New answers tagged antennas
0
I found a reference from the online MIT course 6.013 about the analytical expression of EM wave radiation from a dipole antenna.
"Hertzian dipoles preferentially radiate laterally, with zero radiation along their axis. The electric field strength Eθ varies as sin θ, which yields a circle in a polar plot, as illustrated in Figure 10.2.2(a)"
David ...
0
EM waves as produced by antennas that are themselves visible (one centimeter to one kilometer long) are too low in frequency for your eyes to detect, which means you will never see them- as you point out.
The processes that create visible light photons occur on the size level of atoms and since your eyes do not have the spatial resolution to see an atom, you ...
2
why we have never seen oscillating charges produce em waves, like for instance an antenna.
You can experience this every day by looking at a mirror, which is just a flat polished conductive metallic surface. It is usually built on the back side of a sheet of glass for practical reasons: glass is transparent and it prevents the metallic surface from ...
-2
You can do it with "plasmonic nanoantennas". I quote from this link
Out of all of the devices that emit visible light in a controlled way, plasmonic nanoantennas are the closest to traditional radio antennas. A plasmonic nanoatenna is a nanoscale, precisely shaped metal antenna that has plasma resonances excited in it (bunched-up electron ...
2
at high (radio) frequencies, extension cord ("zip cord") wire has large power losses and is generally unsuitable for transmitting antennas.
telecomm antennas are built so as to produce as much electromagnetic radiation as possible, and point it in the desired direction. They have a wide variety of shapes to do this.
The amateur radio stack exchange ...
1
I think your question might be valid in some non-technical sense. The truth is, the signal does lose energy. As someone previously stated, it is the power of the signal and the impedance of the atmosphere that matter. As the broadcaster of the signal, you have to push enough power through the air to send the signal. If you want to think of these bodies as ...
5
First of all, the photon picture isn't the best way to visualize and model the operation of a radio-frequency antenna. RF antenna engineering is best performed and understood (IMHO) by means of a model in which high frequency AC power flow through a wire creates electromagnetic waves that propagate away into space.
For this to occur, the antenna impedance ...
2
All elements of a circuit will radiate when there is alternating current involved. Typically though, at the frequencies involved, not very much. It is something that circuit designers must be careful of though.
Capacitors themselves typically are encased in some sort of material which will damp the stray radiation, especially if it is a conductive material ...
2
When antennas are placed very close together all bets are off. The near field coupling between antennas elements change the impedance's therefore currents and fields the far field patterns don't follow simple algorithms any more. in fact if they become close enough and fed in phase they asymptote towards a single element with two sources in parallel the far ...
1
You have answered your own question in a sense, by part of your statement that:
"This image shows that the applied voltage is in phase with the current"
this is the definition of the "real" or purely resistive part of the complex impedance. there is no standing wave at the feed point but a travelling wave from the source to the feed point,...
0
A half-wave dipole is fed with the signal at its mid-point.
Consider the current distribution along the antenna. Some current has to flow in and out at the middle, or power cannot be transferred. But at the ends the current has nowhere to flow. So the current distribution along the dipole takes on a distinct "hump" shape, basically half a sine wave....
0
A pi network, or any other impedance tuning circuit used between an antenna and a transmitter, adds or subtracts inductance or capacitance so as to null out the effects of the antenna being nonresonant (mismatched) at the desired frequency and thereby present a purely resistive load impedance to the transmitter.
This is done to prevent power that the ...
0
G'day Jay
A dipole is not a good choice for an under water antenna. The characteristic impedance of water is very low. A dipole has an impedance of approx 73 Ohms at the feed point and has the property of being able match to the impedance of free space, which is 377 Ohms. Also the ends of a thin dipole may have reactive impedance up to several thousand ...
2
An antenna, in general, is NOT a resonator (contra @oliver). The purpose of the antenna is to be a direction dependent impedance transformer that matches the wave impedance of the transmission line connecting the transmitter/receiver to space (air) having impedance $\sqrt{\frac{\mu_0}{\epsilon_0}}=377\Omega$ in all direction. In other words, the apparent ...
0
An antenna is a resonator. If you are not feeding it with its natural frequency, it is not going to oscillate with sufficient amplitude. Of course you can always think about increasing voltage, but usually everything in technology is about efficiency. Imagine you would have to carry a heavy car battery and a high voltage generator with your smartphone in ...
1
Consider an oscillating electric charge (part of an oscillating current). The moving charge introduces a transverse distortion into its (preexisting) radial electric field (strongest when moving most rapidly). As part of a current it also produces a magnetic field (wrapped around the direction of motion), also strongest when the charge is moving most ...
0
I realize that the truly correct answer to this question is perhaps to simply say that if there is a wave solution to Maxwell's equations, then a necessary property of that solution is that the electric and magnetic fields will be at right angles to each other and will be in phase.
I was looking for a more intuitive type understanding. Please tell me if ...
4
Since the times of maximum charge separation (electric field) and maximum current (magnetic field) are out-of-phase in a oscillating dipole, the question makes perfect sense.
The answer is to look at the exact dipole solution:
$${\bf E}=\frac 1{4\pi\epsilon_0}\Big[
\frac{\omega^2}{c^2r}(\hat{\bf r}\times{\bf p})\times\hat{\bf r}+
\big[
\frac 1 {r^3}-\frac{i\...
0
The propagating wave has complex power density (Poynting vector) $\mathbf S = \mathbf E \times \mathbf {\bar H}$. Any phase shift between the fields $\mathbf E $ and $\mathbf H $ is reactive and does not correspond real power. Recall the analogous case of $V\bar I$ from circuit theory.
3
The EM waves act on charges, and the frequency of the movement of that charges is selected by the LC reception circuit of the radio.
But without enough carries present in a good antenna, there is no much signal to select. The effect of the human body is to provide movable charges for the EM waves. When touching the antenna, it increases its role. But even ...
14
Small pocket radios usually have fairly poor antennas in them, because of their small size. But by placing your hand nearby the antenna of that small radio, you are creating a capacitor in which one terminal is the radio chassis and the other is your hand. Any radio frequency signal induced in your skin by radio broadcasts will be capacitively coupled from ...
Top 50 recent answers are included
