4.4. Additional exercises

Exercise 4.7

In figure… the radar-based path-average rainfall intensity \(R\) [mm/h] for a microwave link path is shown. Compute the maximum rainfall intensity for the peak around 23:15 UTC in this figure using the received signal level of the commercial microwave link. As you can see from the figure, \(P_\mathrm{ref} = -46.4\) dBm and \(P_\mathrm{min} = -88.5\) dBm. Use (4.4), \(P_\mathrm{max} = -50.1\) dBm, \(A_\mathrm{a} = 1.30\) dB, \(a = 7.19\) and \(\alpha = 0.334\). Estimate the missing variables from Fig. 4.17 and Fig. 4.25.

Fig. 4.25 The value of the exponent \(b\) [-] as a function of frequency \(f\) [GHz] of the microwave signal for the commercial links used in the study of Overeem et al. (2011)

Answer Exercise 4.7

From figure…: The link length is 7.7 km and the frequency 25.4 GHz.

From Fig. 4.25: A frequency of 25.4 GHz leads to a value of \(b\) of about 0.90.

\[ \langle R \rangle = 0.334 \times 7.19 \times \left( \frac{-46.4 - -88.5 - 1.30}{7.7} \right) ^{0.90} \]

\[ \ldots+ (1-0.334) \times 7.19 \times \left( \frac{-46.4 - -50.1 - 1.30}{7.7} \right)^{0.90} \]

\[ \ldots = 10.77 + 1.68 = 12.5\ \mathrm{mm/h} \]

Exercise 4.8

Compute the semivariance between two raingauges and use Fig. 4.19 to estimate the distance between the two stations.

\(R_1 \: [\text{mm}]\)	\(R_2 \: [\text{mm}]\)
10.0	12.2
3.8	5.1
6.5	2.2
0	0.1
3.8	1.4
12.0	16.5

Remember that: \(Var(X) = \frac{1}{n-1} \cdot \sum_{i=1}^{n} (x_{i}-\mu)^2\).

Answer Exercise 4.8

Compute the difference \(R_1 - R_2\):

\(R_1\)	\(R_2\)	\(R_1 - R_2\)
10.0	12.2	-2.2
3.8	5.1	-1.3
6.5	2.2	4.3
0	0.1	-0.1
3.8	1.4	2.4
12.0	16.5	-4.5

The mean of \(R_1 - R_2\) is:

\[ \mu = \frac{-2.2-1.3+4.3-0.1+2.4-4.5}{6} = -0.23 \]

The variance of \(R_1 - R_2\) is:

\[ Var(X) = \frac{1}{n-1} \cdot \sum_{i=1}^{n} (x_{i}-\mu)^2 \]

\[\begin{split} Var(X) = \frac{1}{5} \cdot [(-2.2--0.23)^2 + (-1.3--0.23)^2 + (4.3--0.23)^2 + (-0.1--0.23)^2 \\ + (2.4--0.23)^2 + (-4.5--0.23)^2] = 10.1 \: \text{mm}^2 \end{split}\]

Using (4.5):

\[ \gamma(d) = \frac{1}{2} Var[R(p_1)-R(p_2)] = \frac{1}{2} \cdot 10.1 = 5.1 \: \text{mm}^2 \]

Read from Fig. 4.19: \(5 \: \text{mm}^2\) corresponds to about 10 km. Of course this type of statistics should be performed with much larger datasets.

Exercise 4.9

Rainfall can be measured by means of (weather) radar.

a) The so-called Marshall-Palmer relation (\(Z=200R^{1.6}\)) is often used to relate radar reflectivity (\(Z\), mm\(^6\)/m\(^{3}\)) and rainfall intensity (\(R\), mm/h). Suppose a weather radar measures a reflectivity of 1000 mm\(^6\)/m\(^{3}\). What is the corresponding rainfall rate? And what is the rainfall rate for \(Z\)=100,000 mm\(^6\)/m\(^{3}\)?

b) Since \(Z\) can vary strongly over orders of magnitude while this is not the case for \(R\), it is common to express \(Z\) in decibels (dBZ). This is a logarithmic scale defined as \(10 \times \log(Z)\), with \(\log\) the logarithm with base 10. Suppose it rains with an intensity of 10 mm/h, what reflectivity (in dBZ) will be measured by the weather radar?

Answer Exercise 4.9a

\[ R = (Z / 200) ^{1 / 1.6} = 5 ^{0.625} = 2.7 \text{ mm/h} \]

\[ R = 500 ^{0.625} = 48.6 \text{ mm/h} \]

Answer Exercise 4.9b

\(Z = 200 R ^{1.6}\), so \(10\log{Z} = 10\log{200} + 16\log{R} = 23 + 16 = 39 \text{ dBZ}\)