Proofreading

_psets/6.md

@@ -147,7 +147,7 @@ $$
 \begin{align*}
 E_\text{max} &= \left( 2 \sqrt{\frac{\mu_0}{\epsilon_0}} \langle P \rangle \right)^\frac{1}{2} \\
 &= \sqrt{2 \cdot 377 \si{\ohm} \cdot \num{8e-5} \si{W/m^2}} \\
-&= 0.24 \si{W/m}
+&= 0.24 \si{V/m}
 \end{align*}
 $$
 
@@ -161,13 +161,16 @@ By Ohm's Law $$V = I (R_r + R_l)$$ (where $$R_r$$ is the radiation resistance, a
 resistance). Thus $$I = V / (R_r + R_l)$$ and the power dissipated in the load resistance is
 
 $$
-P = I^2 R = \frac{V^2 R_l}{(R_r + R_l)^2}
+\begin{align*}
+P &= I^2 R \\
+&= \frac{V^2 R_l}{(R_r + R_l)^2}
+\end{align*}
 $$
 
-This is maximized where its derivative is zero.
+To maximize this we take the derivative
 
 $$
-\frac{d P}{d R_l}  V^2 ( (R_r + R_l)^{-2} - 2 R_l (R_r + R_l)^{-3} )
+\frac{d P}{d R_l} = V^2 ( (R_r + R_l)^{-2} - 2 R_l (R_r + R_l)^{-3} )
 $$
 
 Setting this to zero implies $$R_r + R_l = 2 R_l$$, or
@@ -226,7 +229,7 @@ $$
 A &= \frac{V^2}{4 R} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} \\
 &= \frac{V^2}{4} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}}
 \frac{3}{2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\
-&= \frac{3}{8} \frac{V^2}{E^2_\text{max}} \left( \frac{\lambda}{d} \right)^2
+&= \frac{3}{8 \pi} \frac{V^2}{E^2_\text{max}} \left( \frac{\lambda}{d} \right)^2
 \end{align*}
 $$
 
@@ -239,9 +242,18 @@ $$
 Now we need to relate the voltage and the maximum magnitude of the electric field. The whole point
 of the antenna is to pick up oscillating electric and magnetic fields, and the presence of the
 latter makes voltage path dependent. So let's integrate along our (infinitely thin) antenna. Our
-antenna has an infinitesimal length $$d$$, so the field will be constant across it. Thus $$d$$, $$V
-= E_\text{max} d$$. As such
+antenna has an infinitesimal length $$d$$, so the field will be constant across it. Thus the
+integral is just $$d$$ times the electric field, i.e. $$V = E_\text{max} d$$. As such
+
+$$
+A = \frac{3}{8 \pi} \lambda^2
+$$
+
+Finally we see that the area gain ratio is
 
 $$
-A = \frac{3}{8} \lambda^2
+\begin{align*}
+\frac{A}{G} &= \frac{3}{8 \pi} \lambda^2 \frac{2}{3} \\
+&= \frac{1}{4 \pi} \lambda^2
+\end{align*}
 $$