A half-wave dipole isn't resonant, or is it?

"A half wave dipole is resonant."
"No, it isn't: it actually has an impedance of 73.1 + j42.5 ohms."
"To make it resonant, reduce its length by a little bit."

I decided to figure it out for myself.

Let us consider a center-fed dipole antenna of length $\ell$, where the radius of the conductor is $a$. Let us further assume that the resistivity of the antenna conductor is zero, and that it has an infinitesimal feed gap.

To express the results ahead, we will need the following trigonometric integrals: $$\renewcommand{\Cin}{\:\mathrm{Cin}\;} \renewcommand{\Si}{\:\mathrm{Si}\;} \renewcommand{\Ci}{\:\mathrm{Ci}\;} \Si {x} = \int_0^x\; \frac{\sin t} {t} dt \qquad\qquad \Cin {x} = \int_0^x\; \frac{1-\cos t} {t} dt \qquad\qquad \Ci x = -\int_x^\infty \frac{\cos t}{t} dt = \gamma + \ln x - \Cin x$$ ($\gamma \approx 0.5772...$ is the 'Euler-Mascheroni constant'.)

To express the dimensions in terms of the wavelength $\lambda$, we define:

$$\chi = \frac{2\pi\ell}{\lambda} \qquad\qquad \xi = \frac{4\pi a}{\lambda}$$

For the conditions $a \ll \lambda$, $a \ll \ell$ and $\frac{\lambda}{2}-\varepsilon \lt \ell \lt \frac{\lambda}{2}+\varepsilon$ (for small $\varepsilon$), the current distribution on the dipole is very close to sinusoidal. Using this assumption, the radiation resistance $R_{rad}$ and self-reactance $X_{self}$ of this antenna can be found to be as follows:

$$R_{rad}\left(\chi\right) \approx \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Cin\chi + \left( \Cin\chi- \frac{\Cin 2\chi}{2} \right)\cos\chi - \left( \Si\chi- \frac{\Si 2\chi}{2} \right)\sin\chi \right]$$ $$\begin{align} X_{self}\left(\chi,\xi\right) &\approx \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Si\chi + \left( \Si\chi- \frac{\Si 2\chi}{2} \right)\cos\chi + \left( \Cin\chi-\frac{\Cin 2\chi}{2} - \ln\chi + \ln\xi \right)\sin\chi \right] \\ &= \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Si\chi + \left( \Si\chi-\frac{\Si 2\chi}{2} \right)\cos\chi - \left( \Ci\chi - \frac{\Ci 2\chi}{2} + \ln\sqrt {\chi} - \ln {\xi} \sqrt {\frac {e^\gamma}{2}} \right) \;\sin\chi \right] \\ \end{align}$$

(The derivation for this can be found in section 22.3 in this book.)

For the half wave dipole, $\ell=\lambda/2$, so $\chi=\pi$. Hence, for all $\xi \gt 0$, we have:

$$R_{rad} = \cfrac {\Cin 2\pi} {4\pi} \eta_0 \approx 73.1\;\Omega$$ $$X_{self} = \cfrac {\Si 2\pi} {4\pi} \eta_0 \approx 42.5\;\Omega$$

This tells us a couple of things:

• As $\sin\pi=0$, the impedance of a half-wave dipole is independent of its thickness.
• Since an antenna is resonant when it has zero reactance, the half-wave dipole is not resonant!

Okay, so a dipole is resonant when it is slightly less than $\lambda / 2$ in length. Can we figure out what that length is?

For all other values of $\ell$ close to $\cfrac{\lambda}{2}$, we can use the identity $\csc x + \cot x = \cot\cfrac{x}{2}$ to obtain:

$$R_{rad}\left(\chi\right) \approx \left( \cot \cfrac {\chi} {2} \;\Cin\chi - \cot\chi\cfrac{\Cin 2\chi}{2} - \Si\chi + \cfrac{\Si 2\chi}{2} \right) \cfrac {\eta_0}{\pi} \cot \cfrac{\chi}{2}$$ $$X_{self}\left(\chi,\xi\right) \approx \left( \cot \cfrac {\chi} {2} \;\Si\chi - \cot\chi\cfrac{\Si 2\chi}{2} - \Ci\chi + \cfrac{\Ci 2\chi}{2} - \ln\sqrt {\chi} + \ln {\xi} + \ln \sqrt {\frac {e^\gamma}{2}} \right) \cfrac {\eta_0}{\pi} \cot \cfrac{\chi}{2}$$

The resonant lengths $\ell$ of a dipole for any given conductor radius $a$ correspond to the roots $\chi$ of the following equation for a given $\xi$: $$X_{self}\left(\chi,\xi\right) = 0$$

Since we know that $\chi=\pi$ is not a root, the equation can be re-written as:

$$\cot \cfrac {\chi} {2} \;\Si\chi - \cot\chi\cfrac{\Si 2\chi}{2} - \Ci\chi + \cfrac{\Ci 2\chi}{2} - \ln\sqrt {\chi} + \ln {\xi} + \ln \sqrt {\frac {e^\gamma}{2}} = 0$$

This does not help us analytically determine the resonant value of $\chi$ given $\xi$. It can be done numerically, and like most computations for antennas, this is what is normally done. However, it works well the other way around: to analytically determine the value of $\xi$ for which we have resonance for a given $\chi$.

$$\ln \xi = \cot\chi\cfrac{\Si 2\chi}{2} -\cot \cfrac {\chi} {2} \;\Si\chi + \Ci\chi - \cfrac{\Ci 2\chi}{2} + \ln\sqrt {\chi} - \ln \sqrt {\frac {e^\gamma}{2}}$$

A picture is worth a thousand words. Let us plot $\ln\xi$ against $\chi$ in the range $0 \lt \chi \lt 2\pi$.

The graph tells us (and we can verify analytically) that in the vicinity of $\chi=\pi$, we have:

• As $\chi \to \pi^+$, $\ln\xi \to +\infty$, or equivalently $\xi \to \infty$, and
• As $\chi \to \pi^-$, $\ln\xi \to -\infty$, or equivalently $\xi \to 0^+$

We started with the assumption that $a \ll \lambda$, which implies $\xi \ll 1$, so we need not bother with the $\xi \to \infty$ case. But the $\xi \to 0^+$ case nicely fits in with the $\xi \ll 1$ assumption, and tells us that for small values of $\xi$, resonant dipole antennas have $\chi$ that are somewhat less than $\pi$.

In other words, this confirms that a dipole is resonant when it is slightly less than $\lambda / 2$ in length!

But hang on, can we use this to investigate an infinitesimally thin dipole? Can we say that as $\chi \to \pi^-$ for $\xi \to 0^+$, the half-wave dipole is resonant in the limiting case of the infinitesimal thickness?

Unfortunately not: the approximation breaks down somewhere as the wire gets thinner. It says here (in section 16.3) that an infinitesimally thin, infinitely conductive dipole would actually be resonant for $\ell \approx 0.4857\lambda$, with a corresponding $R_{rad}=67.2\Omega$. Which seems to imply that the expressions for $R_{rad}$ and $X_{self}$ mentioned above hold for $a \ll \lambda$, but not $a \lll \lambda$. How interesting!

I'm still trying to understand exactly why the approximation does not hold for $a \lll \lambda$. The most obvious explanation is that we started out assuming a sinusoidal current distribution which is approximately, but not exactly, true.