Help in a problem in Balanis Antenna theory.

I don't understand the solution of this question at all. The top of the scanned document is the question. Below is the scanned copy of the solution manual.


I don't even know where\[I_{in}=I_0\sin[k(\frac {l}{2}+|z|)]\] comes from.

Or does the question meant cutting the antenna to leave λ/8 from λ/4<z<λ/8 on the position side and λ/8 from -λ/4<z<-λ/8 on the negative side. Essentially making a λ/4 dipole by cutting the middle section out? If so, it should be \[I_{in}=I_0\sin[k(\frac {l}{2}-|z|)]\]

I don't even know where xxx comes from.

Click to expand...

It's just putting in the said geometry (length and feed point) into the general dipole current distribution expression (4-56), page 170.

FvM said:

It's just putting in the said geometry (length and feed point) into the general dipole current distribution expression (4-56), page 170.

Click to expand...

Thanks for the answer. can you explain what the question is asking? Or does the question meant cutting the antenna to leave λ/8 from λ/4<z<λ/8 on the positive side and λ/8 from -λ/4<z<-λ/8 on the negative side?

I understand that the question is asking about the (differential mode only) input impedance of an assymmetrical dipole.

The solution is apparently assuming that the sine shaped current distribution of the symmetrical dipole isn't changed by shifting the feed point. I must confess that I didn't try to verify this prerequisite.

FvM said:

I understand that the question is asking about the (differential mode only) input impedance of an assymmetrical dipole.

The solution is apparently assuming that the sine shaped current distribution of the symmetrical dipole isn't changed by shifting the feed point. I must confess that I didn't try to verify this prerequisite.

Click to expand...

Thanks, that is what I thought, moving the feed point to either z= +λ/8 or z=-λ/8. That is you get an asymmetrical dipole with one side being 3λ/8 and the other side is only λ/8 in length. But the solution is so far out that I really second guess my understanding of the question.

My original thinking was assuming current is the same at the feed point even though it is asymmetrical, just like the symmetrical dipole. The different is finding the vector magnetic potential A for both sides separately with

\[\vec {A}_1 = \frac{\mu}{4\pi}\int^{\frac {3\lambda}{8}}_0 \frac{I_0 \sin[k(\frac {3\lambda}{8}-z')]e^{-jkR}}{R}dz'\]
And
\[\vec {A}_2= \frac{\mu}{4\pi}\int^0_{-\frac {\lambda}{8}} \frac{I_0 \sin[k(\frac {\lambda}{8}+z')]e^{-jkR}}{R}dz'\]

Using these two to get E and H etc. Then add the two to get the complete solution including input impedance. But looking at the solution, it is so far off!!!

When I read the text it is just shifting a fully floating feed point. So one side will be lambda/8, other side will be lambda*3/8. The feed is no long balanced, but it is fully floating, so there is no common mode current that gets into the feed point.

Very simple explanation.
Assuming a very thin dipole, the current is fully dictated by a standing wave pattern (therefore the amplitude versus distance along the dipole has a sinusoidal shape). See it as half wave long resonator that is not operating at its resonant frequency (as there is still some inductive component present).

So if the original feed current was 1A, there will be 1*0.707 = 0.707A at lambda/8 distance from one end. The center feed current does radiate 73 W, therefore Re(Zin) = 73 Ohms.

In the new situation the feed current is 0.707 A, but the distribution remains sinusoidal, so the center current remains 1A, hence the radiated power remains 73W.

Based on P = I^2*R, now Re(Zin) should be 146 Ohms.

You can't use this simple approach for feed points near the ends If you may not assume a sinusoidal current distribution beforehand, it will be very complex to solve.

WimRFP said:

When I read the text it is just shifting a fully floating feed point. So one side will be lambda/8, other side will be lambda*3/8. The feed is no long balanced, but it is fully floating, so there is no common mode current that gets into the feed point.

