Antenna Basics - Page 1
Forum section: electronics / projects

author: vxp036000
author_rank: Regular Contributor
timestamp: April 15, 2012, 06:34:08 pm
content: Gents,One topic I found sorely lacking in my education was the principles and analysis of antenna structures. Sure, I understood Maxwell's equations and could throw some numbers into an EM simulator to get some results, but I still hadn't the slightest clue of how some basic back of the hand type analysis could be done. After spending some time researching the topic, I compiled a few notes that you guys might find interesting (warning: a basic understanding of vector calculus is needed to follow most of this!).I started by introducing the basic procedure for calculating radiated fields and applied it to several common antenna structures. Even if the math is a little advanced, the closed form solutions provided should be useful for calculating radiation patterns, efficiency, etc.*** There is a glaring typo in the loop antenna analysis; can anyone find it?

author: vxp036000
author_rank: Regular Contributor
timestamp: April 23, 2012, 09:52:15 pm
content: To stir the pot some more, I thought I'd throw out a few counter-intuitive conclusions that may be drawn from this sort of analysis.'1. A dipole parallel to the y axis (vertically polarized, as commonly stated) has substantial components of the electric field in the x and z directions as well. Maybe not as vertically polarized as most folks would like to think. It would be interesting to see what these often neglected electric field components do to the radiation pattern (I'll probably have a follow up post on this).2. The radiated electric field decreases linearly with distance. The radiated magnetic field decreases with the square of the distance. This, of course, leads to the well known conclusion that power decreases with the cube of the distance.3. A folded dipole is more efficient than a full wavelength loop (because the radiation resitance is twice that of a full wave loop).

author: Unknown
author_rank: Unknown
timestamp: April 23, 2012, 10:49:40 pm
content: Quote from: vxp036000 on April 23, 2012, 09:52:15 pmSo, some percentage of the energy is radiated off axis and therefore not seen by the receiver (aka. lost) By decreasing off axis radiation, in other words by polarizing the signal more strongly a better efficiency could be obtained.Do I follow correctly?Polarizing the signal more strongly, but how?

author: ejeffrey
author_rank: Super Contributor
timestamp: April 24, 2012, 10:31:48 am
content: Quote from: vxp036000 on April 23, 2012, 09:52:15 pmWell, I am not sure how you are defining your axes -- I call z vertical, and prefer to talk about spherical or cylindrical coordinates. The polarization of a vertical dipole will be vertical in the horizontal plane. Off axis, there will be a radial component but still no azimuthal component, at least in free space. A ground plane breaks the vertical symmetry, so there can be a radial component even at in the plane of the antenna, but still no azimuthal component.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 24, 2012, 05:27:09 pm
content: Actually you've stumbled across a common misconception. Most literature computes the E and H fields from a current driving the dipole and immediately apply the far field approximation. But, what I was alluding to, is that the far field approximation is quite a stretch from reality in many situations. Before applying the far field approximation, you'll notice that the E field has x, y, and z components. Yes, a dipole parallel to the y axis has a dominant E field component in the y direction, which is why we say it is polarized in the y axis. However, the x and z components are not insignificant. Take a look at the notes I attached to the first post and see the expression you end up with for radiated E field before the far field approximation is applied.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 24, 2012, 10:35:50 pm
content: Even in the near field you are still wrong. Your notes are mostly impenetrable to me, and I don't really have any interest in spending time decoding them. Therefore I don't know if your equations are wrong or only your interpretation. Trivial symmetry arguments prove that the electric field cannot have an azimuthal component even in the near field nor can it have a radial component in the horizontal plane.The azimuthal component is easy. Consider the electric field at a point r. Mirror everything about a vertical plane containing the antenna and the point r. Since nothing is changed, the electric field must stay the same. However, the reflection takes the azimuthal field component to its negative value. Therefor the azimuthal component must be zero.To show that in the horizontal plane of the antenna the radial component is zero, consider driving the antenna with a voltage signal of opposite polarity. Because the electric field is linear, the field must also negate. Since you drive the two halves of the dipole with opposite polarity, you get the same effect by swapping the two halves of the dipole: i.e., mirroring across the horizontal plane. This takes points on the horizontal plane to themselves, and it negates the vertical component of the field but not the radial or azimuthal. Therefore, on the horizontal plane, only the vertical component of the field can be nonzero.

