Fundamentals of Cellular and Wireless Networks Lecture ID: ET-IDA-xxx 20.05.2009 , v08 Prof. W. Adi Tutorial-7 Common Mobile Modulation Techniques

} Carson´s rule Problem 1: In US AMPS cellular system FM modulator the maximum deviation frequency fd=12 kHz and the modulated signal bandwidth is fm=4kHz Determine the modulation index Determine the upper bound of the frequency bandwidth For AMPS FM transmission, If SNRin= 10 dB, determine the SNRout of the FM detector. Determine the possible number of duplex channels for a total of 20 MHz available RF bandwidth (Assume channel bandwidth from 2) Compute the number of possible channels and SNRout if the modulation index is doubled. Solution 1: For FM Modulation: f =  f / fm BT  2 (f + 1) fm = 2( f + fm) (upper bound) BT  2  f (lower bound) SNR factor = [SNR]out / [SNR]in  3 (f2 + f3) f =  f / fm = 12 kHz/4 kHz = 3 2. BT  2 (f + 1) fm = 2 (3 + 1) 4 kHz = 32 kHz [SNR]out  [SNR]in x 3 (f2 + f3) = 10 x 3 (32 + 33)= 1080 = 30.33 dB Number of channels 20 000 / 2 x 32 kHz = 625 channels BT  2 (f + 1) fm = 2 (2x3 + 1) 4 kHz = 56 kHz frequency deviation is also doubled  f = f fm = (2x3) 4 = 24 kHz Number of channels 20 000 / 2 x 56 kHz = 178 channels [SNR]out  [SNR]in x 3 (f2 + f3) = 10 x 3 (62 + 63)= 7560 = 38.8 dB } Carson´s rule

„Mobility“ and Doppler Shift BST Vr = v cos  vr   =90º ‘ fc-fd fc fc+fd A v B ‘ v Clasical Doppler Spectrum C V‘r Cos  =0 => vr=0 v . cos  fd = f =  Fading effects due to Doppler shift If the modulated symbol time is Ts = 1/R Where R is the symbol rate then: Fast fading Ts > Tc Channel changs during one symbole time Slow fading Ts < Tc Channel stable during one symbole time Coherence time Tc due to Doppler shift is the time duration over which the channel response is essentially invariant Tc  0.423/fd

Solution 2:  Problem 2: v . cos  f = A Mobile station moving with a speed of 200 Km/h and transmitting a carrier signal of 990 MHz Determine the maximum Doppler frequency shift fd Determine the coherence time Tc What is the minimum symbol rate at which the digital modulated signal needs no equalization (slow fading consideration). At which speed would a symbole be seriously distorted if the modulator is working with a symbol rate of fs =500 symbols/sec Solution 2: 1. Carrier wavelength  = c/f = 3 . 108 m/s / 990 x 106 Hz = 0.3 m maximum for cos =1 => fd = v /  = [200 103/3600] /0.3 = 185 Hz Tc  0.423 / fd= 0.423/185 = 2.28 ms If Ts < Tc then the channel is considerd flat within the symbole time => fs > 1/Tc = 1/ 2.28 x 10-3 = 437 symbole/Second Distortion starts if fs = 1/Tc = 1/ (0.423/ fd) => fd = 0.423 fs substituting in maximum Doppler shift formula v = fd  => v = 0.423 fs  = 0.423 x 500 x 0.3 = 63.45 m/sec = 63.45 10-3 km / (1/3600) = 228.4 km/hour fc-fd fc fc+fd Clasical Doppler Spectrum f = v . cos   Fading effects due to Doppler shift Fast fading Ts > Tc Channel changs during one symbole time Slow fading Ts < Tc Channel stable during one symbole time

Problem 3: Using a digital PSK modulation technique with the following constellation diagram 1. Determine the voltage amplitude for the cosine wave modulation signal 2. Draw the following modulated bit stream 1011010011 1 volt/division

Solution 3: -45o t wave form of cos t +2 A -45o - 2 A 2 45o 2 0 90 180 270 360 t -45o +45o 135o 225o +45o 2.8 v +2 135o +45o A -45o - 2 225o 1 volt/division Voltage amplitude = A A2=22 + 22 => A = 2.8 volt A 2 45o 2 0o 45o  11. 135o 225o -45o