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antenna noise bridge
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by KC8HZM on February 3, 2003
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Hello all,
I am working on an antenna project and want to better understand the concepts behind a noise bridge and how to best use one. I would like to hear of others experiences of using a noise bridge or if anyone could point towards antenna tuning resources, I'd appreciate it.
Thanks for all of your help.
Marten
kc8hzm
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RE: antenna noise bridge
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by WB6BYU on February 3, 2003
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Marten -
First, imagine a "bridge" circuit, such as a common
Wheatstone bridge using resistors. A voltage is
applied across two terminals, and, when the two arms
of the bridge are "balanced" the voltage across two
other terminals of the network is zero.
This is basically the same circuit used in resitive
SWR bridges: the antenna is connected in one arm of
the bridge and transmitter power is applied to the
circuit. A meter reads the resulting balanced (or lack
thereof). If the meter reads zero, the antenna load
matches the calibrated value, typically 50 or 75 ohms.
This bridge makes the measurement at the frequency
of the transmitter - the detector is not frequency
specific.
If one or more parts of the bridge are made variable,
this will vary the load at which the output is a null.
By calibrating these variable elements and adjusting
them for a null, the approximate impedance of the load
can be determined.
A noise generator works on the same approach, but uses
a wide-band RF noise generator for the signal source
(in place of the transmitter). The station receiver
is used for the detector in place of the meter, to
provide frequency selectivity. Thus the measurement
is taken at whatever frequency the receiver is tuned
to. Noise bridges are often made with variable
elements for measuring impedance, but for tuning an
antenna they may be fixed for a 50 ohm load.
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RE: antenna noise bridge
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by KC8HZM on February 3, 2003
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Wow, thanks for that very informative post. I realize that this is some basic electronics here, but how is the total antenna impedance related to the inductive and capacitive reactance of the antenna? Do you simply add the three values, DC resistance, inductive reactance, and capacitive reactance?
I suppose that I need to get my self a good book on RF electronics.
Thanks again,
Marten
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RE: antenna noise bridge
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by WB6BYU on February 4, 2003
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Marten -
A good book will certainly help. The basics should be
covered in the ARRL Handbook and their Antenna Book.
For a bit more detail, they have a book on Antenna
Impedance Matching that covers the Smith Chart and
other useful topics. It gets into a bit of math using
complex numbers: not really a difficult subject, but
can be confusing if they are thrown at you without
warning!
Here is a simple explaination.
First, "DC Resistance" is not the correct term. We
are dealing with "resistance", which means that part
of the impedance that dissipates power (hopefully by
radiation if it is an antenna.) Even without any
reactance, the values at DC and at RF can be quite
different. For example, a common half-wave dipole
looks like an open circuit at DC, but 72 ohms or so
at the resonant frequency. So let's just use the
general term "resistance", knowing that it may be
frequency dependent.
Take a piece of graph paper and put it on the table in
front of you in landscape format. Draw a vertical
axis in the center of the paper and mark it "resistance".
Draw a horizontal axis along the bottom of the paper
and mark it "reactance".
Number both axes starting with zero at the point where
they intersect. Pick a scale factor that puts you
somewhere around 200 ohms at the edge of the paper.
Of course, the bottom axis will have + and - numbers.
Mark the negative side "capacitive" and the positive
side "inductive".
Make a mark at 50 on the resistance axis. This is the
nominal impedance of common coax cables.
Now you can get an idea of the relationship between
resistance and reactance: they are separate axes, and
you can change one without changing the other. The
capacitive and inductive reactances have opposite signs
and cancel each other out in a circuit. (That is, you
just add them together if you have assigned the right
signs to each.)
The term "impedance" refers to the coordinates of a
spot on the graph. Make a mark at resistance = 30 and
reactance = +40. The impedance of this point would be
written "30 + 40j". (The letter "j" is used in
electrical work to represent the square root of -1
when dealing with complex numbers: normal mathematics
uses "i", but that can get confused with current.)
The magnitude of the impedance is the distance from
a point to the origin. Both points on the graph have
the same magnitude, but very different impedances!
