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'[OT] What does 20+j10 mean?'
2007\02\01@153713 by Paul Anderson

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This is a term I've seen when reading about impedance matching
circuits, but thus far I haven't found an explanation of it's meaning.
It has to do with the impedance of the circuit, and it's always of
the form x+jy or x-jy.  Like 20+j10 or 20-j10.  Is it reference to a
complex number?  Any help is greatly appreciated.

--
Paul Anderson
VE3HOP
spam_OUTwackyvorlonTakeThisOuTspamgmail.com
http://www.oldschoolhacker.com
"May the electromotive force be with you."

2007\02\01@155049 by Marcel Birthelmer

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Paul,
yes, it is a complex number. j = sqrt(-1) (like 'i' in mathematics).
In the case of impedances, the imaginary component is the reactance -
this is impedance contributed by inductive or capacitive elements,
whereas the real component is the resistance. So if you have a 10 ohm
resistor and a 1 milli-Henry inductor in series, the impedance at a
frequency of 1KHz is R + j * w * L = R + j * (2 * Pi * 1K) * 1m = 10 +
6.29j .

On a Smith chart (since you're talking about matching), the reactance
circles are the ones that all start on the right real axis and
intersect the outer circle. The ones in the positive half are
inductive, the ones in the negative half are capacitive.

Cheers,
- Marcel

On 2/1/07, Paul Anderson <.....wackyvorlonKILLspamspam@spam@gmail.com> wrote:
{Quote hidden}

> -

2007\02\01@155129 by Shawn Tan

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On Thursday 01 February 2007 20:35, Paul Anderson wrote:
>  It has to do with the impedance of the circuit, and it's always of
> the form x+jy or x-jy.  Like 20+j10 or 20-j10.  Is it reference to a
> complex number?  Any help is greatly appreciated.

yes it is... the imaginary portion of the impedance is affected by capacitance
and inductance..

cheers..

--
with metta,
Shawn Tan

2007\02\01@155218 by PAUL James

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Paul,

Yes, indeed.  It is a complex number.  It relates the magnitude of the
effects of capacitance and inductance with the system impedance.  The j
part of the statement is an imaginary number.  The part to the left of
the sign is the real part of the complex impedance, and the part to the
right of the sign is the imaginary part.


       
Regards,

       
Jim

{Original Message removed}

2007\02\01@155331 by Robert Young

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>From: "Paul Anderson" <.....wackyvorlonKILLspamspam.....gmail.com>
>Reply-To: "Microcontroller discussion list - Public." <EraseMEpiclistspam_OUTspamTakeThisOuTmit.edu>
>To: "Microcontroller discussion list - Public." <piclistspamspam_OUTmit.edu>
>Subject: [OT] What does 20+j10 mean?
>Date: Thu, 1 Feb 2007 15:35:06 -0500
>
>This is a term I've seen when reading about impedance matching
>circuits, but thus far I haven't found an explanation of it's meaning.
>  It has to do with the impedance of the circuit, and it's always of
>the form x+jy or x-jy.  Like 20+j10 or 20-j10.  Is it reference to a
>complex number?  Any help is greatly appreciated.
>

Yes, it is a complex number.  "i" is "taken" to represent current so "j" is
a good alternate.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html

is a good introduction.

Rob


2007\02\01@155821 by John Dammeyer

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Hi Marcel,

Brings up an interesting question.  Should an antenna that is supposed to
have a 50 Ohm impedance be 0+j50 or 50+j0 or 25+j25 or some combination?
Granted this is theoretical but I'm curious.

John Dammeyer


{Quote hidden}

2007\02\01@160215 by Mike Hord

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'j' is what EEs use to refer to the square root of -1, since we were
already using 'i' for some fairly important things.

