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I am a new to the radio design field.
My question is, 'What is so significant about 1/4 wavelength with respect to antennas.'
From the information I have gathered from various sources, it is my impression that a 1/4 wavelength antenna is the most used on account of its performance.
I am looking forward to an answer, elucidating the significance in simple terms that I could understand.
The first part of why a quarter wavelength is special is actually understanding that it's not a quarter wavelength, but a half wavelength.
Consider a quarter-wavelength monopole. If a wavefront originates at the feedpoint, a quarter-cycle later it will have reached the end of the monopole. Here "something interesting" happens, because the antenna ends. That "interesting event" must then propagate back down the antenna to the feedpoint before it can affect the feedpoint impedance. So a quarter-wave element is a half-wavelength round trip.
What then is the "interesting thing" that happens at the end of the antenna? The thing to realize is at the first instant something happens at the feedpoint, the feedpoint doesn't "know" the antenna is going to abruptly end some distance away. Even though the antenna is an open circuit, some current will initially flow, starting a wave propagating down the length of the antenna. When this wave reaches the end of the antenna, another wave starts in the opposite direction.
To develop some intuition for how this works, I suggest reading How does the current know how much to flow, before having seen the resistor? and these excellent animations by Daniel Russell.
The reflections take a half cycle to make their round trip, so by the time the reflection (of opposite polarity) reaches the feedpoint, the feedpoint has also reversed polarity. Thus the current in the reflected wave reinforces the current driven by the feedpoint. This current reinforcement means the feedpoint can drive a larger current with a lower voltage. In other words, it has a low impedance. And this is why an antenna which is an open circuit at DC can be a low impedance around 50 ohms at RF.
Dangerous to start this sequence of answers, but I do.
Any conducting object can be used as antenna. From reciprocity (transmit versus receive) it can be deduced that when power is fed to such an antenna AND there is no loss, that all the power is transmitted. Must be transmitted, since without loss there is no conversion in heat (calories). There are however two important things for this antenna. 1: what is the directivity of the antenna (directivity pattern) and 2: what is the impedance of this antenna. That second aspect, the matching impedance, is a practical problem with limits: when the antenna feedpoint impedance deviates far from real impedance, there is need for complex matching network. And such network is realised with inductors and capacitances and these components do add to losses.
To come back to your question: any resonant antenna is a practical solution to minimise the matching network losses. A resonant quarter wavelength with a dedicated counterpoise or earth system results in a feeding point impedance between 35 and (say) 60 Ohm pure resistive. That is an impedance that can be matched with simple means to most transmitter systems that require 50 Ohm real impedance. Most simple antenna systems are based on quarter wavelength antennas.
And, for next-step or future thoughts: a receiving antenna can be as small as you can imagine, without loss of signal to noise. At least at the antenna output terminal. Of course: noise is added in the amplifier or receiver that is connected to that antenna, so there is a practical limitation to the reduction of the dimensions.
Example: a small telescope antenna used on shortwave for reception of signals with a simple (J-FET) LNA can have a sensitivity (own noise contribution) better than the received atmospheric noise, since the atmospheric noise is relatively high. Transmitting with such small telecope antenna requires series inductance to make impedance real AND impedance matching from that small few-Ohms impedance to the common 50-Ohms wanted load impedance for tranmitters. Also the requirements for the grounding are very different: for reception it is far less important, noise limitation only, than for transmitting, where the efficiency is affected.
PA0FSB
Welcome to Ham SE, @Newbie, and thanks for your question.
If you break a length of wire to create two terminals and apply an AC voltage across the terminals, you will excite an alternating current on the wire. Some wire lengths will accept more power from the generator than other lengths.
The antenna cited in your reference has two terminals but, because Ground has some electrical conductivity, one of the two terminals is actually the Ground under the "$\lambda$\4 radiating element":
The ability of the "antenna" - in this case, the combination of the radiating element and the ground underneath it - to accept power from the generator depends on how much the antenna resists the flow of current. When speaking of AC voltages, we call this Impedance, symbolized by the letter Z and measured in ohms just like resistance for DC. For a given generator voltage, more current flows on the antenna when the Impedance is lower. When more current flows on the antenna, stronger electromagnetic waves are radiated.
The impedance at the terminals - the "feedpoint" - varies with frequency. If we make the antenna 5-meters tall, the impedance seen by the generator will follow the green curve in the graph below as we change the frequency of the generator:
Notice that the antenna has the least resistance to the flow of current - the lowest impedance - at about 14.35-MHz, so this is the frequency at which the most current will flow and the strongest electromagnetic waves will be radiated.
In free space, the length of an electromagnetic wave, $\lambda$, is: $$\lambda = \frac{c}{f}$$ where $c$ is the speed of light (300 million meters/sec) and $f$ is the frequency of the generator (Hz). At $f$=14.35MHz: $$\lambda = \frac{3*10^8}{14.35*10^6} = 20.9-meters$$
So, our 5-m tall antenna is approximately 1/4 of the free-space wavelength of the 20.9-meter signal applied to the antenna terminals.
The reasons the antenna is not exactly $\lambda/4$ are beyond the scope of your question. Antennas like this "quarter wave vertical" are indeed popular, but often for a complex set of reasons - cost, available space, materials on hand, stealth, etc. - not simply its "performance."
In reality, any length/height of a conductor along which r-f current is present can produce e-m radiation into space.
However, some antenna configurations and systems are more effective in generating e-m radiation from a given applied power than others. This is shown in the graphic below by the reduced fields that antenna system produces when radiation resistance is low relative to non-radiating, but resistive elements of the antenna system.
The flow of r-f current in/along a non-radiating resistance produces heat, rather than the radiation of e-m waves.
This explanation might be more intuitive, and thus not completely accurate.
EM fields (the RF someone might be trying to transmit using the antenna) are generated by accelerating charges (electrons, or maybe the lack thereof), over some distance. For a particular transmit frequency, one wants to alter the direction of acceleration at some rate. So, 1/(2*f) later, whatever is driving charges in one direction should probably try be trying to drive them in the opposite direction (for a frequency of f).
Now the EM field driving the electrons can't go faster than the speed of light. So the acceleration of electrons gets to the end of a wire (tip of the antenna) after a certain amount of time, and then gets reflected back. If the distance to and fro is half a cycle, the the charges rushing back get there at the same time the driver is also reversing direction. Thus the two effects reinforce, instead of canceling out. This reinforcement is usually a good thing, as it doesn't heat up the transmitter and/or the feedline as much.
