# Sol - Faculty of Information Engineering & Technology ```Faculty of Information and
Engineering Technology
Dr. Tallal Elshabrawy
Bar Code
Wireless Communication (NETW 701)
Winter 2013
Midterm Examination
Nov. 18th 2013
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 &times; 𝟏 + 𝟓𝟎 &times; 𝟏 + 𝟎. 𝟏 &times; 𝟕𝟓 +&times; 𝟎. 𝟎𝟏 &times; 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 &times; 𝟏 + 𝟓𝟎𝟐 &times; 𝟏 + 𝟎. 𝟏 &times; 𝟕𝟓𝟐 +&times; 𝟎. 𝟎𝟏 &times; 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&lt;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 &lt; 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 &times; 𝟏𝟎𝟖
= 𝟎. 𝟏𝟓 𝒎
𝟐 &times; 𝟏𝟎𝟗
𝒇𝒎 =
𝒗 𝟏𝟐𝟎 &times; 𝟏𝟎𝟎𝟎/𝟑𝟔𝟎𝟎
=
𝟐𝟐𝟐. 𝟐𝟐 𝑯𝒛
𝝀
𝟎. 𝟏𝟓
𝑻𝒄 =
𝟎. 𝟒𝟐𝟑
= 𝟏. 𝟗 𝒎𝒔
𝒇𝒎
𝑻𝑺 = 𝟎. 𝟎𝟐 𝒎𝒔
𝑻𝒄 &gt; 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 &times; 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 &gt; −25) = 𝟎. 𝟏
𝑷𝒓 (𝑷𝑹 (𝒅𝒐 ) + 𝟏𝟎𝒏𝒍𝒐𝒈 (
𝒅𝒐
) + 𝑿𝝈 &gt; −25) = 𝟎. 𝟏
𝒅
𝟏
𝑷𝒓 (𝟏𝟎 𝒍𝒐𝒈(𝟏𝒎𝑾) + 𝟏𝟎 &times; 𝟑. 𝟓 (𝒍𝒐𝒈 ( ) + 𝑿𝝈 &gt; −25) = 𝟎. 𝟏
𝟏𝟎
𝑷𝒓 (𝟎 − 𝟑𝟓 + 𝑿𝝈 &gt; −25) = 𝟎. 𝟏
𝑷𝒓 (𝑿𝝈 &gt; 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 &times; 𝟏𝟎 &times; 𝟏𝟎𝟎.𝟑 (𝟓𝟎 &times; 𝟓𝟎)𝟐
=
(𝟑𝟎𝟎𝟎)𝟒
𝒅𝟒
𝑷𝑹−𝑹𝒆𝒑 = 𝟏. 𝟓𝟒 &times; 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 &times; 𝟏𝟎−𝟒 &times; 𝟏𝟎 &times; 𝟏𝟎𝟎.𝟑
= 𝟒. 𝟖𝟔 &times; 𝟏𝟎−𝟏𝟒 = −𝟏𝟑𝟑. 𝟏𝟑 𝒅𝑩
(𝟒𝝅 &times; 𝟑𝟎𝟎𝟎\𝟎. 𝟏𝟓𝟐 )
6
𝝂 = 𝒉√
𝟐(𝒅𝟏 + 𝒅𝟐 )
𝝀𝒅𝟏 𝒅𝟐
𝝂 = 𝒉√
𝟐 &times; 𝟑𝟎𝟎𝟎
= 𝟑. 𝟓𝟒
𝟎. 𝟏𝟓 &times; 𝟏𝟎𝟎𝟎 &times; 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 &times; 𝟓
= 𝟎𝟎
𝟎. 𝟏𝟐𝟓
𝛉𝑳𝑶𝑺𝟏−𝒓𝟏𝟏 =
𝟐𝝅 &times; 𝟎. 𝟑𝟖𝟓
= 𝟐𝟖. 𝟖𝟎
𝟎. 𝟏𝟐𝟓
𝛉𝑳𝑶𝑺𝟏−𝒓𝟐𝟏 =
𝟐𝝅 &times; 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
&times;
𝟏𝟐𝟎𝛑
𝟒𝝅
10
```