Why is a half wave dipole resonant?

Question

Is the following explanation correct ?

RF energy with a wavelength equal to twice the electrical length of the dipole is applied to the feed point at the center of the half wave dipole.

To understand the following, it's necessary to visualize in your head a number of different things which happen during the time it takes for one cycle of the applied RF energy to occur.

1. The amplitude of the voltage of the RF energy applied at the center feed point which is varying in sinusoidal fashion at it's frequency of oscillation.
2. The waves of RF energy which emanate away from the feed point on each half of the dipole towards the ends of the antenna and which are then reflected and arrive baat the feed point.
3. The change in phase of the waves as they move from center to ends and back again.
4. The effect that the reflection has at the ends of the dipole on the phase of the voltage and current of the waves.
5. The difference in phase between voltage and current of the RF energy applied at the feed point compared to the phase of the voltage and current of the reflected wave which arrive back at the feed point.
6. The standing waves of voltage and current which exist along the length of the antenna.
7. The value of reactance that the difference in phase of the voltage and current of the incident and reflected waves produces at each point on the antenna elements.

A positive half cycle of applied RF energy at the feed point emanates away from the feed point towards one end of the antenna, and a negative half cycle of RF energy at the feed point emanates away from the feed point towards the other end of the antenna. The waves are reflected from the ends of the antenna, arriving back at the feed point in the time it takes for one half of the cycle of the applied AC to occur because the element lengths are exactly 1/4 of the wave length of the applied AC.

At the same time as the positive wave on one side and the negative wave on the other caused by the applied RF energy are traveling from the feed point to the ends and back again, the amplitude of voltage of the applied AC is changing in a sinusoidal fashion, from zero to maximum positive amplitude one side and zero to maximum negative amplitude on the other side, and back again, and so progresses through half of one AC cycle or undergoes a total change in phase of 180 deg.

Between the feed point and the end of each element, the traveling wave changes phase 90 deg in time and space in 1/4 of the time it takes for one cycle of the applied AC to occur. The same phase change occurs in the same amount of time from each end back to the feed point. Excluding what happens at the reflection, this means that both the voltage and the current of the wave changes in phase by 180 deg just because of the fact that the wave has traveled a distance of 1/4 wave length twice.

The reflection occurs at an open circuit, and so the phase of the voltage of the reflected wave does not change, and the phase of the current of the wave is reversed.

The amplitudes spoken about as follows are the independent instantaneous amplitudes of the voltage and current of the incident and reflected waves which change in sinusoidal fashion over the period of 1/2 a cycle of the applied AC.

At the reflection, the amplitude of the positive voltage of the incident wave adds to the amplitude of the positive voltage of the reflected wave to produce double the amplitude of each, and the positive amplitude of the current of the incident wave adds to the negative amplitude of the reversed in phase current of the reflected wave to produce zero current amplitude.

So at the ends of the antenna there is an AC voltage maximum and a current minimum, and an impedance maximum.

When the reflected wave arrives back at the feed point, its voltage is 180 deg out of phase in time with the voltage of the next cycle of the applied ac waveform, and it's current is 360 deg out of phase in time, or in phase, with the current of the next cycle of the applied AC.

The entire process repeats for the negative cycle of the applied RF energy, with everywhere a reversal of polarity of amplitudes.

So at the feed point, the sinusoidal variation of the amplitude of the voltage of the reflected wave is exactly out of phase in time with that of the next cycle of the applied AC, and so they cancel each other out. The sinusoidal variation of the amplitude of the current of the reflected wave is exactly in phase in time with that of the next cycle of the applied AC, and so they add together. The phase of the current of the reflected wave is the same as the phase of the voltage of the next cycle of the applied AC, the voltage and current at the feed point are in phase, there is no reactance present, and the antenna is resonant.

At the inner feed point ends of the antenna elements, there is an AC voltage minimum and an AC current maximum, and an impedance minimum.

At the same time as all that is going on, the incident and reflected waves traveling in opposite directions on the antenna are combining to produce standing waves of voltage and current on the antenna.

The amplitude of the voltage and current of the incident and reflected waves add as they pass each other on the dipole elements to produce a standing wave of voltage and current. The maximum amplitude of the standing wave is the addition of the maximum amplitudes of the incident and reflected waves and this changes at each point along the antenna. At the ends the amplitude of the voltage peak is at a maximum, and the maximum diminishes in sinusoidal fashion as the point along the antenna approaches the center, to arrive at a minimum peak value at the feed point end of each dipole element. At the ends, the amplitude of the current peak is at a minimum, and the maximum increases in a sinusoidal fashion as the point along the antenna approaches the center, to arrive at a maximum peak value at the feed point.

The amplitudes of the standing waves of voltage and current vary in sympathy with the sinusoidal change in amplitude of the applied RF energy at the feed point. The positions of the peaks and troughs of the standing wave remain stationary and fixed in space along the length of the antenna. The standing waves result from the vectorial addition of the amplitudes of voltage and current of the incident and reflected waves and are the actual conditions present on the antenna which can be measured. The original incident and reflected traveling waves cannot exist independently of each other and cannot be measured as their existence is obscured by the vectorial addition of each to the other.

