Faculty of Information and
Engineering Technology
Dr. Tallal Elshabrawy
Bar Code
Wireless Communication (NETW 701)
Winter 2013
Midterm Examination
Nov. 18th 2013
Please read carefully before proceeding.
1. The duration of this exam is 2 hours
2. Only calculators are permitted for this exam
3. This exam booklet contains 14 pages including this one. Two extra sheets of
scratch papers are attached
Good Luck!
Marks:
Problem
Number
Possible Marks
Final Marks
1
2
3
4
5
Total
10
10
10
12.5
17.5
60
Problem 1:
For the power delay profile shown in the figure below,
Pr(τ)
0 dB
-10 dB
-20 dB
-30 dB
50
75
100
τ(nano seconds)
i. Calculate the mean excess delay
𝝉=
∑𝒌 𝑷(𝝉𝒌 )𝝉𝒌 𝟎 × 𝟏 + 𝟓𝟎 × 𝟏 + 𝟎. 𝟏 × 𝟕𝟓 +× 𝟎. 𝟎𝟏 × 𝟏𝟎𝟎
=
= 𝟐𝟕. 𝟕𝟐𝟓 𝒏𝒔
∑𝒌 𝑷(𝝉𝒌 )
𝟏 + 𝟏 + 𝟎. 𝟏 + 𝟎. 𝟎𝟏
ii. Calculate the Excess delay spread (15 dB)
𝑬𝒙𝒄𝒆𝒔𝒔 𝑫𝒆𝒍𝒂𝒚 𝑺𝒑𝒓𝒆𝒂𝒅 𝟏𝟓𝒅𝑩 = 𝟕𝟓 𝒏𝒔
iii. Calculate the rms delay spread
𝝉𝟐 =
∑𝒌 𝑷(𝝉𝒌 )𝝉𝒌 𝟐 𝟎𝟐 × 𝟏 + 𝟓𝟎𝟐 × 𝟏 + 𝟎. 𝟏 × 𝟕𝟓𝟐 +× 𝟎. 𝟎𝟏 × 𝟏𝟎𝟎𝟐
=
= 𝟏𝟒𝟗𝟖. 𝟖
∑𝒌 𝑷(𝝉𝒌 )
𝟏 + 𝟏 + 𝟎. 𝟏 + 𝟎. 𝟎𝟏
𝝈𝝉 = √𝝉𝟐 − (𝝉)𝟐 = 𝟐𝟕. 𝟎𝟐 𝒏𝒔
iv. Calculate the 50% correlation coherence bandwidth
𝑩𝒄 =
𝟏
= 𝟕. 𝟒𝟎𝟏𝟗𝟐 𝑴𝑯𝒛
𝟓𝝈𝝉
2
v. If it is required to send 16 QAM with a bit rate of 200 kbps. What type of
fading will the modulation undergo (For flat fading assume that the 10 x
signal bandwidth<coherence bandwidth)
𝑭𝒐𝒓 𝟏𝟔 𝑸𝑨𝑴 𝒘𝒊𝒕𝒉 𝒃𝒊𝒕 𝒓𝒂𝒕𝒆 𝒐𝒇 𝟐𝟎𝟎 𝑲𝒃𝒑𝒔, 𝒕𝒉𝒆 𝒔𝒚𝒎𝒃𝒐𝒍 𝒓𝒂𝒕𝒆 𝒊𝒔 𝟓𝟎 𝑲𝑺𝒚𝒎𝒃𝒐𝒍𝒔/𝒔
𝑰𝒏 𝒕𝒉𝒆 𝒑𝒂𝒔𝒔 𝒃𝒂𝒏𝒅, 𝒕𝒉𝒆 𝒃𝒂𝒏𝒅𝒘𝒊𝒅𝒕𝒉 𝒕𝒐 𝒕𝒓𝒂𝒏𝒔𝒎𝒊𝒕 𝟓𝟎𝑲𝑺𝒚𝒎𝒃𝒐𝒍𝒔/𝒔 𝒊𝒔 𝑩𝑺 = 𝟓𝟎 𝑲𝑯𝒛
𝑮𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝟏𝟎𝑩𝑺 < 𝐵𝑐, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑚𝑜𝑑𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑤𝑜𝑢𝑙𝑑 𝑢𝑛𝑑𝑒𝑟𝑔𝑜 𝑓𝑙𝑎𝑡 𝑓𝑎𝑑𝑖𝑛𝑔
vi. Given that the mobile is travelling at a speed of 120 Km/hr. Determine
whether the receiver shall experience slow or fast fading given that the
carrier frequency is 2 GHz and the system bandwidth is 10 MHz.
