6
$\begingroup$

Diagrams like the one shown below are often shown to explain antenna theory, but I have always had problem with the concept of voltage being a wave, and because of this the diagrams never make any sense to me.

enter image description here

If the voltage source starts to generate a sine wave at time 0, then the voltage at any point along the transmission line will be:

$$V(x,t) = V_{max}\sin(kx-\omega t)$$

Once the voltage reflects the equation becomes:

$$V(x,t) = V_{max}\sin(kx-\omega t) + V_{max}\sin(kx+\omega t)$$

  • First question: What is the mechanism explaining how this voltage propagates? $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ - A voltage is not a "real" thing, it's measurable, but it is just a concept made up to quantify a complex interaction of coulomb forces between electrons. Voltage is the electrical potential energy between two places a unit charge would feel. Energy is just the potential to do work, which is the integral of force over distance moved in the direction of the force. If there is a 1 volt potential difference between 2 points ($a$ and $b$) within an electric field, that means that a unit charge placed at point $b$ would require 1 joule to be moved to point $a$. That is all voltage is, I don't understand how this can propagate as a wave.

  • Second question: Why does the voltage reflect? $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ - Similar to the reason above, if voltage just describes the amount of work done needed to move 1 coulomb between two places (the work done can be manifested as distance or force, as long as: $\:\:W=\int f \cdot x $), then how can it reflect?

  • Third question: What is the voltage on this line relative to? $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ - Voltage always has to be with respect to something, as it is a potential difference. In closed circuits with an oscillating voltage source you usually treat one side of the voltage source to always be at ground (a constant 0V) and the other side of the voltage source to be oscillating between the max and min values. Because voltage is relative this is exactly the same as treating both sides of the source to be oscillating and the difference between the two always equaling: $V_{max}\sin(\omega t)$. It is just easier to treat one side as ground. In the antenna, what is ground, and what are the voltages relative to?

  • Final question: Why is voltage and current at nodes/ antinodes respectively? $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ - Okay, if I just accept that voltage can be a wave and hence a standing wave in an open transmission line, I still get confused. Why are the points of maximum voltage (antinodes) at the point of zero current (nodes)? At a point of maximum voltage the potential difference is oscillating between its max and min values, so why doesn't the maximum current occur here?


Note: I'm not completely confident with the equations I gave for the value of the voltage at time $t$ and position $x$. Please just edit these if they are not correct as they are not the main concern of my question. I mainly want to know what is happening to the electrons and fields inside the antenna so I can understand why the voltage propagates and why it reflects.

$\endgroup$

2 Answers 2

7
$\begingroup$

(All images in this answer were made by me, for wikipedia! Links here)

Let's start with the simplest question: Voltage relative to what?

transmission line schematic

It's the voltage from one line to the other.

Now let's look at an animation of a transmission line. This one is terminated with an impedance-matched resistor, so we don't have to think about reflections yet.

transmission line animation

The dots represent electrons. The red color is high voltage, and the blue color is low voltage.

What makes the wave flow?

  • The bunched-up electrons repel each other, and the spread-out electrons attract each other. Like a compression wave on a spring (slinky). The amount of attraction or repulsion depends on the capacitance of the line. With a large capacitance, there is less attraction and repulsion, because the other line (which always has the opposite charge) partly cancels out the attractive or repulsive force.

  • The electrons behave as if they had inertia, i.e. they were heavy. They don't really have inertia (the mass of the electron is too small to matter here), but this is the effect of the inductance of the lines. When a force is pushing on the electrons, they accelerate gradually, not instantaneously, as if you were pushing a heavy cart. (Think about what an inductor does.)

So, you can picture it like a compression wave traveling down a spring (slinky). The electrons bunch, which causes repulsion, which makes them spread out, but now they're bunched up somewhere else, and so on.

Next, reflection:

transmission line reflection animation

The top one is a standing wave due to reflection off an open circuit. The bottom is a standing wave due to reflection off a short circuit.

For the top one, the electrons can't move at the end. For the bottom one, they cannot bunch up at the end.

Why does it reflect? It's not so different from a compression wave bouncing off the end of a spring (slinky). Why are the voltage nodes at the current antinodes? I think the answer is clear from the animation.

$\endgroup$
0
2
+100
$\begingroup$

I think you want a handwaving, intuitive explanation. So here is in order from easier to more difficult. Your plot shows not an antenna but an open circuit parallel transmission line that may radiate at its ends and your questions seem to be about the propagation along the line. I know there is something called travelling wave antenna but this is not what you show so let us stick with propagation along the line.

  1. the voltage you are asking about is $between$ the two points of the same $x$ coordinate that an ac voltmeter would measure whose leads are in the plane of that cross-section and are simply connected to the points in question without winding around any of the lines.

  2. the flowing current is measured along the line and since the ends are not connected, open circuit, the current there must be zero at all times, that is just charge conservation. But the voltage is not zero at all times but fluctuates at the rate of the input frequency of the generator. If you think of reducing that frequency almost to zero then the parallel lines are just a capacitor so the end voltage is the same as the generator voltage, so there the voltage has to have non-zero amplitude. As you increase the frequency there will be a phase delay between them but a fluctuating voltage builds up between the ends.

  3. if you think of the charges forming a continuous flow driven by the generator at the left then remember that whatever propagates is propagating at a finite velocity, so that any interaction at further right will happen a bit later then at the left, so "things" clump and dilute, so to speak, resulting in fluctuation not just in time as driven by the generator but in space as well, hence the wave.

  4. the current cannot pass the open circuit where the lines end, so the kinetic energy in the current must go somewhere: the current bounces back, and carries a reflected wave with it.

This is all handwaving and only for visualizing, I do not claim that it will get you to deep understanding of the physics of transmission lines. For that you should study the Telegrapher's Equation and you may start here http://en.wikipedia.org/wiki/Transmission_line

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.