A few days ago I constructed a tape measure type antenna for 2 meters.
It is a 3 element beam and uses a hairpin or beta match.

Ran a return loss on it with the signal generator/spectrum analizer. The
best return loss was about 149 Mhz instead of 146. Thought it may be
because the tape AI used was 1/2 inch wide instead of 1 inch wide. I tried
lenghtning the hair pin form 5 inches to 6 inches. This did not change the
frequency of the return loss to any great ammount , but the RL went from
around 20 or more db to 10 db.

Did a little research on the hair pin match. From what I am getting out of
this, and it is very rough, is that the length of the driven element does
not make much differance in the actual resonate frequency of the antenna.
It is mostly where it is placed on the boom and the other elements that
determin the frequency of the antenna, but the length of the driven element
is determined by the impedance of the transmission line to be matched and
the impedance of the antenna/driven element.

Is that mostly correct or did I miss something in my reading ?

A few days ago I constructed a tape measure type antenna for 2 meters.
It is a 3 element beam and uses a hairpin or beta match.

Ran a return loss on it with the signal generator/spectrum analizer. The
best return loss was about 149 Mhz instead of 146. Thought it may be
because the tape AI used was 1/2 inch wide instead of 1 inch wide. I tried
lenghtning the hair pin form 5 inches to 6 inches. This did not change the
frequency of the return loss to any great ammount , but the RL went from
around 20 or more db to 10 db.

Did a little research on the hair pin match. From what I am getting out of
this, and it is very rough, is that the length of the driven element does
not make much differance in the actual resonate frequency of the antenna.
It is mostly where it is placed on the boom and the other elements that
determin the frequency of the antenna, but the length of the driven element
is determined by the impedance of the transmission line to be matched and
the impedance of the antenna/driven element.

Is that mostly correct or did I miss something in my reading ?

Hmmm. I think it's easier to look at it otherwise.

If you consider the driven element as a dipole in isolation, its
resonant frequency is going to be controlled by the length and the
width (thickness) of the element. Longer elements have lower resonant
frequencies, and for any specific length, thicker/wider elements will
have lower resonant frequencies. My guess is that using 1/2" tape
rather than 1" tape was what made the resonant point somewhat higher
than you had expected.

Adding the reflector and director elements is not going to change the
resonant frequency of the DE very much. What it will tend to do, is
change the radiation resistance (the resistive portion of the
impedance) and thus the impedance at the feedpoint. In a typical
Yagi, the feedpoint impedance is often down in the 25-ohm range.

What the hairpin-match for a Yagi does, is create an L-match which
raises the feedpoint impedance to 50 ohms, compared to what you would
have without such a match but with the DE cut for your desired
resonant frequency.

Starting with a DE length which is resonant, you shorten the DE
slightly. This changes the feedpoint impedance at your desired
resonant frequency from purely resistive (too low) to a slightly lower
resistance in series with a capacitance.

This R + jXc (series) is electrically equivalent to a parallel
combination of R' || jXc' where (in this case) R' is 50 ohms.

You then add a shunt inductance (the hairpin), which appears in
parallel to the capacitive reactance... when the magnitudes of the two
are equal, the combination appears as a very high impedance in
parallel with R' and all you're left with is R', or 50 ohms.

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.
Hence, the reduced return loss. I would not expect this sort of tweak
to affect the resonant frequency, since that's a function of the DE
length and thickness.

So, in your case, what you would want to do is either lengthen the DE
slightly, or use wider measuring tape. Either would bring the DE's
resonant frequency downwards a few MHz. You would then use the
hairpin match (as designed) to bring the feedpoint impedance to 50
ohms. You could try adding a "capacity hat" out at each end of the
DE, which would have a similar effect... but given how difficult
measuring tape is to solder onto, that might be more hassle than just
cutting a new DE and installing it and then just trimming back a bit
at a time.

If you're going to be cutting and trying... you can tune the hairpin
inductance somewhat, without having to actually cut it repeatedly, by
opening and closing the hairpin "loop".

