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antenna lenghts

you can't adjust an antenna for resonance if you have no way to measure (reactance) X. many analyzers provide this function. not to be confused with matching the impedance at the antenna feedpoint to the impedance of the feedline in question.. what's on your mind?
 
freecell, I have read some post on here about putting a ant. on 10ft pole and tuning it. I dont have a ant. anilizer so the only thing I could do is to check the swr on the ant. and mount it on my tower. I have a 10ft pole set up that pivots so I can make adjustments to the ant. I am just experimenting on some old ant. that I have just to learn how to do it. Looks like i am going to buy me an ant. anilizer and learn how to do it the right way :D thanks for the help.
 
And you know, when you get right down to it, the difference of that 'extra' inch or so with that 234 "magic number" isn't really gonna make much difference. Not unless you plan on only using one channel instead of all of them. It'll still radiate every bit of the power that gets to it, even if it isn't resonant. Resonant antennas are nice, but certainly not necessary.
- 'Doc

[another one'a them 'worms']
 
freecell, you forgot about the ... part?

How do you maintain electrical 1/2 wavelength coax with a frequency sweep?

I understand what your thinking is with that, but still? Oh, it's too small to0 matter?

...and I don't use the 1" coupler on that stud. The bottom of the spring is maybe a mm above the sheetmetal. I should measure it again (length and impedance). Now that I think of it, when I first put it on there, I had the 1" coupler, but no spring, so roughly 103.x inches, and it worked good there too!
 
"...the sweep has nothing to do with the line."

How is that? The line is supposed to be "tuned" 1/2 wave. If your sweeping, the tuned line length would have to change with frequency.

"attach a tuned electrical 1/2 wave line to it and look into the other end with an analyzer...then sweep it up the band and you'll find it [resonance]."

But when you find resonance someplace other than what you cut your tuned 1/2 wave line for, the line would no longer be a tuned 1/2 wave line.

"your 103.X" antenna is not at resonance and "it worked good" does not substantiate resonance one way or the other."

There are many factors beyond simple length that affect reactance, so how would you know?

"since i consider a 1.5:1 SWR to be essentially flat especially if the combination of R and j leaves Z somewhere in the proximity of 50 ohms."

Please tell, how do you simplify the complex equation? What is an example combination of R and j that would yield 50 ohms (not R=50 j=0)? What I mean, is R=60 -j10 leaving Z in the proximity of 50 ohms?

"...hmm. Always found it a bit too long. Guess it just depends on who'z doing the measuring."

Or what is doing the measuring?

What would one expect (at least the experts) from the ~103.3" whip mounted top dead center on the roof of your average sedan (the full scale RC)?

Freecell, your saying that there would be high capitive reactance in the CB band and resonance would be somwhere above (F), specifically at 28.577, if my math is correct. And what do we predict R to be at resonance?
 
'C2',
This ine is mine;
"...hmm. Always found it a bit too long. Guess it just depends on who'z doing the measuring."

Or what is doing the measuring?

So I get to rebute/debute/modify/change or make comments, right?

Maybe a better question would have been. "Who, what, where, and when something was measured."? Sounds reasonable to me.

Just never use "jockeying" and "length" in the same sentence. Oops! Wait, that's on another forum... sorry.
- 'Doc
 
the sweep has nothing to do with the line. you know if you would just reread my posts i wouldn't have to do this.

the 1/2 wave line was suggested for repeating the values present at the antenna feedpoint to the analyzer.

the sweep i referred to has to do with determining at what frequency X=0. the 1/2 wave line is not required for this.

there is only 1 real condition to be met to determine resonance. THERE MUST BE EQUAL AMOUNTS OF BOTH Xc and Xl.

"is R=60 -j10 leaving Z in the proximity of 50 ohms?" No....Z =60.828 ohms. your confusion is created here because you don't understand the notation in the condition statement, R=60 -j10....it merely states: resistance is equal to 60 ohms and there is a (+ inductive, - capacitive) capacitive amount of reactance present that equals 10 ohms. in general use the - designator is used to indicate capacitive reactance while the absence of both the - or + designator is understood to be inductive. sorry but it isn't a simple subtraction problem.

here are a couple........

resistance R = 43.910 ohms
reactance X = -25.277 (ohms capacitive)
impedance Z = 50.481 ohms

resistance R = 48.650 ohms
reactance X = 12.943 (ohms inductive)
impedance Z = 50.481 ohms

in neither one of the two instances is the antenna resonant at the frequency where the measurements were taken, so you can see that a 1:1 swr is absolutely meaningless when it comes to determining resonance. that only occurs when X = 0. both of these samples show how an antenna can have a great match and still not be resonant for the frequency at which the swr measurement is taken.

