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Author Topic: Open Wire Reactance and Resistance Calculation  (Read 27276 times)
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aa5wg
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« on: April 11, 2011, 09:07:27 AM »

Hi to all:
i.e. If a center fed doublet and open wire line combination has a final electrical length of 135 degrees (at the ladder line input of the link antenna coupler) how do I calculate the resistance and reactance (capacitive or inductive) at this 135 degree point on the ladder line?
73,
Chuck
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KA3EKH
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« Reply #1 on: April 11, 2011, 09:49:14 AM »

I would use an impedance bridge or an antenna analyzer, not smart enough to do it mathematically. Just curious on how you know its 135 degrees length? Just by the total length? And would open feed line approaching the center of an antenna at a right angle be calculated that way? Think maybe you can figure total length plus feed line for an end fed but center feed would be more complex impedance. But that’s just me, cretin the other senior members of the group will explain all.
Ray F.

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« Reply #2 on: April 11, 2011, 11:15:00 AM »

I think we need to know what % of the 135 degrees is the antenna.  my assumption is 135 deg. is the sum electrical length of one side of the feedline and one side of the dipole.   but with only your information, the feedline could be 130 degrees and one side of the dipole 5 degrees.  we have no way of knowing the geometry of the setup which I think is necessary for a calculation.  BTW I have no idea how to perform the calculation without looking into books I have.  I hope someone else like Walt knows.
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Steve - K4HX
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« Reply #3 on: April 11, 2011, 02:06:53 PM »

Calculate, measure or model the feedpoint impedance of the doublet. It will depend on the length and height above ground. Then plug that number into the feedline calculator link below.

http://fermi.la.asu.edu/w9cf/tran/index.html
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Ralph W3GL
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« Reply #4 on: April 11, 2011, 02:18:49 PM »


   SMITH CHART??? Roll Eyes Roll Eyes Roll Eyes Grin Grin Grin !
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Don
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« Reply #5 on: April 11, 2011, 02:45:47 PM »

And don't omit the small but significant velocity factor of the feeders, plus the end factor correction for the dipole. But for practical real world applications, an approximation, ignoring those minor factors, is good enough.  Those recommended approximate lengths are given in tables on the older ARRL handbooks. Since the figures for precise calculation would hold only for a single frequency in any case, if your application is critical in terms of reactance and resistance, you might just as well go to a single frequency resonant length dipole fed with a flat transmission line, using something low-loss like heliax.

The only case where the delta-frequency error across a ham band becomes significant with balanced resonant feeders is when the feed line consists of a large number of quarter-wavelenths, and the cumulative error caused by a relatively small frequency change moves the feed point at the tuner along a substantial percentage of a quarter-wavelength, in the alternation between a current loop and a voltage loop.
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« Reply #6 on: April 11, 2011, 04:46:20 PM »

" SMITH CHART??? Roll Eyes Roll Eyes Roll Eyes Grin Grin Grin ! "

didn't know Tim had one.


klc
 
 
 
 
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aa5wg
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« Reply #7 on: April 11, 2011, 10:58:13 PM »

Let me try adding some additional information.

The example antenna will be an 80 meter dipole fed with open wire feed line, 450 ohm line.  I took one half of the dipole length and added it to the length of line to come up with a total length and multiplied this length by the velocity factor of .85.  I believe this is the electrical length.  (I don't recall what the actual velocity factor is.)  I want to see how many quarter wavelengths I have.  If the coupler needs help then I will add a capacitive or inductive shunt to bring the system to resonant.  The frequency in use will be 7.15 MHz.

 "One line wire plus one side of the dipole should be a whole-number multiple of a quarter wavelength if you want the system to be resonant."  
(Length in feet = 234/Frequency in MHz) Reference: Understanding Amateur Radio, Copyright 1963, ARRL, page 112.

The total system length is 180 feet.  (67 feet of antenna plus 113 feet of line = 180 feet)  

Then: 234/7.15 MHz = 32.7 feet = rough 1/4 wavelength of 40 meters

Then: 180 feet/32.7 feet = 5.5 quarter wave lengths or an effective system length in degrees of 135 degrees.  This not good because this is not a current loop or node.  The best place to put the link antenna coupler is at a current loop or current node.  This would be current feed or voltage feed.

I think this is a region of high capacitive reactance.  I would need to shunt inductively.  

What is the reactive value, in ohms, for this system 5.5 quarter wavelength length or effective length of 135 degrees?

Chuck

  











 




 
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W2DU
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« Reply #8 on: April 11, 2011, 11:06:15 PM »

Hi Chuck,
I'll take a crack at it. First, for such a calculation the antenna and the feedline need to be considered separately, not as a total length of the combination.

