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Tutorial 5 - Wireless Communications

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Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
Tutorial (5)
Problem (1):
Calculate the RMS delay spread for an HF radio channel for which:
𝑷(𝝉) = 𝟎. πŸ”πœΉ(𝝉) + 𝟎. πŸ‘πœΉ(𝝉 βˆ’ 𝟎. 𝟐) + 𝟎. 𝟏𝜹(𝝉 βˆ’ 𝟎. πŸ’) where 𝝉 is measured in milliseconds.
Determine whether a 5 KHz signal would experience flat fading or frequency selective fading.
πœΜ… =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ (0.6 × 0 + 0.3 × 0.2 + 0.1 × 0.4) × 10βˆ’3
=
= 0.1π‘šπ‘ π‘’π‘
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
0.6 + 0.3 + 0.1
Μ…Μ…Μ…
𝜏2 =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜2 (0.6 × 02 + 0.3 × 0.22 + 0.1 × 0.42 ) × 10βˆ’6
=
= 0.028(π‘šπ‘ π‘’π‘)2
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
0.6 + 0.3 + 0.1
= 0.028 × 10βˆ’6 𝑠 2
𝜎𝜏 = βˆšΜ…Μ…Μ…
𝜏 2 βˆ’ (πœΜ…)2 = (√0.028 βˆ’ (0.1)2 ) × 10βˆ’3 = 0.13416π‘šπ‘ π‘’π‘
∡ 𝐼𝑑 𝑖𝑠 π‘π‘Žπ‘ π‘ π‘π‘Žπ‘›π‘‘ π‘β„Žπ‘Žπ‘›π‘›π‘’π‘™ ∴ 𝐡𝑠 = 𝑅𝑠 =
10𝜎𝜏 = 1.3416π‘šπ‘ π‘’π‘
∴ 𝑇𝑠 < 10𝜎𝜏 β†’βˆ΄ πΉπ‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦ 𝑠𝑒𝑙𝑒𝑐𝑑𝑖𝑣𝑒
1
1
β†’ 𝑇𝑠 =
= 0.2 π‘šπ‘ π‘’π‘
𝑇𝑠
5 × 103
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
Problem (2):
For the power delay profile shown in the figure below,
Pr(Ο„)
0 dB
-10 dB
-20 dB
-30 dB
0
50
75
100
Ο„(nano seconds)
i. Calculate the mean excess delay
πœΜ… =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 10βˆ’1 × 0 + 1 × 50 + 10βˆ’1 × 75 + 10βˆ’2 × 100
=
= 48.347𝑛𝑠𝑒𝑐.
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
10βˆ’1 + 1 + 10βˆ’1 + 10βˆ’2
N.B.: The power must be in Watt to calculate πœΜ…
ii. Calculate the delay spread
2
βˆ’1 ×02 +1×502 +10βˆ’1 ×752 +10βˆ’2 ×1002
βˆ‘ 𝑃(𝜏 )𝜏
10
Μ…Μ…Μ…
𝜏 2 = βˆ‘π‘˜ 𝑃(πœπ‘˜ )π‘˜ =
π‘˜
π‘˜
10βˆ’1 +1+10βˆ’1 +10βˆ’2
= 2613.636(𝑛𝑠𝑒𝑐)2
𝜎𝜏 = βˆšΜ…Μ…Μ…
𝜏 2 βˆ’ (πœΜ…)2 = √2613.636 βˆ’ (48.347)2 = 16.619𝑛𝑠𝑒𝑐
iii. Calculate the 90% correlation coherence bandwidth
𝐡𝐢 =
1
= 1.203𝑀𝐻𝑧
50𝜎𝜏
iv. If it is required to send 16 QAM with a bit rate of 600 kbps. What type of fading will the
modulation undergo (For flat fading assume that the 10 x signal bandwidth<coherence
bandwidth)
