Scilab Textbook Companion for Digital Telephony by J. C. Bellamy Created by Harish Shenoy B.Tech Electronics Engineering NMIMS, MPSTME College Teacher Not decided Cross-Checked by TechPassion May 0, 06 Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/nmeict-intro. This Textbook Companion and Scilab codes written in it can be downloaded from the Textbook Companion Project section at the website http://scilab.in
Book Description Title: Digital Telephony Author: J. C. Bellamy Publisher: Wiley India (P.) Ltd., New Delhi Edition: 3 Year: 000 ISBN: 9788650994
Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular Example of the above book) For example, Exa 3.5 means solved example 3.5 of this book. Sec.3 means a scilab code whose theory is explained in Section.3 of the book.
Contents List of Scilab Codes 4 3 Voice Digitization 5 5 Digital Switching 8 6 Digital Modulation and Radio Systems 3 7 Network Synchronization Control and Management 5 8 Fiber Optic Transmission System 8 9 Digital Mobile Telephony 5 0 Data and Asynchronous Transfer Mode Network 7 Digital Subscriber Access 30 Traffic Analysis 33 3
List of Scilab Codes Exa 3. Exa 3. Exa 3.4 Exa 5. Exa 5. Exa 5.3 Exa 5.5 Exa 6.4 Exa 7. Exa 7.3 Exa 8. Exa 8. Exa 8.3 Exa 8.4 Program to calculate quantization interval and bits needed to encode each sample.................. 5 Program to calculate the minimum bit rate for a PCM encoder must provide for high fidelity......... 6 Program to calculate how many bits per sample can be saved by using DPCM.................. 7 Program to find the idle path in a three stage 89 line switch........................... 8 Program to determine the implementaton complexity of the TS switch....................... 9 Program to determine the implementaton complexity of a 048 channel...................... 0 Program to determine the implementaton complexity of a 307 channel..................... Program to determine system gain of 0Mbps Ghz digital microwave repeater using 4 PSK modulation... 3 Program to determine relative accuracy of maintaining a mutual slip rate ojective of one slip in 0hrs..... 5 Program to determine the minimum and maximum input channel rate accommodated by an M multiplexer 6 Program to determine the loss limit and the multimode dispersion limit of a graded index FOC........ 8 Program to determine the loss limit and the chromatic dispersion limit of a high performance SMF FOC... 9 Program to determine the BDP of SMF system and DS SMF system using DFB LD............... 0 Program to determine the difference in wavelength of two optical signal..................... 4
Exa 8.5 Program to determine the system gain......... Exa 8.6 Program to determine the range of SPE data rates that can be accomodated by the byte stuffing operation.. 3 Exa 9. Progam to determine the probability of maximum interference of a 64 channel CDMA system....... 5 Exa 0. Program to determine the amount of transmission capacity........................... 7 Exa 0.3 Program to determine the probability that the delay of an ATM voice cell.................... 8 Exa. Program to determine the distance limit imposed by the need to echo E bit in a BRI ST interface........ 