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Author Topic: Antenna Analyzer for Open Wire Feeders  (Read 40323 times)
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WD5JKO
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« Reply #25 on: September 20, 2011, 08:26:30 PM »

I have an RF1, the older model. I find it very handy for the price. Measuring C and L at the operating frequency is very handy. Many people put more faith in those lab units with lots of digits that only test at 1 Khz too 0.1%. At 7 Mhz that capacitance they measured might be at the high side of resonance and look like an inductor. Here the Autek's shine.

Still we have to measure right at that SO-239 connector. Just a few inches of leads make a big error when the frequency is say > 5 Mhz. There is no provision to calibrate a line extension like some of the more expensive S11 gizmo's where you cal with a 'open', 'short', and 'X-Ohms resistive load'.

The VA1 looks nice to measure SWR at say 450 ohm level. Still, we take a balanced line to an unbalanced connector. Even holding the unit in hand might unbalance the transmission line a bit. Maybe best to sit it on a piece of wood when doing tests like this.

Measuring parallel line SWR at full strap transmit power requires another approach..

Jim
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aa5wg
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« Reply #26 on: September 20, 2011, 09:54:52 PM »

Jim and all:
I would not have to measure at full 120 watts.  QRP power would be fine if the accuracy is there.

Did my work in the previous replies make any sense to anyone?
Chuck
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Jim, W5JO
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« Reply #27 on: September 20, 2011, 10:11:19 PM »

Nowhere you know why antennas are black art.  Calculate, measure, build and install.  Then fiddle with it until it works.  Another thing, when I started, no one cared what the SWR was.  Just get the florescent buld as bright as possible and operate.

Does knowing the exact impedance help?  In lab circumstance yes.  But when you install, that all changes.
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aa5wg
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« Reply #28 on: September 20, 2011, 10:33:56 PM »

Jim:
You make some excellent points.  Did my calculations make sense?
Chuck
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k4kyv
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Don
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« Reply #29 on: September 21, 2011, 03:21:40 AM »

The graph shows the first quarter wavelength, 2nd, 3rd and 4th quarter wavelengths.  The zero point starts on the far left side of graph and goes to the right to the 1/4 wavelength mark (on the horizontal line of the graph).  Zero to the 1/4 wavelength mark is shown as capacitive reactive.  From the 1/4 wavelenght mark to the 1/2 wavelength point, going right on the graph, shows the reactance as inductive.

This graph is showing odd quarter wavelengths is the starting point of capacitve reactance and even quarter wavelength as the beginning of inductive reactance.  Thus, I think, (jump in here to correct me) my 13.48 quarter wavelength long 15 meter antenna system would be capacitve reactive.

You are right, according to my calculations.  You base the + or - sign of the  reactance  starting at zero  feed line length.  From zero to 1/4λ, the reactance is capacitive. From 1/4/λ to 1/2λ the reactance is inductive, etc.

Your 13.48 quarter wavelengths , ≈ 13.5 quarter λ, has six half wavelengths + another 1.5 quarter wavelengths. When using the chart, you can discount the first 12 quarter λ (six half wavelengths), since the impedance at that point will be the same as the original input impedance. In your case, it is an open circuit. So really, what you  have to deal with is the additional 1.5 quarter wavelengths. The first additional quarter wavelength converts the high-Z apparent load to low-Z, still purely resistive.  So what you are left with is a 1/8λ with a low-Z input, which displays a capacitive reactance as the antenna tuner sees it.  

My 160m dipole has 3/8λ total from the end of the dipole to the end of the feed line at the bottom of the tower.  Each leg of the 160m dipole is 1/8λ, plus the two quarter waves between the feed point of the dipole and the tuner. At the half wavelength point, midway down the tower, the impedance is purely resistive and very high. At the end of the feed line, the additional 1/8λ makes  the apparent impedance capacitive. I could use lumped inductors, but instead simply added another 1/8λ of OWL in my "tennis net".

That clears up the confusion about the odd 1/8λ. Define the starting point as zero; from 0 to 1/4λ, the reactance is capacitive, and alternates each additional quarter λ.
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« Reply #30 on: September 21, 2011, 10:07:35 AM »

Don and everyone:
I think I may be wrong.  Please look at my 3 attachments.  The crude sign wave graph shows the 13.5 quarter wavelength mark in the inductive reactive side.  Thus, I think my 13.5 quarter wavelength antenna would be inductive reactive and not capacitive as I first thought.  And, because it is at the .5 or 1/8 wavelength or 45 degree point it would be 600 ohms reactive.  Please see the other two text attachments to calculate for 600 ohms or see my earlier replies.

