updated the code

26 files changed, 660 insertions, 689 deletions

diff --git a/116/CH10/EX10.1/exa10_1.sce b/116/CH10/EX10.1/exa10_1.sce
index e0dd9a4b2..bfd21704d 100755
--- a/116/CH10/EX10.1/exa10_1.sce
+++ b/116/CH10/EX10.1/exa10_1.sce
@@ -1,42 +1,39 @@
- 

-//Caption:Program to determine the amount of transmission capacity 

- 

-//(a)Assume the link-by-link error control (b) Assume end-to-end error control (c)Repeat the calculation for a bit error probability of 10^-5

- 

-//Example 10.1

- 

-//Page 472

- 

-//(a)With link-by-link 

- 

-frame=1000*10^-8

-

-disp('The expected number of bits of transmission capacity required to retransmit is')

-

-frame*1000

-

-//(b)With end-to-end

- 

-frames=10*10^-5//corrupted frame

-

-disp('The expected number of bits of transmission capacity required is')

-

-frames*1000 

- 

-//(c)With bit error 10^-5

- 

-ans1=1000*10^-5 

- 

-ans1=1000*10^-5*1000

-

-ans=10*10^-2*1000

- 

-//Result

- 

-//(a)0.01 bit/link

- 

-//(b)0.1 bit/link

- 

-//(c)1. 10 bits/link

- 

-//(c)2. 100 bits/link

+
+ 
+//Example 10.1
+ 
+//Page 472
+ 
+//(a)With link-by-link 
+ 
+frame=1000*10^-8
+
+disp('The expected number of bits of transmission capacity required to retransmit is')
+
+frame*1000
+
+//(b)With end-to-end
+ 
+frames=10*10^-5//corrupted frame
+
+disp('The expected number of bits of transmission capacity required is')
+
+frames*1000 
+ 
+//(c)With bit error 10^-5
+ 
+ans1=1000*10^-5 
+ 
+ans1=1000*10^-5*1000
+
+ans=10*10^-2*1000
+ 
+//Result
+ 
+//(a)0.01 bit/link
+ 
+//(b)0.1 bit/link
+ 
+//(c)1. 10 bits/link
+ 
+//(c)2. 100 bits/link
diff --git a/116/CH10/EX10.3/exa10_3.sce b/116/CH10/EX10.3/exa10_3.sce
index 4dc91f9c9..033358474 100755
--- a/116/CH10/EX10.3/exa10_3.sce
+++ b/116/CH10/EX10.3/exa10_3.sce
@@ -1,21 +1,20 @@
-

-//Caption:Program to determine the probability that the delay of an ATM voice cell 

- 

-//Example 10.3

- 

-//Page 488

- 

-disp('Assuming the access link is 90% utilized on average.')

- 

-disp('The queuing theory is provided in Chapter 12. It involves determining the probability that the DSI access queue contains enough cells to represent 10msec of transmission time')

- 

-tm=[(53*8)/(192*8000)]

-

-disp('Therefore, 10 msec delay represents 10/0.276 = 36.2 cell times.')

-

-p=(0.9)*{%e^[-(1-0.9)*36.2]}//Refer to equation 12.25 in chap 12 

- 

-disp("Result")

- 

-disp("P(>10msec) = 2.5% delay will be displayed by more than 10 msec ")

+
+ 
+//Example 10.3
+ 
+//Page 488
+ 
+disp('Assuming the access link is 90% utilized on average.')
+ 
+disp('The queuing theory is provided in Chapter 12. It involves determining the probability that the DSI access queue contains enough cells to represent 10msec of transmission time')
+ 
+tm=[(53*8)/(192*8000)]
+
+disp('Therefore, 10 msec delay represents 10/0.276 = 36.2 cell times.')
+
+p=(0.9)*{%e^[-(1-0.9)*36.2]}//Refer to equation 12.25 in chap 12 
+ 
+disp("Result")
+ 
+disp("P(>10msec) = 2.5% delay will be displayed by more than 10 msec ")
  
\ No newline at end of file
diff --git a/116/CH11/EX11.1/exa11_1.sce b/116/CH11/EX11.1/exa11_1.sce
index 56750b986..45a82df94 100755
--- a/116/CH11/EX11.1/exa11_1.sce
+++ b/116/CH11/EX11.1/exa11_1.sce
@@ -1,30 +1,28 @@
-  

-//Caption:Program to determine the distance limit imposed by the need to echo E bit in a BRI S/T interface

- 

-//Example 11.1

- 

-//Page 501

- 

-//Refer to figure 11.5 on page 500

- 

-disp('By seeing the figure, it can be seen that the minimum delay between a terminal transmitting D bit and receiving it back in the following E bit is seven bit times')

-

-disp('At a 192 kbps data rate the duration of bit is 5.2 usec. Thus, the total round trip propagation time is ')

-

-7*5.2//usec

-

-disp('Assuming no appreciable circuitry delays in the NT,')

-

-c=3*10^8// speed of light

-

-Lmax=(36.4*10^-6)*(1/3)*c

-

-disp('Because round trip propagation involves both direction of transmission')

- 

-Dmax=(1/2)*Lmax

-

-disp("Result")

- 

-disp("Maximum length of wire(Lmax) = 3640 m = 3.64 km")

- 

-disp("Maximum distance(Dmax)= 1820 m = 1.82 km")