30 Exa. Program to determine the theoretical maximum data rate of a prefectly equalized voiceband modem.... 3 Exa. Program to calculate how often do two calls arrive with less than milisec between them............ 33 Exa. Program to calculate the probability that eight or more arrivals occur in an chosen 30 sec............ 34 Exa.3 Program to calculate the probability that a 000 bit data block experiences exaclty 4 errors while being transmitted over a link having error............. 35 Exa.4 Program to calculate the percentage of total traffic carried by first five ckt and traffic carried by all other remaining.......................... 36 Exa.5 Program to calculate how much traffic can the trunk group carry........................ 37 5
Chapter 3 Voice Digitization Scilab code Exa 3. Program to calculate quantization interval and bits needed to encode each sample 3 // Example 3. 4 5 // Page 0 6 7 sqr =30 //SQR=30dB 8 9 q =*0^[ -( sqr -7.78) /0] 0 disp ( Thus 3 q u a n t i z a t i o n i n t e r v a l s a r e r needed f o r each p o l a r i t y f o r a t o t a l o f 6 i n t e r v a l s i n a l l. The number o f b i t z r e q u i r e d a r e d e t e r m i n e d as ) 3 N= log (6) 4 5 // R e s u l t 6 7 // q = 0. 0 7 8 V 8 6
9 //N = 4. 7 = 5 b i t s per sample Scilab code Exa 3. Program to calculate the minimum bit rate for a PCM encoder must provide for high fidelity 3 // Example 3. 4 5 // Page 05 6 7 dr =40 // dynamic range =400dB 8 9 SNR =50 // s i g n a l to n o i s e r a t i o =5 0dB 0 SQR =dr+snr 3 n =[( SQR -.76) /6.0] 4 5 disp ( This can be approximated to 5 b i t s per sample ) 6 7 disp ( Assuming e x c e s s sampling f a c t o r u s i n g D type channel, we c h o o s e sampling r a t e as 48KHz ) 8 9 disp ( T h e r e f o r e r e q u i r e d b i t r a t e i s ) 0 5*48000 3 // R e s u l t 4 5 // 70 kbps 7
Scilab code Exa 3.4 Program to calculate how many bits per sample can be saved by using DPCM 3 // Example 3. 4 4 5 // Page 8 6 7 w =800 // Omega=800Hz 8 9 // x ( t )=A s i n ( p i. wt ), e q u a t i o n f o r s i n e wave with maximum a m p l i t u d e 0 //x ( t )=A( p i ). w. c o s ( p i. wt ), d i f f w. r. t time 3 (* %pi ) *800*(/8000) 4 5 // 0. 6 8 3 a, x ( t ) max 6 7 disp ( s a v i n g s i n the b i t s per sample can be d e t e r m i n e d as ) 8 9 log (/0.68) 0 // R e s u l t 3 // 0. 6 7 b i t s 8
Chapter 5 Digital Switching Scilab code Exa 5. Program to find the idle path in a three stage 89 line switch 3 // Example 5. 4 5 // Page43 6 7 // R e f e r to t a b l e 5. on page36 8 9 disp ( From the t a b l e, s p a c e e x p a n s i o n f a c t o r o f 0. 3 4 i s 0. 0 0. Hence the u t i l i z a t i o n o f each i n t e r s t a g e i s g i v e n by ) 0 0./0.34 3 p = -[( -0.47) ^] // p r o b a b i l i t y t h a t one o f two l i n k s i n s e r i e s i s busy 4 5 disp ( T h e r e f o r e, the e x p e c t e d number o f paths to be t e s t e d are, ) 6 9
