Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

1/4 wavelength antennas

  1. Oct 19, 2013 #1
    Why does a 1/4 wavelength antenna have to be 1/4 wavelength? What property of the wave is responsible for this restriction?
    I've been trying to understand how a 1/4 wavelength antenna works and the best explanation, relatively speaking, was on the following website, http://www.sm0vpo.com/antennas/anten.htm [Broken].

    It still doesn't make any sense. Maybe if I could understand how a 1/4 wavelength antenna works, I could figure out why there is a restriction on the length.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 19, 2013 #2

    meBigGuy

    User Avatar
    Gold Member

    Do you understand how a dipole works? Roughly by making an antenna a half wavelength you minimize reflections and reactive components such that maximum power transfer occurs when it is matched properly. (an oversimplification)

    A quarter wave kind of uses ground (well, a reflection) for the other half of the dipole.

    The wikipedia dipole article describes it pretty well.

    Actually, the article you pointed to describes it well too. If the length is wrong the reflections are not additive. You can go to longer multiples if you want.
     
  4. Oct 19, 2013 #3
    Yes, I presume I understand since all of the videos I watched and articles I read make sense.

    I don't understand how the reflected portion of the signal constitutes the other 1/4 length of a dipole.
     
  5. Oct 19, 2013 #4

    meBigGuy

    User Avatar
    Gold Member

    Realize that only the 1/4 wave section actually radiates. The reflection only serves to make the 1/4 wave antenna non-reactive (not capacitive or inductive) and an efficient radiator.

    The 1/4 wave antenna depends on the ground plane being large enough to actually cause physical reflections of the 1/4 wave conductor. If there is no ground plane it does not work. The article you referenced explains it pretty well. ("The diameter of the tin-plate is assumed to be more than 1/2-wave"). In the wikipedia dipole antenna article they are a little more technical about the quarter wave monopole. This link addresses it also with a bit about the effect of a small ground plane http://www.antenna-theory.com/antennas/monopole.php

    But none of this is a near-field analysis of how the reflections work.

    If you really want to understand you need to work up to this I suppose (I'm not there) http://www.ece.rutgers.edu/~orfanidi/ewa/ch14.pdf. What you ask may well be covered in that book somewhere. http://www.ece.rutgers.edu/~orfanidi/ewa/

    A google search for "near field antenna analysis monopole" yielded a lot of stuff.

    http://ece.wpi.edu/mom/chapter4.pdf seems to get pretty deep.
     
  6. Oct 19, 2013 #5
    I'll be reading those. I wasn't sure if the article for which I provided a link was talking about a physical reflection, like a signal bouncing off of a wall or an electrical signal in a conductive medium at a discontinuity (e.g. what happens when you have impedance mismatch). I didn't know what terms to search for. Thank you.
     
  7. Oct 20, 2013 #6
    I've read through those but still don't intuitively understand what's happening in a monopole antenna.
     
  8. Oct 20, 2013 #7

    meBigGuy

    User Avatar
    Gold Member

    All I can say is that the reflection causes the monopole section to act like there is another monopole section which it interacts with in much the same ways as it would it were a dipole rather than a relection.

    If you think of the combined signal and reflection at an arbitrary single point on a monopole, it is the same as the signal and its symmetrical point on a dipole. The distance from any point on the monopole to the reflector and back is the same as the distance to the symmetrical point on the dipole, so the phases are the same in terms of how the signals interact. That causes it to act the same as a dipole. But you really have to fully understand the dipole to fully understand that (remeber the dipole halves are fed out of phase).
     
  9. Oct 20, 2013 #8

    Baluncore

    User Avatar
    Science Advisor

    A wave travelling along a half wavelength and reflecting from the cut end will get back in time with the next cycle. That is the resonant half dipole situation.

    Now get a drinking straw, assume it is half a wavelength long, cut it in half and place one end against a mirror. It will now look the same length as it did before you cut it. That is why a quarter wave antenna over a ground plane is resonant.

    The antenna length can be different to a half or a quarter wavelength but there will have to be some phase matching network to bring it to resonance.
     
  10. Oct 20, 2013 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    One way to look at is is to consider the currents flowing in the ground due to the fields from the vertical radiator. An antenna mounted at a height above a ground plane will produce currents in the plane which will produce a reflection of the antenna (exactly the same as the currents in a shiny metal reflector will cause a reflection and an optical image which appears to be the same distance 'behind' the reflector as the object. (Mounting an antenna at great height will give you another, virtual, antenna below ground and increase the gain of the antenna in the forward direction: so called Height Gain) The details of the current distribution over the plane are complicated and correspond to the same Maths that are used to describe diffraction - which is another way to 'construct' the visible image.
    For a quarter wave dipole, there are earth currents flowing, induced by the vertical bit and these generate a wave which is the same as would exist if the earth plane were replaced by the other half of the dipole.
    Don't lose sleep if you cannot do the sums for this - the essential thing is to appreciate that reflections in conductors are due to (or can be explained in terms of) induced currents.
     
  11. Oct 21, 2013 #10
    It's very hard to picture that considering the 1/4 wavelength antenna has currents running vertically whereas the ground plane would have ground currents running horizontally due to the orthogonal relationship between the two. I have read about the analogy of looking at "half" of an object in the mirror and it would resemble a whole, however I'm drawing a blank at the science behind it. I don't see how the ground plane mimics the behavior of the other 1/4 wavelength antenna.
     
