## Monday, April 10, 2017

### Solutions To Radio Astronomy Problems

1) Write an expression  for the degree of polarization given in terms of the normalized Stokes parameters..

Solution:

For regular Stokes' parameters we have:

d  =  Ö (Q2  +  U 2 +  V2) /  I

(For 0 < d < 1.)

We have for the normalized Stokes parameters:

0  =   I / S   =   1

1  =   Q / S

2  =   U / S

3  =   V  / S

So through the use of algebra we obtain:

d  =  Ö (s 1 2  +   s 2 2  +   s 3 2  )  /  s 0    =

Ö (s 1 2  +   s 2 2  +   s 3 2  )

2)  Write a matrix equation for a completely unpolarized radio wave using a left circularly polarized wave and a right circularly polarized wave.

Solution:

A completely unpolarized wave requires the result:  C =

½ [1…..0]
[0…..1]

Left circularly polarized wave: has:

½ [1…..j]
[-j…..1]

And:   Right circularly polarized wave: has:

½ [1…..-j]
[j…..1]

Then to get C  we need:

A + B  = C

Where: A  =

1/4 [1…..j]
[-j…..1]

B =

1/4  [1…..-j]
[j…..1]
.
Check by using matrix addition to add and you obtain matrix C

3) The coherency matrix of some individual radio wave is given by:

½ [1…..0]

[0…..0]

Show how the  resultant unpolarized wave's coherency matrix may be obtained by showing the C-matrix for the other wave needed to combine with the individual wave above.

Solution:

The  resultant unpolarized wave's coherency matrix would be:  C=

½ [1…..0]
[0…..1]

The given matrix which needs a complementary matrix to obtain the above is:  A =

½ [1…..0]

[0…..0]

Then matrix subtraction, e.g.   C - A  yields B =

½ [0…..0]

[0…..1]

Which is the matrix needed to obtain C.

4) Four radio waves are detected and analyzed and found to have the characteristics shown below:

a) d = 0

b) d =  ½      AR = 4     and   t=  135 deg

c)  d =  ½      AR = 4     and   t=  - 135 deg

d)  d =  ½      AR = 4     and   t=  45 deg

Find the normalized Stokes parameters and the coherency matrices for these waves.

Solutions:

a) d = 0  denotes a completely unpolarized wave so

Stokes parameters:   [1..0..0..0]

Coherency matrix:  C =

½ [1…..0]

[0…..1]

b)   d =  ½      AR = 4     and   t=  135 deg

We have for the normalized Stokes components:

=   1

1    =  d  cos 2t cose

2    =  d  sin2t cose

3    =  d  sin2e

And:   cose  =   (AR2    -  1) / (AR2   +  1)

=  (42    -  1) / (42   +  1)   =   15/  17

And:   cos 2   =   cos (2 x 135) = cos 270 =  0

sin2t    =   sin (2 x 135) = sin  270 =  -1

Then Stokes parameters for  these characteristics:

[1..0..0....  -1./2]

Coherency matrix:  C =

½ [1…..-j]

[j…..1/2]

(c) is analogous in solution to (b)  except t=  - 135 deg

Then Stokes parameters for  these characteristics:

[1..0..0.... ½]

Coherency matrix: C =

½ [1…..j]

[-j…..½]

d) d =  ½      AR = 4     and   t=  45 deg

cose  =   (AR2    -  1) / (AR2   +  1)

=  (42    -  1) / (42   +  1)   =   15/  17

And:   cos 2   =   cos (2 x 45) = cos 90 =  0

sin2t    =   sin (2 x 45) = sin 90 =  1

Then Stokes parameters for  these characteristics:

[1..0..0.... ½]

Coherency matrix: C =

½ [1…  ..j]

[-j.....½]