initial commit / add all books

2 files changed, 29 insertions, 0 deletions

diff --git a/3542/CH9/EX9.4/9_4.jpg b/3542/CH9/EX9.4/9_4.jpg
new file mode 100644
index 000000000..b7a6894a7
--- /dev/null
+++ b/3542/CH9/EX9.4/9_4.jpg
diff --git a/3542/CH9/EX9.4/Ex9_4.sce b/3542/CH9/EX9.4/Ex9_4.sce
new file mode 100644
index 000000000..afe152911
--- /dev/null
+++ b/3542/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,29 @@
+// Example no 9.4

+// To find a)the time duration of a bit b)the time duration of a slot c)the time duration of a frame d)how long must a user occupying single time slot wait between two successive transmission

+// Page no. 456

+

+clc;

+clear all;

+

+// Given data

+N=8;                                                        // Number of time slots in each frame

+Nb=156.25;                                                  // Number of in each time slot

+DR=270.833*10^3;                                            // Data rate of transmission in channel

+

+// a)To find the time duration of a bit 

+Tb=1/DR;                                                   // The time duration of a bit in sec

+

+// b)To find the time duration of a slot

+Tslot=Nb*Tb;                                              // The time duration of a slot

+

+// c)To find the time duration of a frame

+Tf=N*Tslot;                                               // The time duration of a frame

+

+//d) The waiting time between two successive transmission

+Tw=Tf;                                                   // The arrival time of new frame for its next transmission

+

+// Displaying the result in command window

+printf('\n The time duration of a bit = %0.3f microseconds',Tb*10^6);

+printf('\n The time duration of a slot = %0.3f ms',Tslot*10^3);

+printf('\n The time duration of a frame = %0.3f ms',Tf*10^3);

+printf('\n The arrival time of new frame for its next transmission = %0.3f ms',Tw*10^3);