7 Np =[ -(0.67) ^5]/( -0.67) 8 9 // R e s u l t 0 // Only 3 o f the 5 paths s h o u l d be t e s t e d b e f o r e an i d l e path i s found Scilab code Exa 5. Program to determine the implementaton complexity of the TS switch 3 // Example 5. 4 5 // Page 53 6 7 // R e f e r to f i g u r e 5. 9 on page 5 8 9 N =80 //Number o f l i n k s 0 Nc =4 //Number o f c o n t r o l words 3 Nb =7 //Number o f b i t s per c o n t r o l word 4 5 Nb =5 //Number o f b i t s per c o n t r o l word 6 7 disp ( The number o f c r o s s p o i n t s i n the s p a c e s t a g e i s ) 8 9 Nx=N^ 0 disp ( The t o t a l number o f memory b i t s f o r the s p a c e s t a g e c o n t r o l s t o r e i s ) 3 Nbx =N*Nc*Nb 0
4 5 disp ( The t o t a l number o f memory b i t s f o r the time s t a g e i s ) 6 7 Nbt =(N*Nc *8) +(N*Nc*Nb ) 8 9 disp ( Thus the i m p l e m e n t a t i o n c o m p l e x i t y i s ) 30 3 Cmplx =Nx +[( Nbx + Nbt ) /00] 3 33 // R e s u l t 34 35 // Complexity i s 6784 e q u i v a l e n t c r o s s p o i n t. Scilab code Exa 5.3 Program to determine the implementaton complexity of a 048 channel 3 4 // Example 5. 3 5 6 // Page 56 7 8 k=7 // from e q u a t i o n 5. 4 on page 56 9 0 disp ( Using the v a l u e o f k ) disp ( Using the v a l u e o f k, the number o f c r o s s p o i n t d e t e r m i n e d a r e ) 3 4 *7*6 5 6 disp ( The number o f b i t s o f memory can be d e t e r m i n e d a r e )
7 8 N =[*7* 8*4]+[7* 8*8]+[7* 8*7] 9 0 // R e s u l t // The c o m p o s i t e i m p l e m e n t a t i o n c o m p l e c i t y i s 430 e q u i v a l e n t c r o s s p o i n t s. Scilab code Exa 5.5 Program to determine the implementaton complexity of a 307 channel 3 4 // Example 5. 5 5 6 // Page 6 7 8 9 0 n =3 // b i n a r y w. r. t (N/) ˆ k =7 // d e t e r m i n e d as a b l o c k i n g p r o b a b i l i t y o f 0. 0 0 5 3 4 // R e f e r e q u a t i o n s 5. 8 and 5. 9 5 6 Nx =[*04*7]+[7+(3^) ] 7 8 Nx =[*04*7]+[7*(3^) ] 9 0 Nbx =[*7*8*3*5]+{7*8*3*5} Nbt =[*04*8*8] 3 4 Nbtc =[*04*8*7]
5 6 cmplx =[ Nx +{( Nbx + Nbt + Nbtc ) /00}] 7 8 // R e s u l t 9 30 // Complexity i s 38,854 e q u i v a l e n t c r o s s p o i n t. 3
Chapter 6 Digital Modulation and Radio Systems Scilab code Exa 6.4 Program to determine system gain of 0Mbps Ghz digital microwave repeater using 4 PSK modulation 3 // Example 6. 4 4 5 // Page 35 6 7 // R e f e r to f i g u r e 6. 7 on page 300 8 9 0 disp ( SNR d e t e c t o r i s 3dB h i g h e r than Eb/N0, t h e r e f o r e ) snr =3.7 //SNR=3.7dB 3 4 disp ( S i n c e 4 PSK modulation p r o v i d e s bps /Hz, the s a m p l i n g r a t e i s 5 MHz, which i s Nqyuist r a t e, t h e r e f o r e ) 5 4
6 a =0* log0 (5000000000000) 7 8 a =0* log0 (.3) 9 0 A0=a -3.7-7 -3 - a disp ( At a c a r r i e r f r e q u e n c y o f GHz, the wavelength i s ) 3 4 (3*0^8) /(*0^9) 5 6 FM =6+60+0* log0 (0.5) -5-0* log0 (4* %pi *5*0^4) // Fade Margin can be found by Equation 6. 3 7 8 // R e s u l t 9 //A0 = 6dB 30 // wavelength = 0. 5 m 3 // Fade Margin = 3 8. 5 db 5
Chapter 7 Network Synchronization Control and Management Scilab code Exa 7. Program to determine relative accuracy of maintaining a mutual slip rate ojective of one slip in 0hrs 3 // Example 7. 4 5 // Page 350 6 7 8 df =(/0*60*60) 9 0 df =[/(0*60*60) ] disp ( S i n c e t h e r e a r e 8000 frame per second, the r e l a t i v e a c c u r a c y i s d e t e r m i n e d as ) 3 4 ans =[ df /8000] 5 6 // R e s u l t 7 6
