quarter wave antenna vs quarter wave transmission line - Page 1
Forum section: electronics / beginners

author: rgujju
author_rank: Newbie
timestamp: November 12, 2020, 04:51:11 pm
content: Hello guys,I was reading up on transmission lines and came across this resource they have the following image which looks like a quarter wave antenna to me since the length is λ/4 and terminated in open circuit. The text mentions that due to the λ/4 length the reflected wave will be 90 degrees out of phase and hence cancel each other and hence appear as a short circuit.1) Is this how a λ/4 antenna works as well? Probably not since it would not radiate then.2) What are the differences between an λ/4 antenna and a λ/4 transmission line with open circuit termination?https://resources.altium.com/p/stubs-transmission-lines-what-do-they-do-and-how-do-you-keep-them-happening

author: ejeffrey
author_rank: Super Contributor
timestamp: November 12, 2020, 08:54:08 pm
content: They are related but not the same.A quarter wave transmission line has a return path nearby (much closer than lambda/4) such as the shield of a coax or the ground plane of a microstrip. Therefore it has a small dipole moment and is a poor antenna.A quarter wave antenna is situated well above the ground plane, so the combination of the antenna and it's image reflected across the ground plane have a large dipole moment and are effective radiators.The quarter wave antenna does however make use of a similar impedance transformation from the reflected wave so that it doesn't look like an open circuit but is lower impedance. However it doesn't appear as a short circuit either: because it couples to propagating waves it has a real component of admittance that corresponds to the radiation "loss".

author: rgujju
author_rank: Newbie
timestamp: November 13, 2020, 04:42:03 pm
content: Hi ejeffrey,As per my understanding, in case of the quarter length antenna the reflected wave will be 180 degrees out phase (due to open termination) and hence the resultant waveform will be 0. So how does an antenna radiate then?

author: ejeffrey
author_rank: Super Contributor
timestamp: November 13, 2020, 05:09:53 pm
content: First off, in the transmission line case the reflected wave is only 180 degrees out of phase at the feed point. There is a standing wave with a voltage node at the feed and an anti-node at the open-circuit end. In between you have a sinusoidal variation of the peak voltage. On the other hand, there is a current anti-node at the feed point, so it isn't correct to say that the "signal" is zero there even if the voltage is zero. Since the voltage is zero and power = V*I, the power is zero which is expected for a lossless non-radiating transmission line.For the antenna because there is radiation the reflection is weaker than the incoming wave and therefore does not completely cancel even when there is 180 degrees of phase shift. This results in a non-zero voltage at the input point and thus the ability to deliver power.

author: MathWizard
author_rank: Super Contributor
timestamp: November 15, 2020, 02:09:09 am
content: I know the principles of standing waves, but I don't remember any eqn's for sound, or really know/remember the EM wave equations good enough yet.So is it the case, that's there's some 50 impedance, of "RLC" across some bandwidth.....from the signal conductor, to GND at any given point of the shield?, or just at the ends?And as for the reflection, what does it reflect off of? Is it sort of akin to how waves change speed, direction when crossing between materials, and some amount of energy propigates backwards ?Also, if some coax cable length is X*lamda/4, is it the case that at the endpoint, X*lamda/4, it might seems more L dominated (or R or C) at the end point, and then if you add Lam/4 to the cable, it might seem more R dominates the impedance at the endpoint for example? , and then at if you add Lam/2 to the cable, it might seem more C dominates the imp. at the endpoint ?I'll be learning this stuff by next for sure.

author: gcewing
author_rank: Regular Contributor
timestamp: November 15, 2020, 05:25:42 am
content: Quote from: MathWizard on November 15, 2020, 02:09:09 amIt reflects off the end of the cable.More mathematically, the reflected wave has to be there to satisfy the boundary condition of zero current at the open-circuit end.QuoteYes, it's a lot like that. The same kinds of equations apply.

author: radiolistener
author_rank: Super Contributor
timestamp: November 15, 2020, 08:14:10 am
content: Quote from: rgujju on November 12, 2020, 04:51:11 pmNo. Antenna works just because current in the wire is changed. When di/dt is not zero, the wire radiating EM wave. It happens because electrons in the wire accelerating and decelerating.Antenna length to wavelength relation is not important for antenna. It is important for a feeder and transmitter, because transmitter don't want to get reflected wave back from the antenna. In order to avoid reflected wave from antenna, you're needs to match feeder impedance with antenna impedance. It can be done with matching device or by selection proper antenna size relative to the wavelength.Antenna size in wavelengths just affects impedance at the feeding point. Some antenna sizes have impedance which is more easy to match with the feeder. Some antenna sizes have impedance which is hard to match with feeder. The same with different feeding point. This is why some antenna sizes are useful and some antenna sizes are useless.λ/4 is a half of λ/2 antenna. It needs counterpoise as a second part. Counterpoise is mandatory. In portable transmitter metal case of the transmitter is used as a counterpoise. If you use too small counterpoise it leads to unbalanced load, so you're needs to use RF choke and balun/unun in order to prevent using your feeder as a part of antenna (because it leads to interference and collect noises from your home equipment). Too small counterpoise leads to a high impedance and thus to more complicated impedance match.