Faculty of Computers and Information
Cairo University
Wireless Communication and networks
Sheet 1
[1] Using a transmit power 10𝑚𝑊 and𝑁0 = 2 × 10−9 𝑚𝑊/𝐻𝑧, find the (theoretical) minimum
bandwidth of a channel used to transmit at a rate of 10 Mbps.
Answer:
Givens:
𝑃 = 10𝑚𝑊 = 10 × 103 𝑊 = 𝑆
𝑁𝑜 = 2 × 10−9 𝑚𝑊/𝐻𝑧 = 2 × 10−6 𝑊/𝐻𝑧
𝑅 = 10𝑀𝑏𝑝𝑠 = 10 × 10−6 𝑏𝑝𝑠 = 𝐶
Required to find: B
We have:
𝑆
𝐶 = 𝐵𝑙𝑜𝑔2 (1 + )
𝑁
𝑁 = 𝑁𝑜 𝐵
−2
10
)
2 × 10−9 𝐵
106
0.5 × 107
= 𝑙𝑜𝑔2 (1 +
)
𝐵
𝐵
106 = 𝐵𝑙𝑜𝑔2 (1 +
7
1
0.5 × 10
= 10−6 𝑙𝑜𝑔2 (1 +
)
𝐵
𝐵
1
This is an implicit equation of 𝐵 that can be solved numerically by iteration as follows:
1
1- Start with any value for 𝐵
1
2- Substitute this value in the R.H.S. of the equation to obtain a new value for 𝐵
3. Repeat step 2 using the new value several times until its value becomes fixed
1
𝐵
0.0001
8.96866679319520840592 × 10−6
5.518640063093775711707472999608× 10−6
4.8376001986220801081812825848894× 10−6
4.65466472423368420475356076504× 10−6
4.6012997577022062364108011181411× 10−6
0.5 × 107
10 𝑙𝑜𝑔2 (1 +
)
𝐵
−6
8.96866679319520840592 × 10−6
5.518640063093775711707472999608× 10−6
4.8376001986220801081812825848894× 10−6
4.65466472423368420475356076504× 10−6
4.6012997577022062364108011181411× 10−6
4.5853531049206095997399490922925× 10−6
4.5853531049206095997399490922925× 10−6
4.5805534780986836952595053914688× 10−6
4.5791057536545698013970507307224× 10−6
4.5786687872645699572724793866351× 10−6
4.5785368717769994531873142572448× 10−6
4.5805534780986836952595053914688× 10−6
4.5791057536545698013970507307224× 10−6
4.5786687872645699572724793866351× 10−6
4.5785368717769994531873142572448× 10−6
4.5784970455293417069216918423002× 10−6
B =218.412 KHz
[2] What is the maximum attenuation in a wireless line of sight communication system given that the
transmitter antenna height is 20m and the transmitted power is 5 watts.
Answer:
𝑑 = 3.57√4/3 × 20 = 19.365 𝐾𝑚
4𝜋𝑑 2
4𝜋𝑑 2 59217626406.536
𝐿=(
) =(
) =
𝜆
𝜆
𝜆2
So if 𝜆 was given, we would be able to compute L
[3] If a channel noise temperature is 450 K and the received power is 3.2 nano-watts, estimate the
receiver signal-to-noise ratio for this link, for 10 MHz bandwidth.
Answer:
𝑃𝑟 = 3.2 𝑛𝑎𝑛𝑜𝑊 = 3.2 × 10−9 𝑊 = 𝑆 (𝑆𝑖𝑔𝑛𝑎𝑙 𝑃𝑜𝑤𝑒𝑟)
𝑇 = 450 𝐾
𝐵 = 10 𝑀𝐻𝑧 = 10 × 106 𝐻𝑧
𝑘 = 1.3803 × 10−23 𝐽/𝐾
𝑆𝑁𝑅 =
𝑆
𝑁
𝑆𝑁𝑅 =
𝑆
3.2 × 10−9
=
= 0.0515 × 106
𝑘𝑇𝐵 (1.3803 × 10−23 )(450)(10 × 106 )
𝑁 = 𝑘𝑇𝐵
λ2
[4] A hypothetical isotropic radiator has unity gain and effective aperture 4π for radiation of
wavelengthλ. Calculate the gain of a satellite dish antenna of effective aperture 3.75 square metres at a
frequency of 14 GHz, and estimate the diameter of the parabolic reflector needed to implement such a
dish, given an aperture efficiency factor of 0.65.
Answer:
𝑓 = 14 𝐺𝐻𝑧 = 14 × 109 𝐻𝑧
𝐴𝑒 = 3.75𝑚2
𝑐
3 × 108
=
= 0.0214𝑚
𝑓 14 × 109
To calculate the gain of such antenna
4𝜋𝐴𝑒
4𝜋(3.75)
𝐺= 2 =
= 0.1029 × 106
𝜆
(0.0214)2
𝜆=
Estimating the diameter of similar parabolic antenna with aperture efficiency =0.65
𝐴𝑒 = 0.65𝐴 = 0.65𝜋𝑟 2 = 3.75
𝑟2 =
3.75
= 1.8357
0.65𝜋
𝑟 = 1.3548 𝑚
𝑑 = 2𝑟 = 2.71𝑚
[5] A deep space communication system uses a Cassegrain antenna of diameter 70m at a frequency of
8.45 GHz.
(i) Determine the gain of this dish (in dBi) assuming an aperture efficiency of 80%
(ii) Determine the power received by this dish from a transmission of a satellite having antenna gain 2.2
dBi and transmitter power of 10 W at a distance of 180 million kilometers. Assuming a receiver noise
temperature of 70 K and a receiver bandwidth of 10 Hz, estimate the maximum receiver signal-to-noise
ratio.
