1. The general integral that gives the beam solid angle is equation (2-142) !!

!There is no 𝜙 dependence to the given antenna pattern, so the 𝜙 integral can be done immediately, giving a factor of 2π. The integral is then split up, separating out the three non-zero regions. !

!

!Do the integration. The integral of sine is minus cosine. We can absorb the sign by swapping the limits. !

!

!Evaluating the cosine function. !

!

!Plugging in the numbers, we have !

! !The directivity is then

!

ΩA = F θ ,φ( ) 2 dΩsphere∫∫ = F θ ,φ( ) 2 sinθ dθ

0

π

∫ dφ0

2π

∫

ΩA = 2π sinθ dθ0

π /6

∫ + 13

⎛⎝⎜

⎞⎠⎟2

sinθ dθπ /3

2π /3

∫ + 12

⎛⎝⎜

⎞⎠⎟2

sinθ dθ5π /6

π

∫⎡

⎣⎢

⎤

⎦⎥

ΩA = 2π cos0 − cosπ6

⎛⎝⎜

⎞⎠⎟ +

19cosπ

3− cos 2π

3⎛⎝⎜

⎞⎠⎟ +

14cos 5π

6− cosπ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

ΩA = 2π 1− 32

⎛⎝⎜

⎞⎠⎟+ 1912+ 12

⎛⎝⎜

⎞⎠⎟ +

14

− 32

+1⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2π 5

41− 3

2⎛⎝⎜

⎞⎠⎟+ 19

⎡

⎣⎢⎢

⎤

⎦⎥⎥

ΩA ! 1.750

D = 4πΩA

! 7.179

2. We can start from the expressions in the solution of 6.3-8. For the parasitic element to function as the best possible director, we need to maximize the array factor magnitude in the forward direction !

! !To reach this maximum, the two terms must have the same phase, modulo 2π !

! Evaluating the argument function !

!

!Solve for d and substitute the expression for β !

!

!The shortest (positive) distance is

!

!but integral additions of λ also work. Similarly, for the parasitic element to function as the best possible reflector, we need to maximize the array factor magnitude in the backward direction !

! !Following from above, we find

!

!Solving for d

!

!and the shortest distance is !

!

maximize AF 0( ) = maximize I0 + I1e jβd

arg I1ejβd( ) = arg I0( )+ 2πn

arg I1( )+ arg e jβd( ) = arg I0( )+ 2πn

arg j2

⎛⎝⎜

⎞⎠⎟ + arg e jβd( ) = arg 1( )+ 2πn

π2+ βd = 2πn

d = n − 14

⎛⎝⎜

⎞⎠⎟ λ

d = 34λ

maximize AF π( ) = maximize I0 + I1e− jβd

π2− βd = 2πn

d = 14− n⎛

⎝⎜⎞⎠⎟ λ

d = 14λ

3. The source points we want lie on a circle of radius a in the x-y plane, spaced apart by an angle of π/3. Noting that the first point has an angle of π/6, we subtract 1/2 from the index variable. !

! !

!If this is not obvious, consider Euler’s formula !

! !The radial unit vector has the usual expansion in cartesian coordinates !

! !We just need the dot product to compute the far-field approximate distances !

!

!The expression for the individual distances is also acceptable. Evaluating the trigonometric functions we have !

!

′rn = a cos n − 12

⎛⎝⎜

⎞⎠⎟π3

⎡⎣⎢

⎤⎦⎥x + sin n − 1

2⎛⎝⎜

⎞⎠⎟π3

⎡⎣⎢

⎤⎦⎥y

⎧⎨⎩

⎫⎬⎭

e jψ = cosψ + j sinψ

r = sinθ cosφx + sinθ sinφy + cosθ z

Rn = r − ′rn ⋅ r = r − asinθ cos n − 12

⎛⎝⎜

⎞⎠⎟π3

⎡⎣⎢

⎤⎦⎥cosφ + sin n − 1

2⎛⎝⎜

⎞⎠⎟π3

⎡⎣⎢

⎤⎦⎥sinφ

⎧⎨⎩

⎫⎬⎭

R1 = r − asinθ32cosφ + 1

2sinφ

⎧⎨⎩

⎫⎬⎭

R2 = r − asinθ sinφ{ }

R3 = r − asinθ − 32cosφ + 1

2sinφ

⎧⎨⎩

⎫⎬⎭

R4 = r − asinθ − 32cosφ − 1

2sinφ

⎧⎨⎩

⎫⎬⎭

R5 = r − asinθ −sinφ{ }

R6 = r − asinθ32cosφ − 1

2sinφ

⎧⎨⎩

⎫⎬⎭