𝝀=
𝟑 × 𝟏𝟎𝟖
= 𝟎. 𝟏𝟓 𝒎
𝟐 × 𝟏𝟎𝟗
𝒇𝒎 =
𝒗 𝟏𝟐𝟎 × 𝟏𝟎𝟎𝟎/𝟑𝟔𝟎𝟎
=
𝟐𝟐𝟐. 𝟐𝟐 𝑯𝒛
𝝀
𝟎. 𝟏𝟓
𝑻𝒄 =
𝟎. 𝟒𝟐𝟑
= 𝟏. 𝟗 𝒎𝒔
𝒇𝒎
𝑻𝑺 = 𝟎. 𝟎𝟐 𝒎𝒔
𝑻𝒄 > 𝑻𝑺
𝑺𝒍𝒐𝒘 𝑭𝒂𝒅𝒊𝒏𝒈
3
Problem 2:
Assume communications of a signal of bandwidth 20 MHz to be transmitted over
the 2.4 GHz band within a large city. The Transmitter power is 20 W. Use the Hata
model to calculate the received power at the Receiver within a suburban
environment given that the transmitter height is 35m, the receiver height is 3m
and the distance separation between the transmitter and receiver is 1 Km.
𝑷𝑻 = 𝟐𝟎 𝑾 = 𝟏𝟑. 𝟎𝟏 𝒅𝑩
𝑳𝟓𝟎 (𝑼𝒓𝒃𝒂𝒏) = 𝟔𝟗. 𝟓𝟓 + 𝟐𝟔. 𝟐𝟔𝒍𝒐𝒈(𝒇𝒄 ) − 𝟏𝟑. 𝟖𝟐𝒍𝒐𝒈(𝒉𝒕𝒆 ) − 𝒂(𝒉𝒓𝒆 ) + (𝟒𝟒. 𝟗 − 𝟔. 𝟓𝟓𝒍𝒐𝒈(𝒉𝒕𝒆 ) )𝒍𝒐𝒈(𝒅)
𝒇𝒄 = 𝟐𝟒𝟎𝟎 𝑴𝑯𝒛
𝒉𝒕𝒆 = 𝟑𝟓 𝒎
𝒉𝒓𝒆 = 𝟑 𝒎
𝒅 = 𝟏 𝑲𝒎
𝑳𝟓𝟎 (𝑼𝒓𝒃𝒂𝒏) = 𝟔𝟗. 𝟓𝟓 + 𝟐𝟔. 𝟐𝟔𝒍𝒐𝒈(𝟐𝟒𝟎𝟎 ) − 𝟏𝟑. 𝟖𝟐𝒍𝒐𝒈(𝟑𝟓 ) − 𝒂(𝒉𝒓𝒆 ) + (𝟒𝟒. 𝟗 − 𝟔. 𝟓𝟓𝒍𝒐𝒈(𝟑𝟓) )𝒍𝒐𝒈(𝟏)
𝟐
𝒂(𝒉𝒓𝒆 ) = 𝟑. 𝟐(𝒍𝒐𝒈(𝟏𝟏. 𝟕𝟓𝒉𝒓𝒆 )) − 𝟒. 𝟗𝟕 𝒅𝑩
𝒇𝒐𝒓 𝒇𝒄 ≥ 𝟑𝟎𝟎𝑴𝑯𝒛
𝟐
𝒂(𝒉𝒓𝒆 ) = 𝟑. 𝟐(𝒍𝒐𝒈(𝟏𝟏. 𝟕𝟓 × 𝟑)) − 𝟒. 𝟗𝟕 𝒅𝑩
𝒂(𝒉𝒓𝒆 ) = 𝟐. 𝟔𝟗 𝒅𝑩
𝑳𝟓𝟎 (𝑼𝒓𝒃𝒂𝒏) = 𝟔𝟗. 𝟓𝟓 + 𝟖𝟖. 𝟕𝟔 − 𝟐𝟏. 𝟑𝟒 + 𝟐. 𝟔𝟗 + 𝟎 = 𝟏𝟑𝟗. 𝟔𝟔 𝒅𝑩
𝒇𝒄 𝟐
𝑳𝟓𝟎 (𝑺𝒖𝒃𝒖𝒓𝒃𝒂𝒏) = 𝑳𝟓𝟎 (𝑼𝒓𝒃𝒂𝒏) − 𝟐 [𝒍𝒐𝒈 ( )] − 𝟓. 𝟒
𝟐𝟖
𝟐𝟒𝟎𝟎 𝟐
𝑳𝟓𝟎 (𝑺𝒖𝒃𝒖𝒓𝒃𝒂𝒏) = 𝑳𝟓𝟎 (𝑼𝒓𝒃𝒂𝒏) − 𝟐 [𝒍𝒐𝒈 (
)] − 𝟓. 𝟒 = 𝟏𝟑𝟓. 𝟔 − 𝟔. 𝟖𝟕𝟒 − 𝟓. 𝟒 = 𝟏𝟐𝟔. 𝟕𝟖 𝒅𝑩
𝟐𝟖
𝑷𝑹 = 𝟏𝟑. 𝟎𝟏 − 𝟏𝟐𝟔. 𝟕𝟖 = −𝟏𝟏𝟑. 𝟕𝟕 𝒅𝑩 = −𝟖𝟑. 𝟕𝟕 𝒅𝑩𝒎 = 𝟒. 𝟐 𝒑𝑾
