# reflection coefficient for complex characteristic impedance

• posted

Hi,

I understand that when the load (ZL) matches the characteristic impedance (Zo), the reflection coefficient is zero using the following equation:

reflection coefficient = (ZL-Zo)/(ZL+Zo)

If the characteristic impedance is a complex value (e.g. 50+j10), should the load be 50-j10 (conjugate match) in order to have the reflection coefficient equal to zero? However, according to the reflection coefficient equation, a conjugate matched load won't give a zero reflection coefficient.

Thanks a lot for your help! Philip

• posted

Hi,

I understand that when the load (ZL) matches the characteristic impedance (Zo), the reflection coefficient is zero using the following equation:

reflection coefficient = (ZL-Zo)/(ZL+Zo)

If the characteristic impedance is a complex value (e.g. 50+j10), should the load be 50-j10 (conjugate match) in order to have the reflection coefficient equal to zero? However, according to the reflection coefficient equation, a conjugate matched load won't give a zero reflection coefficient.

Thanks a lot for your help! Philip

• posted

You answered your own question: "ZL matches Zo" works the same with real or complex numbers. Under what conditions will the numerator go to zero? ZL = Zo.

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Modulo a complex conjugation.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal```
• posted

I've not studied complex modulo, could you explain the rationale behind that?

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Thanks for the replies!

As I stated in my original message, if the characteristic impedance is

50+j10, and the load is a complex conjugate (50-j10. Then the reflection coefficient is not zero.

reflection coefficient = (ZL-Zo)/(ZL+Zo) = (50+j10-50+j10)/(50+j10+50- j10) = j20/100

However, if ZL=Zo, then the reflection coefficient is equal to zero but I thought for complex impedance, we need to conjugate in order to match?

Philip

• posted

Conjugate matching means that the load needs to be the complex conjugate of the source impedance. In other words, you match an inductive source using a capacitive load, and vice versa.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal```
• posted

-

An imaginary reflection coefficient, does that mean the reflected wave is 90 degrees out of phase wrt the input?

George H.

• posted

e

OK, but why would you want to do that?

George H.

t
• posted

The reflection coefficient is (well can be) used to measure an impedance. It does not give you the complex conjugate. Once you have assertained the impedance (from the reflection coeff) you can work out the conjugate and go ahead with the matching.

• posted

Because conjugate matching *does* provide you with maximum power transfer, which is often desirable when dealing with tiny signals from antennas, sensors, etc. (I.e., when there's so very little to begin with, getting as much of it as possible makes a lot of sense.)

(But don't let anyone tell you that conjugate matching is always what you watch -- many university courses will tend to suggest this, and it just isn't true. Once you have "reasonable" power levels/SNRs running around your circuit, it's often easier to just worry about voltage or current gain from stage to stage rather than power gain. There's a lot to be said for close-to-zero impedance voltage sources and close-to-infinite-impedance current sources, after all...)

---Joel

• posted

We had a professor who taught (!) that all transmitters must be impedance matched, so with a 50ohms antenna the source resistance must be also 50ohms. Which, of course, is wrong. I told him that this would result in some explosions and possibly a major fire ...

```--
Regards, Joerg

http://www.analogconsultants.com/```
• posted

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Thanks Joel, I guess I'll have to work through the math to convince myself of that.

George H.

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• posted

Philip,

here is a link to a flash program I made to show how a reflected wave is developed on a transmission line and to give insight into VSWR and the reflection coefficient.

• posted

It's not a proper proof, but you can pretty much convince yourself of it by arguing that the reactive (complex) elements don't produce or consume any energy themselves, therefore even with them in the picture you couldn't obtain

*more* power (...in the steady-state case, at least...) than if there weren't any reactive elements and one obvious way to get rid of the reactive elements is for them to "cancel themselves out" by having conjugate reactances.

The mathematical proof does show up in various engineering books and it is what you'd expect: Write an equation for the power delivered to the load and set the first derivative to zero and solve. It should be a good exercise... :-)

---Joel

• posted

Ah, but in which direction do you take the derivative? Since we are on the complex plane, after all :)

Tim

```--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms```
• posted

Since you wrote the expression for power delivered to the load, you isolate the real part of it (we don't need no stinkin' reactive power... VARs? What's that?) and now it's no longer a matter of complex analysis. :-)

I wish I'd taken a proper complex analysis course (from the math department) back in college... you get all these little bits and pieces of it through engineering school, but it was understandably often rather light on details. You were required to take "modern physics" (fun with tunneling and wave equations, leading to semiconductor theory) from the physics department, which, although they considered it a "service course," I still think was better than if the engineering department had been teaching it as apparently happens at many a school.

---Joel

• posted

They also told us that reactive elements such as capacitors could be ingored in dissipation calcs. So, I built another shortwave amp of the big foot type. One fine day, after a few hours of heavy use ... *PHOOMP* ... SPLAT. Thanks to a rather tall front panel the shrapnel cone was nearly vertical so nothing lodged itself in my body. One of the hockey-puck capacitors was literally gone, except for the severed screw terminals.

```--
Regards, Joerg

http://www.analogconsultants.com/```
• posted

According to me, if the load is 50-j10 and the line has Z0=50-j10, you will have reflection coeff=ro=0, which means no reflections from the load into the line. Let's go to the other end of the line: ro is still zero there, so we'll see an impedance of 50-j10. At this point, if we want, we can conjugate match using a source with Z=50+j10 and get maximum power transfer.

```--
RoV - IW3IPD
http://digilander.libero.it/rvise/```
• posted

Clearly you went to the wrong store for your capacitor -- you're supposed to get them from the place that also provides the frictionless pullies and massless springs; then all will be well.

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