reflection coefficient for complex characteristic impedance
Philip Lee
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
Philip Lee
Tim Williams
news:95283f62-7e81-483b...@y11g2000yqm.googlegroups.com...
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
Phil Hobbs
> "Philip Lee"<fei...@gmail.com> wrote in message
> news:95283f62-7e81-483b...@y11g2000yqm.googlegroups.com...
>> 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.
>
> 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
>
Modulo a complex conjugation.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net
Tim Williams
news:4CA11A45...@electrooptical.net...
> Modulo a complex conjugation.
I've not studied complex modulo, could you explain the rationale behind
that?
Philip Lee
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
Phil Hobbs
> "Phil Hobbs"<pcdhSpamM...@electrooptical.net> wrote in message
> news:4CA11A45...@electrooptical.net...
>> Modulo a complex conjugation.
>
> I've not studied complex modulo, could you explain the rationale behind
> that?
>
> Tim
>
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.
George Herold
An imaginary reflection coefficient, does that mean the reflected wave
is 90 degrees out of phase wrt the input?
George H.
George Herold
> >news:4CA11A45...@electrooptical.net...
> >> Modulo a complex conjugation.
>
> > I've not studied complex modulo, could you explain the rationale behind
> > that?
>
> > Tim
>
> 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.
OK, but why would you want to do that?
George H.
neddie
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.
Joel Koltner
news:ab8d7275-d9c5-437b...@i3g2000yql.googlegroups.com...
On Sep 28, 12:58 am, Phil Hobbs
<pcdhSpamMeSensel...@electrooptical.net> wrote:
>> 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.
>OK, but why would you want to do that?
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
Joerg
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/
"gmail" domain blocked because of excessive spam.
Use another domain or send PM.
George Herold
wrote:
> "George Herold" <gher...@teachspin.com> wrote in message
>
> news:ab8d7275-d9c5-437b...@i3g2000yql.googlegroups.com...
> On Sep 28, 12:58 am, Phil Hobbs
>
> <pcdhSpamMeSensel...@electrooptical.net> wrote:
> >> 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.
> >OK, but why would you want to do that?
>
> 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.)
Thanks Joel, I guess I'll have to work through the math to convince
myself of that.
George H.
brent
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.
Joel Koltner
>myself of that.
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
Tim Williams
news:RHqoo.196418$mN7....@en-nntp-08.dc1.easynews.com...
> The mathematical proof does show up in various engineering books and it
> is
> what you'd expect: Write an equation for the power delivered to the load
> and
> set the first derivative to zero and solve. It should be a good
> exercise...
Ah, but in which direction do you take the derivative? Since we are on
the complex plane, after all :)
Joel Koltner
news:i7tebn$r02$1...@news.eternal-september.org...
> Ah, but in which direction do you take the derivative? Since we are on the
> complex plane, after all :)
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
Joerg
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.
RoV
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/
Joel Koltner
news:8geuaf...@mid.individual.net...
> 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*
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.
Joerg
And only to be used in a zero-energy house. Since I can't turn on the
amp because of energy outflow the cap would not have been destroyed.
George Herold
wrote:
> "Tim Williams" <tmoran...@charter.net> wrote in message
Hmm, I never liked courses from the math department. Too short on
practical examples. My favorite math course in college was taught by
an engineer. The text still sits on my shelf (and gets used).
"Advanced Engineering Mathematics" by C.R. Wylie.
Still, I wish I had a better understanding of integration in the
complex plane. The theory of residues and all that. I've been wading
through Penrose's tome, "The Road to Reality". I’m in chapter 7 where
he talks about complex integration and I find my understanding is
lacking.
George H.
Joel Koltner
>Still, I wish I had a better understanding of integration in the
>complex plane. The theory of residues and all that.
Exactly -- I've been told the results of, e.g., Cauchy's Residue Theorem and
used it, but it would be nice to go through the derivation someday as well.
Granted, there's nothing stopping me from finding and cracking open an
appropriate book if I really wanted to do that, so apparently I'm only
motivated to the level of, "I'd sign up for a course where then I'll show up
and probably learn something" rather than to the level of actually tracking
down the material myself. :-)
> I've been wading
>through Penrose's tome, "The Road to Reality". I’m in chapter 7 where
>he talks about complex integration and I find my understanding is
>lacking.
Here's an alternative when your brain is starting to boil? -->
http://www.amazon.com/Logicomix-Search-Truth-Apostolos-Doxiadis/dp/1596914521
---Joel
ehsjr
> Joel Koltner wrote:
>
>>"Joerg" <inv...@invalid.invalid> wrote in message
>>news:8geuaf...@mid.individual.net...