+
+//Example 11.1
+ 
+//Page 501
+ 
+//Refer to figure 11.5 on page 500
+ 
+disp('By seeing the figure, it can be seen that the minimum delay between a terminal transmitting D bit and receiving it back in the following E bit is seven bit times')
+
+disp('At a 192 kbps data rate the duration of bit is 5.2 usec. Thus, the total round trip propagation time is ')
+
+7*5.2//usec
+
+disp('Assuming no appreciable circuitry delays in the NT,')
+
+c=3*10^8// speed of light
+
+Lmax=(36.4*10^-6)*(1/3)*c
+
+disp('Because round trip propagation involves both direction of transmission')
+ 
+Dmax=(1/2)*Lmax
+
+disp("Result")
+ 
+disp("Maximum length of wire(Lmax) = 3640 m = 3.64 km")
+ 
+disp("Maximum distance(Dmax)= 1820 m = 1.82 km")
diff --git a/116/CH11/EX11.2/exa11_2.sce b/116/CH11/EX11.2/exa11_2.sce
index 665583926..f8a703198 100755
--- a/116/CH11/EX11.2/exa11_2.sce
+++ b/116/CH11/EX11.2/exa11_2.sce
@@ -1,19 +1,17 @@
-  

-//Caption:Program to determine the theoretical maximum data rate of a prefectly equalized voiceband modem

- 

-//Example 11.2

- 

-//Page 513

- 

-disp('The signal-to-quantizing-noise ratio(SQR) is given in chap3 to be on the order of 36dB, which corresponds to power ratio of 3981.')

- 

-disp('Using this value in Shannon theorem for the theoretical capacity of a channel yield,') 

- 

-SNR=3981

-

-C=3100*[log2(1+SNR)]

-

-disp("Result")

- 

-disp("data rate = 37 kbps")

-

+   
+//Example 11.2
+ 
+//Page 513
+ 
+disp('The signal-to-quantizing-noise ratio(SQR) is given in chap3 to be on the order of 36dB, which corresponds to power ratio of 3981.')
+ 
+disp('Using this value in Shannon theorem for the theoretical capacity of a channel yield,') 
+ 
+SNR=3981
+
+C=3100*[log2(1+SNR)]
+
+disp("Result")
+ 
+disp("data rate = 37 kbps")
+
diff --git a/116/CH12/EX12.1/exa12_1.sce b/116/CH12/EX12.1/exa12_1.sce
index 1032ecdda..b604ac666 100755
--- a/116/CH12/EX12.1/exa12_1.sce
+++ b/116/CH12/EX12.1/exa12_1.sce
@@ -1,23 +1,21 @@
-

-//Caption:Program to calculate how often do two calls arrive with less than 0.01 sec between them

- 

-//Example 12.1

- 

-//Page 524

- 

-disp('The average arrival rate is')

-

-lam=(3600/10000)//arrivals per sec

-

-disp('From equation 12.2, the probability on arrival in 0.01-sec interval is')//equation on page 524

-

-P0=(%e^-0.0278)

-

-disp('Thus 2.7% arrivals occur withnin 0.01 sec of the pervious arrival. Since the arrival rate is 2.78 arrivals per second, the rate of occurrence of intervarrival time less than 0.01 sec is')

-

-2.78*0.027

-

-disp("Result")

- 

-disp("0.075 times/sec")

+
+//Example 12.1
+ 
+//Page 524
+ 
+disp('The average arrival rate is')
+
+lam=(3600/10000)//arrivals per sec
+
+disp('From equation 12.2, the probability on arrival in 0.01-sec interval is')//equation on page 524
+
+P0=(%e^-0.0278)
+
+disp('Thus 2.7% arrivals occur withnin 0.01 sec of the pervious arrival. Since the arrival rate is 2.78 arrivals per second, the rate of occurrence of intervarrival time less than 0.01 sec is')
+
+2.78*0.027
+
+disp("Result")
+ 
+disp("0.075 times/sec")
  
\ No newline at end of file
diff --git a/116/CH12/EX12.2/exa12_2.sce b/116/CH12/EX12.2/exa12_2.sce
index 638ae0ef0..e5e5dd433 100755
--- a/116/CH12/EX12.2/exa12_2.sce
+++ b/116/CH12/EX12.2/exa12_2.sce
@@ -1,35 +1,33 @@
-

-//Caption:Program to calculate the probability that eight or more arrivals occur in an chosen 30 sec

- 

-//Example 12.2

- 

-//Page 526

- 

-disp('The average number of arrivals in a 30 sec interval is,')

-

-lamt=4*(30/60)

-

-disp('The probability of eight or more arrivals is,')

-

-P0=1

-

-P1=[(2^1)/(1)]

-

-P2=[(2^2)/(1*2)]

-

-P3=[(2^3)/(1*2*3)]

-

-P4=[(2^4)/(1*2*3*4)]

-

-P5=[(2^5)/(1*2*3*4*5)]

-

-P6=[(2^6)/(1*2*3*4*5*6)]

-

-P7=[(2^7)/(1*2*3*4*5*6*7)]

-

-i=1-{(%e^-2)*[P0+P1+P2+P3+P4+P5+P6+P7]}

-

-disp("Result")

- 

-disp("P(2) = 0.0011")

+
+//Example 12.2
+ 
+//Page 526
+ 
+disp('The average number of arrivals in a 30 sec interval is,')
+
+lamt=4*(30/60)
+
+disp('The probability of eight or more arrivals is,')
+
+P0=1
+
+P1=[(2^1)/(1)]
+
+P2=[(2^2)/(1*2)]
+
+P3=[(2^3)/(1*2*3)]
+
+P4=[(2^4)/(1*2*3*4)]
+
+P5=[(2^5)/(1*2*3*4*5)]
+
+P6=[(2^6)/(1*2*3*4*5*6)]
+
+P7=[(2^7)/(1*2*3*4*5*6*7)]
+
+i=1-{(%e^-2)*[P0+P1+P2+P3+P4+P5+P6+P7]}
+
+disp("Result")
+ 
+disp("P(2) = 0.0011")
  
\ No newline at end of file
diff --git a/116/CH12/EX12.3/exa12_3.sce b/116/CH12/EX12.3/exa12_3.sce
index 742b5f27b..2044e15e8 100755
--- a/116/CH12/EX12.3/exa12_3.sce
+++ b/116/CH12/EX12.3/exa12_3.sce
@@ -1,19 +1,18 @@
-  

