Lectures Notes
Written By
Professor Dr. Saad Saffah
1
(
)
(
)

 
 
 1  2
 1  2
E  H  dS = −  J  E dv −    E dv −    H dv


t 2 
t 2 
▪ The cross product in the left-hand side can be written as following and known as
instantaneous Poynting vector
  
P = EH
▪ UPW power transmission, apply Poynting theorem to UPW. WE can easily show that
the time-averaged power density in a plane wave is



*
1
Pave = Re Es  H s
2

▪ The power through a surface is then


P =  Pave  dS
2
▪ If we have a plane wave traveling in lossy media, the spatial part of wave is:

−z − jz j 
Es = E x 0 e e e a x
▪ The intrinsic impedance can be complex and can be written in the form of
 = e
j 
▪ The magnetic field is then


E x 0 −z − jz j − j 
1
H s = a p  Es =
e e e e ay


Then The Poynting vector can be expressed as:
2

1 (Exo ) − 2z

Pave =
e
cos az
2 
3
▪ The polarization of a UPW describes the shape and locus
of the tip of E-vector at a given point in space as function
of time

j y jt − jkz 
jx jt − jkz 
E t , z = E xo e e e
a x + E yo e e e
ay
( )

j (t − kz + y ) 
j (t − kz +x ) 
E (t , z ) = E xo e
a x + E yo e
ay
▪ If we observe the E-field at a fix point, say… z = 0 , we have:
(
)



j ( t + y ) 
j ( t +x ) 
E (t ,0 ) = E xo e
a x + E yo e
a y = E xo cos( t +  x )a x + E yo cos  t +  y a y
▪ In an x-y plane, the above equation represent parametric equations describing various
shape of circle, ellipse, or straight line. These all depends on the phase in two axis and
amplitudes. If we choose a proper reference point for the time,
▪ we can set
x = 0, and  = y - x , so

j ( t ) 
j ( t + ) 
E (t ,0) = E xo e
a x + E yo e
ay
4
5
If the phases x and y are in phase or
completely out of phase
x =  y
x =  y  
(
)
in − phase
(
)
out − of − phase



E (z, t ) = E xo a x + E yo a y e j ( t −kz )



E (z, t ) = E xo a x − E yo a y e j ( t −kz )
6
▪ If Exo = Eyo  a, and the phase different between
two E-field components are
 =  y − x = 

2
▪ Here we have two situations:
A- Left-handed circular (LHC) polarization
 =  y − x = +

2
Right-handed circular (RHC) polarization
 =  y − x = −

2
7
▪ In most general cases, we have Ex0  Ey0, and
elliptically polarized.
  0, , or  /2. The wave is said to be
8
9
10
11