Am I correct in thinking the open circuit lines (top attachment figure 3-31) could be used for a center fed zepp and a short-circuited lines (top attachment figure 3-30) could be used for a closed loop fed with ladder line?

What do you think?
Chuck


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ke7trp
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« Reply #31 on: September 21, 2011, 03:22:29 PM »

So you can bandswitch it?

C


c
I would like to be able to know the reactance at the coupler and line juction when the total antenna system length is at a mid-way point between a current loop and voltage loop at this junction.  I can then add the correct amount of reactance (shunt or series) to resonant the antenna system.

Chuck
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aa5wg
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« Reply #32 on: September 21, 2011, 09:02:43 PM »

c:
The extra shunt or series reactance (capacitive or inductive) would be mounted on or next to the link coupler.  If needed, either one could be plugged into the circuit.  Band switching would be a snap.

Would anyone care to look at my above work to see if it is correct please?
Thank you,
Chuck
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k4kyv
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« Reply #33 on: September 22, 2011, 12:13:07 AM »

I'm still thinking about it.

From your diagram, from zero all the way to 12/4λ, you have 6 complete half-wavelengths (or 3 full-wavelengths), so that at that point you are still exactly where you started - at a high Z nonreactive point, so you can ignore those 3λ to the left.

The 13th quarter wavelength section converts that hi-Z resistive point to a low-z resistive point. You could replace the first 13 quarter-wavelengths of wire with a low-value non-inductive resistor. Following that, the additional 1/8λ starts out at a low-Z purely resistive point and terminates midway between a low-Z and high-Z resistive point. The additional 1/8λ according to the chart, shows that a less than one quarter-wavelength short-circuited line (and a low Z compared to the Zo impedance of the line has the same reactive sign as a short-circuit) would have an inductive reactance at the input terminals.  Huh

I'm going to have to study that a little more, because I see a contradiction, or else I remember wrong. I was comparing the shorted-out OWL section under discussion to a grounded vertical - capacitive at less that 1/4λ, so you use a loading coil to bring the shortened length to resonance.  When I operated in the apartment in Cambridge, I used an end-fed zepp running across the street with a slightly less than 1/4λ feeder, and used a variable inductor in series with the feeders to bring the shortened stub to resonance. But I see that the diagram suggests that a less than 1/4λ shorted stub would be inductive, not capacitive.

We have to be overlooking something simple and obvious here. Confusion over whether the low-Z point is at the generator or at the load end of the stub?
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« Reply #34 on: September 22, 2011, 08:58:07 AM »

Jim and all:
I would not have to measure at full 120 watts.  QRP power would be fine if the accuracy is there.

Chuck,

QRP measurements are fine for tune in and adjusting. But If you are trying to troubleshoot a problem, sometimes it wont rear it's ugly head at flea power levels. If something is zorched, or arcing over, you want to make as many of your tests as you can at Full power. If something is intermittant, a lot can change as you QRO
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aa5wg
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« Reply #35 on: September 22, 2011, 08:58:47 AM »

Don:
Do you agree that the 13.5 quarter wavelength long 15 meter antenna systyem is (1) inductively reactive and (2) at 600 ohms or 135 degrees past the high impedance point which is 45 degrees (1/8th wavelength) past the peak current point or low impedance point?   My crude quarter wave chart is showing current.  
Thank you for your help.
Chuck
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aa5wg
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« Reply #36 on: September 22, 2011, 09:00:10 AM »

Slab Bacon:
I agree.  Testing at higher power sometimes provides a different picture.  Very good tip.
Thanks.
Chuck
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k4kyv
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Don
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« Reply #37 on: September 22, 2011, 11:55:57 AM »

Don:
Do you agree that the 13.5 quarter wavelength long 15 meter antenna systyem is (1) inductively reactive and (2) at 600 ohms or 135 degrees past the high impedance point which is 45 degrees (1/8th wavelength) past the peak current point or low impedance point?   My crude quarter wave chart is showing current.  
Thank you for your help.
Chuck

It still seems like it should be capacitive, but the chart you scanned says inductive. I have several editions of ARRL antenna book, dating from pre-WW2 to just a few years ago, as well as other texts, such as the Laport Antenna book and Radio antenna book. I need to pull out some of those where I can study the chart and read the entire text, and see what I am missing or where the confusion lies.