-//Caption:Program to calculate the probability that a 1000 bit data block experiences exaclty 4 errors while being transmitted over a link having 10^-5 error rate

- 

-//Example 12.3

- 

-//Page 527

- 

-disp('Assuming inpendenterror, we can obtain the probability of exactly 4 errors directly from the Poisson distribution. The average number of errors is,')

-

-lamt=[(10^3)*(10^-5)]

-

-disp('Thus,')

-

-P4={[(0.01^4)/(1*2*3*4)]*%e^-0.01}

-

-disp("Result")

- 

-disp("P(4) = 4.125*10^-10")

+
+ 
+//Example 12.3
+ 
+//Page 527
+ 
+disp('Assuming inpendenterror, we can obtain the probability of exactly 4 errors directly from the Poisson distribution. The average number of errors is,')
+
+lamt=[(10^3)*(10^-5)]
+
+disp('Thus,')
+
+P4={[(0.01^4)/(1*2*3*4)]*%e^-0.01}
+
+disp("Result")
+ 
+disp("P(4) = 4.125*10^-10")
  
\ No newline at end of file
diff --git a/116/CH12/EX12.4/exa12_4.sce b/116/CH12/EX12.4/exa12_4.sce
index eff220a3f..7ddacf887 100755
--- a/116/CH12/EX12.4/exa12_4.sce
+++ b/116/CH12/EX12.4/exa12_4.sce
@@ -1,34 +1,33 @@
-  

-//Caption:Program to calculate the percentage of total traffic carried by first five ckt and traffic carried by all other remaining

- 

-//Example 12.4

-

-//Page 529

- 

-disp('The Traffic intensity of system is,')

-

-A=1*2

-

-disp('The raffic intensity carried by i active ckt is exactly i erlangs. Hence the traffic carried by 1st 5 ckt is,')

-

-P1=[(1*2^1)/(1)]

- 

-P2=[(2*2^2)/(1*2)]

- 

-P3=[(3*2^3)/(1*2*3)]

-

-P4=[(4*2^4)/(1*2*3*4)]

-

-P5=[(5*2^5)/(1*2*3*4*5)]

-

-A5={(%e^-2)*[P1+P2+P3+P4+P5]}

-

-disp('All of remaining ckts carry,')

-

-Ar=2-1.89

-

-disp("Result")

- 

-disp("A(5) = 1.89 erlangs")

- 

-disp("A(remaining) = 0.11 erlangs")

+  
+
+//Example 12.4
+
+//Page 529
+ 
+disp('The Traffic intensity of system is,')
+
+A=1*2
+
+disp('The raffic intensity carried by i active ckt is exactly i erlangs. Hence the traffic carried by 1st 5 ckt is,')
+
+P1=[(1*2^1)/(1)]
+ 
+P2=[(2*2^2)/(1*2)]
+ 
+P3=[(3*2^3)/(1*2*3)]
+
+P4=[(4*2^4)/(1*2*3*4)]
+
+P5=[(5*2^5)/(1*2*3*4*5)]
+
+A5={(%e^-2)*[P1+P2+P3+P4+P5]}
+
+disp('All of remaining ckts carry,')
+
+Ar=2-1.89
+
+disp("Result")
+ 
+disp("A(5) = 1.89 erlangs")
+ 
+disp("A(remaining) = 0.11 erlangs")
diff --git a/116/CH12/EX12.5/exa12_5.sce b/116/CH12/EX12.5/exa12_5.sce
index 1c77bc6db..26994b43c 100755
--- a/116/CH12/EX12.5/exa12_5.sce
+++ b/116/CH12/EX12.5/exa12_5.sce
@@ -1,27 +1,26 @@
-  

-//Caption:Program to calculate how much traffic can the trunk group carry

- 

-//Example 5.5

- 

-//Page 534

- 

-//Refer figure 12.5 on page 533

- 

-disp('From fig, it can be that the output circuit utilization for B=0.1 and N=24 is 0.8.')

-

-N=24

-

-op=0.8

- 

-disp('Thus the carried traffic intensity is ')

-

-N*op

-

-disp('Since the blocking probability is 0.1, the maximum level of offered traaffic is,')

-

-A=[19.2/(1-0.1)]

-

-disp("Result")

-

-disp("A = 21.3 erlangs")

+
+ 
+//Example 5.5
+ 
+//Page 534
+ 
+//Refer figure 12.5 on page 533
+ 
+disp('From fig, it can be that the output circuit utilization for B=0.1 and N=24 is 0.8.')
+
+N=24
+
+op=0.8
+ 
+disp('Thus the carried traffic intensity is ')
+
+N*op
+
+disp('Since the blocking probability is 0.1, the maximum level of offered traaffic is,')
+
+A=[19.2/(1-0.1)]
+
+disp("Result")
+
+disp("A = 21.3 erlangs")
  
\ No newline at end of file
diff --git a/116/CH3/EX3.1/exa3_1.sce b/116/CH3/EX3.1/exa3_1.sce
index 6dafadd7c..e2f8746de 100755
--- a/116/CH3/EX3.1/exa3_1.sce
+++ b/116/CH3/EX3.1/exa3_1.sce
@@ -1,19 +1,19 @@
-//Caption:Program to calculate quantization interval and bits needed to encode each sample

- 

-//Example 3.1

- 

-//Page 101

- 

-sqr=30//SQR=30dB

- 

-q=1*10^[-(sqr-7.78)/20] 

- 

-disp('Thus 13 quantization intervals arer needed for each polarity for a total of 26 intervals in all. The number of bitz required are determined as')

- 

-N=log2(26)