Very simple explanation.
Assuming a very thin dipole, the current is fully dictated by a standing wave pattern (therefore the amplitude versus distance along the dipole has a sinusoidal shape). See it as half wave long resonator that is not operating at its resonant frequency (as there is still some inductive component present).

So if the original feed current was 1A, there will be 1*0.707 = 0.707A at lambda/8 distance from one end. The center feed current does radiate 73 W, therefore Re(Zin) = 73 Ohms.

In the new situation the feed current is 0.707 A, but the distribution remains sinusoidal, so the center current remains 1A, hence the radiated power remains 73W.

Based on P = I^2*R, now Re(Zin) should be 146 Ohms.

You can't use this simple approach for feed points near the ends If you may not assume a sinusoidal current distribution beforehand, it will be very complex to solve.

Click to expand...

I still don't understand how can you establish a continuous standing wave along the uneven dipole. The two sides are not connected together and reflection doesn't travel through. I can see each side establish it's own current signal as I posted in the last post.

EDIT: I went back and look at the equations for various length. The current signal makes sense on the long side as input current is 0.707 of max current and the max is still at the original mid point. But I still don't get the short side is 0.707 of the max current.

Thanks

In an exact analysis, the sinoidal current distribution reveals as a simplification, also for the center fed dipole, because it ignores the electrical field of the feeding voltage. But the involved error is apparently small for thin wire antennas with a small feed gap.

I still don't understand how can you establish a continuous standing wave along the uneven dipole. The two sides are not connected together and reflection doesn't travel through. I can see each side establish it's own current signal as I posted in the last post.

Click to expand...

I don't see how the assymmetrical case would fundamentally differ from the symmetrical one. The halves are "unconnected" in both cases.

FvM said:

In an exact analysis, the sinoidal current distribution reveals as a simplification, also for the center fed dipole, because it ignores the electrical field of the feeding voltage. But the involved error is apparently small for thin wire antennas with a small feed gap.

Click to expand...

Sorry, I don't follow this part. Please explain. Thanks

You can see the two conductors as a transmission line (therefore a standing wave pattern can exist). See the center fed dipole as two somewhat longer then quarter wave transmission lines (open terminated) with the source between the two transmission lines. The loss due to radiation you can see as two resistors left and right of the source (3*36.5 Ohms). You can shift the source as in the question. Leave the loss in place (73 Ohms resistor in the middle, don't move them together with the source).

Now you can use a Smith chart or other means to determine the reflection coefficient when moving the source away from the center. From the reflection coefficient you can determine the current. Based on the impedance versus frequency curve for the center fed dipole, you can deteremine the appearent characteristic impedance of the antenna conductors. For a very thin dipole, think of Zc = 400 Ohms. Zc will be the reference impedance (so 73 Ohms will be 0.18 Ohms on a normalized chart).

It is because of that the reflection coefficient belonging to 73 Ohms is way off the center of the chart, you get a strong standing wave pattern. From this you can also see that there is still a traveling wave component as the reflection coefficient isn't 1, so the sinusoidal current distribution is an approximation.

If you would do this for a very thick dipole (Zc drops), you will see that the sinusoidal current distribution no longer holds as the standing wave component (that causes the sinusoidal distribution) reduces.

When you move further from the center, the feed point impedance increases, hence the voltage across the gap. This gives distortion of the E-field becuase there is a capacitive path across the source. More voltage across this capacitance gives more displacement current. But I think this is way beyond the question.

You can imagine the standing wave pattern also via a mechanical experiment. If you tension a wire between two fixed rigid points you can get it into resonance. A small movement near one end results in large movement in the middle. the same happens in the half wave dipole antenna.

If you would use a very light but thick rope (wool?), you will notice that you can't get a nice sinusoidal resonance due to insufficient weight/m and air friction.

A standing wave pattern can also be based on current. Your source provides the curent, so current can pass from the short conductor to the long one (and vice versa).