In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ

--
David Ryeburn

To send e-mail, change "netz" to "net"

On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


No, David Platt is correct. You are assuming the effective capacitance
is in parallel. The antenna equivalent circuit is a *series* RC. Put in
too much *shunt* inductance and the combination looks inductive at the
feed point.

John KD5YI

On 9/25/2013 4:27 AM, John S wrote:
On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


No, David Platt is correct. You are assuming the effective capacitance
is in parallel. The antenna equivalent circuit is a *series* RC. Put in
too much *shunt* inductance and the combination looks inductive at the
feed point.

John KD5YI

I'm wrong. Please disregard.

Sorry.

John

On Tue, 24 Sep 2013 10:38:17 -0400, "Ralph Mowery"
wrote:

A few days ago I constructed a tape measure type antenna for 2 meters.
It is a 3 element beam and uses a hairpin or beta match.

Ran a return loss on it with the signal generator/spectrum analizer. The
best return loss was about 149 Mhz instead of 146. Thought it may be
because the tape AI used was 1/2 inch wide instead of 1 inch wide. I tried
lenghtning the hair pin form 5 inches to 6 inches. This did not change the
frequency of the return loss to any great ammount , but the RL went from
around 20 or more db to 10 db.

Did a little research on the hair pin match. From what I am getting out of
this, and it is very rough, is that the length of the driven element does
not make much differance in the actual resonate frequency of the antenna.
It is mostly where it is placed on the boom and the other elements that
determin the frequency of the antenna, but the length of the driven element
is determined by the impedance of the transmission line to be matched and
the impedance of the antenna/driven element.

Is that mostly correct or did I miss something in my reading ?



Look there before you waste too much time:
https://sites.google.com/site/tapemeasureantenna/


w.

On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


Actually, thinking more about this, I believe a hairpin is a shorted
transmission line. So, using a Smith Chart, I investigated a 25-j25 load
and played with the chart to see what happened.

It turns out that the feed impedance is indeed inductive if the stub is
too long. So, I have now reverted to agreeing with Mr. Platt.

I think that the stub (hairpin) will have no effect if its length is a
quarter wave. Shorter than that, it becomes inductive. Very short and it
is highly inductive. Longer is less inductive. See where this is going?

Cheers es 73,
John

On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive.
Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent function that increases from zero at zero length up to a maximum on the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance is proportional to the inductive reactance so - very short and it is slightly inductive (low reactance). Longer is more inductive (up to 1/4WL).

What you say above seems to be referring to capacitive reactance in open-circuit stubs where very short is highly reactive and longer is less reactive..

On 9/26/2013 7:11 AM, W5DXP wrote:
On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive. Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent
function that increases from zero at zero length up to a maximum on
the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL).
The equivalent inductance is proportional to the inductive reactance
so - very short and it is slightly inductive (low reactance). Longer
is more inductive (up to 1/4WL).

What you say above seems to be referring to capacitive reactance in
open-circuit stubs where very short is highly reactive and longer is
less reactive.


Yep, you're right. I was only noticing that the length of the line was
longer, but that is a shorter stub. My mistake again.

In article ,
W5DXP wrote:

On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive.
Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent function
that increases from zero at zero length up to a maximum on the Smith Chart as
the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance
is proportional to the inductive reactance so - very short and it is slightly
inductive (low reactance). Longer is more inductive (up to 1/4WL).

Agreed. So the inductive susceptance of a short shorted stub will be
high, while the inductive susceptance of a longer (but less than a
quarter wavelength) shorted stub will be low. Put that longer stub in
parallel with a dipole a little bit shorter than a half wave long (and
thus with a small amount of capacitive reactance and a large amount of
capacitive susceptance) and the net result comes out capacitive. This is
what I meant when I said that a shorted stub a bit too long to resonate
the (too short) dipole makes the whole thing will come out capacitive,
whereas if the stub is a bit too short for resonance, the whole thing
will come out inductive.

David, VE7EZM and AF7BZ

--
David Ryeburn

To send e-mail, change "netz" to "net"