- 

-//Result

- 

-//q = 0.078 V

-

+
+ 
+//Example 3.1
+ 
+//Page 101
+ 
+sqr=30//SQR=30dB
+ 
+q=1*10^[-(sqr-7.78)/20] 
+ 
+disp('Thus 13 quantization intervals arer needed for each polarity for a total of 26 intervals in all. The number of bitz required are determined as')
+ 
+N=log2(26)
+ 
+//Result
+ 
+//q = 0.078 V
+
 //N = 4.7 = 5 bits per sample 
\ No newline at end of file
diff --git a/116/CH3/EX3.2/exa3_2.sce b/116/CH3/EX3.2/exa3_2.sce
index f2c233f69..8953c29e4 100755
--- a/116/CH3/EX3.2/exa3_2.sce
+++ b/116/CH3/EX3.2/exa3_2.sce
@@ -1,26 +1,26 @@
-//Caption:Program to calculate the minimum bit rate for a PCM encoder must provide for high fidelity

- 

-//Example 3.2

- 

-//Page 105

- 

-dr=40//dynamic range=400dB

- 

-SNR=50//signal to noise ratio =5 0dB

- 

-SQR=dr+SNR 

- 

-n=[(SQR-1.76)/6.02] 

- 

-disp('This can be approximated to 15 bits per sample')

- 

-disp('Assuming excess sampling factor using D-type channel, we choose sampling rate as 48KHz')  

- 

-disp('Therefore required bit rate is')

- 

-15*48000 

- 

-//Result

-

-//720kbps

+
+ 
+//Example 3.2
+ 
+//Page 105
+ 
+dr=40//dynamic range=400dB
+ 
+SNR=50//signal to noise ratio =5 0dB
+ 
+SQR=dr+SNR 
+ 
+n=[(SQR-1.76)/6.02] 
+ 
+disp('This can be approximated to 15 bits per sample')
+ 
+disp('Assuming excess sampling factor using D-type channel, we choose sampling rate as 48KHz')  
+ 
+disp('Therefore required bit rate is')
+ 
+15*48000 
+ 
+//Result
+
+//720kbps
  
\ No newline at end of file
diff --git a/116/CH3/EX3.4/exa3_4.sce b/116/CH3/EX3.4/exa3_4.sce
index eb0c422f7..def0f0f09 100755
--- a/116/CH3/EX3.4/exa3_4.sce
+++ b/116/CH3/EX3.4/exa3_4.sce
@@ -1,23 +1,23 @@
-//Caption:Program to calculate how many bits per sample can be saved by using DPCM

- 

-//Example 3.4

- 

-//Page 128

- 

-w=800//Omega=800Hz

- 

-//x(t)=A sin(2pi.wt), equation for sine wave with maximum amplitude

- 

-//x'(t)=A(2pi).w.cos(2pi.wt), diff w.r.t time

- 

-(2*%pi)*800*(1/8000)

- 

-//0.62831*a, x'(t)max

- 

-disp('savings in the bits per sample can be determined as ')

- 

-log2(1/0.628)

-  

-//Result

- 

+
+ 
+//Example 3.4
+ 
+//Page 128
+ 
+w=800//Omega=800Hz
+ 
+//x(t)=A sin(2pi.wt), equation for sine wave with maximum amplitude
+ 
+//x'(t)=A(2pi).w.cos(2pi.wt), diff w.r.t time
+ 
+(2*%pi)*800*(1/8000)
+ 
+//0.62831*a, x'(t)max
+ 
+disp('savings in the bits per sample can be determined as ')
+ 
+log2(1/0.628)
+  
+//Result
+ 
 //0.67 bits
\ No newline at end of file
diff --git a/116/CH5/EX5.1/exa5_1.sce b/116/CH5/EX5.1/exa5_1.sce
index ab46c0d28..e1c2ef58c 100755
--- a/116/CH5/EX5.1/exa5_1.sce
+++ b/116/CH5/EX5.1/exa5_1.sce
@@ -1,21 +1,21 @@
-//Caption:Program to find the idle path in a three stage 8192 line switch

- 

-//Example 5.1

- 

-//Page243

- 

-//Refer to table 5.2 on page236

- 

-disp('From the table, space expansion factor of 0.234 is 0.002. Hence the utilization of each interstage is given by')

- 

-0.1/0.234

-

-p=1-[(1-0.427)^2]// probability that one of two links in series is busy

-

-disp('Therefore, the expected number of paths to be tested are,')

- 

-Np=[1-(0.672)^15]/(1-0.672)

- 

-//Result

- 

+
+ 
+//Example 5.1
+ 
+//Page243
+ 
+//Refer to table 5.2 on page236
+ 
+disp('From the table, space expansion factor of 0.234 is 0.002. Hence the utilization of each interstage is given by')
+ 
+0.1/0.234
+
+p=1-[(1-0.427)^2]// probability that one of two links in series is busy
+
+disp('Therefore, the expected number of paths to be tested are,')
+ 
+Np=[1-(0.672)^15]/(1-0.672)
+ 
+//Result
+ 
 //Only 3 of the 15 paths should be tested before an idle path is found
\ No newline at end of file
diff --git a/116/CH5/EX5.2/exa5_2.sce b/116/CH5/EX5.2/exa5_2.sce
index 065ab7a3b..91fc8d0cf 100755
--- a/116/CH5/EX5.2/exa5_2.sce
+++ b/116/CH5/EX5.2/exa5_2.sce
@@ -1,36 +1,36 @@
-//Caption:Program to determine the implementaton complexity of the TS switch

- 

-//Example 5.2

-

-//Page 253

-

-//Refer to figure 5.19 on page 252

- 

-N=80//Number of links  

- 

-Nc=24//Number of control words 

- 

-Nb1=7//Number of bits per control word 

- 

-Nb2=5//Number of bits per control word

-

-disp('The number of crosspoints in the space stage is')

- 

-Nx=N^2

- 

-disp('The total number of memory bits for the space stage control store is')

- 

-Nbx=N*Nc*Nb1

- 

-disp('The total number of memory bits for the time stage is')

- 

-Nbt=(N*Nc*8)+(N*Nc*Nb2)

- 

-disp('Thus the implementation complexity is ')

- 

-Cmplx=Nx+[(Nbx+Nbt)/100]

-

-//Result

- 

-//Complexity is 6784 equivalent crosspoint.

