The Complex Amplitude: Making Calculations Easier
If you worked through problem 3 of the last section you realize how cumbersome it is to keep
track of phase relationships using trig functions. Since we will have to do a lot of this sort of
thing in practical computations in optics, it is important to find an easier method. Luckily, there
is one is available.
The method is based on the properties of complex numbers. If you need a refresher on these,
please look over Appendix B now. We will assume a familiarity with this material in the
following discussion.
The property that makes complex numbers so very useful in physics is the equation
jθ
(1) e = cosθ + j sinθ
j
so that cos = Re(e ). This allows an oscillating electric field to be written as the real part of a
complex number:
(2) E0 cos(ω t ) = E0 Re( e jωt ) = E0 Re( e − jωt )
Physicists generally prefer to use - t in the exponent while engineers tend to prefer + t. We will
stick with the physicist’s convention.
More generally, we can include phases by using
(3)
E0 cos(ω t + φ ) = Re( E0 e − j (ω t +φ ) )
= Re( E0 e − jφ e− j (ωt ) ) = Re( E 0 e − j ( ωt ) )
where E0 = E0 e-j is called the complex amplitude. The utility of this notation stems from the
fact that the energy density is proportional to the square of the electric field:
uE = ½ 0 (E0 cos( t + ))2 = ½ 0 E02 cos2( t + ) , so that
(4)
uE =
ε0 E02
2
cos (ω t + φ ) =
2
ε0 E02
4
=
ε0
4
E0
2
where E denotes the modulus of E. In words, the energy density is proportional to the squared
modulus of the complex amplitude.
Another important property is that the complex amplitude of a superposition is the sum of the
individual complex amplitudes. For example, if the net electric field can be written as
(5) E ( t ) = E1 cos(ω t + φ1 ) + E2 cos(ω t + φ2 )
Then
E ( t ) = Re( E1 e− j (ω t +φ1 ) ) + Re( E2 e − j ( ω t +φ2 ) )
(6)
= Re( E1 e − j (ω t +φ1 ) + E2 e − j (ω t +φ2 ) ) = Re(( E1 e − j φ1 + E2 e − jφ2 ) e − jω t )
= Re(( E1 + E 2 ) e − jω t )
so that, if we write E(t) = Re(Ee-j t), we have
(7)
E = E1 e − jφ1 + E2 e − jφ2 = E1 + E 2
4.1
Combining this with eq. (4) gives
4 uE
2
= E = E1 + E 2 = ( E1 * + E 2 *)( E1 + E 2 )
2
ε0
2
2
2
2
= E 1 + E 2 + ( E 1 * E 2 + E 2 * E1 )
= E1 + E 2 + 2 Re( E1 * E 2 )
(8)
= E12 + E22 + 2 Re( E1 E2 e − j ( φ2 − φ1 ) )
= E12 + E22 + 2 E1 E2 cos(φ2 − φ1 )
For the special case where E1 = E2 E0 , this can be rewritten with the help of the identity
1 + cos = 2 cos2( /2)
to get equation (5) of the last section.
The time-averaged energy density is given directly in terms of the complex amplitudes without
the need for performing a further time average.
solved problem: Find the average electric field energy density for the three-generator problem
using complex amplitudes.
solution: Define E p =
V0 − j φp
where p = 1, 2, or 3.
e
d
= E1 + E 2 + E 3 = ( E1 * + E 2 * + E 3 *)( E1 + E 2 + E 3 )
4 uE
2
ε0
= E1 + E 2 + E 3 + 2 Re ( E1 * E 2 + E1 * E 3 + E 2 * E 3 )
2
2
2
2
V 
=  0  [ 3 + 2 Re( e j ( φ1 −φ2 ) + e j ( φ1 −φ3 ) + e j ( φ2 −φ3 ) )]
d
2
V 
=  0  [ 3 + 2 cos(φ1 − φ2 ) + 2 cos(φ1 − φ3 ) + 2 cos(φ2 − φ3 ) ]
d
This is the same as the answer found for problem 3 of the last section, obtained here with
4.2
considerably less labor.
More generally, for any number of superposed electric fields oscillating at the same frequency,
we have
(9)
E ( t ) = ∑ E p cos(ω t + φp ) ⇒ E = ∑ E p e
p
− j φp
p
and
4 uE
ε0
(10)
2
=
∑E
= ∑Ep +
2
p
p
p
∑E



=  ∑E p *  ∑E q 
 p
 q

p
* Eq
p,q
p ≠q
Problem 1: Show that equation (10) implies that
uE


ε0 

=  ∑ E p 2 + 2 ∑ E p Eq cos(φp − φq )
4 p

distinct pairs


( p, q )
4.3
= ∑Ep
p