The Short Dipole Antenna Loss Resistance

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How do we derive the equation for the loss resistance of a short dipole antenna? We'll try to do that on this page.

Let's start with the equation for the d.c. (0 Hz) resistance (R) of a material with resistivity (), length L and cross sectional area A:
resistance of a block


The reason this equation is only valid at d.c. is because at higher frequencies, the current does not flow uniformly through the cross sectional area A. As a result, the loss resistance increases, because the current is actually flowing through a smaller area than the full cross section. This concept is known as skin depth, and is discussed on the skin depth page.

If we want to calculate the power dissipated as a result of the finite-conductivity (or the resistivity) of a material with electric current flowing through it, then we can use equation [2]:

power lossed through a resistor


In equation [2], R is the total resistance, and I is the current that flows through the material. For the case of the short dipole antenna, we have a current that drops off linearly with distance:

current distribution on short dipole antenna

Figure 1. Current Distribution on a Short Dipole Antenna.

To calculate the total power lost from this current, we use integration:

total power lost through resistance


In the above equation, dR is the resistance of a small section of the antenna. We can approximate the current along a small section as constant, and so we can rewrite dR as:

differential loss resistance


In equation [4], we rewrote the cross sectional area as the skin depth times the circumference of the wire (2*pi*a). Does that make sense? If not, think about it for a while and prove it to yourself.

We can rewrite/shift the electric current distribution of I(x) into:

current for short dipole


The reason Equation [5] was rewritten this way is in order to simply integrate the first section, and then multiply the result by 2. [We can avoid doing 2 integrals, because they will both produce the same value for loss.]

The power dissipation equation then becomes:

starting loss resistance derivation


We can substitute Equation [4] and I(x) to get:

continuing loss resistance derivation


Then we figure out the total dissipate power as:

almost done with loss resistance derivation


Now we've gotten somewhere. If we substitute in the skin depth:

adding in skin depth


Euqation [8] then becomes:

finishing power dissipation


If we set the power dissipated equal to an imaginary loss resistance (as if this  was a simple ac circuit with a current flowing through of magnitude I0), we get:

finishing power dissipation


Now, I'm missing a factor of a square root of 2 in there [sqrt(2)], and honestly I don't know where it comes from.  But all the other terms are right, and you can see all the ideas behind how it is derived.

Hopefully you understand why the loss increases with higher frequency (skin depth), why the loss is different than if we were measuring a block of material, and why the result is proportional to the square root of resistivity, instead of resistivity to the first power.

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