It would seem to me that if you took an 1/8-wave stub and shorted out one end or placed a low value resistor across that end, that the opposite end would show a capacitive reactance. That's what your 13 quarter waves (or any number of odd quarter-wavelengths for that matter) is the equivalent of: a short across the OWL (if the entire length of wire is a parallel conductor), or a low-value resistor (if the first quarter-wave is pulled apart and stretched out horizontally to form a half-wave dipole, 1/4λ per leg).

600Ω sounds reasonable, but I haven't tried to calculate the actual value of reactance or resistance at that point, or look it up in the charts. Exact figures would depend on the impedance at the feed point of the antenna (which would vary somewhat depending on height above ground, inverted vee versus true horizontal, the presence of nearby objects and antennas, etc).

Regarding the reactance (inductive or capacitive?), either I missing something fundamental, or we are each talking about two different things and don't realise it.  Let me review my ARRL book with those some charts you scanned and maybe some of the other references and see if that helps clear up where the contradiction is.

Interesting discussion in any case.  That's the way we learn things. And sometimes the "books" lack clarity, are inadvertently misleading, and occasionally, just plain wrong.
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« Reply #38 on: September 22, 2011, 01:41:12 PM »

Don:
The universal reactace chart,I scanned with the two formulas, gave me 600 ohms reactance for the 1/8th wavelength or 45 degree point on the curve.
This is interesting because the feed line impedance, i.e. 600 ohms, would always effectively be the same at the 45 degree point along that crude current curve drawing, i.e. 600 ohms reactance at 45 degrees = 600 ohm impedance feed line.  The same goes for 300 ohm, 450 ohm etc.

I am going to use a variable capacitor or tapped inductor instead of a stub line or extra feed line to counter the reactance found at this 45 degree point.

I too, will dig into some other text material.

Thank you.
Chuck

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k4kyv
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« Reply #39 on: September 22, 2011, 02:33:54 PM »

But what is the resistive component at the 45˚ point? I don't think 600 ohms of reactance at the tuner would mean anything in regards to the line impedance Zo, since the line is running with standing waves, and the 600Ω reactive point exists only at the end of the transmission line where you are feeding it, and at 1/2λ intervals up towards the antenna, and the tuner is cancelling out that reactance anyway. On second thought, the resistive component would be irrelevant for the same reason; is not constant, but varying along the line in sinusoidal fashion, too.

This looks like a situation where conjugate match theory would be helpful. But I would probably just use trial-and-error and a field strength meter to see what works best.  That would most likely be quicker. Nevertheless, it's still nice to know why something works, rather than flying by the seat of your trousers the way Deforest did when he invented the triode tube.
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« Reply #40 on: September 22, 2011, 07:17:38 PM »

Don:
I agree.
It would like be nice to know how to predict the component values at these 45 degree, 1/8th wavelength points, for open wire lines.
This brings me back to my first question.  I wonder where can I find an antenna analyzer designed and built specifically for open wire lines?
Chuck
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« Reply #41 on: September 22, 2011, 08:43:28 PM »

Don, if I may horn in, I'd like to comment on the reactances obtained with xmsn-line stubs.

A stub shorted at one end with lengths less than 1/4 wl will be inductive at the opposite end. As length increases from 1/4 wl to 1/2 wl the reactance will be capacitive.

A stub open on  one end with lengths less than 1/4 wl will be capacitive at the opposite end. As length increases from 1/4 wl  to 1/2 wl the reactance will be inductive.

Hope this helps.

PS--I should also have said that at length 1/8 wl, or 45°, the reactance X = Zo of the line.

Walt
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« Reply #42 on: September 22, 2011, 09:25:03 PM »

Walt:
Thank you for the conformation the 45 degree point is reactive with same Zo of the line, i.e. 600 ohms.

If you have time would take a look at my reply #30 of this post.  The last attachment is a crude current graph of my 15 meter total antenna system length.

It shows at the far right that this system is inductive reactive at the 45 degree point.  

Is this correct (inductive reactive) or do I need to gather further information?

Chuck
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k4kyv
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« Reply #43 on: September 22, 2011, 09:50:29 PM »

Don, if I may horn in, I'd like to comment on the reactances obtained with xmsn-line stubs.

A stub shorted at one end with lengths less than 1/4 wl will be inductive at the opposite end. As length increases from 1/4 wl to 1/2 wl the reactance will be capacitive.

A stub open on  one end with lengths less than 1/4 wl will be capacitive at the opposite end. As length increases from 1/4 wl  to 1/2 wl the reactance will be inductive.