author: gregariz
author_rank: Frequent Contributor
timestamp: April 24, 2012, 10:56:01 pm
content: Quote from: vxp036000 on April 23, 2012, 09:52:15 pm2. is not correct. Both E and H fall off exponentially with distance. The higher order component makeup however is mixed in the near field and dependent on the antenna type. In the far field antenna type is not a factor.3. is not theoretically correct. Folding the dipole does allow for a higher radiation resistance however this is not an indicator of efficiency. Having more gain is not an indicator of efficiency. A matching problem does arrise for electrically small loops (much less than a full wave) where the resistive losses start to become large in relation to radiation resistance. In the real world people have overcome that by using superconducting elements, but this is usually not practical. Full wave loops however provide a higher radiation resistance that is more practically matchable. You'll see them used in Quad array antenna's and they are very good.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 25, 2012, 01:42:31 am
content: Instead of relying on my own derivations, I can also refer you to page 283 of Balanis "Advanced Engineering Electromagnetics". There is also considerable laboratory evidence that suggest otherwise. I stand by my statement that the E field has x, y, and z components. The flaw in your reasoning is that the analogy is not quite valid. I think the math speaks for itself, but conceptually it is quite simple.The radiated electric field is determined by the current distribution along the antenna element. The current distribution along a half wave dipole is very close to sinusoidal, peaking at the feedpoint in the center. Draw a normal line to the current distibution at every point along the dipole. Guess what, it points in every direction. In the far field, the E field component parallel to the antenna element dominates. Hence, we approximate the E field radiation as a plane wave. But in the near field (i.e., in the case of an MRI system), the wavefront is nowhere near planar (again, because the current distribution is not constant along the antenna length) and this means that the E field has x, y, and z components.Quote from: ejeffrey on April 24, 2012, 10:35:50 pm

author: vxp036000
author_rank: Regular Contributor
timestamp: April 25, 2012, 02:00:57 am
content: E and H do not decrease exponentially with distance. It is certainly true that the E and H fields have exponential terms, but notice that the exponential terms of the E field are multiplied by a term 1 / R and the H field exponential terms are multiplied by a 1 / R^2 term. This is the linear and inverse square decrease that I was referring to. The exponential terms are of the form e^(j R), which means the phase changes but the amplitude of the exponential term is constant. Therefore, the 1 / R from the E field and 1 / R^2 for the H field is what determines the decrease in amplitude with distance.Radiation resistance absolutely is a measure of efficiency. The radiation efficiency is given by Prad / Ptotal or Prad / (Presistive + Prad). Substituting, we have (I ^2 Rrad) / (I ^2 Rresistive + I ^2 Rrad), which reduces to Rrad / (Rresistive + Rrad ). So we see that the radiation resistance directly effects radiation efficiency. Higher Rrad increases radiation efficiency and lower Rrad decreases radiation efficiency. Note that radiation efficiency is unrelated to gain, I'm not sure where you were going with that. In fact, I can demonstrate that a high gain antenna array is considerably less efficient than a folded dipole (why?).Quote from: gregariz on April 24, 2012, 10:56:01 pm

author: gregariz
author_rank: Frequent Contributor
timestamp: April 25, 2012, 05:31:41 am
content: 1/r and 1/r2 are both exponential functions. 1/r is not a linear functionRadiation resistance can be used as part of the overall calculation for efficiency but it does not define it as you suggested. Resistive losses are necessary to a complete calculation. I can tell by looking at your math that you probably have a science background rather than an engineering. The physical realities of the problem are not captured in your math. For example you need to consider your power transfer into the antenna. The benefits of raising the radiation resistance of an antenna are usually offset by the losses you incur making the impedance transformation to the antenna and as I explained by going to a superconducting element those losses have been negated so raising the radiation resistance will not do you any good. In the case of MRI it gets even worse because your efficiency is dampened by your effective Q which is determined by your loading ( which by the way puts you in the far field pretty quickly)But I have no wish to further this discussion. I built all these antenna types when I was a kid. The Quad was the best performing HF antenna i ever built. If you take the time to build each of these antenna types you'll realise what practical limitations exist due to finite conductors, matching, losses and ground location and quality. You will not see significant differences in efficiency between any of fundamental antenna types using standard construction techniques.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 25, 2012, 07:45:50 am
content: Quote from: gregariz on April 25, 2012, 05:31:41 amExponential means e^(a*r) and nothing else. The technical term for r^a is a power law. It is common and colloquially understood to say that 1/r "falls off linearly with distance" and 1/r^2 "falls quadratically with distance". To be perfectly clear is is better to simply call them "1/r" and "inverse square law", but calling them exponential is both wrong and confusing.

author: vk6zgo
author_rank: Super Contributor
timestamp: April 25, 2012, 11:07:56 am
content: The Inverse Square Law originates in classic optics,& depends on the radiating source being small (in theory,a point source with zero dimensions.),compared to the distance from that source (in wavelengths), to the place at which the observation is being carried out.In the real world,with antennas of practical proportions,the near field is a mess,which I will leave to those of you with the mathematical ability to analyse.Theefficiencyof an antenna depends upon the ratio of R(rad) to R(loss).If R(loss) is small compared to R(rad),the efficiency will be higher,no matter what the absolute values of either.In practice real antennas are not in free space,& are affected by their distance from the Earth's surface,the constitution of that surface (sandy,muddy,salt water,etc) its smoothness ,& the effects of other natural or man made artifacts in the vicinity.