  12. Oct 21, 2013 #11

    meBigGuy

    User Avatar
    Gold Member

    The reflected signal from the ground plane is exactly the same as what a radiator would generate. The monopole cannot tell the difference. It cannot tell whether it is talking to another monopole or its own reflection. If you want to fully understand it you will have to trace paths and see it is the same. Look at the EM waves that cross over the center plane for a dipole (a plane orthogonol to the antenna at the feed) and see they are the same as a reflected wave at that plane.
     
  13. Oct 21, 2013 #12

    meBigGuy

    User Avatar
    Gold Member

    http://www.aa5tb.com/dipole01.gif

    If you put a reflector in the middle of that antenna (and fed it correctly) the current and voltage patterns would look identical. If you want the mathematical proof or derivation of that, I can't help you other than the reference material I already posted.
     
  14. Oct 21, 2013 #13
    I've seen that graphical illustration of a dipole and I understand it. I just don't understand how the monopole antenna has the same current/voltage curve. What happens at the ground plane that causes the same behavior?
     
  15. Oct 21, 2013 #14

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Two ways of thinking about this problem, which may help. (It could be a visualisation matter). I remember 'feeling' that the E fields everywhere around the monopole must be vertical but they are not.
    1. Remember. If you are looking at the antenna from anywhere other than right down on the plane of the earth plane, you will see a component of both the vertical currents in the monopole and the horizontal currents in the earth plane.
    2. Take a small element of the vertical wire. This is a 'short' radiator and the E field will only be vertical in the horizontal direction. The E field lines will go out of the top of the element, round and into the bottom of the element (i.e. not all vertical). So the elemental E fields will actually have a component in the plane of the earth plane if they have originated from anywhere above the plane. Hence, you can get an induced a current which is at right angles to the direction of the originating current.
     
  16. Oct 21, 2013 #15
    So if we suppose that the ground plane is actually parallel to the ground and the monopole itself is vertically oriented, then the E field that is due to the monopole mainly looks like it does in the following picture: http://web.ncf.ca/ch865/graphics/EFldChargedCylinder.jpeg i.e. the E field is "parallel" to the ground plane. Because the E field is "parallel" to the ground plane it will induce current in the ground plane and they will be in phase with the current in the antenna itself. Is my analysis correct?

    Why does the ground plane have to exist? Why wouldn't the antenna radiate (at the same frequency) if there was no ground plane?
     
  17. Oct 21, 2013 #16

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Sorry, that picture is hard to understand. There is no E field actually at the plane - which is why the currents are induced (boundary conditions apply), any waves arriving just have to be reflected. The induced E fields are symmetrical in direction about the vertical (laws of reflection - in optics - same thing).
    The ground plane doesn't "have to exist" but its presence causes two things. It gets the feed point impedance to half the value of the dipole. It also defines a path for currents in the other half (the screen or other one of the pair) of the feeder to flow into. If you just stick a coax in the air, with a length of the inner sticking, out every Amp that flows in the exposed bit needs to flow down the outside of the screen. This gives an uncertain impedance and an uncertain radiation pattern. (Kirchoff 1 applies at the drive point, as ever)
    And of course, actually, the other reason for currents flowing in the earth plane is that the earth current from the feeder has to flow somewhere.
     
  18. Oct 21, 2013 #17
    http://farside.ph.utexas.edu/teaching/302l/lectures/img1298.png
    This concept is what you're talking about, correct?

    So the ground is necessary to ensure that whatever amount of current flows through the antenna has a return path?
     
  19. Oct 21, 2013 #18

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    That's the basics of reflection - in terms of 'rays'. But the same idea of symmetry applies when you draw the wave fronts.
    Something will take the currents. Using a ground plane defines where they go - and fixes the impedance.
    I have a feeling that you are trying to get into this topic at a level that's too deep for your basic knowledge (a common problem). I suggest you do some more reading round (in receive mode) rather than trying to do it by Q and A. Your Qs are not always too relevant, I think.
     
  20. Oct 21, 2013 #19
    You're probably right. What do you suggest I read?
     
  21. Oct 22, 2013 #20

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    There are loads of Google hits when you search google with monopole antenna theory etc. but I couldn't find anything that I'd say was just what you want. You really need a book on antenna and propagation theory. I have had many books available to me in the past but no more, I'm afraid, so I couldn't be sure if any book I could name would actually have an explanation in detail about how a ground plane works. Kraus has a well known classic book on antennae - written when your granny was a girl, no doubt - but it's a good source for generally reliable stuff about such things. You need something old and dusty, found in the corner of a s/h book shop, perhaps.

    It's normally taken as 'a given' that an image is formed but it should suffice to say that a reflection of any frequency of radiation should be expected to form an image in the same way - due to the currents (no details needed) induced in it. If you can believe it works optically (you can see it does) then you should be able to accept it works at MF, too and produces an image in the same way. The image and the monopole go together to form the overall pattern. In the case of a vertical monopole, the virtual currents in the virtual image are the right way to enhance the field in the horizontal direction. However, it may be interesting to note that a horizontal radiator, over a ground plane, produces an image in which the virtual currents (which are in the opposite direction) serve to cancel the radiated field in the horizontal direction. HF antenna arrays deliberately use HP for this reason because they need to fire upwards to the sky and suppress the 'groundwave'.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 1/4 wavelength antennas
  1. 1/4 lambda stub (Replies: 9)

Loading...