# Tutorial 5 - Wireless Communications ```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. 𝟔𝜹(𝝉) + 𝟎. 𝟑𝜹(𝝉 − 𝟎. 𝟐) + 𝟎. 𝟏𝜹(𝝉 − 𝟎. 𝟒) where 𝝉 is measured in milliseconds.
Determine whether a 5 KHz signal would experience flat fading or frequency selective fading.
𝜏̅ =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 (0.6 × 0 + 0.3 × 0.2 + 0.1 × 0.4) × 10−3
=
= 0.1𝑚𝑠𝑒𝑐
∑𝑘 𝑃(𝜏𝑘 )
0.6 + 0.3 + 0.1
̅̅̅
𝜏2 =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘2 (0.6 × 02 + 0.3 × 0.22 + 0.1 × 0.42 ) × 10−6
=
= 0.028(𝑚𝑠𝑒𝑐)2
∑𝑘 𝑃(𝜏𝑘 )
0.6 + 0.3 + 0.1
= 0.028 × 10−6 𝑠 2
𝜎𝜏 = √̅̅̅
𝜏 2 − (𝜏̅)2 = (√0.028 − (0.1)2 ) × 10−3 = 0.13416𝑚𝑠𝑒𝑐
∵ 𝐼𝑡 𝑖𝑠 𝑝𝑎𝑠𝑠𝑏𝑎𝑛𝑑 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 ∴ 𝐵𝑠 = 𝑅𝑠 =
10𝜎𝜏 = 1.3416𝑚𝑠𝑒𝑐
∴ 𝑇𝑠 < 10𝜎𝜏 →∴ 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑒
1
1
→ 𝑇𝑠 =
= 0.2 𝑚𝑠𝑒𝑐
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(τ)
0 dB
-10 dB
-20 dB
-30 dB
0
50
75
100
τ(nano seconds)
i. Calculate the mean excess delay
𝜏̅ =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 10−1 × 0 + 1 × 50 + 10−1 × 75 + 10−2 × 100
=
= 48.347𝑛𝑠𝑒𝑐.
∑𝑘 𝑃(𝜏𝑘 )
10−1 + 1 + 10−1 + 10−2
N.B.: The power must be in Watt to calculate 𝜏̅
2
−1 ×02 +1×502 +10−1 ×752 +10−2 ×1002
∑ 𝑃(𝜏 )𝜏
10
̅̅̅
𝜏 2 = ∑𝑘 𝑃(𝜏𝑘 )𝑘 =
𝑘
𝑘
10−1 +1+10−1 +10−2
= 2613.636(𝑛𝑠𝑒𝑐)2
𝜎𝜏 = √̅̅̅
𝜏 2 − (𝜏̅)2 = √2613.636 − (48.347)2 = 16.619𝑛𝑠𝑒𝑐
iii. Calculate the 90% correlation coherence bandwidth
𝐵𝐶 =
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 = 𝑚𝑇𝑏
∴ 𝑅𝑠 =
(𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑖𝑡𝑠 𝑝𝑒𝑟 𝑠𝑦𝑚𝑏𝑜𝑙)
𝑅𝑏 600 × 103
=
= 150 × 103 𝑠𝑦𝑚𝑏𝑜𝑙𝑠 /𝑠𝑒𝑐
𝑚
4
(∵ 16𝑄𝐴𝑀 ∴ 𝑙𝑜𝑔2 16 = 4𝑏𝑖𝑡𝑠/𝑠𝑦𝑚𝑏𝑜𝑙)
∵ 𝑄𝐴𝑀 𝑚𝑜𝑑𝑢𝑙𝑎𝑡𝑖𝑜𝑛 (𝑆𝑖𝑛𝑒 𝑎𝑛𝑑 𝐶𝑜𝑠𝑖𝑛𝑒) ∴ 𝑃𝑎𝑠𝑠𝑏𝑎𝑛𝑑 𝑠𝑖𝑔𝑛𝑎𝑙