author: gregariz
author_rank: Frequent Contributor
timestamp: April 25, 2012, 08:03:51 pm
content: Quote from: ejeffrey on April 25, 2012, 07:45:50 amI've never heard anyone refer to it in that way when talking about E&H field components. Exponential does not mean euler base and simply refers to a function with an exponent not equal to one which would be a linear function.but just to make everyone happy simply convert between bases so that 1/r = exp^(-ln(r))Edit: BTW, I wasnt going to comment because I dont think its right to be here educating people but the OP is correct to some degree in point 1 but he is wrong on an account the the radiation integral is not consistant. There is a tangential component when a projection is made. Anyone who has done any physical optics will have run across it and it was documented a lifetime ago before the war. It is also the reason that there are some exceptions to reciprocity and something to be mindful of if you are radiating into a lossy/conductive media. In the comms sphere however with identical antenna's at large distances away there is no significant tangential component received at the other end, simply due to geometry so the polarization is more or less pure.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 25, 2012, 09:41:33 pm
content: Quote from: gregariz on April 25, 2012, 08:03:51 pmIt isn't the base that matters (you can fold that into the constant 'a'). What makes an exponential function exponential is that the variable is in the exponent. x^2 is a power law not an exponential.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 25, 2012, 10:40:48 pm
content: I think we are in violent agreement at least on the E field decreasing with 1 / R and the H field 1 / R^2. We also agree on resistive losses, you'll notice that I included them in my definition of efficiency, Rrad / (Rrad + Rresistive), where Rrad is radiation resistance and Rresistive is the resistance that dissipates power lost as heat. So I'm not sure why this expression would be incorrect.As far as transfering power to the antenna, it is certainly true that losses in an impedance matching network can more than make up for the increase in antenna efficiency for large values of radiation resistance. However, consider the case of a yagi antenna with a driven half wave dipole. The radiation efficiency of this antenna is somewhat lower than that of a single isolated dipole because of the low radiation resistance. So, we use a folded dipole for the driven element to increase the radiation resistance, improving efficiency and with a perhaps more significant benefit of making the antenna easier to match to a 50 ohm feed.The loading on the antenna does not change the distance at which near / far field components dominant. The wavelength is the sole factor in determining how far away the far field approximation becomes reasonably close to reality. Loading the antenna absolutely will, however, change the radiation pattern and radiation resistance.For most situations, the far field approximation that you're describing is perfectly valid. But there are some common applications in which the near field is dominant and the analogy falls apart. I think this is part of the reason why misconceptions seem so abundant in antenna theory; there are greatly simplified analogies that are valid most of the time, but people like to apply these analogies to situations in which the assumptions they are based upon are no longer valid. This is why I fall back to relying on Maxwell's equations and let the math lead to the most accurate model.Quote from: gregariz on April 25, 2012, 05:31:41 am

author: vxp036000
author_rank: Regular Contributor
timestamp: April 26, 2012, 01:04:05 am
content: I'm curious as to what you mean by the radiation integral isn't constant. I calculated the radiated magnetic vector potential by integrating the current density, using A = (mu / ( 4 pi ) ) integral ( J (x) (1 / R) e^( - j beta R) dx ), where J is the current density along the antenna, assuming the antenna is parallel to the x axis. From the magnetic vector potential, it is straight-forward to find the radiated magnetic field using H = (1 / mu) del X A. I then calculated the radiated E field from the H field. What part of this is invalid?Quote from: gregariz on April 25, 2012, 08:03:51 pm