"your saying that there would be high capitive reactance in the CB band and resonance would be somwhere above (F), specifically at 28.577, if my math is correct."

i have no clue what you're talking about here or where you got "the math." all i stated is that at 103.3" the antenna is NOT resonant at 27.185 and that resonance would occur further up the band. the only possible way you could have come up with that frequency is to model an antenna that exact length in a real installation and measure equal amounts of Xc and Xl at that frequency. RESONANCE ONLY OCCURS WHEN X = 0.
 
If you know the R and X values and you have a calculator with a square root function, it's fairly straightforward using the old Pythagorean theorem. Linear graph paper might help.

Plot R along the horizontal axis and X along the vertical axis, using any convenient scale.

Connect the ends of the two values -- you have just created an impedance vector. Now, solve.

In Freecell's first example, R=43.91 and X=-25.28 (rounded).

Square both values: 1928 and 640. Added together = 2568.

Now calculate the square root of 2568 = 50.67 ohms.

The ol' "square root of the sum of the squares" trick.
 
Thanks Beetle, that is the simple answer I was looking for.

freecell, it just seemed that you were saying you needed to use a tuned 1/2 wave feed to accurately measure the impedance at the other end, or else some other length would induce some impedance transformation. That is more than just an implication:

"if you want the values of resistance and reactance present at the feedpoint repeated at the opposite end of "a line" terminating with the analyzer then an electrically tuned 1/2 wave line will be required to make that happen."

Is this following statement not contradictory?

"the sweep i referred to has to do with determining at what frequency X=0. the 1/2 wave line is not required for this."

Are you implying that if there is no reactance at the antenna that X = 0 no matter what length of feedline?

You also said that the resonant length of the antenna should be somewhere around 108" You posted the figure 246? Is not the math 246/27.185 = 108.6 inches, where you claim resonance (AKA X=0) would be?

I just extrapolate that to my length of 103.3" and leave F as the unknown. That yields the supposed frequency of resonance using your 246 value. Our debate was over 234 (to account for the VF of metal) vs. 246.

"there is only 1 real condition to be met to determine resonance. THERE MUST BE EQUAL AMOUNTS OF BOTH Xc and Xl."

I don't know if I believe that. If I take a 7 foot whip and a nice tuner to add the conjugate match, is the 7 foot whip now resonant on the CB band?
 
No........

resonance occurs (at any given frequency) when values of inductive and capacitive reactance are equal, cancelling each other out leaving (reactance) X equal to = 0. whether you believe that or not is irrelevant. that's the definition of resonance.

Beetle,

thanks for the math lesson.
 
A wise one told me that the antenna would not be resonant, but the system would.

And I asked two other engineers how to simplify a complex impedance into a simple number, like 60 ohms, and I just could not get the answer. I thought it was the sum of squares, but was not completely sure. It makes a little sense considering that constant impedances plot circles on a smith chart.

Now, what the heck is going on here?

For Z = 1.3-j0.7, find SWR:

SWR = (1 + |Gama| ) / (1 - |Gama| ) where Gama = (Zl – Zo) / ( Zl + Zo)

Gama = 50 * (1.3-j0.7) – 50 / 50 * (1.3-j0.7) + 50 = 0.2042 + j 0.2422 = 0.3168 Ang 50°

SWR = 1+ 0.3168 / 1 – 0.3168 ~= 1.9 <-- wrong answer?
 
"I thought it was the sum of squares........"
that's because you're not paying attention........

it's the SQUARE ROOT of the SUM of the SQUARES.

given: 1.3 -j0.7, find Z.

1.3 squared = 1.69
0.7 squared = 0.49

1.69 + 0.49 = 2.18

the SQUARE ROOT of 2.18 = 1.476

Z = 1.476 ohms........impedance is a combination of resistance and reactance. 1.3 -j0.7 represents the values of resistance and reactance and IMPEDANCE is DERIVED from applying (Pythagoreans Theorum) the required calculations to the values of resistance and reactance before IMPEDANCE can be determined.

and for whatever it's worth:

formula.gif


SWR is equal to 1 + the square root of the reflected power divided by the forward power, DIVIDED BY 1 - the square root of the reflected power divided by the forward power.

wise, huh? engineers, huh? yeah, 1.9 is the WRONG answer. try 33.8:1 VSWR at the load for 50 ohm line.
 
Excuse me for being lazy...I was paying attention and meant to say root of the sum of the squares.

I also forgot to mention that the *complex* impedance was not normalized, so the VSWR would be 1.476:1.

my fault there.

Given that, does anyone know what the engineer was calculating there or where it went wrong?

Unless, I think maybe freecell, your eluding to it in the image?

:(
 
"I also forgot to mention that the *complex* impedance was not normalized, so the VSWR would be 1.476:1."

no....1.476 was the value for Z solved after performing the proper calculations to: 1.3 -j0.7....if you want to refer to that as normalized to differentiate between that and the complex version that's fine. the number has nothing to do with the VSWR. i told you what that was in my last post. re-read it.
 

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