The AM Forum

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

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.

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.

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

   SMITH CHART??? ::) ::) ::) ;D ;D ;D !

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.

" SMITH CHART??? ::) ::) ::) ;D ;D ;D ! "

didn't know Tim had one.


klc
 
 
 
 

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

  











 




 

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

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

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.

The current maxima would occur at about 165 and 230 feet.

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.

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

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

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

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.  ;)

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.

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.

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

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

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.

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

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

Steve in his prior post is correct and identification of the antenna terminal impedance vs. frequency, for example, is key. With that accomplished, there are a number of interactive Smith Chart programs which are easy to learn to solve the original post. See the site shown below.
 
http://sss-mag.com/smith.html

EZNEC DEMO will handle the impedance problem, or try any of the NEC freeware programs like 4NEC2. However, EZNEC in its DEMO form is really easy to learn and quite FUN and addictive! The DEMO version will readily handle most wire arrays.

Then after finding the antenna Z (from a WIRE MODEL) use the Smith Chart utilities to calculate the Z looking into the feed line or any other matching system. As Walt highlighted, this can be accomplished via math calculation. However, the REAL BEAUTY of the INTERACTIVE Smith Chart utilities is you obtain a quick intuition and feel for what reactive elements will accomplish and how the antenna impedance gets modified. I wrote one of these routines in MathCAD in 2001 and it appeared in Applied Microwaves. It is located at the site mentioned above. However, there are far more powerful and intuitive Smith Chart utilities available as shareware and I will try to post one or more of them here.

Alan W4AMV

If you use feet for the cable length, the graph will be in feet. And yes, multiply the numbers by 100 (at least for this example).

See the image below. Notice this image is displaying ONLY current. You'll see a current peak at around 0.16. This means the current peak is 16 feet back towards the antenna from the tuner end of your 113 foot long feedline. So, if you cut off 16 feet or made the feedline 97 feet long from the start, you'd be at a current peak (or close enough).

W4AMV mentioned some modeling programs. 4Nec2 will model the feedlines too and do stub and matching section calculations.

Alan, Steve and all:

Thank you for the information regarding Smith Charts.  I will look for one of these programs and give it a go.

Steve, the calculated input impedance at the antenna coupler and feed line is found to be 83.48 ohms +j371.24.  This +j371.24 inductive reactance translates to a 8.3 uH coil.  This is for the 113 foot feed line example.

Question:  Can I use this 8.3 uH coil as a single shunt inductor at the input of the line at the antenna coupler?
               And/or, could I divide 8.3 uH by 2 and place a 4.15 uH coil in series with each leg of the ladder line at the line input to antenna coupler?

Thank you.
Chuck

Hi Chuck. Glad to help, what FUN! I need clarification on your question. The calculated input Z at the antenna coupler (which is at the end of the feed line??) is 83.48 ohms + j 371.24 ohms. Is that correct ? If that is the case, then the problem is to match this to 50 ohms ? If that is the case,  then you calculated the SERIES impedance at a given frequency and found that the equivalent L is +j371.25 ohms. Is that correct ? If so, then you only need to add a series CAPACITIVE reactance at THAT calculated frequency. That reactance is -j 371. Now I would have a real value remaining of 83 ohms. Not a bad match to 50. So... you can see, I hope, why I have a problem with your QUESTION. Let me know where I might be amiss.

Hi Chuck,

Glad you found the stub-matching material of useful value.

The 1606A and B are identical except for cosmetics--the A is black and the B is gray. Oh yeah, I forgot--the knobs on two are of different style.

On my Z-matching material in Reflections, as you know, I couldn't attach those files in BAS or EXE. Don't know why they're not allowed as attachments. So here's what I'm gonna do--I'm gonna email them to you. In the form you'll receive them you can use them directly on receiving them. Since you have an edition of Reflections the explanation of how to use the programs is explained in Chapter 15, along with the print-out of the programs. You'll find it interesting that when using the hyperbolic program on the PC vs the hand calculator with reverse-polish notation, the answers come out identically to at least 10 significant figures. They sorta validate each other, hey?

Chuck, you didn't say whether you're the same Chuck Pool I corresponded with several years ago.

Walt

Hi Walt:
Thank you for the information on the 1606A and 1601B.
Yes sir, I am the same guy who corresponded with you several years ago.
I enjoy reading your books.  I have three of them.
I look forward to receive your email and giving the software a go. 
Thanking you.
Chuck

Alan:
The calculated input Z at the antenna coupler is at the end of the feed line and it is 83.48 ohms +j371.24.  Yes this is correct.  The challenge is to match this to 50 resistive ohms, yes.  I did not calculate the SERIES impedance at the given frequency of 7.15 MHz.  At least I don't think so.  I have not had any good sleep for a couple of days and I think I goofed.  The calculated impedance was found to be 83.24 ohms +j371.24 (inductive) thus I should add -J371.24 of capacitive reactance to cancel the found inductive value of +j371.24.  This translates to an approximate capacitor of 60 pf.

Question:  Can I shunt this 60pf capacitor at the input end of the line at the tuner?  And/or could I also divide this 60 into two 30 pf capacitors and place one capacitor in series with each side of the line?  This would take place at the line input at the tuner.

Did I get it right?  I hope so.
Chuck

Hi Chuck. Yes, you almost on the track. The impedance between the terminals is SERIES in form. That is to say, IMPEDANCE is by definition a SERIES network. Therefore, to accomplish what you desire, you need to add a SERIES element. If you wish, add twice the required C in each arm of the feed line. Or, of course you could add a 1:1 balun. However, recognize a balun is not perfect and the impedance that you initially had will be slightly modified by the reactance of the balun. In any case, to your question, 120 pF added in series with each arm is desired. If you wanted to add a parallel element, you are going to have to find a PARALLEL network which has the same frequency response as this SERIES IMPEDANCE network.  And it exists! However, that is a topic for another thread! Hope this helps.

Alan:
The link antenna coupler I use can switch between series or parallel configuration.  Series for low Z line input and parallel for high Z line input.
I calculated 60 pf worth of shunt capacitance is needed to cancel the inductive reactance.  Or, I could use 120 pf worth of capacitance in each leg of the input line for series tune.  Is this correct?
Chuck

The impedance is 83.48 ohms +j371.24.  That is a low Z line input, so the link antenna coupler will need to be in series configuration. Adding 120 pf in series with each leg would be equivalent to reducing the total capacitance in the main tuned circuit, since you are effectively placing the series tuning capacitor(s) in the ATU in series with the 120 pf capacitors you inserted into each leg of the balanced feed line to cancel out the +j371.24  .  You should be able tune out the reactance with nothing more than the main tuning capacitor, without the need for an additional set of 120 pf capacitors.  The main tuning capacitor would simply resonate at a lower capacitance setting than it would if the 83.48 ohms were purely resistive.

If the transmitter is picky about working into something close to a 50Ω load, you can transform the 83.48Ω to 50Ω, either by physically adjusting the link if it is set up for variable coupling by swinging the link in and out of the coil or else by rotating it variometer style, or by placing a variable capacitor in series with the link.  See a late 50s or early 60s ARRL handbook for figuring out how to determine the size of the series variable capacitor.

Sure, yes Don and Chuck. You're matcher is far more flexible than I had in mind!

Don, Alan , Walt and all:
Give me a couple of days please before I ask a follow up question.  I do appreciate all of your help!
Please check back because I value your advice very much!
Chuck

Hi to all:
I am still have a couple of questions.  But, I need to finish some reading.  I will check back as soon as I can.  Thank you for your help.
Chuck