∴ 𝐵𝑠 = 𝑅𝑠 = 150 × 103 𝐻𝑧
10𝐵𝑠 = 1500 𝑘𝐻𝑧 = 1.5𝑀𝐻𝑧
∵ 𝐵𝑐 = 1.203𝑀𝐻𝑧
∴ 10𝐵𝑠 > 𝐵𝑐
∴ 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑎𝑑𝑖𝑛𝑔
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(τ)
0 dB
-10 dB
-20 dB
-30 dB
20
60
100
τ(nano seconds)
Faculty of Information Engineering & Technology
The Communications Department
Course: Wireless Communications [NETW 701]
Dr. Tallal Elshabrawy
i. Calculate the mean excess delay
𝜏̅ =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 0 × 1 + 20 × 0.1 + 60 × 0.1 + 100 × 0.01
=
= 7.438𝑛𝑠𝑒𝑐
∑𝑘 𝑃(𝜏𝑘 )
1 + 0.1 + 0.1 + 0.01
2
2 ×1+202 ×0.1+602 ×0.1+1002 ×0.01
∑ 𝑃(𝜏 )𝜏
0
̅̅̅
𝜏 2 = ∑𝑘 𝑃(𝜏𝑘 )𝑘 =
𝑘
𝑘
1+0.1+0.1+0.01
= 413.22(𝑛𝑠𝑒𝑐)2
𝜎𝜏 = √̅̅̅
𝜏 2 − (𝜏̅)2 = 18.918 𝑛𝑠𝑒𝑐
iii. Calculate the 90% correlation coherence bandwidth
𝐵𝐶 =
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)
∵ 𝑝𝑎𝑠𝑠𝑏𝑎𝑛𝑑
∴ 𝐵𝑠 = 𝑅𝑠 =
𝑅𝑏
𝑚
10𝐵𝑠 < 𝐵𝑐
10 × (
𝑅𝑏
) < 𝐵𝑐
𝑚
𝑚
1
>
10𝑅𝑏 𝐵𝑐
𝑚>
10𝑅𝑏
𝐵𝑐
𝑚 > 5.675
𝑚𝑚𝑖𝑛 = 6 𝑏𝑖𝑡𝑠 𝑝𝑒𝑟 𝑠𝑦𝑚𝑏𝑜𝑙 → 𝑀 = 2𝑚 = 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
i.
Find the range of values required for the rms delay spread of the channel such that the
𝑓𝑜𝑟 𝑓𝑙𝑎𝑡 𝑓𝑎𝑑𝑖𝑛𝑔: 𝑇𝑠 > 10𝜎𝜏
∵ 𝐵𝑃𝑆𝐾
∴ 𝑀 = 2𝑠𝑦𝑚𝑏𝑜𝑙𝑠
𝑇𝑠 = 𝑚𝑇𝑏 = 𝑇𝑏 =
10𝜇 > 10𝜎𝜏
∴ 𝑚 = log 2 𝑀 = log 2 2 = 1𝑏𝑖𝑡 /𝑠𝑦𝑚𝑏𝑜𝑙
1
= 10 𝜇𝑠
𝑅𝑏
1𝜇 > 𝜎𝜏
0 ≤ 𝜎𝜏 < 1 × 10−6
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
𝑓𝑚
𝑣 = 30 ×
𝜆=
1000
25
=
𝑚/𝑠
60 × 60
3
𝑐
3 × 108
3
=
=
𝑚
9
𝑓𝑐 5.8 × 10
58
𝑓𝑚 = 𝑓𝑑,𝑚𝑎𝑥
𝑣
= =
𝜆
25
3
3
58
= 161.11 𝐻𝑧
0.423
𝑇𝑐 =
= 2.626 × 10−3 𝑠 = 2.626 𝑚𝑠
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 → 𝜃 = 0° Doppler shift is positive
𝑓 = 𝑓𝑐 + 𝑓𝑑
𝑓𝑑 =
𝑣
𝑐𝑜𝑠(𝜃)
𝜆
𝑓𝑑,𝑚𝑎𝑥 =
𝑣
= 161.11 𝐻𝑧
𝜆
𝑓 = 5.8 × 109 + 161.11 = 5.800000161 𝐻𝑧
iv.