8 // Hence the c l o c k must be a c c u r a t e to. 7 p a r t s i n 0 ˆ 9. Scilab code Exa 7.3 Program to determine the minimum and maximum input channel rate accommodated by an M multiplexer 3 // Example 7. 3 4 5 // Page 354 6 7 disp ( The maximum i n f o r m a t i o n r a t e per c h a n n e l i s d e t e r m i n e d as ) 8 9 Imax =[(6.3*88) /76] 0 disp ( The minimum i n f o r m a t i o n r a t e per c h a n n e l i s d e t e r m i n e d as ) 3 Imin =[(6.3*87) /76] 4 5 disp ( S i n c e t h e r e a r e t h r e e p o s s i b l e c o m b i n a t i o n s o f two e r r o r s i n the C b i t s, the p r o b a b i l i t y o f m i s i n t e r p r e t i n g an S b i t i s ) 6 7 3*(0^ -6) ^ 8 9 76/6.3 // d u r a t i o n o f each master frame 0 [(3*0^ -) /(86*0^ -6) ] 3 // R e s u l t 4 5 // 0.06 0ˆ 6 m i s f r a m e s per second 7
8
Chapter 8 Fiber Optic Transmission System Scilab code Exa 8. Program to determine the loss limit and the multimode dispersion limit of a graded index FOC // Caption : Program to d e t e r m i n e the l o s s l i m i t and the multimode d i s p e r s i o n l i m i t o f a graded i n d e x FOC 3 4 // Example 8. 5 6 // Page 388 7 8 // R e f e r to f i g u r e 8. on page 385 9 0 Pin =4 // i n p u t power = 4dB 3 A=3 // a t t e n u a t i o n 4 5 LL =( Pin /A) // Loss Limit 6 9
7 disp ( Using Gbps km as t y p i c a l BDP o f graded i n d e x multimode f i b e r, the multimode d i s p e r s i o n d i s t a n c e i s d e t e r m i n e d as ) 8 9 Dl =(000/90) // D i s p e r s i o n l i m i t 0 // R e s u l t 3 // Loss Limit = 4 km 4 5 // D i s p e r s i o n Limit =. km Scilab code Exa 8. Program to determine the loss limit and the chromatic dispersion limit of a high performance SMF FOC 3 // Example 8. 4 5 // Page 389 6 7 // R e f e r f i g u r e 8. on page 385 8 9 disp ( The a t t e n u a t i o n o f s i n g l e mode f i b r e o p e r a t i n g at 300nm i s a p p r o x i m a t e l y 0. 3 5 db/km. Thus, ) 0 Pin =4 // i n p u t power = 4dB 3 A =0.35 4 5 LL =( Pin /A) // Loss Limit 6 7 disp ( Using 50 Gbps km as BDP o f a s i l i c a s i n g l e mode f i b e r, the c h r o m a t i c d i s p e r s i o n l i m i t i s d e t e r m i n e d as ) 0
8 9 Cd =(50000/47) // Chromatic d i s p e r s i o n l i m i t 0 // R e s u l t 3 // Loss Limit = 0 km 4 5 // Chromatic D i s p e r s i o n Limit = 5 9 9. 5 = 600 km Scilab code Exa 8.3 Program to determine the BDP of SMF system and DS SMF system using DFB LD 3 4 // Example 8. 3 5 6 // Page 393 7 8 // R e f e r to t a b l e 8. on page 39, a l s o to f i g u r e 8. 6 on page 39 9 0 smf =6 smf =6 // d i s p e r s i o n co e f f i c i e n t o f SMF at 550nm 3 4 sw =0.4 // s p e c t r a l width o f the s o u r c e 5 6 BDP =[50/( smf *sw)] // assuming l i n e code as NRZ 7 8 disp ( The BDP o f the DS SMF system i s d e t e r m i n e d as ) 9 0 smf =3.5 // d i s p e r s i o n co e f f i c i e n t o f DS SMF at 550nm
