Problem 3.1  
A centre-fed Hertzian dipole is excited by a current I0 = 50 A. If the dipole is   λ/50   in length, determine  
the maximum radiated power density at a distance of 1 km

$$ P_{rad} = \frac{1}{2} \lvert I_o \lvert^2 R_r \\ R_r = 80 \pi^2 (\frac{1}{\lambda})^2 =80 \pi^2 (\frac{1}{50})^2 = 0.316 \ \Omega \\ \Rightarrow P_{rad} = \frac{1}{2} \lvert 50 \lvert^2 R_r = 395.78 \ W $$

$$ D = \frac{4 \pi r^2 S_{max}}{P_{rad}} \Rightarrow S_{max} = \frac{D \cdot P_{rad}}{4 \pi r^2} \\ D_{HerztianDipole} = \frac{3}{2} \Rightarrow S_{max} = 47.1 \cdot 10^{-6} \mu W / m^2 $$

A 50 cm long centre-fed dipole directed along the   z-direction radiates into air and is excited by a 1 MHz  
source. Assuming the current amplitude is   I0   = 20 A, determine:  
a) The power density radiated at   2 km   along the broadside of the antenna pattern.

>>> import math
>>> D = 1.5
>>> R_r = lambda l, lamb: 80 * math.pow(math.pi * (l / lamb), 2)
>>> length = 50e-2
>>> freq = 1e6
>>> lamb = 3e8 / freq
>>> I_o = 20
>>> R_r(length, lamb)
0.002193245422464302
>>> P_rad = lambda I_o, R_r: (1/2) * math.pow(I_o, 2) * R_r
>>> P_rad(I_o, R_r(length, lamb))
0.43864908449286044
>>>mP_rad = P_rad(I_o, R_r(length, lamb))
>>> S_max = lambda D, P_rad, r: (D * P_rad) / (4 * math.pi * math.pow(r, 2))
>>> D_HD = 3/2
>>> S_max(D_HD, mP_rad, 2e3)
1.3089969389957473e-08

$$ R_r = 0.002193245422464302 \ \Omega \\ P_{rad} = 0.43864908449286044 \ W \\ S_{max} = 1.3089969389957473e-08 $$

b) What is the power density radiated at   2 km   in the direction of   θ   = 45◦?

$$ S(\theta, \phi) = S_{max} \cdot \sin^2 \theta $$

>>> S_max(D_HD, mP_rad, 2e3)*math.pow(math.sin(45 * math.pi / 180), 2)
6.544984694978735e-09

c) The fraction of the total power radiated within the section between   θ   = 85◦   and   θ   = 95◦