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authorpriyanka2015-06-24 15:03:17 +0530
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+// Example 6.2, page no-230
+clear
+clc
+t=20 //TDMA frame length in ms
+lc=352 //length of carrier and clock recovery frequency in bits
+lu1=48 //length of unique word in bits
+lo=510 //length of order wire channel in bits
+lm= 256 //length of management channel in bits
+lt=320 // length of transmit timming channel in bits
+ls1=24 // length of service channel in bits
+gt=64 // Guard time in bits
+rb=2 // reference burst
+br=90*10^6 //burst bit rate 90Mbps
+dr= 64*10^3 //data rate 64 kbps
+bfr=br*t*10^-3
+lr=lc+lu1+lo+lm+lt
+tb=lc+lu1+lo+ls1
+tob=(lr*rb)+(tb*t)+((t+rb)*gt)
+feff=(bfr-tob)*100/bfr
+feff=ceil(feff*100)/100
+vsb=dr*t*10^-3
+x=bfr*feff/100
+printf("The number of bits in a frame for a voice sub-burst is %d\n\n The total no of bits available in a frame for carrying traffic is %d\n\n Maximum no of PCM voice channels in a frame is %d channels",vsb,x,x/vsb)