Hope this helps.

PS--I should also have said that at length 1/8 wl, or 45°, the reactance X = Zo of the line.

Walt

Walt,

Thanks for your input.  Maybe you can clear up something for me.

I based my supposition regarding the capacitive reactance of the < 1/4λ stub shorted at one end, on the behaviour of the familiar vertical antenna working against ground.  A 1/4λ vertical will be purely resistive @  approximately 36 ohms.  A vertical shorter than 1/4λ will have capacitive reactance, and you cancel out the C reactance by adding a loading coil in series at the base of the antenna. The inductive reactance of the coil cancels the capacitive reactance of the antenna, and the antenna is tuned to  resonance. Of  course, we know the resistive component is now less than 36 ohms, decreasing rapidly as the wire length decreases, but nevertheless the reactance is  always capacitive and we use a series inductance to tune it.

So, why does a quarter wave or less of parallel line, shorted at one end, behave differently (I'm talking about inductive vs capacitive reactance) from a single wire shorted to earth? Again, we can agree that the single vertical wire radiates while the parallel line, which has a corresponding wire of opposite polarity in close vicinity, doesn't.

One difference in my analogy compared to the vertical antenna is that the loading coil is inserted at the base (the shorted end) of the antenna, while with the stub, the reactance is measured (and tuned out with the ATU) at the opposite (unshorted end).  Could that have something to do with the apparent (to me) paradox? However, the loading coil can be attached to the top of the vertical (the unshorted end), if there is a capacitive top hat or something similar for the loading coil to work into, serving as a terminal to complete the circuit back to earth.

So would a 1/8λ vertical wire shorted to earth, which could be tuned to resonance by inserting a series coil at the base, display inductive reactance at the top (if the measurement is taken with no  loading coil at the base to resonate the wire)?

Don
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« Reply #44 on: September 22, 2011, 10:22:20 PM »

Good question, Don, but right now it's too late to consider further. I'll definitely get back with you tomorrow.

Walt
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« Reply #45 on: September 23, 2011, 11:45:37 AM »

A stub shorted at one end with lengths less than 1/4 wl will be inductive at the opposite end. As length increases from 1/4 wl to 1/2 wl the reactance will be capacitive.

A stub open on  one end with lengths less than 1/4 wl will be capacitive at the opposite end. As length increases from 1/4 wl  to 1/2 wl the reactance will be inductive.

Walt

I've seen this work out in the field for antennas.

At one time (1987) I had seven 75M  dipoles at 120' high, EACH fed with open wire line. They were all broadside facing NE/SW with 130' of open wire (1/2 wavelength) coming to the ground for each  dipole.  They were positioned just like a Yagi, spaced about 50' apart.

To tune them I had the driven element connected to a receiver and a beacon out in the woods.  I made all parasitic dipoles a little short, like about 120' each. By moving a shorting wire along the bottom of each feedline I was able to reflect back a small amount of inductance to the top junction of these parasitic dipoles and tune the array for maximum f-b or forward gain. It worked really FB using relays to move the shorting stubs for NE or SW coverage.

Earlier I had the same array up at 60', and used 1/4 wave long open wire feeders. By using 250pf variable caps on each end of the openwire, I was able reflect back inductance and tune the elements in the same way.


The drawback was, believe it or not, all that openwire loss added up and I could never get the same gain as using just simple pretuned Yagi elements.  The Yagi elements were as low as 20 ohms at the center once tuned, so the mismatch loss added up over seven elements.  I thought it would be much less, but alas, it was a factor.

T
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« Reply #46 on: September 23, 2011, 12:42:28 PM »

Addressing reactance in lines and antennas.

Let’s first consider dipoles and verticals over ground. At this point we’ll consider all wires and conductors to be lossless—zero resistance. We’ll begin with a center-fed half-wavelength dipole, each half of which is 90 degrees in length. Current leaving the center encounters 90 degrees in travelling to the far end, sees an open circuit, developing a 180-degree phase change, then encounters an additional 90 degrees on return to the center. Thus, during one round trip from center and return, the current has traveled 360 degrees, returning precisely in phase with the next cycle of current (and voltage) arriving at the center from the source. Consequently, no reactance has been developed, and the dipole is in resonance.

However, because power is lost during return to the center, due to energy radiated by the dipole, the magnitude of the returning (reflected) current is somewhat less than at the beginning. Consequently, the difference in magnitudes of the source and reflected current determine the terminal resistance of the resonant dipole.