In the case of a value x+jy, x is the resistive portion of the
impedance and y is the reactive component.  The reactance of
a component in a single-frequency circuit (this sort of shorthand
only works if you are evaluating an AC circuit) is
Z=wI for inductors, where w (which should be a lower case omega)
is the frequency in radians/sec (2*pi*Hz) OR
Z=-1/(wC) for capacitors.

Inductive loads have a positive value while capactive loads have
a negative value.  Currents and voltages can also be expressed
this way.  There are other ways of doing it, too- a magnitude and
an angle is also common, where the "real" portion of the x+jy
form makes up the abscissa leg of a right triangle, the "imaginary"
component forms the ordinate leg, the magnitude is the length
of the hypotenuse, and the angle is, the angle formed by the
hypotenuse and the abscissa.

If all this sounds confusing, that's because it is.  This is ususally
at least a half semester or more of college, and that's for the
people who don't need to use it.  If you plan on making your living
in a higher-power AC field, you'll have this ground in with a mortar
and pestle.  I haven't thought of it in five or six years, so if I got
any of it wrong, well, oops.

Mike H.


> This is a term I've seen when reading about impedance matching
> circuits, but thus far I haven't found an explanation of it's meaning.
>  It has to do with the impedance of the circuit, and it's always of
> the form x+jy or x-jy.  Like 20+j10 or 20-j10.  Is it reference to a
> complex number?  Any help is greatly appreciated.

2007\02\01@161942 by Alanis, Cristo jesus n/a

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The antenna impedance is measured as a real part only at resonance
frequency, that is 50+j0 ohm, in case that there is a complex part it
tells you that there is an inductive or capacitive part, or in simple
words the antenna is operating at a different frequency and you have to
compensate...

       
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> {Original Message removed}

2007\02\01@162540 by Thomas C. Sefranek

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A "proper" antenna resonant is purely resistive...
So 50+j0 would represent a 50 ohm antenna.

BTW MOST antennas are NOT 50 ohms!

 *
 |  __O    Thomas C. Sefranek  RemoveMEWA1RHPTakeThisOuTspamARRL.NET
 |_-\<,_   Amateur Radio Operator: WA1RHP
 (*)/ (*)  Bicycle mobile on 145.41MHz PL74.4

ARRL Instructor, Technical Specialist, VE Contact.
hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

{Original Message removed}

2007\02\01@170252 by Nate Duehr

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On 2/1/07, Thomas C. Sefranek <spamBeGonetcsspamBeGonespamcmcorp.com> wrote:

> http://www.harvardrepeater.org

Hey Thomas... was just looking over your repeater info.  On your
autopatch page you have a warning that the patch will dial silently.

You can program you S-Com to send a series of three "dits" (or any
other normal message in CW or voice if you have the voice option) in
CW when an Autopatch is dialed, so the person knows their command was
received and the controller is acting on it.

If you do the audio routing correctly on the message, it won't mute or
otherwise bother the audio from the phone line at all.  We've done
that for years on our autopatches.

Nice to know if your DTMF was received and that the controller is
acting on it, before the phone starts ringing.

Nate WY0X

2007\02\01@171754 by Richard Prosser

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John,
50+0j Ideally - the antennae should present a resistive load as it is
the real power you are trying to transmit.

RP

On 02/02/07, John Dammeyer <TakeThisOuTjohndEraseMEspamspam_OUTautoartisans.com> wrote:
{Quote hidden}

>

2007\02\01@173519 by Harold Hallikainen

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For some discussion on impedance in rectangular and polar notation, see
http://sujan.hallikainen.org/rw/theory/theory12.html and
http://sujan.hallikainen.org/rw/theory/theory13.html

Harold


--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2007\02\01@175254 by John Dammeyer

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face
Hi Thomas.

Great text art!  Love the bicycle with the antenna.

