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LAB-4 Wireless Networks Course Instructor: Dr.Amina SaleemLab Conducted by Miss Moonerah Al-EidiQuestion No 1:The room temperature is usually taken to be T=17oC or 290K. At this temperature find the thermal noise power density.Solution:No=KTWattsHzWhere K=1.3810-23 is the Boltzman constant. And No is the thermal noise power density per Hertz or bandwidth.No=KTWattsHz=(1.38×10-23×290=4×10-21WattsHz=-204dBW/HzThe expression for dBW=10log10(Power in watts/1Watt).dBW is an absolute level of power relative to 1W which selected as a reference and defined to be 0dBwThe thermal noise is called white noise as it is constant over the entire bandwidth. So given a receiver with the above noise temperature and a 10MHz bandwidth, the thermal noise at the receiver’s output is given by:No=KTBNdB=-228.6+10log294+10log107N=-133.9dBWQuestion No 2:Consider a voice channel being used via modem to transmit digital data. Assuming a bandwidth of 3100 Hz. Then the Nyquist capacity is given by the following considering M=2C=2B=6200bpsIf we use M-Ary Signalling where the total number of levels are M=8 then the channel capacity is eaual to 18600 b/s over a bandwidth of 3100Hz.C=6200×3=18600 bits/secondQuestion No 365722541021000Prove the following expression for the Signal energy/bit to Noise power per Hz or the ratio EbNoThe relationship between the EbNo and the SNR can be represented as -7620035687000The relationship between the EbNo and the spectral efficiency can be found out as Question No 4:Given EbNo=8.4dB is required to achieve a bit error rate of 10-4.If the effective noise temperature is 290k and the data rate is 2400bps, what received signal is required.8.4=SdBW-10log2400+228.6dBW-10log290=-161.8dBWQuestion No 5:Find the minimum Eb/No required to achieve a spectral efficiency of 6bps/Hz.EbNo=1626-1=10.5=10.21dBQuestion No 6:If the maximum distance between two antennas for LOS transmission is 41Km. Find the length of the transmitting antenna if the length of the receiving antenna is 10 m. 41=3.57(Kh1+4/3×10)h1=46.2mQuestion No 7:Following Figure gives us the relationship between the free space loss with distance and the frequency. What can you interpret between the relationship of the loss with increasing distance and increasing frequencies.Solution:We can see from the following curve, that the free space loss increases with increasing distance from the transmitter and it is higher for higher frequencies.0-444500 ................
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