# Chapter 11 The Transmission of binary data in communication systems¶

## Example 11.1 Page no 392¶

In [2]:
#Given
t=0.0016
No_words=256.0
bits_word = 12.0

#Calculation
tword= t/No_words
tbit = tword/bits_word
bps =1/tbit

#Result
print"(a) The time duration of the word ",tword*10**8,"microsecond"
print"(b) The time duration of the one bit is ",round(tbit*10**8,4),"microseconds"
print"(c) The speed of transmission is ",bps/10**5,"kbps"

(a) The time duration of the word  625.0 microsecond
(b) The time duration of the one bit is  52.0833 microseconds
(c) The speed of transmission is  19.2 kbps


## Example 11.2 Page no 400¶

In [5]:
#Given
B=12.5*10**3
SN_dB= 25

#Calculation
import math
C_th = 2*B
SN=316.2
C =B*3.32*log10(SN+1)
N= 2**(C/(2.0*B))

#Result
print"(a) The maximum theorotical data rate is ",C_th/10**3,"kbps"
print"(b) The maximum theorotical capacity of channel is ",round(C/10**3,1),"Kbps"
print"(c) The number of levels needed to acheive maximum speed are ",round(N,2)

(a) The maximum theorotical data rate is  25.0 kbps
(b) The maximum theorotical capacity of channel is  103.8 Kbps
(c) The number of levels needed to acheive maximum speed are  17.78


## Example 11.3 Page no 430¶

In [3]:
#Given
block =512
packets =8
BER = 2*10**-4

#Calculation
avg_errors = block*packets*8*BER

print"Average number of errors are ",avg_errors

Average number of errors are  6.5536