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
∡ 𝑇𝑠 = π‘šπ‘‡π‘
∴ 𝑅𝑠 =
(π‘€β„Žπ‘’π‘Ÿπ‘’ π‘š 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑏𝑖𝑑𝑠 π‘π‘’π‘Ÿ π‘ π‘¦π‘šπ‘π‘œπ‘™)
𝑅𝑏 600 × 103
=
= 150 × 103 π‘ π‘¦π‘šπ‘π‘œπ‘™π‘  /𝑠𝑒𝑐
π‘š
4
(∡ 16𝑄𝐴𝑀 ∴ π‘™π‘œπ‘”2 16 = 4𝑏𝑖𝑑𝑠/π‘ π‘¦π‘šπ‘π‘œπ‘™)
∡ 𝑄𝐴𝑀 π‘šπ‘œπ‘‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› (𝑆𝑖𝑛𝑒 π‘Žπ‘›π‘‘ πΆπ‘œπ‘ π‘–π‘›π‘’) ∴ π‘ƒπ‘Žπ‘ π‘ π‘π‘Žπ‘›π‘‘ π‘ π‘–π‘”π‘›π‘Žπ‘™
∴ 𝐡𝑠 = 𝑅𝑠 = 150 × 103 𝐻𝑧
10𝐡𝑠 = 1500 π‘˜π»π‘§ = 1.5𝑀𝐻𝑧
∡ 𝐡𝑐 = 1.203𝑀𝐻𝑧
∴ 10𝐡𝑠 > 𝐡𝑐
∴ πΉπ‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦ 𝑠𝑒𝑙𝑒𝑐𝑑𝑖𝑣𝑒 π‘“π‘Žπ‘‘π‘–π‘›π‘”
N.B: When it is required to know the type of fading, you must state in the solution whether it
is flat or frequency selective and also, whether it is slow or fast fading. But, in this problem, v
is not given, so it is not possible to get Tc and know whether it is slow or fast fading.
Problem (3):
For the power delay profile shown in the figure below,
Pr(Ο„)
0 dB
-10 dB
-20 dB
-30 dB
20
60
100
Ο„(nano seconds)
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
i. Calculate the mean excess delay
πœΜ… =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 0 × 1 + 20 × 0.1 + 60 × 0.1 + 100 × 0.01
=
= 7.438𝑛𝑠𝑒𝑐
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
1 + 0.1 + 0.1 + 0.01
ii. Calculate the delay spread
2
2 ×1+202 ×0.1+602 ×0.1+1002 ×0.01
βˆ‘ 𝑃(𝜏 )𝜏
0
Μ…Μ…Μ…
𝜏 2 = βˆ‘π‘˜ 𝑃(πœπ‘˜ )π‘˜ =
π‘˜
π‘˜
1+0.1+0.1+0.01
= 413.22(𝑛𝑠𝑒𝑐)2
𝜎𝜏 = βˆšΜ…Μ…Μ…
𝜏 2 βˆ’ (πœΜ…)2 = 18.918 𝑛𝑠𝑒𝑐
iii. Calculate the 90% correlation coherence bandwidth
𝐡𝐢 =
1
= 1.0572 𝑀𝐻𝑧
50𝜎𝜏
iv. Given that an M-QAM modulation scheme would be adopted, determine the minimum
value of M such that flat fading communication is achieved given that the required bit rate
is 600 kbps. (For flat fading assume that 10 x signal bandwidth<coherence bandwidth)
∡ π‘π‘Žπ‘ π‘ π‘π‘Žπ‘›π‘‘
∴ 𝐡𝑠 = 𝑅𝑠 =
𝑅𝑏
π‘š
10𝐡𝑠 < 𝐡𝑐
10 × (
𝑅𝑏
) < 𝐡𝑐
π‘š
π‘š
1
>
10𝑅𝑏 𝐡𝑐
π‘š>
10𝑅𝑏
𝐡𝑐
π‘š > 5.675
π‘šπ‘šπ‘–π‘› = 6 𝑏𝑖𝑑𝑠 π‘π‘’π‘Ÿ π‘ π‘¦π‘šπ‘π‘œπ‘™ β†’ 𝑀 = 2π‘š = 64 𝑄𝐴𝑀
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
Problem (4):
If a baseband binary message with a bit rate Rb = 100kbps is modulated by an RF carrier using
BPSK, answer the following:
i.