Answer:
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 70𝑚
𝑟 = 35𝑚
9
𝑓 = 8.45𝐺𝐻𝑧 = 8.45 × 10 𝐻𝑧
𝑐
3 × 108
𝜆= =
= 0.0355𝑚
𝑓 8.45 × 109
i)
𝐴𝑒 = 80%𝐴 = 0.8𝐴 = 0.8𝜋𝑟 2 = 0.8𝜋(35)2 = 3080
4𝜋𝐴𝑒 4𝜋(3080)
𝐺= 2 =
= 0.0307 × 109
𝜆
(0.0355)2
ii) 𝑇 = 70𝐾
𝑃𝑡 = 10𝑊
𝑑 = 180𝑚𝑖𝑙𝑙𝑖𝑜𝑛𝐾𝑚 = 180 × 109 𝑚
𝐵 = 10𝐻𝑧
𝑃𝑡 (4𝜋𝑑)2
=
𝑃𝑟 𝐺𝑡 𝐺𝑟 𝜆2
2.2
𝐺 = 2.2𝑑𝐵𝑖 ≫ 𝐺 = 10 10 = 1.6596
𝑃𝑡 𝐺𝑡 𝐺𝑟 𝜆2 (10)(1.6596)2 (0.0355)2
≫ 𝑃𝑟 =
=
= 6.777 × 10−27 𝑊
(4𝜋𝑑)2
(4𝜋180 × 109 )2
𝑆𝑁𝑅 =
𝑆
𝑁
𝑆𝑁𝑅 =
𝑃𝑟
6.777 × 10−27
=
= 7.01 × 10−7
𝑘𝑇𝐵 (1.3803 × 10−23 )(70)(10)
𝑁 = 𝑘𝑇𝐵
[6] Compute the required diameter of a parabolic antenna needed to receive a signal at a power of 2 mW
when the transmitted power is 3 watts and the transmission distance is 100 Km in the following cases:
(i) The transmitted signal frequency is 320 KHz
(ii) The transmitted signal frequency is 5 GHz
Answer:
i)
𝑓 = 320𝐾𝐻𝑧 = 320 × 103 𝐻𝑧
𝑑 = 100𝐾𝑚 = 100 × 103 𝑚
8
𝑐
3 × 10
𝜆= =
= 937.5𝑚
𝑓 320 × 103
𝑃𝑟 = 2𝑚𝑊 = 2 × 10−3 𝑊
𝑃𝑡 = 3𝑊
Assume gain at transmitter = gain at receiver
𝑃𝑡 (4𝜋𝑑)2
𝑃𝑟 (4𝜋𝑑)2 (2 × 10−3 )(4𝜋100 × 103 )2
=
≫
𝐺
𝐺
=
=
= 1198.77 ≫ 𝐺 = 34.62
𝑡 𝑟
𝑃𝑟 𝐺𝑡 𝐺𝑟 𝜆2
𝑃𝑡 𝜆2
3(937.5)2
𝐺=
4𝜋𝐴𝑒
4𝜋(𝐴𝑒)
=
= 34.62
2
𝜆
(937.5)2
(937.5)2 34.62
𝐴𝑒 =
= 2.42 × 106
4𝜋
𝐴𝑒 = 0.65𝐴 = 0.65𝜋𝑟 2 = 2.42 × 106
2.42 × 106
𝑟2 =
= 1.848 × 106
𝑟 = 1.088 × 103 𝑚
0.65𝜋
𝑑 = 2𝑟 = 544.24𝑚
ii) Same as i) but with 𝑓 = 5𝐺𝐻𝑧 = 5 × 109 𝐻𝑧
[7] If a receiver is moved with a constant speed of 10 km/hour away from a transmitter radiating at a
frequency of 8.43 GHz.
(i) What is the Doppler shift in the frequency at the receiver side?
(ii) Write an expression for the received signal (as a function of time and distance) if the transmitter
sends a pure cosine signal. Assume free space noise free channel.
Answer:
i)
ii)
∆𝑓 = −
𝑣𝑡𝑟
𝑓
𝐶
=−
10×1000/(60×60)
3×108
× 8.43 × 109 = −78 𝐻𝑧
Since there will be an attenuation L that is function of distance, the amplitude will
2
change by a factor of √𝐿 , There will be also a time delay of
shift of
2𝜋𝑓𝑑
𝑐
=
2𝜋𝑑
𝜆
𝑑
𝑐
(leading to a phase
) and a frequency shift of -78 Hz, The signal will be:
2
𝑑
√ 𝑐𝑜𝑠 (2𝜋(8.43 × 109 − 78) (𝑡 − ))
𝐿
𝑐