However, consider what would happen if there were zero radiation. This condition will occur if we fold the dipole at its center, thus forming a quarter-wavelength open-circuit stub. Radiation is canceled because the radiation from each leg of the stub is equal and opposite.  Consequently, with lossless conductors, the open-circuited stub presents zero resistance at the opposite end, a virtual short circuit. (With real conductors the resistance will be very small.)

We’ll now consider a quarter-wavelength dipole, one-eighth wavelength on each side. One eighth wavelength is 45 degrees.  In this case the current encounters 90 degrees of travel during the round trip, plus the 180-degree phase shift at the open circuit at the far end of the dipole, for a total of 270 degrees of phase change. Thus the returning current arrives at the center 90 degrees earlier than the current (and voltage) arriving from the source. We know that when current leads voltage the result is a capacitive reactance, the condition prevailing in this case.

If we now fold the quarter-wavelength dipole at the center we again get an  open-circuit stub, this time one of one-eighth wavelength, or 45 degrees. Because the electrical lengths have not changed while folding this dipole into a stub, the open-circuit stub less than a quarter-wave in length, the reactance of the stub is capacitive.

 The following are two equations that determine the reactance of a transmission-line stub shorter than a quarter-wavelength :

XL = jZo tan length in degrees.

XC = -jZo cot length in degrees.

For  line lengths between 1/4wl and  1/2wl the signs of the reactances reverse.

I hope this essay helps in understanding the relation concerning reactance appearing on lines and antennas.

Walt



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« Reply #47 on: September 23, 2011, 01:10:10 PM »

I figured it out!

On several occasions in years past, I have used half wave end-fed zepps with a quarter wave of tuned feeder, and open-wire fed dipoles with a half wave of open wire tuned feeder, as a single band antenna for 75/80m. The transmitter had a link-coupled final.  I was able to use the antenna all the way across the band by cutting the line slightly short for the top end of the band, and placing a rotary inductor in series with one of the feeders. As I moved down the band, I would roll in additional inductance and the antenna would  load up just as it would with a full-fledged link coupled tuner. As I QSYed down, I added in more inductance to tune out the additional capacitance as the frequency was lowered. The same phenomenon exists with the ≤ quarter-wave vertical as discussed in the provious post.

The only way for this to work is, if a shorted stub, less than 1/4λ shows inductive reactance, then an open stub the same length must show capacitive reactance, just as Walt pointed out.  And that's exactly what the charts say.  

The end-fed zepp feed line is like an open stub, since one feeder floats free and the other is attached to a high voltage point. The dipole with the half-wave feeder can be thought of as a quarter wave feeder feeding another quarter wave feeder, which feeds the low-Z dipole.  The high voltage point is right at the middle of the feed line, so that the bottom half of the feed line could be thought of as a 1/4λ open stub, feeding the voltage point.

The charts show that the reactance at the input terminals of on open-circuited line, as a function of line length in wavelengths, is the opposite sign (inductive vs capacitive) from that of a short-circuited line.  So it depends on which end of the line you insert the capacitance or inductance.  In the case of the ≤ quarter-wave vertical and the zepp feeders, the inductance is placed near the shorted end, and at that point, you are looking at an open-circuited line.  But in the case of the antenna/feed line combination with the 13.5 quarter wavelengths, the first 13 quarter-waves could be  replaced by a short-circuit or by a low value resistor, and the  remaining 1/8λ, from the tuner's viewpoint, is a short-circuited line, therefore it shows inductive reactance.

What I have been overlooking is the fact that the same line, with a short circuit or a resistor attached to one end, is an open-circuited line from the resistor's (or shorting bar's) viewpoint, but looking back from the opposite end of that same line, one sees a short-circuited line. I was simply thinking short line = loading coil = capacitive reactance, and long line = series capacitor = inductive reactance.  Of course, this is true only from the perspective of the appropriate end of the line.

The grounded 1/8λ vertical indeed does show capacitive reactance at its base. But if you went up in a hot-air balloon or helicopter and measured the reactance by touching the probe to the tip-top, that same antenna would show inductive reactance.

Senior moment in action.
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« Reply #48 on: September 24, 2011, 11:52:12 AM »

On my post #46 (yesterday) I forgot to include the minus sign with the jCOT term in the equation for XC. I am correcting that error in #46.

Walt
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« Reply #49 on: September 24, 2011, 06:48:57 PM »

I am double checking to make sure my graph in reply #30 is correct.  Is it correct?  Don, Walt and everyone your replies were very helpful.
Thank you.

Chuck

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AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
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