Expert: Philip A. Stahl - 2/28/2008

Question
QUESTION: Help! Can you show me how to do at least one of these problems?

Three radio waves from different objects in space have the following characteristics, where d is the degree of polarization and AR denotes the axial ratio of polarization ellipse. Find the normalized Stokes parameters and the coherency matrix for each:

i) d = 1/2,   AR= 4,  tau (angle) = 135 deg
ii) d = 1 , AR = -4,  tau = pi/4
iii) d = 1, AR = -1

Thanks!

ANSWER: Hello,

Well, I am not going to work any of those problems for you, but will set up another example and full solution, which you can then choose to model your own solutions on.

For this case, we will consider a wave with d = 0 (unpolarized).

We know the axial ratio is a function of left and right electric polarization amplitudes (E_R and E_L) so:

AR = (E_L + E_R)/ (E_L – E_R)  = cot (Z)

Obviously, if the wave is unpolarized then the amplitudes, E_R =  E_L,= 0 so: AR = 0

In general the normalized Stokes parameters will always be a combination of  contributions such that:

S [s_i] = S[s] +  S{s’]

Where [s_i] is as before (see previous problem on received spectral power):

(so)
(s1)
(s2)
(s3)

and s =

[1- d]
[0]
[0]
[0]

and s’ =

[d]
[d cos(2Z) cos(2t)]
[d cos(2Z)sin(2t)]
[d sin(2Z)]

where the latter column vector (matrix) discloses the contribution for partial polarization such that:

cos (2Z) = (AR^2 – 1)/ (AR^2 + 1)

and:  sin(2Z) = 2AR/ (AR^2 + 1)


Now – since d = 0 for an unpolarized wave (and hence NO partial polarization occurs)
Then the second column matrix term drops out from the S eqn.above  – then we have for the normalized Stokes parameters in this case:

(so)
(s1)
(s2)
(s3)

=

[1]
[0]
[0]
[0]


Note that the three last elements of the matrix are 0, confirming that AR = 0 as we suspected

Now the coherency matrix is obtained via the new elements, such that:

[C] =

(s11…..s12)
(s21….s22)

and:

s11 = ½ (s0 + s1)

s12 = ½ (s2 + js3)

s21 = 1/2 (s2 – js3)

s22 =  ½( s0 – s1)

Thus, substituting the quantities into the matrix [C] we have, [C] =
(1/2 ……..0)
(0 ………1/2)

or:

½    [1…..0]
    [0…..1]

Now, it is also possible to write the final result as the sum of a left-circularly polarized wave and a right one, since the terms in j will cancel. Thus:

[C] =

¼  x

(1 ….j)
(-j….1)

+   ¼  x

(1 ….-j)
(j….1)

And doing the matrix addition, produces the same result as before, [C]=

(1/2 ……..0)
(0 ………1/2)

Given this example, you ought to be able to work out the 3 problems you list. If you still encounter difficulties, get back to me and I will provide more hints.  Note again, that for nearly all practical cosmic wave cases there will be partial polarization and the Stokes function S[s_i] will reduce to the 2nd matrix terms in Z, t.














---------- FOLLOW-UP ----------

QUESTION: Ok, I worked the last problem (part iii) and can you tell me if I have it done correctly? Then I will do the other two!

d = 1, AR = -1

so cos(2Z) = (AR^2 – 1)/ (AR^2 + 1) =  (1 - 1)/ (1 + 1)= 0

sin(2Z)=  2AR/ (AR^2 + 1) = 2(-1)/ [(-1)^2 + 1]

= -2/ 2 = -1

so the wave is right-hand polarized and Z = -90 degrees

The normalized Stokes parameters are then:

s0 =1

s1=  d cos(2Z)cos (2t) = 0,

s2 = d cos(2Z) sin (2t) = 0

s3 = d sin (2Z) = 1(-1) = -1

so:

s0= 1,  s1 = 0, s2 = 0,  s3 = -1

The coherence matrix elements are:

s11 = ½(s0 + s1) = ½(1 + 0) = ½

s12 = ½(s2 + js3) = 1/2(j (-1)) = -j/2

s21= ½ (s2 – j s3) = ½(j) = j/2

s22 = ½(s0 – s1) = ½(1) = ½

So the coherence matrix is:

[1/2…..-j/2]
[j/2……1/2]


Answer
Hello,

Very good! You have solved part (iii) correctly. Just bear in mind that the coherency matrix (also called a "Jones matrix" in some optical applications) is more often written:

C =

1/2 [1…..-j]
....[j….. 1]

Also, remember that j, -j denote unit imaginary numbers (otherwise also written i, -i).

Good work!