+
+ 
+//Example 5.2
+
+//Page 253
+
+//Refer to figure 5.19 on page 252
+ 
+N=80//Number of links  
+ 
+Nc=24//Number of control words 
+ 
+Nb1=7//Number of bits per control word 
+ 
+Nb2=5//Number of bits per control word
+
+disp('The number of crosspoints in the space stage is')
+ 
+Nx=N^2
+ 
+disp('The total number of memory bits for the space stage control store is')
+ 
+Nbx=N*Nc*Nb1
+ 
+disp('The total number of memory bits for the time stage is')
+ 
+Nbt=(N*Nc*8)+(N*Nc*Nb2)
+ 
+disp('Thus the implementation complexity is ')
+ 
+Cmplx=Nx+[(Nbx+Nbt)/100]
+
+//Result
+ 
+//Complexity is 6784 equivalent crosspoint.
  
\ No newline at end of file
diff --git a/116/CH5/EX5.3/exa5_3.sce b/116/CH5/EX5.3/exa5_3.sce
index 63592b8f8..6c501c2c0 100755
--- a/116/CH5/EX5.3/exa5_3.sce
+++ b/116/CH5/EX5.3/exa5_3.sce
@@ -1,23 +1,23 @@
-

-//Caption:Program to determine the implementaton complexity of a 2048 channel

- 

-//Example 5.3

- 

-//Page 256

- 

-k=7//from equation 5.14 on page 256

- 

-disp('Using the value of k')

- 

-disp('Using the value of k, the number of crosspoint determined are')

- 

-2*7*16

-

-disp('The number of bits of memory can be determined are')

- 

-N=[2*7*128*4]+[7*128*8]+[7*128*7]

-

-//Result

- 

-//The composite implementation complecity is 430 equivalent crosspoints.

+
+
+ 
+//Example 5.3
+ 
+//Page 256
+ 
+k=7//from equation 5.14 on page 256
+ 
+disp('Using the value of k')
+ 
+disp('Using the value of k, the number of crosspoint determined are')
+ 
+2*7*16
+
+disp('The number of bits of memory can be determined are')
+ 
+N=[2*7*128*4]+[7*128*8]+[7*128*7]
+
+//Result
+ 
+//The composite implementation complecity is 430 equivalent crosspoints.
  
\ No newline at end of file
diff --git a/116/CH5/EX5.5/exa5_5.sce b/116/CH5/EX5.5/exa5_5.sce
index 6385f0ade..3c720dd34 100755
--- a/116/CH5/EX5.5/exa5_5.sce
+++ b/116/CH5/EX5.5/exa5_5.sce
@@ -1,31 +1,31 @@
-

-//Caption:Program to determine the implementaton complexity of a 131,072 channel

-

-//Example 5.5

- 

-//Page 261

- 

-disp('The space switch can be designed bt take the first space stage. A value of 32 is chosen as a convenient binary number.')

-

-n=32//binary w.r.t (N/2)^2

-

-k=27//determined as a blocking probability of 0.0015

-

-//Refer equations 5.18 and 5.19

- 

-Nx=[2*1024*27]+[27+(32^2)]

-

-Nx=[2*1024*27]+[27*(32^2)]

-

-Nbx=[2*27*128*32*5]+{27*128*32*5}

-

-Nbt=[2*1024*128*8]

- 

-Nbtc=[2*1024*128*7]

- 

-cmplx=[Nx+{(Nbx+Nbt+Nbtc)/100}]

- 

-//Result

-

-//Complexity is 138,854 equivalent crosspoint.

+
+
+
+//Example 5.5
+ 
+//Page 261
+ 
+
+
+n=32//binary w.r.t (N/2)^2
+
+k=27//determined as a blocking probability of 0.0015
+
+//Refer equations 5.18 and 5.19
+ 
+Nx=[2*1024*27]+[27+(32^2)]
+
+Nx=[2*1024*27]+[27*(32^2)]
+
+Nbx=[2*27*128*32*5]+{27*128*32*5}
+
+Nbt=[2*1024*128*8]
+ 
+Nbtc=[2*1024*128*7]
+ 
+cmplx=[Nx+{(Nbx+Nbt+Nbtc)/100}]
+ 
+//Result
+
+//Complexity is 138,854 equivalent crosspoint.
  
\ No newline at end of file
diff --git a/116/CH6/EX6.4/exa6_4.sce b/116/CH6/EX6.4/exa6_4.sce
index 314e151dd..31bf871f9 100755
--- a/116/CH6/EX6.4/exa6_4.sce
+++ b/116/CH6/EX6.4/exa6_4.sce
@@ -1,35 +1,31 @@
-

-//Caption: Program to determine system gain of 10Mbps, 2Ghz digital microwave repeater using 4 PSK modulation

- 

-//Example 6.4

- 

-//Page 325

- 

-//Refer to figure 6.17 on page 300

- 

-disp('From figure, Eb/N0 for 4 psk modulation can be determined 10.7 dB.')