author: vxp036000
author_rank: Regular Contributor
timestamp: April 28, 2012, 02:51:32 pm
content: So I was re-visited my near field calculations in preparation for writing some Matlab code to plot the radiation pattern of a dipole including the often neglected electric field components. Lo and behold, I found that I overlooked a term in differentiating the scalar potential. So, although none of my conclusions are altered by this slight error, here is a more accurate analysis.For those that remain convinced that a vertically polarized dipole has no horizontal electric field component, I encourage you to perform a simple experiment. Connect an RF transmitter to a horizontally polarized dipole antenna and place it several hundred feet away from a receiver with a vertically polarized dipole antenna. The receiver will detect the transmitter just fine, so clearly some of the radiation is vertically polarized. My model explains why this is so.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 28, 2012, 09:15:55 pm
content: If I plug x=0 into your equations, Ey and Ez appear to both vanish. In the plane of the antenna, the field is perfectly vertically polarized, exactly as I have been saying.If I set y=0, Ey vanishes, and if I substitute z=0, Ez vanishes. More generally, Ey is proportional to y and Ez is proportional to z, and in the same way. That is, the field in the azimuthal direction is zero. Exactly as I said.When your equations disagree with the symmetry of the problem, your equations are almost every time wrong. In this case, the dipole antenna is rotationally symmetric, and symmetric under reflection along with reversing the phase of the drive.The major caveat to this is that sometimes our notation doesn't manifest symmetries directly. One of the most notable examples is the magnetic field. Why don't the same symmetry arguments I used for the electric field apply to the magnetic field? They do, the problem is that our notation of the magnetic field requires us to break symmetry by picking a handedness. We use the right-hand convention, but we could equally well define magnetic fields to be left handed without changing any measurable quantities. That means that when considering mirror symmetries, we should switch to left handed fields. That is inconvenient, so instead we just remember to flip the magnetic field after a mirror symmetry.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 28, 2012, 09:44:09 pm
content: You bring up an interesting point. However, when I move away from x = 0, Ey and Ez are present. These are the non-zero field components I refer to. So if we are referring to the E field right at the antenna, you are correct, the Ey and Ez components are zero. But moving away from the antennathere are Ey and Ez components, which I interpret to mean that the field is no longer solely polarized in the Ex direction. Does this make sense or am I still misunderstanding you?Quote from: ejeffrey on April 28, 2012, 09:15:55 pm

author: Construct
author_rank: Contributor
timestamp: April 28, 2012, 09:46:02 pm
content: Quote from: vxp036000 on April 28, 2012, 02:51:32 pmThis is a very poor test. It won't really tell you anything about the polarization purity of an electrically thin dipole antenna. When operating near the earth and/or buildings, polarization tends to get twisted in surprising ways due to reflections, etc. Furthermore, with any non-ideal antenna you're going to have some radiation coming off of your cable and feed structure, which is going to be randomly polarized.For some perspective, at my previous employer we would attach 50-Ohm loads to our antenna ports for setting up indoor test links. The links still worked just fine within a reasonable distance for two reasons: 1) modern radios are extraordinarily sensitive and 2) RF energy is very good at leaking out all over the place.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 28, 2012, 10:17:38 pm
content: Quote from: vxp036000 on April 28, 2012, 09:44:09 pmI think you are still not understanding me. According to the notation of your notes, the dipole is oriented in the "x" direction. I am saying that in the entire y-z plane (all points where x=0), the electric field is purely in the 'x' direction. Furthermore, I am saying that everywhere in space there is no azimuthal component of the field -- that is, the only components are vertical, or pointing to/from the antenna, never side-to-side.So yes, the general form of the field, written in rectangular coordinates, has x,y,and z components. However, they behave in exactly the way everyone expects. At 'ground' level the field is perfectly vertical. At other elevations, you want to tip your receive antenna somewhat towards or away from the transmitter.Once again, I say that polar coordinates would be a better choice here. You would immediately see that Ephi=0, and the overall form of the equations would be simpler.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 28, 2012, 10:23:29 pm
content: I completely agree with this. It's possible that I was using the term polarization incorrectly. Yes, polar coordinates would probably make more sense, but I did it in rectangular form just to show that it could be done.So what you're saying is that if I have a transmitter and receiver with dipoles vertically oriented and offset by some amount vertically, then I would need to rotate the antennas off the vertical axes to maximize the received field. Correct?Quote from: ejeffrey on April 28, 2012, 10:17:38 pm

author: ejeffrey
author_rank: Super Contributor
timestamp: April 28, 2012, 11:01:07 pm
content: Quote from: vxp036000 on April 28, 2012, 10:23:29 pmYes, I am saying that you tilt your receiver towards the transmitter (or away) but not to the side. And yes, at least part of the problem is the polarization nomenclature. Horizontal and vertical polarization are both orthogonal to the direction of propagation (from the transmitter to you). A vertical dipole does not have any horizontal polarization anywhere. The third field component is the longitudinal polarization which is indeed usually neglected in the far field because it is quite small but non-zero, except on axis where it vanishes.One more thing: E and H both fall off as 1/r in the far field (there are faster decaying components, but those are the traveling wave components). Your equations look right, so I think you just misread them. In fact they have a constant ratio: E/B=c. This is necessary so that the power properly falls off as 1/r^2 which it must by conservation of energy.

author: vxp036000
author_rank: Regular Contributor
timestamp: April 29, 2012, 01:37:51 am
content: You're right, the H field decreases with 1 / R in the far field. Not sure how I missed that. Off to generate some Matlab plots including the faster decaying components...Quote from: ejeffrey on April 28, 2012, 11:01:07 pm