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+//Exa 21.3
+//To find-
+// a) The bandwidth of a subchannel,
+// b) modulation efﬁciency,
+// c) user symbol rate,
+// d) user data rate if the information bits are encoded with a rate of 3/4,
+// e) time utilization efﬁciency of the system.
+
+clc;
+clear all;
+
+InfoSc=48;//Information subcarriers
+SyncSc=4;//synchronization subcarriers
+ReservedSc=12;//Reserved subcarriers
+Symrate=250;  //ksps(kilosymbols per second)
+BW=20;  ///in MHz
+Grdt=800; //Guard time in nsec
+
+//solution
+TotalSc=InfoSc+SyncSc+ReservedSc;//Total subcarriers
+BW_Sch=BW*10^6/TotalSc;//BW of subchannel
+Mod_eff=Symrate*10^3/(BW_Sch);//Modulation efficiency
+User_txrate=InfoSc*Symrate*10^3;
+User_bitsymbol=4;  //16-QPSK is used
+disp("From table 21.7 For modulation scheme as 16-QAM and coding rate =3/4 then User data rate will be 36Mbps");
+User_DR=36; //Mbps
+Sym_Dur=1/(Symrate*10^3);
+TimeUti=Sym_Dur/(Sym_Dur+(Grdt/10^9));
+
+printf(' The bandwidth of subchannel is %.1f kHz\n',BW_Sch/10^3);
+printf(' Modulation efficiency is %.1f symbols/sec/Hz \n',Mod_eff);
+printf(' User symbol rate is %d Msps \n',User_txrate/10^6);
+printf(' Time Utilization efficiency is %.2f \n',TimeUti);