-

-//Using equation 3.24 from Appendix C

- 

-disp('SNR detector is 3dB higher than Eb/N0, therefore')

-

-snr=13.7//SNR=13.7dB

-

-disp('Since 4 PSK modulation provides 2bps/Hz, the sampling rate is 5 MHz, which is Nqyuist rate, therefore')

-

-a1=10*log10(125000000000000)

-

-a2=10*log10(1.3)

- 

-A0=a1-13.7-7-3-a2

-

-disp('At a carrier frequency of 2GHz, the wavelength is')

-

-(3*10^8)/(2*10^9)

- 

-FM=116+60+20*log10(0.15)-5-20*log10(4*%pi*5*10^4)//Fade Margin can be found by Equation 6.31

-

-//Result

-//A0 = 116dB

-//wavelength = 0.15 m

+
+
+//Example 6.4
+ 
+//Page 325
+ 
+//Refer to figure 6.17 on page 300
+ 
+
+disp('SNR detector is 3dB higher than Eb/N0, therefore')
+
+snr=13.7//SNR=13.7dB
+
+disp('Since 4 PSK modulation provides 2bps/Hz, the sampling rate is 5 MHz, which is Nqyuist rate, therefore')
+
+a1=10*log10(125000000000000)
+
+a2=10*log10(1.3)
+ 
+A0=a1-13.7-7-3-a2
+
+disp('At a carrier frequency of 2GHz, the wavelength is')
+
+(3*10^8)/(2*10^9)
+ 
+FM=116+60+20*log10(0.15)-5-20*log10(4*%pi*5*10^4)//Fade Margin can be found by Equation 6.31
+
+//Result
+//A0 = 116dB
+//wavelength = 0.15 m
 //Fade Margin = 38.5 dB
\ No newline at end of file
diff --git a/116/CH7/EX7.2/exa7_2.sce b/116/CH7/EX7.2/exa7_2.sce
index 26ef7ae20..693554624 100755
--- a/116/CH7/EX7.2/exa7_2.sce
+++ b/116/CH7/EX7.2/exa7_2.sce
@@ -1,20 +1,18 @@
-

-//Caption:Program to determine relative accuracy of maintaining a mutual slip rate ojective of one slip in 20hrs

-

-//Example 7.2

- 

-//Page 350

- 

-disp('The slip rate objective implies that thee frame rate produced by one clock can be different than the frame rate produced by the other clock by no more then')

- 

-dF=(1/20*60*60)

-

-dF=[1/(20*60*60)]

- 

-disp('Since there are 8000 frame per second, the relative accuracy is determined as')

-

-ans=[dF/8000]

- 

-//Result

- 

+
+
+//Example 7.2
+ 
+//Page 350
+ 
+
+dF=(1/20*60*60)
+
+dF=[1/(20*60*60)]
+ 
+disp('Since there are 8000 frame per second, the relative accuracy is determined as')
+
+ans=[dF/8000]
+ 
+//Result
+ 
 //Hence the clock must be accurate to 1.7 parts in 10^9.
\ No newline at end of file
diff --git a/116/CH7/EX7.3/exa7_3.sce b/116/CH7/EX7.3/exa7_3.sce
index 775b3085c..d9527962a 100755
--- a/116/CH7/EX7.3/exa7_3.sce
+++ b/116/CH7/EX7.3/exa7_3.sce
@@ -1,27 +1,26 @@
-  

-//Caption:Program to determine the minimum and maximum input channel rate accommodated by an M12 multiplexer

- 

-//Example 7.3

- 

-//Page 354

- 

-disp("The maximum information rate per channel is determined as")

- 

-Imax=[(6.312*288)/1176]

-

-disp('The minimum information rate per channel is determined as')

- 

-Imin=[(6.312*287)/1176]

-

-disp('Since there are three possible combinations of two errors in the C bits, the probability of misinterpreting an S bit is')

-

-3*(10^-6)^2

-

-1176/6.312//duration of each master frame

-

-[(3*10^-12)/(186*10^-6)]

-

-//Result

- 

-//0.016*10^-6 misframes per second

+  
+
+//Example 7.3
+ 
+//Page 354
+ 
+disp("The maximum information rate per channel is determined as")
+ 
+Imax=[(6.312*288)/1176]
+
+disp('The minimum information rate per channel is determined as')
+ 
+Imin=[(6.312*287)/1176]
+
+disp('Since there are three possible combinations of two errors in the C bits, the probability of misinterpreting an S bit is')
+
+3*(10^-6)^2
+
+1176/6.312//duration of each master frame
+
+[(3*10^-12)/(186*10^-6)]
+
+//Result
+ 
+//0.016*10^-6 misframes per second
  
\ No newline at end of file
diff --git a/116/CH8/EX8.1/exa8_1.sce b/116/CH8/EX8.1/exa8_1.sce
index 89f07d16f..a672daf52 100755
--- a/116/CH8/EX8.1/exa8_1.sce
+++ b/116/CH8/EX8.1/exa8_1.sce
@@ -1,27 +1,26 @@
-  

-//Caption:Program to determine the loss limit and the multimode dispersion limit of a graded index FOC

- 

-//Example 8.1

-

-//Page 388

-

-//Refer to figure 8.2 on page 385

-

-Pin=42//input power = 42dB

-

-disp('The attenuation of a multimode fiber operating at 820nm is approximately 3db/km. Thus,')

-

-A=3//attenuation

-

-LL=(Pin/A)//Loss Limit

-

-disp('Using 2 Gbps-km as typical BDP of graded index multimode fiber, the multimode dispersion distance is determined as')

- 

-Dl=(2000/90)//Dispersion limit  

- 

-//Result

- 

-//Loss Limit = 14 km

- 

-//Dispersion Limit = 22.2 km

+  
+//Caption:Program to determine the loss limit and the multimode dispersion limit of a graded index FOC
+ 
+//Example 8.1
+
+//Page 388
+
+//Refer to figure 8.2 on page 385
+
+Pin=42//input power = 42dB
+
+
+A=3//attenuation
+
+LL=(Pin/A)//Loss Limit
+
+disp('Using 2 Gbps-km as typical BDP of graded index multimode fiber, the multimode dispersion distance is determined as')
+ 
+Dl=(2000/90)//Dispersion limit  
+ 
+//Result
+ 
+//Loss Limit = 14 km
+ 
+//Dispersion Limit = 22.2 km
  
