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Faculty of Computers and Information Cairo University Wireless Communication and networks Sheet 1 [1] Using a transmit power 10ππ andπ0 = 2 × 10−9 ππ/π»π§, find the (theoretical) minimum bandwidth of a channel used to transmit at a rate of 10 Mbps. Answer: Givens: π = 10ππ = 10 × 103 π = π ππ = 2 × 10−9 ππ/π»π§ = 2 × 10−6 π/π»π§ π = 10ππππ = 10 × 10−6 πππ = πΆ Required to find: B We have: π πΆ = π΅πππ2 (1 + ) π π = ππ π΅ −2 10 ) 2 × 10−9 π΅ 106 0.5 × 107 = πππ2 (1 + ) π΅ π΅ 106 = π΅πππ2 (1 + 7 1 0.5 × 10 = 10−6 πππ2 (1 + ) π΅ π΅ 1 This is an implicit equation of π΅ that can be solved numerically by iteration as follows: 1 1- Start with any value for π΅ 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 × 10−6 5.518640063093775711707472999608× 10−6 4.8376001986220801081812825848894× 10−6 4.65466472423368420475356076504× 10−6 4.6012997577022062364108011181411× 10−6 0.5 × 107 10 πππ2 (1 + ) π΅ −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: π = 3.57√4/3 × 20 = 19.365 πΎπ 4ππ 2 4ππ 2 59217626406.536 πΏ=( ) =( ) = π π π2 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 receiver signal-to-noise ratio for this link, for 10 MHz bandwidth. Answer: ππ = 3.2 πππππ = 3.2 × 10−9 π = π (ππππππ πππ€ππ) π = 450 πΎ π΅ = 10 ππ»π§ = 10 × 106 π»π§ π = 1.3803 × 10−23 π½/πΎ πππ = π π πππ = π 3.2 × 10−9 = = 0.0515 × 106 πππ΅ (1.3803 × 10−23 )(450)(10 × 106 ) π = πππ΅ 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: π = 14 πΊπ»π§ = 14 × 109 π»π§ π΄π = 3.75π2 π 3 × 108 = = 0.0214π π 14 × 109 To calculate the gain of such antenna 4ππ΄π 4π(3.75) πΊ= 2 = = 0.1029 × 106 π (0.0214)2 π= Estimating the diameter of similar parabolic antenna with aperture efficiency =0.65 π΄π = 0.65π΄ = 0.65ππ 2 = 3.75 π2 = 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. Answer: ππππππ‘ππ = 70π π = 35π 9 π = 8.45πΊπ»π§ = 8.45 × 10 π»π§ π 3 × 108 π= = = 0.0355π π 8.45 × 109 i) π΄π = 80%π΄ = 0.8π΄ = 0.8ππ 2 = 0.8π(35)2 = 3080 4ππ΄π 4π(3080) πΊ= 2 = = 0.0307 × 109 π (0.0355)2 ii) π = 70πΎ ππ‘ = 10π π = 180ππππππππΎπ = 180 × 109 π π΅ = 10π»π§ ππ‘ (4ππ)2 = ππ πΊπ‘ πΊπ π2 2.2 πΊ = 2.2ππ΅π β« πΊ = 10 10 = 1.6596 ππ‘ πΊπ‘ πΊπ π2 (10)(1.6596)2 (0.0355)2 β« ππ = = = 6.777 × 10−27 π (4ππ)2 (4π180 × 109 )2 πππ = π π πππ = ππ 6.777 × 10−27 = = 7.01 × 10−7 πππ΅ (1.3803 × 10−23 )(70)(10) π = πππ΅ [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) π = 320πΎπ»π§ = 320 × 103 π»π§ π = 100πΎπ = 100 × 103 π 8 π 3 × 10 π= = = 937.5π π 320 × 103 ππ = 2ππ = 2 × 10−3 π ππ‘ = 3π Assume gain at transmitter = gain at receiver ππ‘ (4ππ)2 ππ (4ππ)2 (2 × 10−3 )(4π100 × 103 )2 = β« πΊ πΊ = = = 1198.77 β« πΊ = 34.62 π‘ π ππ πΊπ‘ πΊπ π2 ππ‘ π2 3(937.5)2 πΊ= 4ππ΄π 4π(π΄π) = = 34.62 2 π (937.5)2 (937.5)2 34.62 π΄π = = 2.42 × 106 4π π΄π = 0.65π΄ = 0.65ππ 2 = 2.42 × 106 2.42 × 106 π2 = = 1.848 × 106 π = 1.088 × 103 π 0.65π π = 2π = 544.24π ii) Same as i) but with π = 5πΊπ»π§ = 5 × 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. Answer: i) ii) βπ = − π£π‘π π πΆ =− 10×1000/(60×60) 3×108 × 8.43 × 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 √πΏ , There will be also a time delay of shift of 2πππ π = 2ππ π π π (leading to a phase ) and a frequency shift of -78 Hz, The signal will be: 2 π √ πππ (2π(8.43 × 109 − 78) (π‘ − )) πΏ π