of 1 mm? Would you buy one? I think we would need to run the numbers in your example to see if that would happen.

Again, note that I'm talking about my opinion right now, since I'm still looking deeper. I plan to run the numbers and read a little more.

Last edited: Apr 21, 2012
18. ### WBahn Thread Starter Moderator

Mar 31, 2012
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I personally don't know this. By that I simply mean that the only derivation I have made (lately) of the transmission line equations is from a lumped element model of a transmission line that then takes the limit as the element size goes to zero. So I can see that there might be something that it is not taking into account. I think I went through the development from a truly fields perspective a quarter century ago, but I don't remember for sure and certainly don't recall if there were any simplifying assumptions that went into going from the fundamental results to the transmission line equations.

Actually. I think I just recalled something that may shed a lot of light on things. I think the analysis used is contingent on the mode of propagation being TEM waves. I seem to recall that lossy transmission lines are not strictly TEM-mode propagation. I'll have to track down something on that, but if that is the case, then it makes more sense that the classic transmission line equations are bounded pretty tightly to transmission lines that are, at most, only slightly lossy which, as you have pointed out, means that the reactive component of the characteristic impedance is driven into the basement.

19. ### studiot AAC Fanatic!

Nov 9, 2007
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I thought the negative sign referred to a reverse travelling wave.

20. ### WBahn Thread Starter Moderator

Mar 31, 2012
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No, not when it is a negative normalized resistance (which isn't a resistance at all and I think using the word "resistance" is what causes so much of the confusion, but I can't really think of a better word to use -- so it is really important to always use the term "normalized resistance").

For a sine wave of the form:

v(x,t) = Vsin(wt-bx+phi)

The w gives the phase dependence on time and the b gives the phase dependence on distance. If w and b are of the same sign, then you have a wave travelling in the +x direction while, if they have opposite sign, then you have a wave travelling in the -x direction. The way to look at this, so that you get it right even if the equation you are looking at has a + sign where mine above has a - sign, is to ask what has to happen to x if t is made slightly larger. For the equation above (assuming positive w and b), a slightly larger (more positive) t increases the phase a bit and therefore you need to increase x (make it more positive) in order to subtract the phase back off and keep the same amplitude. Thus, as t increases the waveform moves to the right. If they are of the same sign.