# Wireless Communication and networks Sheet 1

```Faculty of Computers and Information
Cairo University
Wireless Communication and networks
Sheet 1
[1] Using a transmit power 10ππ andπ0 = 2 &times; 10−9 ππ/π»π§, find the (theoretical) minimum
bandwidth of a channel used to transmit at a rate of 10 Mbps.
Givens:
π = 10ππ = 10 &times; 103 π = π
ππ = 2 &times; 10−9 ππ/π»π§ = 2 &times; 10−6 π/π»π§
π = 10ππππ  = 10 &times; 10−6 πππ  = πΆ
Required to find: B
We have:
π
πΆ = π΅πππ2 (1 + )
π
π = ππ π΅
−2
10
)
2 &times; 10−9 π΅
106
0.5 &times; 107
= πππ2 (1 +
)
π΅
π΅
106 = π΅πππ2 (1 +
7
1
0.5 &times; 10
= 10−6 πππ2 (1 +
)
π΅
π΅
1
This is an implicit equation of π΅ that can be solved numerically by iteration as follows:
1
1
2- Substitute this value in the R.H.S. of the equation to obtain a new value for π΅
3. Repeat step 2 using the new value several times until its value becomes fixed
1
π΅
0.0001
8.96866679319520840592 &times; 10−6
5.518640063093775711707472999608&times; 10−6
4.8376001986220801081812825848894&times; 10−6
4.65466472423368420475356076504&times; 10−6
4.6012997577022062364108011181411&times; 10−6
0.5 &times; 107
10 πππ2 (1 +
)
π΅
−6
8.96866679319520840592 &times; 10−6
5.518640063093775711707472999608&times; 10−6
4.8376001986220801081812825848894&times; 10−6
4.65466472423368420475356076504&times; 10−6
4.6012997577022062364108011181411&times; 10−6
4.5853531049206095997399490922925&times; 10−6
4.5853531049206095997399490922925&times; 10−6
4.5805534780986836952595053914688&times; 10−6
4.5791057536545698013970507307224&times; 10−6
4.5786687872645699572724793866351&times; 10−6
4.5785368717769994531873142572448&times; 10−6
4.5805534780986836952595053914688&times; 10−6
4.5791057536545698013970507307224&times; 10−6
4.5786687872645699572724793866351&times; 10−6
4.5785368717769994531873142572448&times; 10−6
4.5784970455293417069216918423002&times; 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.
π = 3.57√4/3 &times; 20 = 19.365 πΎπ
4ππ 2
4ππ 2 59217626406.536
πΏ=(
) =(
) =
π
π
π2
So if π 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
ππ = 3.2 πππππ = 3.2 &times; 10−9 π = π (ππππππ πππ€ππ)
π = 450 πΎ
π΅ = 10 ππ»π§ = 10 &times; 106 π»π§
π = 1.3803 &times; 10−23 π½/πΎ
πππ =
π
π
πππ =
π
3.2 &times; 10−9
=
= 0.0515 &times; 106
πππ΅ (1.3803 &times; 10−23 )(450)(10 &times; 106 )
π = πππ΅
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.
π = 14 πΊπ»π§ = 14 &times; 109 π»π§
π΄π = 3.75π2
π
3 &times; 108
=
= 0.0214π
π 14 &times; 109
To calculate the gain of such antenna
4ππ΄π
4π(3.75)
πΊ= 2 =
= 0.1029 &times; 106
π
(0.0214)2
π=
Estimating the diameter of similar parabolic antenna with aperture efficiency =0.65
π΄π = 0.65π΄ = 0.65ππ 2 = 3.75
π2 =
3.75
= 1.8357
0.65π
π = 1.3548 π
π = 2π = 2.71π
[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.
ππππππ‘ππ = 70π
π = 35π
9
π = 8.45πΊπ»π§ = 8.45 &times; 10 π»π§
π
3 &times; 108
π= =
= 0.0355π
π 8.45 &times; 109
i)
π΄π = 80%π΄ = 0.8π΄ = 0.8ππ 2 = 0.8π(35)2 = 3080
4ππ΄π 4π(3080)
πΊ= 2 =
= 0.0307 &times; 109
π
(0.0355)2
ii) π = 70πΎ
ππ‘ = 10π
π = 180ππππππππΎπ = 180 &times; 109 π
π΅ = 10π»π§
ππ‘ (4ππ)2
=
ππ πΊπ‘ πΊπ π2
2.2
πΊ = 2.2ππ΅π β« πΊ = 10 10 = 1.6596
ππ‘ πΊπ‘ πΊπ π2 (10)(1.6596)2 (0.0355)2
β« ππ =
=
= 6.777 &times; 10−27 π
(4ππ)2
(4π180 &times; 109 )2
πππ =
π
π
πππ =
ππ
6.777 &times; 10−27
=
= 7.01 &times; 10−7
πππ΅ (1.3803 &times; 10−23 )(70)(10)
π = πππ΅
[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
i)
π = 320πΎπ»π§ = 320 &times; 103 π»π§
π = 100πΎπ = 100 &times; 103 π
8
π
3 &times; 10
π= =
= 937.5π
π 320 &times; 103
ππ = 2ππ = 2 &times; 10−3 π
ππ‘ = 3π
Assume gain at transmitter = gain at receiver
ππ‘ (4ππ)2
ππ (4ππ)2 (2 &times; 10−3 )(4π100 &times; 103 )2
=
β«
πΊ
πΊ
=
=
= 1198.77 β« πΊ = 34.62
π‘ π
ππ πΊπ‘ πΊπ π2
ππ‘ π2
3(937.5)2
πΊ=
4ππ΄π
4π(π΄π)
=
= 34.62
2
π
(937.5)2
(937.5)2 34.62
π΄π =
= 2.42 &times; 106
4π
π΄π = 0.65π΄ = 0.65ππ 2 = 2.42 &times; 106
2.42 &times; 106
π2 =
= 1.848 &times; 106
π = 1.088 &times; 103 π
0.65π
π = 2π = 544.24π
ii) Same as i) but with π = 5πΊπ»π§ = 5 &times; 109 π»π§
[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.
i)
ii)
βπ = −
π£π‘π
π
πΆ
=−
10&times;1000/(60&times;60)
3&times;108
&times; 8.43 &times; 109 = −78 π»π§
Since there will be an attenuation L that is function of distance, the amplitude will
2
change by a factor of √πΏ , There will be also a time delay of
shift of
2πππ
π
=
2ππ
π
π
π