4
Problem 3:
Assume that local average signal strength field measurements were made inside a
building, and post processing revealed that the measured data fit a distantdependent mean power law model having a log-normal distribution about the
mean. Assume the mean power law was found to be Pr(d) 𝛼 d-3.5. If a signal of
1mW was received at do=1m from the transmitter, and at a distance of 10m, 10%
of the measurements were stronger than -25 dBm, define the standard deviation,
σ, for the path loss model at d=10 m.
𝑷𝒓 (𝑷𝑹 > −25) = 𝟎. 𝟏
𝑷𝒓 (𝑷𝑹 (𝒅𝒐 ) + 𝟏𝟎𝒏𝒍𝒐𝒈 (
𝒅𝒐
) + 𝑿𝝈 > −25) = 𝟎. 𝟏
𝒅
𝟏
𝑷𝒓 (𝟏𝟎 𝒍𝒐𝒈(𝟏𝒎𝑾) + 𝟏𝟎 × 𝟑. 𝟓 (𝒍𝒐𝒈 ( ) + 𝑿𝝈 > −25) = 𝟎. 𝟏
𝟏𝟎
𝑷𝒓 (𝟎 − 𝟑𝟓 + 𝑿𝝈 > −25) = 𝟎. 𝟏
𝑷𝒓 (𝑿𝝈 > 10) = 𝟎. 𝟏
𝟏𝟎
𝑸 ( ) = 𝟎. 𝟏
𝝈
From Q-Table: 𝑸(𝟏. 𝟐𝟗) = 𝟎. 𝟏
𝟏𝟎
= 𝟏. 𝟐𝟗
𝝈
𝝈 = 𝟕. 𝟕𝟔 𝒅𝑩
5
Problem 4:
The figure below shows a system where one obstacle is between the transmitter
and receiver. A repeater is inserted in the middle such that it amplifies the power
by 10 dB before retransmitting. If PT = 10W, GT = 10 dB, GR = 3 dB, GT-Rep = 10
dB, GR-Rep = 3 dB, compute the received power. (fc= 2 GHz, Assume between the
transmitter and the repeater that the two-ray model approximation is valid.