>>
>>>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*
>>
>>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.
>>
>
>
> And only to be used in a zero-energy house. Since I can't turn on the
> amp because of energy outflow the cap would not have been destroyed.
>
No problem. Wrap a few turns of your antenna around the cap.
Be sure to go _counter_ clockwise. (Ask me how I know.)
And use magnets, plenty of them, positioned _just so_. That way,
the energy you put into your antenna will fill up the cap and thus
power the amp, with plenty left over to radiate around the world.
The counterclockwise antenna-magno-restrictive-field will hold the
cap together, preventing the dreaded phoomp.
Warning: You must get the cap from the Place Joel is talking
about.
Ed
Tim Williams
> Still, I wish I had a better understanding of integration in the
> complex plane. The theory of residues and all that. I've been wading
> through Penrose's tome, "The Road to Reality". I’m in chapter 7 where
> he talks about complex integration and I find my understanding is
> lacking.
It's one of the more remarkable proofs in mathematics. The complex plane
is a plane, of course, so you do path integrals, and you get the subject
of path independence. The remarkable part is, you get this very specific
unit when you integrate around a pole, regardless of what distance you did
it at.
This makes sense mathematically, but intuitively, it's like taking the
highway from a restaraunt to a bar, and suddenly realizing your position
is off by pi/2 miles from when you took the surface streets last time.
On the real line, of course, you can either integrate a function, or you
can't. Functions that you can integrate don't have poles in the range
(you can't integrate 1/x from -1 to 1 -- how do you know how much infinity
you've integrated in each direction?), and have a finite number of holes
(points where the limit exists but the function is undefined, like (x^2 +
2x + 1) / (x + 1) has a hole at x = -1; infinite holes are excluded, which
takes care of some pathological cases).
It makes sense to me to think of a flow field, with rotational flow (curl,
if you're in 3D). Complex numbers can be "implemented" by 2x2 matrices,
at some expense in redundancy (since half the numbers in the matrix are
"unused" in a certain sense). For instance, i ==> [0, 1; -1, 0] is also a
90 degree rotation on a 2D vector. Now, when you integrate around
something tangential, regardless of what distance you do it at, you get
the total amount of 'circulation'. That makes sense, because the rotation
weaker at a distance, but it's a proportionally longer path. If you're
thinking of an apparently one-dimensional function, there is no rotation,
and you miss the point. So the trick is, the phasor representation of
that function rotates about the pole. It's not even a pole -- yes okay,
it sticks up towards infinity, but it's more like... a tornado than a
pole. That's a far better name for it!
George Herold
wrote:
> "George Herold" <gher...@teachspin.com> wrote in message
>
> news:c4081e08-dcf2-47bb...@w19g2000yqb.googlegroups.com...
>
> >Still, I wish I had a better understanding of integration in the
> >complex plane. The theory of residues and all that.
>
> Exactly -- I've been told the results of, e.g., Cauchy's Residue Theorem and
> used it, but it would be nice to go through the derivation someday as well.
> Granted, there's nothing stopping me from finding and cracking open an
> appropriate book if I really wanted to do that, so apparently I'm only
> motivated to the level of, "I'd sign up for a course where then I'll show up
> and probably learn something" rather than to the level of actually tracking
> down the material myself. :-)
Hmm I bet there's some nice lecture video's on the web.
>
> > I've been wading
> >through Penrose's tome, "The Road to Reality". I’m in chapter 7 where
> >he talks about complex integration and I find my understanding is
> >lacking.
>
> Here's an alternative when your brain is starting to boil? -->http://www.amazon.com/Logicomix-Search-Truth-Apostolos-Doxiadis/dp/15...
Oh that looks fun, I put Penrose aside for a few days and read some
trashy novels.
>
> ---Joel
Oh BTW the conjugate matching is pretty easy to see. I made a voltage
source with complex source impedance Z driving a load with complex
impedance L. Then the power in the load is something like Vin^2 * (L/Z
+L)^2 * 1/(L). (I need to be a little careful about the 1/(L) term, I
think this should be the magnitude, but I'm not sure.) Anyway the
term L/Z+L has a maximum when the two complex terms cancel. ie
conjugate matching.
George H.
George Herold
> "George Herold" <gher...@teachspin.com> wrote in message
Thanks Tim, You might enjoy Penrose's book. Lots of math in it. I
got mine for only $15, for a 1000+ page book that's a lot of words per
dollar.
George H.