The voltage standing wave pattern will show a jump (disconinuity) at the feedpoint. When the dipole is very thin, the voltage allong the antenna conductors is high (especially at the extremeties of the conductors) and the jump is relatively small compared to the voltage at the ends.

I know the voltage concept is problematic as it depends on the path you take, but see it as if the antenna is above a ground plane (antenna parallel to ground plane), then you can define the voltage between the antenna and the ground plane. Note that the ground plane is only there for the voltage concept, not for calculating radiation pattern and input impedance. So see it as a transparant groundplane for radiation.

WimRFP said:

You can see the two conductors as a transmission line (therefore a standing wave pattern can exist). See the center fed dipole as two somewhat longer then quarter wave transmission lines (open terminated) with the source between the two transmission lines. The loss due to radiation you can see as two resistors left and right of the source (3*36.5 Ohms). You can shift the source as in the question. Leave the loss in place (73 Ohms resistor in the middle, don't move them together with the source).
Where is the 3*36.5Ω come from? I thought the Radiation Resistance change with the length ( in λ) of the wire. So as you move away from the center, the Radiation Resistance increase on the longer side and decrease on the shorter side. I do see the total Radiation Resistance remain approx 73Ω

Now you can use a Smith chart or other means to determine the reflection coefficient when moving the source away from the center. From the reflection coefficient you can determine the current. Based on the impedance versus frequency curve for the center fed dipole, you can deteremine the appearent characteristic impedance of the antenna conductors. For a very thin dipole, think of Zc = 400 Ohms. Zc will be the reference impedance (so 73 Ohms will be 0.18 Ohms on a normalized chart).
I never quite get how to calculate the reactance( the j part) of the input impedance. I know in open end tx line, \[Z_{in}=-jZ_{line} \cot (kl)\]. But I still don't know how to calculate \[Z_{line}\] which is the characteristic impedance of the dipole wire. I know this is approximation as there is the thickness of the wire and the feed gap to consider.

It is because of that the reflection coefficient belonging to 73 Ohms is way off the center of the chart, you get a strong standing wave pattern. From this you can also see that there is still a traveling wave component as the reflection coefficient isn't 1, so the sinusoidal current distribution is an approximation.

If you would do this for a very thick dipole (Zc drops), you will see that the sinusoidal current distribution no longer holds as the standing wave component (that causes the sinusoidal distribution) reduces.

When you move further from the center, the feed point impedance increases, hence the voltage across the gap. This gives distortion of the E-field becuase there is a capacitive path across the source. More voltage across this capacitance gives more displacement current. But I think this is way beyond the question.

You can imagine the standing wave pattern also via a mechanical experiment. If you tension a wire between two fixed rigid points you can get it into resonance. A small movement near one end results in large movement in the middle. the same happens in the half wave dipole antenna.

If you would use a very light but thick rope (wool?), you will notice that you can't get a nice sinusoidal resonance due to insufficient weight/m and air friction.

Click to expand...

I understand the peak current is higher than the input current on the longer side. I just don't quite follow why the the input current of the shorter side is the same. I want to see whether my reasoning is correct of not. As the image below, both sides of the dipole is driven by the same Vin source which is like the two half of the dipole are in series. So the current source to one side is EXACTLY the same as the current sink from the other side as shown in the diagram.


I don't think the answer in the solution manual is that accurate. The reason is the calculation of the E given in equation (4.60) in page 172 is based on symmetric dipole. If you work out the steps, there is a major simplification using the fact of integration of an odd function with symmetrical limit is ZERO and half of the long equation is eliminated. This will not work if the limit is not symmetrical as in this case. So you cannot use the solutions that is derived from (4.60).....Which the solution comes from. I can scan my notes on the exact step that I worked out as it's too long to type it out in Latex.

But if everything is only an approximation, I guess using the exact same current as a symmetric dipole is just as good a guess as anything. Is this the case?