BDP =[50/( smf *sw)] // assuming l i n e code as NRZ 3 4 // R e s u l t 5 6 //BDP = 39 Gbps=km (SMF) 7 8 //BDP = 79 Gbps km (DS SMF) Scilab code Exa 8.4 Program to determine the difference in wavelength of two optical signal 3 // Example 8. 4 4 5 // Page 40 6 7 c =3*0^8 // speed o f l i g h t 8 9 wl =500*0^ -9 // wavlength =500nm 0 f =[(3*0^8) /wl] 3 disp ( Thus the upper and l o w e r f r e q u e n c i e s a r e d e t e r m i n e d as 00,00 and 99,999 GHz r e s p e c t i v e l y. The c o r r e s p o n d i n g w a v e l e n g t h s a r e ) 4 5 lam =[ c /(99999*0^9) ] 6 7 lam =[ c /(0000*0^9) ] 8 9 // R e s u l t 0 // The d i f f e r e n c e i n w a v e l e n g h t s i s 0. 0 5nm
Scilab code Exa 8.5 Program to determine the system gain 3 // Example 8. 5 4 5 // Page 405 6 7 // R e f e r to t a b l e 8. and f i g u r e 8. 8 on page 394 8 9 dr =565 // data r a t e 0 wl =550*0^ -9 // wavelength 3 disp ( The use o f 5B6B l i n e code i m p l i e s the l i n e data r a t e i s, ) 4 5 565*(6/5) 6 7 // 678 Mbps 8 9 disp ( The r e c e i v e r s e n s i t i v i t y f o r 678 Mbps i s d e t e r m i n e d from f i g 8. 8 or t a b l e 8. as ) 0 rsen = -34.5 3 A=( -5 - rsen ) // system g a i n 4 5 BDP =[500/(7*0.4) ] 6 7 BDPs =[73.6/0.678] 8 9 lossp =(0.+0.) *(65) 30 3
3 lossm =A- lossp 3 33 // R e s u l t 34 35 // System g a i n = 9. 5 db 36 37 //BDP = 7 3. 6 Gbps 38 39 //BDP s p a c i n g = 09 km 40 4 // Path Loss = 6 db 4 43 // Loss Margin = 3. 5 db Scilab code Exa 8.6 Program to determine the range of SPE data rates that can be accomodated by the byte stuffing operation // Example 8. 6 3 4 // Page 45 5 6 frames =4*9*87 // Four SPE f r a m e s 7 8 rate =8* frames *000 // normal r a t e SPE 9 0 Rmin =8*33*000 // minimum SPE r a t e 3 4 disp ( When n e g a t i v e byte s t u f f i n g i s used to accomodate a f a s t incoming SPE r a t e, 333 b y t e s o f data a r e t r a n s m i t t e d i n f o u r f r a m e s. Thus, the h i g h e s t s l i p r a t e i s ) 5 4
6 Rmax =8*333*000 //maximum SPE r a t e 7 8 // R e s u l t 9 0 // Normal SPE r a t e = 5 0. Mbps //Minimum SPE r a t e = 5 0. 0 9 6 Mbps 3 4 //Maximum SPE r a t e = 5 0. 8 Mbps 5
Chapter 9 Digital Mobile Telephony Scilab code Exa 9. Progam to determine the probability of maximum interference of a 64 channel CDMA system 3 // Example 9. 4 5 // Page 447 6 7 disp ( The p r o b a b i l i t y o f 63 d e s t r u c t i v e i n t e r f e r e r s i s merely the p r o b a b i l i t y o f o c c u r e n c e o f 63 e q u a l l y l i k e l y b i n a r y e v e nts, ) 8 9 Pmax =(0.5) ^63 //maximum p r o b a b i l i t y 0 disp ( The v a l u e o f a d e s i r e d r e c e i v e s i g n a l i s the a u t o c o r r e l a t i o n o f a codeword with i t s e l f and can t h e r e f o r e be r e p r e s e n t e d as a v a l u e o f 6 4. ) 3 disp ( The mean and v a r i e n c e o f a sum o f 63 such v a r i a b l e a r e 0 and 63, r e s p e c t i v e l y. The s i g n a l to i n t e r f e r e n c e r a t i o i s now d e t e r m i n e d as, ) 4 6
5 a =[(64^) /63] 6 7 SIR =0* log0 (a) 8 9 // R e s u l t 0 // S i g n a l to i n t e r f e r e n c e r a t i o = 8. db 7
Chapter 0 Data and Asynchronous Transfer Mode Network Scilab code Exa 0. Program to determine the amount of transmission capacity 3 // Example 0. 4 5 // Page 47 6 7 // ( a ) With l i n k by l i n k 8 9 frame =000*0^ -8 0 disp ( The e x p e c t e d number o f b i t s o f t r a n s m i s s i o n c a p a c i t y r e q u i r e d to r e t r a n s m i t i s ) 3 frame *000 4 5 // ( b ) With end to end 6 7 frames =0*0^ -5 // c o r r u p t e d frame 8
