EE/Ae 157a
Homework #5
Due Date: December 3, 2015
The microwave temperature of a soil surface covered by a vegetation canopy can be written as
T  1  i  Tg e  Tc 1  e   iTc 1  e  e
where Tg and Tc are the temperatures of the soil and the canopy, respectively. The subscript i
refers to the polarization (either horizontal or vertical). The optical depth  of the canopy is a
function of the vegetation water content according to
  bWc
The constant b is a vegetation opacity coefficient that is determined experimentally and has a
value of approximately 0.1 at L band.
For dawn (6am local time) orbit overpasses, the soil and vegetation temperatures are
approximately equal to the surface air temperature, and the brightness temperature can be
written:
T  Ta 1  i e 2 
where Ta is the surface air temperature. Assume that the soil has a dielectric constant of 15.
Calculate the expected change in microwave temperature for a 5% change in soil dielectric
constant, and plot this as a function of the vegetation water content.
Solution:
By far the simplest way to do this problem is to calculate the brightness temperature for the
two cases of the dielectric constant directly from the equation
T  Ta 1  i e 2 
A 5% change in the dielectric constant means the dielectric constant changes from 15 to 15.75.
The reflection coefficient for horizontal polarization is
𝜌ℎ (𝜃) = |
𝑐𝑜𝑠𝜃 − √𝜖 − 𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠𝜃 + √𝜖 − 𝑠𝑖𝑛2 𝜃
2
|
And for vertical polarization, it is
𝜌𝑣 (𝜃) = |
𝜖𝑐𝑜𝑠𝜃 − √𝜖 − 𝑠𝑖𝑛2 𝜃
𝜖𝑐𝑜𝑠𝜃 + √𝜖 − 𝑠𝑖𝑛2 𝜃
2
|
Here, 𝜖 is the dielectric constant of the surface. We need to calculate these for theta equal to
zero and 45 degrees. The results are
𝜌ℎ (0)
0.3476
0.3570
𝜖
15.00
15.75
𝜌ℎ (45)
0.4717
0.4808
𝜌𝑣 (0)
0.3476
0.3570
To calculate the percent change in the brightness temperature, we note that
100
∆𝑇
𝑇(𝜖 = 15) − 𝑇(𝜖 = 15.75)
= 100 {
}
𝑇
𝑇(𝜖 = 15)
Using the expression for the brightness temperature, this can be written as
∆𝑇 𝑒 −2𝑏𝑊𝑐 (𝜌(𝜖 = 15.75, 𝜃) − 𝜌(𝜖 = 15. , 𝜃))
100
=
𝑇
1 − 𝜌(𝜖 = 15. , 𝜃)𝑒 −2𝑏𝑊𝑐
So, for 𝜃 = 0, we find
100
∆𝑇
0.0094𝑒 −2𝑏𝑊𝑐
=
𝑇
1 − 0.3476𝑒 −2𝑏𝑊𝑐
And for 𝜃 = 45, there are two cases:
100
∆𝑇ℎ
0.0091𝑒 −2𝑏𝑊𝑐
=
𝑇ℎ
1 − 0.4717𝑒 −2𝑏𝑊𝑐
100
∆𝑇𝑣
0.0087𝑒 −2𝑏𝑊𝑐
=
𝑇𝑣
1 − 0.2225𝑒 −2𝑏𝑊𝑐
The plot of these changes is shown on the next page.
𝜌𝑣 (45)
0.2225
0.2311
2.0
Percent Change in Brightness Temperature
1.8
1.6
1.4
1.2
DTV0
1.0
DTV45
DTH45
0.8
DTh0
0.6
0.4
0.2
0.0
0
1
2
3
Vegetation Water Content
4
5