Let's consider terminal impedance of the antenna first. For this calculation I suggest using EZNEC, inputting the wire length, diameter and the desired height above ground.

Now for the feedline. If you're going to use a manufactured line obtain the characteristic impedance (Zo) and the velocity factor (V) from the mfgrs specifications. If you're going to fabricate your own feedline with say, #12 or #14 bare wire with 6" spacing Zo will be almost exactly 600 ohms. If the same wires are separated around 2" the Zo will decrease to around 500 ohms (I'll perform the calculations of this for you tomorrow, or perhaps I'll dig out the equation for it.)

I have two programs appearing in Reflections for obtaining the input impedance of the feedline when terminated in any load impedance--one is HP-2 for HP calculators using reverse-polish notation, the other is Program 7, LINTRNSM, using hyperbolic notation. These programs appear in Chapter 15 in all three editions of Reflections. I'm going to attach the programs, but to use the hyperbolic version you'll need the explanation for using that also appears in Chapter 15, so I'm also going to attach that Chapter for your convenience.

LINTRNSM.BAS runs on GWBASIC.EXE. To run LINTRNSM you must first bring up GWBASIC.EXE. I tried to attach the BAS and EXE files but discovered these files cannot be attached. So guess we're stuck with the HP-2 program, using a calculator with reverse-polish notation. Sorry about that.       

If you have any questions you can reach me at walt@w2du.com.

Walt

* CHAP15.doc (69 KB - downloaded 438 times.)
* CHAP15HP.doc (45 KB - downloaded 422 times.)
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aa5wg
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« Reply #9 on: April 11, 2011, 11:21:58 PM »

Hi Walt:
Thank you for jumping in.  I will dig out some books, including yours, and try to find this information. 
I will check back.
73,
Chuck
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Steve - K4HX
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« Reply #10 on: April 11, 2011, 11:35:46 PM »

Using NEC, I modeled your antenna at a height of 60 feet over "average" ground (Rel. dielectric constant 13.000, conductivity: 0.00500 mhos/meter).

The feedpoint impedance at 7.15 MHz is 5098.200 + i 1227.5.

Using the W9CF applet and 180 feet of 450 Ohm line (VF: 0.95, matched loss: 0.19dB/100 ft), the impedance at the end is 108.3 + i 432 Ohms. The tuner will see an inductive load.

The total loss in the feedline is 1.74 dB or about 33 watts lost in the feedline if you put 100 watts into it.

I'd be more concerned with the feedline loss than what the load at the end happens to be.
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Steve - K4HX
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« Reply #11 on: April 11, 2011, 11:53:57 PM »

The current maxima would occur at about 165 and 230 feet.
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k4kyv
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Don
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« Reply #12 on: April 12, 2011, 12:40:42 AM »

First, for such a calculation the antenna and the feedline need to be considered separately, not as a total length of the combination...

Now for the feedline. If you're going to use a manufactured line obtain the characteristic impedance (Zo) and the velocity factor (V) from the mfgrs specifications. If you're going to fabricate your own feedline with say, #12 or #14 bare wire with 6" spacing Zo will be almost exactly 600 ohms. If the same wires are separated around 2" the Zo will decrease to around 500 ohms...

If you are mainly concerned with resonant lengths and hitting a current loop or node with a balanced tuner, I'd think you could just as well consider the total length of feed line plus one leg of the dipole if you are using open wire line fabricated with single conductor wires and spreaders, since the velocity factor of the feed line will be almost identical to that of the antenna wire.  But with manufactured twin lead like window line or solid dielectric ribbon, the velocity factor is enough to cause a quarter wavelength of feed line to be substantially shorter than a quarter wavelength of antenna, so it might be well to consider the antenna and feed line separately.

I have found it possible to compensate for small discrepancies in total length by adjusting the variable capacitor of a balanced tuner, but when the tuner is looking into something close to the midpoint between a maximum voltage point and a maximum current point, it is impossible to efficiently couple to the load without externally tuning out the reactance.  Tapping down on the ATU coil without externally cancelling the reactance throws too much stray reactance into the tuned circuit to allow the balanced tuner to work properly. My experience with tapping uncompensated feeders down on the coil and tuning back to resonance with the main tuning capacitor, was flash-over in the capacitor even at low power, and very poor transfer efficiency through the tuner.

