Question

4) At the rate of 10 bps binary data are transmitted in a channel. Power spectral density of the noise at the receiver input 10 (W/Hz). Find the avarage carrier power required to maintain a minumum avarage probability of error 10 for modulations which are given below. a) Coherent binary PSK b) DPSK c) QPSK

Answer 1

(9) For Coherent binary psk Pe = Q( 2 E 2, Eb No ( 6 ) For DPSK (Binary Case) Eb exp(- PB = A enp 2 M=4 (0) For Opsk a Es Sm No 2Q M Ps RBO PR Ps 2

ca BPSK PB = 10 ( 2 EG No From Q function table Q (0) > 10-4 =) x 2 4.75 키 atb 4.75 Eb= 4-75 x No 4.737 -10 given No 10 A No = 2X10 2 - 0 =) Eb = 4.75 X 2X10 2. 2.256 X10 3-9 EXT. We Know that power (p) , tbt Rb =) los bit interval -9 P = 2.256 X10 -17 2-2568 10 Watt ti 108 ib) 6 LO =) Eb = exp(- Eb ln (2.10 6) X. No xen(Qx1039) 2.624 X10 2.624X109 x 1.22.624 X10 2x1010 9 Eba -17 7 = Eb XT6 = 2.624 8109 108

Es No sin Cm Pe. > Opsk Pg= a (e de son (*)] = 10 Es > symbol energy Eb bit energy For QPSK, Es= 2 Eb Esz (10gm) Eb M = 4 for QPSK Pse Q2-x26 =10 NO Vz From & function teable 265 4.75: No 4.75" X No 2- -0 7 Eb = 4.75 x 2x 10 2 १ Eb2 2.256 x 10 -17 Walts 7) P3 2 Eb Tb = 2.25.6 x 10