Despite this, the difference in phase of the voltage and current which results from addition of the incident and reflected traveling waves at any point along the lengths of the dipole elements determines the amount of reactance present at each point.

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Answer 1

I can't follow your explanation to the end, so I can't say if it's correct or not. Please enjoy a simpler explanation:

Consider not a dipole, but a quarter-wave section of balanced transmission line. One end is an open circuit, and the other end is the "feedpoint".

Now say you apply a 1 volt step to that feedpoint. It's not yet "known" the end is open, so a wave travels down the transmission line, with voltage and current in proportion with the transmission line's characteristic impedance. Let's say it's a 300 ohm line, so 1/300 = 3.3 mA is the current your voltage source will have to supply.

Some time later, the wavefront corresponding to the start of the voltage step reaches the end of the line. But wait, the circuit is open! The open at the end of the line needs to "tell" the voltage source that it has sent too much current. The voltage source initially thought 3.3 mA was required to make 1 V, but actually only 0 mA is required because the impedance of an open circuit is infinite.

It does this by initiating a 2nd wavefront, superimposed on the first one. This 2nd wavefront is 3.3 mA in the opposite direction. The superposition of these two waves is 0 mA, the "right" amount of current.

This 2nd wavefront arrives at the voltage source some time later, and this is the first time the voltage source has learned anything about what's at the end of the transmission line. The -3.3 mA opposite wave will also be accompanied by 1 V, again due to the characteristic impedance of the line. When this reaches the voltage source it will reduce the current its producing to hold its output voltage to 1 V rather than allowing voltage to rise to 2 V. This sets off another superimposed wave (but smaller this time) that goes through the same process, and this process repeats until the current through the voltage source approaches 0 mA.

Impedance describes the relationship between current and voltage. One way to measure it is to attach a voltage source (fixed, known voltage) and then measure the current. The real part of impedance (resistance) tells us how much current to expect. The complex part of impedance (reactance) tells us when to expect the current. If peak current coincides with peak voltage, reactance is zero and the dipole is resonant.

Now consider the current that was observed in response to the voltage step in the example above. Ignore the 3.3 mA that initially flows when the step occurs, because with AC analysis we are concerned with the equilibrium condition at a single frequency, not the transient analysis.

• It takes 90 degrees for the wavefront to reach the end of the transmission line, because it's a 1/4 wave long.
• The current of the reflected wave is inverted, so that's another 180 degrees.
• It takes another 90 degrees for the reflected wave to reach the feedpoint.

90 + 180 + 90 = 360 degrees, and since in AC analysis we assume sinusoidal inputs with no start or end, 360 is equivalent to 0 degrees. In other words, voltage and current are in phase, and reactance is zero.

From here it's just a small step to a resonant dipole. Cut the transmission line down the middle and pull the conductors apart (and possibly make some minor adjustment to the length to account for the altered velocity factor) and you've got yourself a dipole.

The only difference is that in the case of an ideal 1/4 wave open, transmission line, 100% of the energy sent into the line ends up reflected back at the voltage source with current in phase, meaning with each reflection the voltage source must produce more current to maintain its output voltage. Thus the current approaches infinity: the voltage source sees a short circuit.

But in a dipole some of the energy is lost to radiation, and so the voltage source sees a low but non-zero impedance in the neighborhood of 75 ohms, depending on wire diameter, interaction with the ground, resistive losses, etc.

Answer 2

Here's another slant on the standing waves of voltage and current in a resonant dipole antenna, cut to a half-wavelength considering the velocity factor.

When such an antenna has been energized at its resonant frequency, the current reaches the end of the antenna at the end of a half-cycle, with the voltage at a maximum and the current at zero.

The current now reverses direction for the next half-cycle. The voltage and current follow a sinusoidal waveform, the current touching maximum and the voltage zero midway through, till the end of the half-cycle at the other end of the antenna.

The process repeats with the antenna current doing a to and fro traverse every cycle.

The fixed voltage and current values along the antenna are represented by the standing waves which reverse polarity every half cycle.

Electromagnetic radiation/reception from the antenna is a result of these standing waves.

Answer 3

You don't need standing waves or a sinusoidal input to find the resonant frequency of a dipole.

Zap the feed point of a conductive dipole with a very short (much much less than one period) voltage impulse. That voltage pulse will travel to one end (due to limiting speed of light in some dielectric) in finite time, then bounce to the other end at that same speed, then travel back to the feed point (its second pass). How long does that full round trip of the dipole take? Call that one period of the resonant frequency.

All you need is a fast storage oscilloscope and a fast single shot pulse generator.

If off-center fed, you will see two mid-period negative voltage passes, but the same positive impulse for one full period.

If the total antenna resistance is low enough, that round trip will repeat until it decays away (partially radiated as EM if the radiation resistance is high enough, the rest as heat).

Linear superposition says your transmit waveform is just the arithmetic sum of a bunch of these impulses tracing out the same shape.

Answer 4

Here is another very simple answer.

There is only one condition which determines when a half wave dipole can be resonant, and that's when the electrical lengths of the dipole elements are each precisely 1/4 the wavelength of the applied RF energy.

A half wave dipole is not resonant because the capacitive and inductive reactances cancel out, because at resonance there is no reactance of any kind present because the AC voltage and current at the feed point are in phase.