Output power curve of an antenna

Question

My approach is that a radio wave is made from zillion of accelerated electrons which emit zillions of photons. Since the current flows in a open circuit, the antenna generator works against something like a capacitor. So I want to know, is my imagination comparable with the radio specialists knowledge? Could it be that the power output of an antenna often looks like below?

The red line is the ideal sine curve. The axis to the right is the time, the other axis is the power output. The blue line is the curve in reality, which one can get from an antenna generator.

From theoretical assumptions only, I suppose that the following points improve the curve closer to a sine

1. to some extend a longer antenna,
2. a lower frequency of the antenna generator,
3. a capacitor on the end of the antenna rod.

The background is my answer to the question “Why are longwave radio towers so tall?”.

Answer 1

It's important to be clear about definitions.

Power in this context is usually defined as the product of root-mean-square (RMS) voltage and current. Assuming transmission of an unmodulated carrier, power is constant over time because RMS current and voltage are constant over time, by the definition of RMS.

It makes no sense to think about power "between" the peaks and valleys of current or voltage. All electromagnetic radiation oscillates at some frequency, and thus must necessarily cross zero periodically. If the voltage and current didn't cross zero, there could be no oscillation, and thus no radiation.

Instead of power, what about just voltage? Again assuming only an unmodulated carrier, voltage is exactly a sine wave. This holds true at any antenna length, at any frequency, and even if there are capacitors or inductors as part of the antenna.

I'm guessing by your drawing you're thinking the antenna looks somehow like a capacitor to the transmitter. But it does not: typically the antenna is tuned so that it looks like a resistor, because the antenna is tuned to be resonant. Resonance is by definition a zero reactance, so the antenna impedance must be resistive.

What if the antenna isn't resonant? Well, in that case we can just forget about antennas and just ask what happens if an AC voltage source is connected to a capacitor:

^{simulate this circuit – Schematic created using CircuitLab}

The answer is the voltage across the capacitor is still sinusoidal. It has to be, by definition of how voltage sources work. The voltage across C1 is equal to the voltage across V1, by Kirchoff's voltage law.

What about current? Also sinusoidal. The current through a capacitor is given by the equation:

$$ I(t) = C {\mathrm d V(t) \over \mathrm d t} $$

where $t$ is time, $C$ is capacitance, and $V$ is voltage. If $V(t)$ is some sinusoidal function, then by basic calculus it should be clear $I(t)$ is another sinusoid, shifted 90 degrees in phase and scaled in amplitude by a factor of $C$.

The only time the voltage between any two points in any antenna, or the current through any segment of an antenna is not a sinusoid is when the signal being transmitted is not a sinusoid. Even then, in practice it's usually pretty close to a sinusoid, especially if we're only looking at a few cycles since any modulation which would modify the sinusoid is usually slow compared to the period of the oscillation.

By Fourier analysis we can see that anything which isn't pretty close to a sine wave will contain energy on many frequencies. When a transmitter is transmitting on unintended frequencies that's called interference, and a good part of an RF engineer's job is avoiding precisely that, since it tends to make the legitimate users of those unintended frequencies upset.

Answer 2

Firstly, I think someone may or may not agree with the concepts proposed here, but that doesn't make it a bad question. The OP has a background in physics and is asking how comparable are his observations to that of experienced hams, this forum would seem an ideal place to ask such questions, but it also means he looks at things a little bit differently than the rest of us mere mortals.

If I understood correctly

the model is looking at the 'electron mass' (for lack of a better phrase) of the wire. In a given cycle, the voltage source will try push in x electrons into the wire, and then take back the same number of electrons back in the -ve half. Because the voltage source is a sine wave the power plot also resembles a sine wave. But what happens if the wire is not long enough, i.e. it physically does not carry said number of electrons. The voltage (and power) will peak too soon, much like when a capacitor is charged it will not accept anymore electrons. The OP added a capacitor, here the capacitor is akin to increasing the length of the wire, i.e. it provides it with more "electron mass", and things are back in phase again. But too much "mass" and we start getting back-currents... here I would think the power peaks too late (?)

What makes this analogy hard to grasp is that in practice, if the antenna is too short for the desired frequency we get a capacitive component in the impedance, and we correct it by adding inductance... in the OP model the solution is too add more "mass" to the wire, he does it by adding a capacitor.

I would argue we could add "mass" by making the wire thicker, then in practice you'd be sacrificing gain for bandwidth.. as indeed happens with real antennas.

I hope I got that right... but the point is, if it's a bad model, please someone explain why? As a noob, I am always searching for interesting analogies to help me understand complex concepts, antenna theory is a complex concept! Much like how water pipe analogies help beginners learn DC concepts.. I know the hams in this forum long ago stopped thinking in terms of water pipes..

Edit: Many are pointing out that inductance was not mentioned in the model. At the lower frequencies, such as provided by the DC motor source, inductance would have very little effect and capacitance would dominate in his observations.

Question for the OP, is this all theoretical or are you getting these values from an actual setup ?

Answer 3

After some research I’m able to answer my - honestly not well formulated - question.

In my question I’m using a AC-motor to push and pull electrons inside a wire. Any wire is a resistor and some amount of the kinetic energy of the moved electrons gets dissipated into heat. Having a long enough wire, all the power from the DC-generator will be used for heating the wire. The sketch of the power consumption is a sine, where the negative part is folded up (image 1):

The setup with only a wire is a transmitter. Because any acceleration of charges is accompanied by the emission of photons. The interaction of all these photons produces a radiation. This radiation is modulated in the sense that the number of emitted photons follows the frequency from the antenna generator. To prevent such a radiation coaxial cables were invented.

Suppose that open circuits were unknown, it would be possible to build a real working antenna from the wire. What has to be done is to shield one half of the wire. Building a tower from concrete with a lot of metal in it and providing the wire first on the outside and then inside the tower, one dissipate the radiation in the inside to heat and the outside cable is a real working radio transmitter. Adding some modulation to the frequency or changing the power, one could transmit informations.

Such a device loses half of the transmitted energy. Now the blue graph in the sketch shows the theoretically emitted energy and the red graph alludes to the real emitted energy (image 2).

An open circuit will don it with more efficiency. But only cutting and and reducing the length of the wire, the number of moved by the generator electrons get reduced. The generator works into void and the emitted energy is very low. The black graph represents energy dissipation into heat (image 3).

Adding a small capacitor to the end of the open wire we get a better result. For some time the electrons are able to flow into the capacitor and the current in the wire flows for a longer time. By this we reducing the heat losses and transmitting more power. Making the capacitor big enough the circuit is tuned to its resonant frequency and the heat losses are reduced to a minimum.

That’s not all. The electrons in the charged capacitor at the end of the half wave are working against the generator and this is the reason why the emitted energy is shifted to the left in the sketch (image 4).

Folding the capacitor to an antenna rod, one get a real working transmitter (touch the sketch, if it’s not animated).

Source

What, if the length of the antenna rod does not match the needed capacity to be tuned? Like in low frequency transmissions. It’s simply looks like image 3.

What I has had to asked: “Was such a dis-tuned transmitter built, was the received power measured and does the result looks like in image 3. Sorry that I was sloppy.