We can extend the same concept to SWR: it is a function
of the distance from the actual impedance to that of
the feedline. In this case, both getting the reactance
closer to zero, and the resistance closer to the nominal
value, will decrease the SWR. If you plotted the points
with an SWR of 2 : 1 on your chart, they would include
100 + 0j and 25 + 0j, plus intermediate resistance
values with both inductive and capacitive reactances.
Here is where the Smith Chart comes in handy: the grid
is distorted so that all points with the same SWR fall
on a circle around the origin. And it makes other
calculations much simpler to solve graphically also.
For example, to calculate the total resistance of two
resistors in series you simply add the values, but for
two in parallel you have to take the reciprical of the
sum of the recipricals. This same rule applies to
any sort of impedances, and calculations can get rather
tiring if you have to take a lot of recipricals of
complex numbers. So when there are elements in parallel
we do the math in reciprical units (admittence,
conductance, and susceptance.) The Smith Chart allows
us to convert back and forth quite easily between the
series and parallel forms.
Well, that has gotten us far beyond "simple", but it
should give you an idea of what you are getting
yourself into. I encourage you to explore it further,
taking it in small enough steps that you can understand
one before taking the next one.
Good luck! - Dale WB6BYU
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RE: antenna noise bridge
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by KC8HZM on February 4, 2003
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Wow, thank you for that very informative post. That is exactly what I was looking for. It's a big help. I have the 1979 Amateur Handbook that I'm poring over. I also am getting the ARRL Antenna book through inter-library loan.
Antennas are quite remarkable things. It's hard to find good literature that isn't either far too oversimplified or too advanced.
I'm familiar with imaginary numbers, but too see them applied is something new to me.
Also, admittence and susceptance are new terms to me. I'd heard of the 'mho' unit of conductance, but not the other two.
So am I correct in saying that an arc plotted on this graph centered at the origin and going through 50 ohms from purely inductive reactance (0-50j) to purely capacitive reactance (0+50j) would all match 1:1 with a transmitter designed for a 50 ohm load?
Again, thanks for your help in clearing this up,
Marten
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RE: antenna noise bridge
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by WB6BYU on February 5, 2003
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"So am I correct in saying that an arc plotted on this
graph centered at the origin and going through 50 ohms
from purely inductive reactance (0-50j) to purely
capacitive reactance (0+50j) would all match 1:1 with a
transmitter designed for a 50 ohm load?
"
Absolutely NOT!
All these points would have the same magnitude of
impedance, but that is not the same as saying they are
the same impedance.
The transmitter (and/or the coax) is designed for
(50 + 0j), and won't deliver any power to (0 +/- 50j).
Note that power can only be disipated in a resistance,
not in reactance, so a pure reactance can not absorb
power from the transmitter. At the extremes, the SWR
on a 50 ohm coax cable will be infinite, even though the magnitude of the impedance is 50 ohms. This is why
you have to visualize the impedance world in two
dimensions, because any one-dimensional value doesn't
tell you the whole story.
The closer to (50 + 0j) the load is, the better the SWR. Well, this is a bit of a generalization on
rectangular graph paper, but on a Smith Chart the
distance from the (50 + 0j) at the center of the chart
to the point representing the actual load impedance
directly scales to the SWR.
Oh, and the units for conductance have been changed
from the old "mho" to Siemens, abbreviated "S".
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RE: antenna noise bridge
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by KC8HZM on February 6, 2003
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Thanks for clearing that up, getting some of the basics here sorted out. Impedance is seems to be a fairly complex issue, but I'm getting a handle on it.
This makes sense, that is how a gamma T match works then, by adding capacitance to "cancel" out the inductance of a dipole. (?)
Furthermore, that could explain the two readouts on a noise bridge. So while a noise bridge won't tell you exactly what the antenna's resonant frequency is, but rather the complex value of the impedance. I'm I getting this more or less correct?
Again, thanks for taking the time to explain this so well,
Marten
kc8hzm
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RE: antenna noise bridge
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by WB6BYU on February 7, 2003
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Yes, it really is a "complex" issue, since looking at
it from the perspective of complex numbers actually
makes it much simpler...
