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Example 6.9
Find the resultant field of equal-amplitude N-waves, each phase-shifted from
the preceding by the same ∆φ.

Solution: The problem consists of computing an expression of the following
kind:

(
˜
˜
˜
E ˜ = E + E +…+ E = E ( +1 e j∆φ + e j2∆φ +…+ e jN− ) 1 ∆φ ) (6.78)
tot . 1 2 n 0
We have encountered such an expression previously. This sum is that corre-
sponding to the sum of a geometric series. Computing this sum, the modulus
square of the resultant phasor is

2 (1− e jN ∆φ ) (1− e − jN ∆φ )
˜
E = E 2
tot . 0 (1− e j∆φ ) (1− e j − ∆φ )
(6.79)
 1− cos(N∆φ )  sin (N∆φ / )2 
2
= E 0   1− cos(∆φ )   = E 0   sin (∆φ / )2  
2
2
2
Because the source is the same for each of the components, the modulus of
each phasor is related to the source amplitude by E = E source /N. It is usually
0
as function of the source field that the results are expressed.

In-Class Exercises

Pb. 6.44 Plot the normalized square modulus of the resultant of N-waves as
a function of ∆φ for different values of N (5, 50, and 500) over the interval –π
< ∆φ < π.
Pb. 6.45 Find the dependence of the central peak value of Eq. (6.79) on N.
Pb. 6.46 Find the phase shift that corresponds to the position of the first
minimum of Eq. (6.79).
Pb. 6.47 Find in Eq. (6.79) the relative height of the first maximum (i.e., the
one following the central maximum) to that of the central maximum as a
function of N.
Pb. 6.48 In an antenna array with the field representing N aligned, equally
spaced individual antennae excited by the same source is given by Eq. (6.78).
If the line connecting the point of observation to the center of the array is
making an angle θ with the antenna array, the phase shift is ∆φ = 2 π dcos( θ),
λ

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