With my 160m 1/4λ dipole, I tuned out the reactance presented by the odd number of eighth wavelengths by switching in an additional 1/8 wave-length of feed line to bring a current node right to the tuner, allowing parallel tuning to be used all the way across the band, and the variable capacitor in the tuner easily compensates for the extra reactance that creeps in as I tune towards either end of the band.  Instead of that extra length of feedline, I could have inserted an inductor in series with each feeder or shunted an inductance across the feeders. But I believe I would have had to vary the external inductance as I tuned across the band, in addition to adjusting the variable capacitor in the parallel-tuned circuit, because fixed inductances (or capacitances) external to the tuner would have thrown in more reactance with a given change in frequency than the fixed length of extra feed line does.  Walt, I'd be interested in your opinion on this.

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aa5wg
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« Reply #13 on: April 12, 2011, 08:40:46 AM »

Hi Don, Steve and all:
Thank you for your input.  Steve, the line length is 113 feet not 180 feet.  The system length is 180 feet, that is, 1/2 of antenna length plus line.

Don, I agree with your approach in cancelling the system reactive component when at a 1/8 wavelength point.  This point or spot along the system length is not at a current loop or node but in between.  

I want to experiment by cancelling the reactive component (at this in between current loop/node spot) with added series inductors in each side of the line or with a single shunt inductor.  Or, if needed two series capacitors in line with each side of the ladder line or a shunt capacitor could be used if the system reactance is inductive.  

I am looking for a way to calculate, in advance, how much reactance is needed (ball park numbers) to cancel the system reactive component via the two series in line components or the shunt component.  

I did some digging and found in the 1949 ARRL Antenna Book, page116 - 121 information on how to calculate in advance this system reactive component.  However, I have to use their tiny graphs to get the job done.  I prefer a software version of the below information I found.

Does anyone know of a software version of the below ARRL information?

Page 116 from above reference states: "However, it is possible to determine in advance the approximate circuit conditions that will exist, and in many cases this will save a good deal of time in experimental work.  The input impedance of the transmission line can be determined if the characteristic impedance of the line, its electrical length, and the standing-wave are known."

The above antenna book goes on to say, "The only requirement is that the antenna should be operated at or quite close to resonance so that it presents an essentially resistive load to the line."  Does this mean my highly reactive experiment would not work with their approach?

There is more if you want me to quote it.

There you have it.  I would like to do what Don is doing, adding series or shunt reactance to the system to cancel unwanted reactance.   But, I would like to predict in advance these series or shunt values, capacitive or inductive.    

Chuck

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aa5wg
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« Reply #14 on: April 12, 2011, 09:48:54 AM »

Hi to all:
I am still digging and found some information in the antenna manual, by Woodrow Smith formerly Editor, Radio Magazine..., 1948, pages 257 - 263.  I will review this material tonight along with Walt's information.

Chuck
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« Reply #15 on: April 12, 2011, 01:00:21 PM »

With complex loads a bridge or antenna analyzer is easier and faster! Then trial and error with what you have on hand to match the load tells you way more than some software's ideal match, but that’s just me and its well know I don’t have the mental capabilities of the more senior members.
Ray F.
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flintstone mop
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« Reply #16 on: April 12, 2011, 02:18:13 PM »

With complex loads a bridge or antenna analyzer is easier and faster! Then trial and error with what you have on hand to match the load tells you way more than some software's ideal match, but that’s just me and its well know I don’t have the mental capabilities of the more senior members.
Ray F.

You are not alone...I have never been a number cruncher.........I'm terrible with math.........The MFJ analyzer does tell many things by watching the meters. Capacitive and inductive reactance....all that good stuff. And Rs meter should be close to a single digits.
Das all I know.
Fred
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Steve - K4HX
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« Reply #17 on: April 12, 2011, 02:56:02 PM »

Quote
I am looking for a way to calculate, in advance, how much reactance is needed (ball park numbers) to cancel the system reactive component via the two series in line components or the shunt component. 


I gave you a way in my previous posts. Have your tried it? Put 113 feet into the applet and see what you get. Use the Force Luke.  Wink
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« Reply #18 on: April 12, 2011, 03:06:33 PM »

Quote
With complex loads a bridge or antenna analyzer is easier and faster!

