Worked example 7.3.1: The feed of an S-band air-traffic-control radar

$f_{co,10}=\frac{c}{2a}=\frac{3\times 10^8}{0.14428}=2.079\mathrm{GHz}$ and $s=\sqrt{1-(2.079/2.8{)}^2}=0.670$. With $\lambda =10.71\mathrm{cm}$:

$3.1$ times below the breakdown field.

$R_s=13.8\mathrm{m\Omega }$ at $2.8\mathrm{GHz}$, $2b/a=0.944$ and $(f_c/f{)}^2=0.551$:

Over $40\mathrm{m}$ the loss is $0.85\mathrm{dB}$, so $0.823$ of the power, $823\mathrm{kW}$ peak, arrives. The average power is $1\mathrm{MW}\times 10^{-6}\mathrm{s}\times 1000{\mathrm{s}}^{-1}=1\mathrm{kW}$, of which $17.7\%$, that is $177\mathrm{W}$, heats the guide.

$E_{\max }=1.5MV/m$:

a quarter of the $9.82\mathrm{MW}$ at the full breakdown field.

$(3/0.958{)}^2=9.8$: the breakdown power is $9.8$ times the transmitted $1\mathrm{MW}$, in agreement with $9.82\mathrm{MW}$. Units: $E^{2}ab/Z$ is $(V/m{)}^2{\mathrm{m}}^2/\Omega =\mathrm{W}$.