\ No newline at end of file
diff --git a/116/CH8/EX8.2/exa8_2.sce b/116/CH8/EX8.2/exa8_2.sce
index de9ce5d18..33ccb2a76 100755
--- a/116/CH8/EX8.2/exa8_2.sce
+++ b/116/CH8/EX8.2/exa8_2.sce
@@ -1,27 +1,26 @@
-  

-//Caption:Program to determine the loss limit and the chromatic dispersion limit of a high performance SMF FOC

- 

-//Example 8.2

- 

-//Page 389

- 

-//Refer figure 8.2 on page 385

- 

-disp('The attenuation of single-mode fibre operating at 1300nm is approximately 0.35dB/km. Thus,')

-

-Pin=42//input power = 42dB

-

-A=0.35

- 

-LL=(Pin/A)//Loss Limit

-

-disp('Using 250 Gbps-km as BDP of a silica single-mode fiber, the chromatic dispersion limit is determined as')

-

-Cd=(250000/417)//Chromatic dispersion limit

- 

-//Result

- 

-//Loss Limit = 120 km

- 

-//Chromatic Dispersion Limit = 599.52 = 600 km

+
+ 
+//Example 8.2
+ 
+//Page 389
+ 
+//Refer figure 8.2 on page 385
+ 
+disp('The attenuation of single-mode fibre operating at 1300nm is approximately 0.35dB/km. Thus,')
+
+Pin=42//input power = 42dB
+
+A=0.35
+ 
+LL=(Pin/A)//Loss Limit
+
+disp('Using 250 Gbps-km as BDP of a silica single-mode fiber, the chromatic dispersion limit is determined as')
+
+Cd=(250000/417)//Chromatic dispersion limit
+ 
+//Result
+ 
+//Loss Limit = 120 km
+ 
+//Chromatic Dispersion Limit = 599.52 = 600 km
  
\ No newline at end of file
diff --git a/116/CH8/EX8.3/exa8_3.sce b/116/CH8/EX8.3/exa8_3.sce
index 871086e69..64197ad3c 100755
--- a/116/CH8/EX8.3/exa8_3.sce
+++ b/116/CH8/EX8.3/exa8_3.sce
@@ -1,29 +1,29 @@
-  

-//Caption:Program to determine the BDP of SMF system and DS SMF system using DFB LD 

- 

-//Example 8.3

- 

-//Page 393

- 

-//Refer to table 8.1 on page 392, also to figure 8.6 on page 391

- 

-smf=16

-

-smf=16//dispersion co-efficient of SMF at 1550nm 

- 

-sw=0.4//spectral width of the source

- 

-BDP=[250/(smf*sw)]//assuming line code as NRZ

-

-disp('The BDP of the DS SMF system is determined as')

-

-smf=3.5//dispersion co-efficient of DS SMF at 1550nm

-

-BDP=[250/(smf*sw)]//assuming line code as NRZ 

-

-//Result

- 

-//BDP = 39 Gbps=km (SMF)

- 

-//BDP = 179 Gbps-km (DS SMF)

+  
+
+ 
+//Example 8.3
+ 
+//Page 393
+ 
+//Refer to table 8.1 on page 392, also to figure 8.6 on page 391
+ 
+smf=16
+
+smf=16//dispersion co-efficient of SMF at 1550nm 
+ 
+sw=0.4//spectral width of the source
+ 
+BDP=[250/(smf*sw)]//assuming line code as NRZ
+
+disp('The BDP of the DS SMF system is determined as')
+
+smf=3.5//dispersion co-efficient of DS SMF at 1550nm
+
+BDP=[250/(smf*sw)]//assuming line code as NRZ 
+
+//Result
+ 
+//BDP = 39 Gbps=km (SMF)
+ 
+//BDP = 179 Gbps-km (DS SMF)
  
\ No newline at end of file
diff --git a/116/CH8/EX8.4/exa8_4.sce b/116/CH8/EX8.4/exa8_4.sce
index 4eba55ae7..b7b936590 100755
--- a/116/CH8/EX8.4/exa8_4.sce
+++ b/116/CH8/EX8.4/exa8_4.sce
@@ -1,23 +1,22 @@
- 

-//Caption:Program to determine the difference in wavelength of two optical signal

- 

-//Example 8.4

- 

-//Page 402

- 

-c=3*10^8//speed of light 

- 

-wl=1500*10^-9//wavlength =1500nm  

- 

-f=[(3*10^8)/wl]  

- 

-disp('Thus the upper and lower frequencies are determined as 200,001 and 199,999 GHz respectively. The corresponding wavelengths are')

- 

-lam1=[c/(199999*10^9)]

- 

-lam2=[c/(200001*10^9)]

-

-//Result

-

-//The difference in wavelenghts is 0.015nm

+
+ 
+//Example 8.4
+ 
+//Page 402
+ 
+c=3*10^8//speed of light 
+ 
+wl=1500*10^-9//wavlength =1500nm  
+ 
+f=[(3*10^8)/wl]  
+ 
+disp('Thus the upper and lower frequencies are determined as 200,001 and 199,999 GHz respectively. The corresponding wavelengths are')
+ 
+lam1=[c/(199999*10^9)]
+ 
+lam2=[c/(200001*10^9)]
+
+//Result
+
+//The difference in wavelenghts is 0.015nm
  
\ No newline at end of file
diff --git a/116/CH8/EX8.5/exa8_5.sce b/116/CH8/EX8.5/exa8_5.sce
index 8fa8e320d..107eeb6ab 100755
--- a/116/CH8/EX8.5/exa8_5.sce
+++ b/116/CH8/EX8.5/exa8_5.sce
@@ -1,45 +1,44 @@
-  

