Problem-1Problem 2problem3Problem-4

Question

Robert · Accepted Answer

Problem-1
ANSWER 
(a)  
Formula for calculating  
P(e) = Q(√
   
  
 )  1.647 so we get, 
√
   
  
  > 1.647 on solving for Eb/No we get 
Eb/No > 1.647
2/2 = 1.356  
Minimum Eb/No = 1.356   Answer  
(b)  
We are given that  
spectral efficiency G = 1chip/sec/Hz, Bandwidth (Bw) = 3 MHz,  Chip energy Ec=No, 1kbit/user, 
and P(e) 1.356 Ec  
So, 
N > 1.356  
So minimum processing  gain will be greater 1.356 chip / bit  or since chips are produced in 
integers so we can also write as 2 chips/bit. 
Number of users  
For calculating  maximum no of users we need to know its maximum processing zone so , 
Data rate for single user: 1 K bits/sec 
Maximum processing gain: 3 M/1K=3000 chips/bit  ≥ N 
Thus 3000  ≥N  ≥2 
Since an orthogonal code can be used in synchronous system, no multi access interference. 
Thus, the number of users is equal to the number of codes. We can choose N =3000 and the 
total number of users is also 3000.
Problem 2
ANSWER 
(a)  
Formula for calculating  
P(e) = Q(√
   
  
 )  2.3 so we get, 
√
   
  
  > 2.

Problem-1 Problem 2 problem3 Problem-4

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