That is indeed what the two scales on some noise
bridges tell you - the R and X values of the complex
impedance at the frequency to which the receiver is
tuned. (Some of the older noise bridges only had one
knob for "resistance".)
The gamma and Tee matches are a bit different: at some
point along an antenna element there will be an
impedance that matches the transmission line, and the
impedance should be perfectly resistive at that point.
However, this may be a significant portion of a wave-
length from the ground connection (to the middle of the
element). The gamma rod connects taps the feedline
on to the element at the desired spot, but the rod
and the element combine to form a transmission line
shorted at the end where the rod connects to the
element. A shorted short length of transmission line
has inductance: the capacitor cancels this inductance.
(Or you could shorten the element to introduce a
capacitive reactance at the tap point, but a variable
capacitor is usually easier to adjust.)
However, viewed with complex numbers and converting
between the series and parallel forms, the standard
matching networks suddenly make much more sense as to
why they can transform one impedance to another.
In fact, the two-color Smith Chart makes calculation of
"L" matching networks trivial.
You seem well on your way to understanding impedance,
so the next step is to try to apply it! Calculate some
impedance transformations. For example, if you measure
an impedance with your noise bridge, you should be able
to calculate a network to match it to 50 ohms. Or
you could set your tuner to match it, measure or
estimate the capacitance/inductance values the tuner
uses to do so, and calculate the resulting impedance
(which should be close to 50 ohms.)
Good luck!
- Dale wb6byu
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RE: antenna noise bridge
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by KC8HZM on February 9, 2003
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I went out and bought a study guide for the extra class exam. It gets into impedance and also has some Smith charts which make more sense now that I can see what they look like.
OK, to take an example. Say I measure my antenna impedance to be 34 + j20 ohms at 7 Mhz (I just made that up). That means that the antenna has an inductive reactance component to its impedance. I need to cancel that out.
Xc = 1/(2*pi*f*C)
20 = 1/(2*pi*7*10^6*C)
C = 1.14 * 10^-9
Looking at most of the matching networks, the capacitor is placed in parallel and the inductor in series. So could a 1 nano-farad capacitor be placed in parallel with the antenna to cancel out the inductive reactance?
Would a resistor be used to bring the total impedance up to 50 -j0 ohms?
Guess I'm full of questions, thanks again!
Marten
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RE: antenna noise bridge
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by WB6BYU on February 10, 2003
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"Standard" notation is the series form: if you put a
1.1nF capacitor in series with your load, it would
cancel the reactance and you would have (34 + 0j) for
the impedance. The SWR would be just under 1.5 : 1.
To see what happens when you put a capacitor in
parallel with the load, convert the impedance to the
parallel (admittence) form. The sample Smith chart
I found is pretty small, but it looks like this would
be (0.02 - 0.03j)S. What a stroke of luck! (You did
a good job of choosing an example!) All we have to do
is to add a capacitor with a susceptance of 0.03S in
parallel with the original load, and it should give a
good match. This would be about 1500nf if I did the
math right in my head.
The "two color" Smith chart makes this quite simple.
It has the impedance coordinates in black and the
admittence coordinates superimposed in red. (You can
do the same thing by copying a Smith chart onto an
overhead transparancy and overlaying it on the
original with the axis reversed.) Then the critical
curves are those for R = 50 ohms and 0.02 S. From
any point on the chart you should be able to follow a
curve to one of these axes, then follow that curve to
the origin. (Which curve you use will depend on the
location of the impedance on the chart, and will
determine whether you have the shunt element or the
series element on the load end.) In both cases, the
distance along the curve corresponds to adding a
reactance in series or parallel.
In general, high impedance loads are matched with the
shunt element across the load, and low impeance loads
with the shunt element across the generator. The
series and shunt elements can each be of either sign
(capacitor or inductor). Usually you will find that
the Smith chart will lead to one of the configurations
as being the most likely for a particular load.
Good luck!
- Dale WB6BYU
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