Assume that all power received at the receiver is coming from the repeater)
h2 =75m
ht =50m
hr =50m
hrepeater =50m
1 Km
2 Km
1 Km
2 Km
𝑷𝑻 = 𝟏𝟎𝑾, 𝑮𝑻 = 𝟏𝟎𝒅𝑩, 𝑮𝑹 = 𝟑𝒅𝑩 , 𝑮𝑻−𝑹𝒆𝒑 = 𝟏𝟎𝒅𝑩, 𝑮𝑹−𝑹𝒆𝒑 = 𝟑𝒅𝑩
𝒇𝒄 = 𝟐 𝑮𝑯𝒛
𝝀 = 𝟎. 𝟏𝟓 𝒎
𝑷𝑹−𝑹𝒆𝒑 =
𝑷𝑻 𝑮𝑻 𝑮𝑹−𝑹𝒆𝒑 (𝒉𝒕 𝒉𝒓 )𝟐 𝟏𝟎 × 𝟏𝟎 × 𝟏𝟎𝟎.𝟑 (𝟓𝟎 × 𝟓𝟎)𝟐
=
(𝟑𝟎𝟎𝟎)𝟒
𝒅𝟒
𝑷𝑹−𝑹𝒆𝒑 = 𝟏. 𝟓𝟒 × 𝟏𝟎−𝟓 = −𝟒𝟖. 𝟏𝟑 𝒅𝑩
𝑷𝑻−𝑹𝒆𝒑 = 𝑷𝑹−𝑹𝒆𝒑 + 𝟏𝟎 = −𝟑𝟖. 𝟏𝟑 𝒅𝑩
𝑷𝑹−𝑳𝑶𝑺 =
𝑷𝑻−𝑹𝒆𝒑 𝑮𝑻−𝑹𝒆𝒑 𝑮𝑹
(𝟒𝝅𝒅\𝝀𝟐 )
𝑷𝑹−𝑳𝑶𝑺 =
𝟏. 𝟓𝟒 × 𝟏𝟎−𝟒 × 𝟏𝟎 × 𝟏𝟎𝟎.𝟑
= 𝟒. 𝟖𝟔 × 𝟏𝟎−𝟏𝟒 = −𝟏𝟑𝟑. 𝟏𝟑 𝒅𝑩
(𝟒𝝅 × 𝟑𝟎𝟎𝟎\𝟎. 𝟏𝟓𝟐 )
6
𝝂 = 𝒉√
𝟐(𝒅𝟏 + 𝒅𝟐 )
𝝀𝒅𝟏 𝒅𝟐
𝝂 = 𝒉√
𝟐 × 𝟑𝟎𝟎𝟎
= 𝟑. 𝟓𝟒
𝟎. 𝟏𝟓 × 𝟏𝟎𝟎𝟎 × 𝟐𝟎𝟎𝟎
𝑮𝑫𝒊𝒇𝒇 = −𝟐𝟑 𝒅𝑩
𝑷𝑹 = 𝑷𝑹−𝑳𝑶𝑺 − 𝑮𝑫𝒊𝒇𝒇 = −𝟏𝟑𝟑. 𝟏𝟑 − 𝟐𝟑 = −𝟏𝟓𝟔. 𝟏𝟑 𝒅𝑩
7
Problem 5:
Given a corridor as shown below where the walls behave as perfect reflectors and
the electric field changes in the z-direction (perpendicular to the x-y plane). Two
transmitters are installed as shown that are set to be 15 m apart and the receiver
is 5 m away from the transmitter Tx1. The carrier frequency is 2.4 GHz
Er11
1m
2m
Er21
1m
Tx1
Rx
ELOS1
1m
ELOS2
Tx2
5m
Er22
Er12
y
x
15 m
Derive an exact expression for the total amount of received power at the receiver
while considering the line-of-sight ray, rays that are reflected once before
reaching the receiver (Do not make any approximations).