8 9 disp ( The e x p e c t e d number o f b i t s o f t r a n s m i s s i o n c a p a c i t y r e q u i r e d i s ) 0 frames *000 3 // ( c ) With b i t e r r o r 0ˆ 5 4 5 ans =000*0^ -5 6 7 ans =000*0^ -5*000 8 9 ans =0*0^ -*000 30 3 // R e s u l t 3 33 // ( a ) 0. 0 b i t / l i n k 34 35 // ( b ) 0. b i t / l i n k 36 37 // ( c ). 0 b i t s / l i n k 38 39 // ( c ). 00 b i t s / l i n k Scilab code Exa 0.3 Program to determine the probability that the delay of an ATM voice cell 3 // Example 0. 3 4 5 // Page 488 6 7 disp ( Assuming the a c c e s s l i n k i s 90% u t i l i z e d on a v e r a g e. ) 9
8 9 disp ( The queuing t h e o r y i s p r o v i d e d i n Chapter. I t i n v o l v e s d e t e r m i n i n g the p r o b a b i l i t y t h a t the DSI a c c e s s queue c o n t a i n s enough c e l l s to r e p r e s e n t 0 msec o f t r a n s m i s s i o n time ) 0 tm =[(53*8) /(9*8000) ] 3 disp ( T h e r e f o r e, 0 msec d e l a y r e p r e s e n t s 0 / 0. 7 6 = 3 6. c e l l t i m e s. ) 4 5 p =(0.9) *{ %e ^[ -( -0.9) *36.]} // R e f e r to e q u a t i o n. 5 i n chap 6 7 disp ( R e s u l t ) 8 9 disp ( P(& gt ; 0 msec ) =. 5% d e l a y w i l l be d i s p l a y e d by more than 0 msec ) 30
Chapter Digital Subscriber Access Scilab code Exa. Program to determine the distance limit imposed by the need to echo E bit in a BRI ST interface // Example. 3 4 // Page 50 5 6 // R e f e r to f i g u r e. 5 on page 500 7 8 disp ( By s e e i n g the f i g u r e, i t can be s e e n t h a t the minimum d e l a y between a t e r m i n a l t r a n s m i t t i n g D b i t and r e c e i v i n g i t back i n the f o l l o w i n g E b i t i s s e v e n b i t t i m e s ) 9 0 disp ( At a 9 kbps data r a t e the d u r a t i o n o f b i t i s 5. u s e c. Thus, the t o t a l round t r i p p r o p a g a t i o n time i s ) 7*5. // u s e c 3 4 disp ( Assuming no a p p r e c i a b l e c i r c u i t r y d e l a y s i n the NT, ) 3
5 6 c =3*0^8 // speed o f l i g h t 7 8 Lmax =(36.4*0^ -6) *(/3) * c 9 0 disp ( Because round t r i p p r o p a g a t i o n i n v o l v e s both d i r e c t i o n o f t r a n s m i s s i o n ) Dmax =(/) * Lmax 3 4 disp ( R e s u l t ) 5 6 disp ( Maximum l e n g t h o f w i r e ( Lmax ) = 3640 m = 3. 6 4 km ) 7 8 disp ( Maximum d i s t a n c e (Dmax)= 80 m =. 8 km ) Scilab code Exa. Program to determine the theoretical maximum data rate of a prefectly equalized voiceband modem // Example. 3 4 // Page 53 5 6 disp ( The s i g n a l to q u a n t i z i n g n o i s e r a t i o (SQR) i s g i v e n i n chap3 to be on the o r d e r o f 36dB, which c o r r e s p o n d s to power r a t i o o f 3 9 8. ) 7 8 disp ( Using t h i s v a l u e i n Shannon theorem f o r the t h e o r e t i c a l c a p a c i t y o f a c h a n n e l y i e l d, ) 9 0 SNR =398 C =300*[ log (+ SNR )] 3
3 4 disp ( R e s u l t ) 5 6 disp ( data r a t e = 37 kbps ) 33