Is the channel fast or slow fading?
∵ 𝑇𝑠 ≪ 𝑇𝑐 → 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 𝑲𝒎/𝒉𝒓
𝜆=
𝑐
3 × 108
2
=
=
𝑚
6
𝑓𝑐 1950 × 10
13
𝑣 = 1×
1000
5
=
𝑚/𝑠
60 × 60 18
𝑣
𝑐𝑜𝑠(𝜃)
𝜆
5
𝑣 18
𝑓𝑑,𝑚𝑎𝑥 = = 2 = 1.81 𝐻𝑧
𝜆
𝑓𝑑 =
13
𝑓𝑚𝑖𝑛 = 𝑓𝑐 − 𝑓𝑑,𝑚𝑎𝑥 = 1950 × 106 − 1.81 = 1949.9999998 𝑀𝐻𝑧
𝑓𝑚𝑎𝑥 = 𝑓𝑐 + 𝑓𝑑,𝑚𝑎𝑥 = 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 𝑲𝒎/𝒉𝒓
𝜆=
𝑐
3 × 108
2
=
=
𝑚
6
𝑓𝑐 1950 × 10
13
𝑣 = 100 ×
1000
500
=
𝑚/𝑠
60 × 60
18
𝑣
𝑐𝑜𝑠(𝜃)
𝜆
500
𝑣
𝑓𝑑,𝑚𝑎𝑥 = = 18
2 = 181 𝐻𝑧
𝜆
𝑓𝑑 =
13
𝑓𝑚𝑖𝑛 = 𝑓𝑐 − 𝑓𝑑,𝑚𝑎𝑥 = 1950 × 106 − 181 = 1949.999819 𝑀𝐻𝑧
𝑓𝑚𝑎𝑥 = 𝑓𝑐 + 𝑓𝑑,𝑚𝑎𝑥 = 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
(μs)
(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
𝜏̅ =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 0 × 0.0501 + 0.2 × 1 + 0.5 × 0.631 + 1.6 × 0.251 + 2.3 × 0.158 + 5 × 0.1
=
∑𝑘 𝑃(𝜏𝑘 )
0.0501 + 1 + 0.631 + 0.251 + 0.158 + 0.1
𝜏̅ =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 1.7805
=
= 0.6742 𝜇𝑠
∑𝑘 𝑃(𝜏𝑘 )
2.641
̅̅̅
𝜏2 =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 2 0 × 0.0501 + 0.22 × 1 + 0.52 × 0.631 + 1.62 × 0.251 + 2.32 × 0.158 + 52 × 0.1
=
∑𝑘 𝑃(𝜏𝑘 )
0.0501 + 1 + 0.631 + 0.251 + 0.158 + 0.1
̅̅̅
𝜏2 =
∑𝑘 𝑃(𝜏𝑘 )𝜏𝑘 2 4.1761
=
= 1.5813 (𝜇𝑠)2
∑𝑘 𝑃(𝜏𝑘 )
2.641
𝜎𝜏 = √̅̅̅
𝜏 2 − (𝜏̅)2 = 1.0615 𝜇𝑠
Using 90% Correlation Bandwidth
5×106
1
𝐵𝐶 = 50𝜎 = 18.84 𝐾𝐻𝑧  18.84×103 = 265.39 ≈ 265 𝐹𝐷𝑀𝐴 𝐶ℎ𝑎𝑛𝑛𝑒𝑙𝑠
𝜏
Using 50% Correlation Bandwidth
5×106
1
𝐵𝐶 = 5𝜎 = 188.4 𝐾𝐻𝑧  188.4×103 = 26.539 ≈ 26 𝐹𝐷𝑀𝐴 𝐶ℎ𝑎𝑛𝑛𝑒𝑙𝑠
𝜏
```