𝑨𝒔 𝒔𝒉𝒐𝒘𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒇𝒊𝒈𝒖𝒓𝒆 𝒕𝒉𝒆𝒓𝒆 𝒘𝒊𝒍𝒍 𝒃𝒆 𝟔 𝒓𝒂𝒚𝒔;
− 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒐𝒇 𝒔𝒊𝒈𝒉𝒕 𝒇𝒓𝒐𝒎 𝑻𝒙𝟏 𝒘𝒊𝒕𝒉 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒊𝒆𝒍𝒅 𝑬𝑳𝑶𝑺𝟏
− 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒐𝒇 𝒔𝒊𝒈𝒉𝒕 𝒇𝒓𝒐𝒎 𝑻𝒙𝟐 𝒘𝒊𝒕𝒉 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒊𝒆𝒍𝒅 𝑬𝑳𝑶𝑺𝟐
− 𝒕𝒘𝒐 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅 𝒓𝒂𝒚𝒔 𝒅𝒖𝒆 𝒕𝒐 𝑻𝒙𝟏 𝒘𝒊𝒕𝒉 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒊𝒆𝒍𝒅𝒔 𝑬𝒓𝟏𝟏 , 𝑬𝒓𝟏𝟐
− 𝒕𝒘𝒐 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅 𝒓𝒂𝒚𝒔 𝒅𝒖𝒆 𝒕𝒐 𝑻𝒙𝟐 𝒘𝒊𝒕𝒉 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒊𝒆𝒍𝒅𝒔 𝑬𝒓𝟐𝟏 , 𝑬𝒓𝟐𝟐
𝑬𝑻𝑶𝑻 = 𝑬𝑳𝑶𝑺𝟏 + 𝑬𝑳𝑶𝑺𝟐 + 𝟐𝑬𝒓𝟏𝟏 + 𝟐𝑬𝒓𝟐𝟏 (𝑨𝒍𝒍 𝒐𝒇 𝒕𝒉𝒆𝒎 𝒉𝒂𝒗𝒆 𝒕𝒉𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒊𝒆𝒍𝒅 𝒊𝒏 𝒕𝒉𝒆 𝒛 − 𝒅𝒊𝒓. )
8
𝑭𝒐𝒓 𝑳𝑶𝑺𝟏 :
𝒅𝑳𝑶𝑺𝟏 = 𝟓 𝒎
𝑭𝒐𝒓 𝑳𝑶𝑺𝟐 :
𝒅𝑳𝑶𝑺𝟐 = 𝟏𝟎 𝒎
𝑭𝒐𝒓 𝑬𝒓𝟏𝟏 :
𝒅𝟏𝟏 = 𝟐√𝟏𝟐 + 𝟐. 𝟓𝟐 = 𝟓. 𝟑𝟖𝟓 𝒎