Thanks for the explanation.

Hello Alan, I made error, it should be 2*36.5 instead of 3*36.5, sorry for that. I hope it didn't confuse you much. See it as all loss due to radiation is concentrated in a single resistor (or two resistors in case of center fed)

If you reduce the total length (both sides), the Re(Zin) reduces as a given center feed current produces less far field as you have less total moving charge (or less current*length product).

However here, you only change the feed point position and given a very thin half wave antenna, the current distribution remains more or less the same.

your drawing is correct, it is what I had in mind.
The source is assumed to have small size (that means << 0.25lambda) and free from everything (so floating). What goes into the source, must go out.

Your comment on accuracy is right, it is an approximation as exact solution is difficult I think. When (wire thickness)/lambda is very small (say: 0.001), Zc of the wire is large. As long as Zc >> 73 Ohms, standing wave behavior dictates the current distribution. If you have access to a simulator, just give it a try and look to the current density. In case of thick dipoles, there is large deviation.

The assumption becomes better when there is ground plane close by (say at < 0.1lambda). The ground plane reduces the total radiated power given a certain feed current (due to the negative image). This results in a strong reduction of radiation resistance for the center fed half wave dipole. I used off-center feed in a directional UHF RFID label design (dipole over ground plane) to raise the feed point impedance. Off-center fed antennas are frequently used in the amateur world as they show reasonable impedance at harmonics of the fundamental frequency.

half-wave resonating patch antennas have also a reasonable sinusoidal current distribution and offset feed is frequently used to find a 50 Ohms feed point. Note that in a patch antenna the feed goes via a "probe feed" that only connects with the patch, so it is a voltage feed.

The Im(Zin) component.
I think it would scale with same factor, but that is just a guess. From a practical standpoint I know the it does not scale, but this may be due to capacitance across the feedpoint.

If I really need to know it, I would calculate it based on a transmission line approach. In a practical case you want Im(Zin)=0, and that is done by changing the antenna length a bit (in your case, reduce the length to make it zero).

One disadvantage of off-center fed antennas is their significant common mode voltage at the feedpoint (a floating source does not exist in many cases). Reason is that you are no longer in a voltage minimum. This common mode voltage will cause a common mode current in the feed cable, unless you use a very good common mode suppression.

I'm also interested to know the deviation from ideal current distribution for the offset feed case.

"everything is only an approximation" has been my basic conclusion in post #8 and applies both to symmetrical and asymmetrical feed. The question is how much stronger the deviation from ideal model is in the offset feed case.

The other question is if your alternative approach presented in post #5 to calculate both antenna halves separately gives a basically correct solution or introduces other, possibly more severe errors. Seriously speaking, I don't know.

If you want to know how it turns out in practice, you need to take the thickness/lambda ratio into account, as this really matters. So if you follow an approach that doesn't do this, it has no use to continue with that approach.

To be honest I don't have the mathematical skills to find an exact solution for a practical situation.

If I can I will avoid an off-center series feed (is Alan's case) as common mode issues can be really problematic. I do use off-center voltage feed frequenty where the source is connected directly to the antenna conductor without cutting the conductor. An example is the PIFA (quarter wave radiator), probe fed patch antenna, or an off-center fed half wave slot antenna. In cases where the ground isn't close to the antenna, the deviation is significant. Even with my simple transmission line approach there is deviation, especially when the voltage feed is near one edge, or near the middle. It only gives me a first shot in the good direction. Simulation and/or measurement does the fine tuning.

It is not just the feed point position, but also the length needs to change to get the antenna into resonance (assuming that you start from a center fed resonant antenna).

In case of wide band dipoles (thick radiators), I don't even make a calculation on the off-center feedpoint, simulation with educated guess gives the fastest result.

Knowing the theoretical case (sinusoidal distribution) is good as this drastically reduces the number of shots to hit the target.