Find the range of values required for the rms delay spread of the channel such that the
received signal is a flat fading signal.
π‘“π‘œπ‘Ÿ π‘“π‘™π‘Žπ‘‘ π‘“π‘Žπ‘‘π‘–π‘›π‘”: 𝑇𝑠 > 10𝜎𝜏
∡ 𝐡𝑃𝑆𝐾
∴ 𝑀 = 2π‘ π‘¦π‘šπ‘π‘œπ‘™π‘ 
𝑇𝑠 = π‘šπ‘‡π‘ = 𝑇𝑏 =
10πœ‡ > 10𝜎𝜏
∴ π‘š = log 2 𝑀 = log 2 2 = 1𝑏𝑖𝑑 /π‘ π‘¦π‘šπ‘π‘œπ‘™
1
= 10 πœ‡π‘ 
𝑅𝑏
1πœ‡ > 𝜎𝜏
0 ≀ 𝜎𝜏 < 1 × 10βˆ’6
ii.
If the modulation carrier frequency is 5.8GHz, what is the coherence time of the
channel, assuming a vehicle speed of 30km/hr?
𝑇𝑐 =
0.423
π‘“π‘š
𝑣 = 30 ×
πœ†=
1000
25
=
π‘š/𝑠
60 × 60
3
𝑐
3 × 108
3
=
=
π‘š
9
𝑓𝑐 5.8 × 10
58
π‘“π‘š = 𝑓𝑑,π‘šπ‘Žπ‘₯
𝑣
= =
πœ†
25
3
3
58
= 161.11 𝐻𝑧
0.423
𝑇𝑐 =
= 2.626 × 10βˆ’3 𝑠 = 2.626 π‘šπ‘ 
161.11
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
iii.
Calculate the received carrier frequency if the receiver is moving towards the
transmitter.
∡The receiver is moving towards the transmitter β†’ πœƒ = 0° Doppler shift is positive
𝑓 = 𝑓𝑐 + 𝑓𝑑
𝑓𝑑 =
𝑣
π‘π‘œπ‘ (πœƒ)
πœ†
𝑓𝑑,π‘šπ‘Žπ‘₯ =
𝑣
= 161.11 𝐻𝑧
πœ†
𝑓 = 5.8 × 109 + 161.11 = 5.800000161 𝐻𝑧
iv.
Is the channel fast or slow fading?
∡ 𝑇𝑠 β‰ͺ 𝑇𝑐 β†’ Slow fading
Problem (5):
Determine the maximum and minimum spectral frequencies received from a stationary GSM
transmitter that has a center frequency of exactly 1950MHz, assuming that the receiver is
traveling at speeds of: a) 1Km/hr; b) 100Km/hr.