𝑭𝒐𝒓 𝑬𝒓𝟐𝟏 :
𝒅𝟐𝟏 = 𝟐√𝟏𝟐 + 𝟓𝟐 = 𝟏𝟎. 𝟏𝟗𝟖 𝒎
𝚫𝑳𝑶𝑺𝟏−𝑳𝑶𝑺𝟐 = 𝟏𝟎 − 𝟓 = 𝟓 𝒎
𝚫𝑳𝑶𝑺𝟏−𝒓𝟏𝟏 = 𝟓. 𝟑𝟖𝟓 − 𝟓 = 𝟎. 𝟑𝟖𝟓 𝒎
𝚫𝑳𝑶𝑺𝟏−𝒓𝟐𝟏 = 𝟏𝟎. 𝟏𝟗𝟖 − 𝟓 = 𝟓. 𝟏𝟗𝟖 𝒎
𝛉𝑳𝑶𝑺𝟏−𝑳𝑶𝑺𝟐 =
𝟐𝝅 × 𝟓
= 𝟎𝟎
𝟎. 𝟏𝟐𝟓
𝛉𝑳𝑶𝑺𝟏−𝒓𝟏𝟏 =
𝟐𝝅 × 𝟎. 𝟑𝟖𝟓
= 𝟐𝟖. 𝟖𝟎
𝟎. 𝟏𝟐𝟓
𝛉𝑳𝑶𝑺𝟏−𝒓𝟐𝟏 =
𝟐𝝅 × 𝟓. 𝟏𝟗𝟖
= 𝟐𝟏𝟎. 𝟐𝟒𝟎
𝟎. 𝟏𝟐𝟓
9
𝑭𝒐𝒓 𝑬𝒓𝟏𝟏 , 𝛉𝑳𝑶𝑺𝟏−𝒓𝟏𝟏 𝒅𝒖𝒆 𝒕𝒐 𝒏𝒐𝒓𝒎𝒂𝒍 𝒊𝒏𝒄𝒊𝒅𝒆𝒏𝒄𝒆 𝒉𝒂𝒔 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍 𝒑𝒉𝒂𝒔𝒆 𝒔𝒉𝒊𝒇𝒕 𝒐𝒇 𝟏𝟖𝟎𝟎 ,
𝛉𝑳𝑶𝑺𝟏−𝒓𝟏𝟏 = 𝟐𝟎𝟖. 𝟖𝟎
𝑭𝒐𝒓 𝑬𝒓𝟐𝟏 , 𝛉𝑳𝑶𝑺𝟏−𝒓𝟐𝟏 𝒅𝒖𝒆 𝒕𝒐 𝒏𝒐𝒓𝒎𝒂𝒍 𝒊𝒏𝒄𝒊𝒅𝒆𝒏𝒄𝒆 𝒉𝒂𝒔 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍 𝒑𝒉𝒂𝒔𝒆 𝒔𝒉𝒊𝒇𝒕 𝒐𝒇 𝟏𝟖𝟎𝟎 ,
𝛉𝑳𝑶𝑺𝟏−𝒓𝟐𝟏 = 𝟑𝟎. 𝟐𝟒𝟎
𝟐𝑬𝟎 𝒅𝟎
𝒅𝟐𝟏
𝑬𝟎 𝒅𝟎
𝒅𝑳𝑶𝑺𝟏
𝑬𝟎 𝒅𝟎
𝒅𝑳𝑶𝑺𝟐
𝟐𝑬𝟎 𝒅𝟎
𝒅𝟏𝟏
𝐄𝑻𝑶𝑻 =
𝑬𝟎 𝒅𝟎 𝟐𝑬𝟎 𝒅𝟎
𝟐𝑬𝟎 𝒅𝟎
𝑬𝟎 𝒅𝟎
(
+
𝒄𝒐𝒔(𝟐𝟎𝟖. 𝟖) +
𝒄𝒐𝒔(𝟑𝟎. 𝟐𝟒) +
)
𝒅𝑳𝑶𝑺𝟏
𝒅𝟏𝟏
𝒅𝟐𝟏
𝒅𝑳𝑶𝑺𝟐
√
𝟐𝑬𝟎 𝒅𝟎
𝟐𝑬𝟎 𝒅𝟎
+(
𝒔𝒊𝒏(𝟐𝟎𝟖. 𝟖) +
𝒔𝒊𝒏(𝟑𝟎. 𝟐𝟒))
𝒅𝟏𝟏
𝒅𝟐𝟏
𝟐
𝐄𝑻𝑶𝑻
𝟐
𝟐
𝟏 𝟐𝒄𝒐𝒔(𝟐𝟎𝟖. 𝟖) 𝟐𝒄𝒐𝒔(𝟑𝟎. 𝟐𝟒)
𝟏
𝟐𝒔𝒊𝒏(𝟐𝟎𝟖. 𝟖) 𝟐𝒔𝒊𝒏(𝟑𝟎. 𝟐𝟒)
= 𝑬𝟎 𝒅𝟎 √( +
+
+ ) +(
+
)
𝟓
𝟓. 𝟑𝟖𝟓
𝟏𝟎. 𝟏𝟗𝟖
𝟏𝟎
𝟓. 𝟑𝟖𝟓
𝟏𝟎. 𝟏𝟗𝟖
𝟐
𝐄𝑻𝑶𝑻 = 𝟎. 𝟏𝟔𝟒𝟖𝑬𝟎 𝒅𝟎
𝑷𝑹 =
(𝟏𝟔𝟒𝟖𝑬𝟎 𝒅𝟎 )𝟐 𝑮𝑹 𝝀𝟐
×
𝟏𝟐𝟎𝛑
𝟒𝝅
10