WimRFP said:

The assumption becomes better when there is ground plane close by (say at < 0.1lambda). The ground plane reduces the total radiated power given a certain feed current (due to the negative image). This results in a strong reduction of radiation resistance for the center fed half wave dipole. I used off-center feed in a directional UHF RFID label design (dipole over ground plane) to raise the feed point impedance. Off-center fed antennas are frequently used in the amateur world as they show reasonable impedance at harmonics of the fundamental frequency.

The Im(Zin) component.
I think it would scale with same factor, but that is just a guess. From a practical standpoint I know the it does not scale, but this may be due to capacitance across the feedpoint.

One disadvantage of off-center fed antennas is their significant common mode voltage at the feedpoint (a floating source does not exist in many cases). Reason is that you are no longer in a voltage minimum. This common mode voltage will cause a common mode current in the feed cable, unless you use a very good common mode suppression.

Click to expand...

Thanks for the reply, that really helps. The important thing I got out is it's an approximation as there are other factors like feed gap and thickness of wire all come into play. The fact the the input current on both sides ( and the feed point) are forced to be the same force the current into almost a continuous standing wave. This make the current assumption for symmetrical feed a good approximation in this case.

One non related question in your comment about dipole close to a ground plane. I thought the effect is opposite as the Radiation Resistance increases when close to the ground plane. This this discussed in P187 to p192 with Fig. 4-18 showing the Rr vs h from ground. Also Prad double from 0<θ<90 deg when the dipole is very close to the ground.this is shown by (4.102) in p 190 if substitute kh->0 and use numerical expansion of sine and cosine. But this is not relevant in this subject, just a question.

Again, thanks for the detail explanation.

Alan

- - - Updated - - -

FvM said:

I'm also interested to know the deviation from ideal current distribution for the offset feed case.

"everything is only an approximation" has been my basic conclusion in post #8 and applies both to symmetrical and asymmetrical feed. The question is how much stronger the deviation from ideal model is in the offset feed case.

The other question is if your alternative approach presented in post #5 to calculate both antenna halves separately gives a basically correct solution or introduces other, possibly more severe errors. Seriously speaking, I don't know.

Click to expand...

I kept forgetting that the current is always an approximation from the beginning. I just kept taking it literally. Thanks for your time in explaining all these. I am going to work through this problem and see whether I have any more question.

Thanks.

Alan

- - - Updated - - -

I am looking at the solution as shown in post #1 above. I don't think the equation of Iin is correct. It should be \[I_{in}=I_0 \sin[k(\frac {l}{2}-|z|)]\] derived from (4.60) in page 172. Also length should be \[l=\frac {\lambda}{2}\] instead of \[l=\frac {\lambda}{4}\]. The third line should be \[I_{in}=I_0 \sin[k(\frac {\lambda}{4}-\frac {\lambda}{8})]\]. That will match up to the rest of the solution. Am I correct?

Hello Alan,

I had in my mind a dipole parallel to a ground plane. Your figure is for a dipole perpendicular to the infinite ground plane. Sorry for causing confusion (I should improve my English).

The parallel to ground plane situation is shown in figure. 4.29. Such situations can be treated with an image dipole with out-of-phase (180 degrees) feed (that you get for free). The center-fed radiation resistance drops significantly, but the characteristic impedance Zc of the conductors reduces only slowly, so Rrad(center fed)/Zc reduces. This results in lots of standing waves, hence lots of stored (oscillating) energy and high resonator Q factor (so small usefull bandwidth), and good sinusiodal current distribution.


I hope you have access to a simulator to play with some situations. As you are working on dipoles, you can use a "demo/student" version. It would be nice to have somebody around you that knows some software as this reduces the learning curve.

Simulation is not a substitute for knowledge, but animated current views and static current distribution views will enable you to gain knowledge faster. You can also measure current and voltage distributions, but this is not easy. Errors are easily made.