And you're quite right.  I've done smith charts on the rubber duckies and
there is no way they are 50+j0.  At the high frequencies it's amazing how
hard it is to get rid of that j factor.  Best I was ever able to do was what
I've got posted at:

http://www.autoartisans.com/images/WORM-433-2.03.jpg

John Dammeyer


{Quote hidden}

> {Original Message removed}

2007\02\01@195910 by Paul Anderson

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On 2/1/07, PAUL James <RemoveMEJames.PaulEraseMEspamEraseMEcolibrys.com> wrote:
> The j
> part of the statement is an imaginary number.  The part to the left of
> the sign is the real part of the complex impedance, and the part to the
> right of the sign is the imaginary part.
>
I really appreciate everyone's help, and I think I'm starting to
understand how the math works:)  I'm just going to run by some things,
and see if I'm on the right track.

Now, the math for impedance mimics  ohm's law - Z=V/I.  Since we're
dealing with a wave, you can express the voltage as V*cos(wt).  I
often lags voltage, so it is cos(wt-phi) (I think it's phi), with phi
being the lag in phase between the two.  The equation becomes
Z=V*cos(wt)/I*cos(wt-phi)

The Euler relation says that e^jx = cos x + j sin x...  So we can turn
cos(wt) into e^jwt, cos(wt - phi) becomes e^jwt-phi.  Z =
V*e^jwt/I*e^jwt-phi..  The jwt exponent divides out, leaving us with
(V/I)e^-jphi.  Hmm.  I'm not sure I'm right on that.  Since R = V/I,
we could substitute it in place over (V/I).  So we have R * e^-jphi?

Does that sound like I'm on the right track?


--
Paul Anderson
VE3HOP
RemoveMEwackyvorlonspam_OUTspamKILLspamgmail.com
http://www.oldschoolhacker.com
"May the electromotive force be with you."

2007\02\02@120648 by Marcel Birthelmer

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> Now, the math for impedance mimics  ohm's law - Z=V/I.  Since we're
> dealing with a wave, you can express the voltage as V*cos(wt).  I
> often lags voltage, so it is cos(wt-phi) (I think it's phi), with phi
> being the lag in phase between the two.  The equation becomes
> Z=V*cos(wt)/I*cos(wt-phi)
>
> The Euler relation says that e^jx = cos x + j sin x...  So we can turn
> cos(wt) into e^jwt, cos(wt - phi) becomes e^jwt-phi.  Z =
> V*e^jwt/I*e^jwt-phi..  The jwt exponent divides out, leaving us with
> (V/I)e^-jphi.  Hmm.  I'm not sure I'm right on that.  Since R = V/I,
> we could substitute it in place over (V/I).  So we have R * e^-jphi?
>
> Does that sound like I'm on the right track?

Sort of... you dropped the imaginary component, though. And in any
case, I'd say you're probably better off just using a trig expansion
of cos(a+b) = cos(a)cos(b) - sin(a)sin(b).

Anyway, this entire mess has been rather formalized in what are called
phasors. a Phasor is a voltage/current at a certain frequency, along
with a phase shift. You can look those up to get an idea of how to
deal with phase shifts etc. very conveniently.

- Marcel

2007\02\02@125347 by David VanHorn

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On 2/1/07, Thomas C. Sefranek <RemoveMEtcsTakeThisOuTspamspamcmcorp.com> wrote:
>
> A "proper" antenna resonant is purely resistive...
> So 50+j0 would represent a 50 ohm antenna.
>
> BTW MOST antennas are NOT 50 ohms!


The antenna becomes completely resistive (X+j0) at resonance.
What X is, depends on the antenna.

Thing is, the coupling loss between 50 ohms and X is relatively small, until
you get to rather large or small values of X.

If X is 100, then the loss is relatively small.

http://www.eznec.com/misc/food_for_thought/Food%20for%20thought%20-%20Forward%20and%20Reverse%20Power.txt

Most people miss this point, and butcher their antennas till they get close
to 1-1 VSWR, and assume that this is "properly tuned"..  FAR better to let
the antenna be resonant, and deal with the mismatch using a line section, or
some other efficient means, IF it's even big enough to worry about.

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