diff --git a/_psets/6.md b/_psets/6.md index bb1422a5e27be9cea04ddff7665ce79a4c0c4610..611e574f3d70881c0adfc9d216df46466a34df9e 100644 --- a/_psets/6.md +++ b/_psets/6.md @@ -79,3 +79,67 @@ $$ {:.question} For an infinitesimal dipole antenna, what are the gain and the area, and what is their ratio? + +From the text we know that + +$$ +\langle |P| \rangle += \frac{I_0^2 k^2 d^2}{32 \pi^2 r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \sin^2 \theta +$$ + +and + +$$ +W = \frac{I_0^2 \pi}{3} \sqrt{\frac{\mu_0}{\epsilon_0}} \left( \frac{d}{\lambda} \right)^2 +$$ + +Thus + +$$ +\begin{align*} +G &= \max_{\theta, \phi} \frac{\langle |P(r = 1, \theta, \phi)| \rangle}{W/4 \pi} \\ +&= 4 \pi \frac{I_0^2 k^2 d^2}{32 \pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}} +\frac{3}{I_0^2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\ +&= \frac{(2 \pi \lambda^{-1})^2}{8 \pi^2} 3 \lambda^2 \\ +&= \frac{3}{2} +\end{align*} +$$ + +since $$k = 2 \pi / \lambda$$. + +To find the area, recall that the maximum power received is $$V^2 / 8 R$$. This will be equal to +the area times the Poynting vector. Thus + +$$ +\begin{align*} +\frac{V^2}{8 R} &= \langle | P | \rangle A \\ +&= \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E^2_\text{max} A +\end{align*} +$$ + +Solving for $$A$$, + +$$ +\begin{align*} +A &= \frac{V^2}{4 R} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} \\ +&= \frac{V^2}{4} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} +\frac{3}{2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\ +&= \frac{3}{8} \frac{V^2}{E^2_\text{max}} \left( \frac{\lambda}{d} \right)^2 +\end{align*} +$$ + +since for an infinitesimal dipole radiator + +$$ +R = \frac{2 \pi}{3} \sqrt{\frac{\mu_0}{\epsilon_0}} \left( \frac{d}{\lambda} \right)^2 +$$ + +Now we need to relate the voltage and the maximum magnitude of the electric field. The whole point +of the antenna is to pick up oscillating electric and magnetic fields, and the presence of the +latter makes voltage path dependent. So let's integrate along our (infinitely thin) antenna. Our +antenna has an infinitesimal length $$d$$, so the field will be constant across it. Thus $$d$$, $$V += E_\text{max} d$$. As such + +$$ +A = \frac{3}{8} \lambda^2 +$$