In a nice RF quiet field that may be true. But with medium to strong ambient RF fields such as radio/TV stations and the like, they can have you chasing your tail in a hurry. My MFJ analyzer is useless below 7 Mhz at home. At the cabin where things are quiet it helps out quite a bit above 160M. At 160M it still has a tendency to show eveything as inductive when in fact those real low dipoles are quite capacitive.
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KA3EKH
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« Reply #19 on: April 12, 2011, 04:59:41 PM »

Antenna analyzers are complex and can be shaky; I don’t have one but do have a couple impedance bridges and have never had issues generating power to feed the network under test. Have done this in broadcasting for years and only recently started using a bridge for Ham stuff.  Suppose I contribute to the global noise floor by driving signals thru the bridge out to the antenna or load under test but as my wife will tell you I'm not the sharpest tool in the shed.  I do use a more modern HP Network Analyzer for things at higher frequencies where the shape and connections of the bridge would be an issue but 0.54 to 7.5 range I love to use the old Delta.
Ray F.
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« Reply #20 on: April 12, 2011, 05:48:04 PM »

Chuck, I was too groggy at mid-night last night to recall stub-matching equations that will match a mismatched line, leaving no reflections between the stub and the source.

Consider the load on the line to be a higher impedance than the line impedance Zo. The place for the stub on the line is then determined by rho, the voltage reflection coefficient determined from the SWR. The distance from the load to the stub in electrical degrees is the (arc cos rho)/2.  rho = (SWR -1)/(SWR +1)

The impedance of the stub (or a lumped inductance) is found by this equation:

                                 1/(sqrt SWR/SWR - 1).

When normalized to a 50-ohm Zo with a mismatch of 3:1, this equation equals 1.1547, and the normalized line impedance is thus 1 - 1.1547 at the matching point. When denormalized the line impedance at the matching point is 50 - j57.7 ohms.

As an example, consider a 50-ohm line terminated with Z = 150 + j0 ohms, resulting in a 3:1 SWR between the stub and the load termination. The 3:1 SWR yields a reflection coefficient rho = 0.5. Taking the arc cos 0.5 we get 60°, (which appears on the periphery of the Smith Chart at the matching point.) However, those degrees are reflection degrees, which must be divided by 2 to get electrical degrees. Thus the stub is located 30° from the load toward the source. The SWR on the line from the stub toward the source is 1:1.

Hope this helps.

Walt
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aa5wg
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« Reply #21 on: April 12, 2011, 09:35:38 PM »

Steve:
I tried the calculator and it is a slick tool.  How did you arrive at 165 and 230 feet for current maximum?

Thank you for showing me this calculator.  Do you know of another calculator that does not require the use of NEC or similar programs?
Chuck
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Steve - K4HX
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« Reply #22 on: April 12, 2011, 10:11:24 PM »

If you look at the graph below the area where you enter the numbers, you can select different items to be displayed in a drop-down menu. Select current. The graph is set up so your tuner would be at the left side and the antenna feedpoint at the far right. You can see the current peaks and then look at the horizontal axis to see how far down the line the peak occurs.

I think for any calculator or a Smith Chart, you have to know the feedpoint impedance. Everything down the feedline back towards the transmitter stems from this. If your antenna is (or will be) relatively flat and at a height of 60 feet, the feedpoint impedance will be pretty close to the numbers from my model. So, plug them into the W9CF applet and play around. Select the 600 Ohm line option. Notice how much less loss there is compared to the 450 window line.

Another good one to graph is impedance. It shows, R, X and Z. Here you can really see the "flat spots" on the line where differences in length or frequency will only cause small changes in the impedance. This means you wouldn't need to retune every time you QSY 10 kHz. You'll also notice these impedance flat spots correspond to current maxima.

If you plug in 185 for the feedline length, you'll see the impedance at the tuner end is about 55 Ohms resistive and some small inductance. You could probably just hook up a good balun and skip the tuner all together.

No matter what, it's a cinch, a center-fed full wave dipole will have a very high impedance and be highly inductive. If your antenna is different from what I modeled, let me know. I'll gladly model your specifics.

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aa5wg
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« Reply #23 on: April 12, 2011, 10:25:56 PM »

Steve:
I see the horizontal axis current peaks are at .16, .82, .146.  Do I multiply by 100 to arrive at 16, 82 and 146?  Are these numbers in feet i.e. 16 feet, 82 feet and 146 feet for the current loops?  At first I thought these numbers were in wavelengths.

Thank you.
Chuck
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« Reply #24 on: April 12, 2011, 10:33:29 PM »

Walt:
Thank you for your information on stub matching.  I find this material very interesting.  I will have to get the software program you listed and look for a General Radio 1606-A or 1606B.  I don't know the differenced in the 1606A vs. the 1606B.  Sir, do you know?

At present, I am re-reading material in your books to refresh my thinking.

Link antenna couplers and ladder line are an enjoyable topic for me.  Predicting the resistance and reactance at the input of the line is an enjoyable challenge.  

Thank you.
Chuck
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