𝒗 = 𝟏 π‘²π’Ž/𝒉𝒓
πœ†=
𝑐
3 × 108
2
=
=
π‘š
6
𝑓𝑐 1950 × 10
13
𝑣 = 1×
1000
5
=
π‘š/𝑠
60 × 60 18
𝑣
π‘π‘œπ‘ (πœƒ)
πœ†
5
𝑣 18
𝑓𝑑,π‘šπ‘Žπ‘₯ = = 2 = 1.81 𝐻𝑧
πœ†
𝑓𝑑 =
13
π‘“π‘šπ‘–π‘› = 𝑓𝑐 βˆ’ 𝑓𝑑,π‘šπ‘Žπ‘₯ = 1950 × 106 βˆ’ 1.81 = 1949.9999998 𝑀𝐻𝑧
π‘“π‘šπ‘Žπ‘₯ = 𝑓𝑐 + 𝑓𝑑,π‘šπ‘Žπ‘₯ = 1950 × 106 + 1.81 = 1950.000002 𝑀𝐻𝑧
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
𝒗 = 𝟏𝟎𝟎 π‘²π’Ž/𝒉𝒓
πœ†=
𝑐
3 × 108
2
=
=
π‘š
6
𝑓𝑐 1950 × 10
13
𝑣 = 100 ×
1000
500
=
π‘š/𝑠
60 × 60
18
𝑣
π‘π‘œπ‘ (πœƒ)
πœ†
500
𝑣
𝑓𝑑,π‘šπ‘Žπ‘₯ = = 18
2 = 181 𝐻𝑧
πœ†
𝑓𝑑 =
13
π‘“π‘šπ‘–π‘› = 𝑓𝑐 βˆ’ 𝑓𝑑,π‘šπ‘Žπ‘₯ = 1950 × 106 βˆ’ 181 = 1949.999819 𝑀𝐻𝑧
π‘“π‘šπ‘Žπ‘₯ = 𝑓𝑐 + 𝑓𝑑,π‘šπ‘Žπ‘₯ = 1950 × 106 + 181 = 1950.000181 𝑀𝐻𝑧
Problem (6):
The TU-6 Channel (Typical Urban 6 Path Channel) is a popular multipath channel model used in
wireless telecommunications. The power delay profile information of the TU-6 Channel is in the
table shown below:
Multipath
Delay
Power
Power
Number
(ΞΌs)
(dB)
(Lin)
1
0
-3
0.501
2
0.2
0
1
3
0.5
-2
0.631
4
1.6
-6
0.251
5
2.3
-8
0.158
6
5.0
-10
0.1
Suppose that the total system bandwidth is 5 MHz and the carrier frequency is 2 GHz, how many
FDMA channels could be supported such that each channel exhibits a flat fading like behavior
For flat fading like behavior, each FDMA channel must be at maximum equal to the coherence
bandwidth. Let’s calculate the coherence bandwidth
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
πœΜ… =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 0 × 0.0501 + 0.2 × 1 + 0.5 × 0.631 + 1.6 × 0.251 + 2.3 × 0.158 + 5 × 0.1
=
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
0.0501 + 1 + 0.631 + 0.251 + 0.158 + 0.1
πœΜ… =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 1.7805
=
= 0.6742 πœ‡π‘ 
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
2.641
Μ…Μ…Μ…
𝜏2 =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 2 0 × 0.0501 + 0.22 × 1 + 0.52 × 0.631 + 1.62 × 0.251 + 2.32 × 0.158 + 52 × 0.1
=
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
0.0501 + 1 + 0.631 + 0.251 + 0.158 + 0.1
Μ…Μ…Μ…
𝜏2 =
βˆ‘π‘˜ 𝑃(πœπ‘˜ )πœπ‘˜ 2 4.1761
=
= 1.5813 (πœ‡π‘ )2
βˆ‘π‘˜ 𝑃(πœπ‘˜ )
2.641
𝜎𝜏 = βˆšΜ…Μ…Μ…
𝜏 2 βˆ’ (πœΜ…)2 = 1.0615 πœ‡π‘ 
Using 90% Correlation Bandwidth
5×106
1
𝐡𝐢 = 50𝜎 = 18.84 𝐾𝐻𝑧  18.84×103 = 265.39 β‰ˆ 265 𝐹𝐷𝑀𝐴 πΆβ„Žπ‘Žπ‘›π‘›π‘’π‘™π‘ 
𝜏
Using 50% Correlation Bandwidth
5×106
1
𝐡𝐢 = 5𝜎 = 188.4 𝐾𝐻𝑧  188.4×103 = 26.539 β‰ˆ 26 𝐹𝐷𝑀𝐴 πΆβ„Žπ‘Žπ‘›π‘›π‘’π‘™π‘ 
𝜏
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