-//Caption:Program to determine the system gain

- 

-//Example 8.5

- 

-//Page 405

- 

-//Refer to table 8.2 and figure 8.8 on page 394

- 

-dr=565//data rate

-

-wl=1550*10^-9//wavelength

-

-disp('The use of 5B6B line code implies the line data rate is,')

-

-565*(6/5)

-

-//678Mbps

- 

-disp('The receiver sensitivity for 678 Mbps is determined from fig 8.8 or table 8.2 as ')

- 

-rsen=-34.5

-

-A=(-5-rsen)//system gain 

- 

-BDP=[500/(17*0.4)]

-

-BDPs=[73.6/0.678]

- 

-lossp=(0.2+0.2)*(65)

- 

-lossm=A-lossp

- 

-//Result

- 

-//System gain = 29.5 dB

- 

-//BDP = 73.6 Gbps

- 

-//BDP spacing = 109 km

- 

-//Path Loss = 26 dB

- 

-//Loss Margin = 3.5 dB

+
+ 
+//Example 8.5
+ 
+//Page 405
+ 
+//Refer to table 8.2 and figure 8.8 on page 394
+ 
+dr=565//data rate
+
+wl=1550*10^-9//wavelength
+
+disp('The use of 5B6B line code implies the line data rate is,')
+
+565*(6/5)
+
+//678Mbps
+ 
+disp('The receiver sensitivity for 678 Mbps is determined from fig 8.8 or table 8.2 as ')
+ 
+rsen=-34.5
+
+A=(-5-rsen)//system gain 
+ 
+BDP=[500/(17*0.4)]
+
+BDPs=[73.6/0.678]
+ 
+lossp=(0.2+0.2)*(65)
+ 
+lossm=A-lossp
+ 
+//Result
+ 
+//System gain = 29.5 dB
+ 
+//BDP = 73.6 Gbps
+ 
+//BDP spacing = 109 km
+ 
+//Path Loss = 26 dB
+ 
+//Loss Margin = 3.5 dB
  
\ No newline at end of file
diff --git a/116/CH8/EX8.6/exa8_6.sce b/116/CH8/EX8.6/exa8_6.sce
index 55809510b..9aceca8ad 100755
--- a/116/CH8/EX8.6/exa8_6.sce
+++ b/116/CH8/EX8.6/exa8_6.sce
@@ -1,26 +1,24 @@
-  

-//Caption:Program to determine the range of SPE data rates that can be accomodated by the byte stuffing operation

- 

-//Example 8.6

-

-//Page 415

- 

-frames=4*9*87//Four SPE frames 

-

-rate=8*frames*2000//normal rate SPE

-

-disp('When positive byte stuffing is used to accomodate a slow incoming SPE rate, 3131 bytes of data are transmitted in four frames. Thus, the lowest slip rate is')

- 

-Rmin=8*3131*2000//minimum SPE rate

-

-disp('When negative byte stuffing is used to accomodate a fast incoming SPE rate, 3133 bytes of data are transmitted in four frames. Thus, the highest slip rate is')

-

-Rmax=8*3133*2000//maximum SPE rate

-  

-//Result

-

-//Normal SPE rate = 50.112 Mbps 

-

-//Minimum SPE rate = 50.096 Mbps

-

+
+//Example 8.6
+
+//Page 415
+ 
+frames=4*9*87//Four SPE frames 
+
+rate=8*frames*2000//normal rate SPE
+
+
+ 
+Rmin=8*3131*2000//minimum SPE rate
+
+disp('When negative byte stuffing is used to accomodate a fast incoming SPE rate, 3133 bytes of data are transmitted in four frames. Thus, the highest slip rate is')
+
+Rmax=8*3133*2000//maximum SPE rate
+  
+//Result
+
+//Normal SPE rate = 50.112 Mbps 
+
+//Minimum SPE rate = 50.096 Mbps
+
 //Maximum SPE rate = 50.128 Mbps
\ No newline at end of file
diff --git a/116/CH9/EX9.1/exa9_1.sce b/116/CH9/EX9.1/exa9_1.sce
index b53df56f5..45d170984 100755
--- a/116/CH9/EX9.1/exa9_1.sce
+++ b/116/CH9/EX9.1/exa9_1.sce
@@ -1,23 +1,22 @@
-  

-//Caption:Progam to determine the probability of maximum interference of a 64 channel CDMA system

- 

-//Example 9.1

- 

-//Page 447

- 

-disp('The probability of 63 destructive interferers is merely the probability of occurence of 63 equally likely binary events,')

- 

-Pmax=(0.5)^63//maximum probability

-

-disp('The value of a desired receive signal is the autocorrelation of a codeword with itself and can therefore be represented as a value of 64. ')

-

-disp('The mean and varience of a sum of 63 such variable are 0 and 63, respectively. The signal-to-interference ratio is now determined as,')

-

-a=[(64^2)/63]

-

-SIR=10*log10(a)

-

-//Result

- 

-//Signal to interference ratio = 18.1 dB

+  
+ 
+//Example 9.1
+ 
+//Page 447
+ 
+disp('The probability of 63 destructive interferers is merely the probability of occurence of 63 equally likely binary events,')
+ 
+Pmax=(0.5)^63//maximum probability
+
+disp('The value of a desired receive signal is the autocorrelation of a codeword with itself and can therefore be represented as a value of 64. ')
+
+disp('The mean and varience of a sum of 63 such variable are 0 and 63, respectively. The signal-to-interference ratio is now determined as,')
+
+a=[(64^2)/63]
+
+SIR=10*log10(a)
+
+//Result
+ 
+//Signal to interference ratio = 18.1 dB
  
\ No newline at end of file