Chapter Traffic Analysis Scilab code Exa. Program to calculate how often do two calls arrive with less than milisec between them // Example. 3 4 // Page 54 5 6 disp ( The a v e r a g e a r r i v a l r a t e i s ) 7 8 lam =(3600/0000) // a r r i v a l s per s e c 9 0 disp ( From e q u a t i o n., the p r o b a b i l i t y on a r r i v a l i n 0.0 s e c i n t e r v a l i s ) // e q u a t i o n on page 54 P0 =( %e ^ -0.078) 3 4 disp ( Thus. 7% a r r i v a l s o c c u r w i t h n i n 0. 0 s e c o f the p e r v i o u s a r r i v a l. S i n c e the a r r i v a l r a t e i s. 7 8 a r r i v a l s per second, the r a t e o f o c c u r r e n c e o f i n t e r v a r r i v a l time l e s s than 0. 0 s e c i s ) 5 6.78*0.07 34
7 8 disp ( R e s u l t ) 9 0 disp ( 0. 0 7 5 t i m e s / s e c ) Scilab code Exa. Program to calculate the probability that eight or more arrivals occur in an chosen 30 sec // Example. 3 4 // Page 56 5 6 disp ( The a v e r a g e number o f a r r i v a l s i n a 30 s e c i n t e r v a l i s, ) 7 8 lamt =4*(30/60) 9 0 disp ( The p r o b a b i l i t y o f e i g h t or more a r r i v a l s i s, ) P0 = 3 4 P =[(^) /() ] 5 6 P =[(^) /(*) ] 7 8 P3 =[(^3) /(**3) ] 9 0 P4 =[(^4) /(**3*4) ] P5 =[(^5) /(**3*4*5) ] 3 4 P6 =[(^6) /(**3*4*5*6) ] 5 35
6 P7 =[(^7) /(**3*4*5*6*7) ] 7 8 i= -{( %e ^ -) *[ P0+P+P+P3+P4+P5+P6+P7 ]} 9 30 disp ( R e s u l t ) 3 3 disp ( P ( ) = 0. 0 0 ) Scilab code Exa.3 Program to calculate the probability that a 000 bit data block experiences exaclty 4 errors while being transmitted over a link having error 3 // Example. 3 4 5 // Page 57 6 7 disp ( Assuming i n p e n d e n t e r r o r, we can o b t a i n the p r o b a b i l i t y o f e x a c t l y 4 e r r o r s d i r e c t l y from the P o i s s o n d i s t r i b u t i o n. The a v e r a g e number o f e r r o r s i s, ) 8 9 lamt =[(0^3) *(0^ -5) ] 0 disp ( Thus, ) 3 P4 ={[(0.0^4) /(**3*4) ]* %e ^ -0.0} 4 5 disp ( R e s u l t ) 6 7 disp ( P ( 4 ) = 4.5 0ˆ 0 ) 36
Scilab code Exa.4 Program to calculate the percentage of total traffic carried by first five ckt and traffic carried by all other remaining 3 // Example. 4 4 5 // Page 59 6 7 disp ( The T r a f f i c i n t e n s i t y o f system i s, ) 8 9 A =* 0 disp ( The r a f f i c i n t e n s i t y c a r r i e d by i a c t i v e c k t i s e x a c t l y i e r l a n g s. Hence the t r a f f i c c a r r i e d by s t 5 c k t i s, ) 3 P =[(*^) /() ] 4 5 P =[(*^) /(*) ] 6 7 P3 =[(3*^3) /(**3) ] 8 9 P4 =[(4*^4) /(**3*4) ] 0 P5 =[(5*^5) /(**3*4*5) ] 3 A5 ={( %e ^ -) *[ P+P+P3+P4+P5 ]} 4 5 disp ( A l l o f r e m a i n i n g c k t s c a r r y, ) 6 7 Ar = -.89 8 9 disp ( R e s u l t ) 30 3 disp ( A( 5 ) =. 8 9 e r l a n g s ) 3 33 disp ( A( r e m a i n i n g ) = 0. e r l a n g s ) 37
Scilab code Exa.5 Program to calculate how much traffic can the trunk group carry 3 // Example 5. 5 4 5 // Page 534 6 7 // R e f e r f i g u r e. 5 on page 533 8 9 disp ( From f i g, i t can be t h a t the output c i r c u i t u t i l i z a t i o n f o r B=0. and N=4 i s 0. 8. ) 0 N =4 3 op =0.8 4 5 disp ( Thus the c a r r i e d t r a f f i c i n t e n s i t y i s ) 6 7 N*op 8 9 disp ( S i n c e the b l o c k i n g p r o b a b i l i t y i s 0., the maximum l e v e l o f o f f e r e d t r a a f f i c i s, ) 0 A =[9./( -0.) ] 3 disp ( R e s u l t ) 4 5 disp ( A =. 3 e r l a n g s ) 38