Text: Wireless Communication Andrea Molisch. Problem-1
Problem-2
Problem-3
Problem-4
Problem-5
Problem-6

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Text: Wireless Communication Andrea Molisch. Problem-1Problem-2Problem-3Problem-4Problem-5Problem-6

David · Accepted Answer

Text: Wireless Communication Andrea Molisch. Problem-1
Answer: 
a)At the input we need to consider both Man-made noise and the thermal noise. 
Thermal Noise = k*To= -174 dBm/Hz 
Since, noise powers cannot be added in dB , we convert them to mW/Hz. 
Thermal noise: No= 10
-(174/10)
mW/Hz = 4.00 x 10
-18
 mW/Hz 
Man-made noise: Nmm= No, dBm/Hz +10 dB = -164 dBm/Hz = 10^(-164/10 )mW/Hz  =  3.981*10
-17
mW/Hz 
Adding the two noises, 4.00 x 10
-18
+ 3.981*10
-17
mW/Hz = 4.3981*10
-17
mW/Hz = -163.5676 dBm/Hz 
BandWidth: 10 * log10(2 * (10^6)) = 63.0103 dB 
Input Noise: Nin = No + Nmm + 10log10(B) = -163.5676+63.0103 = -100.5573 dBm 
Carrier to Noise ratio: (CNRin) = Cin –Nin = (-85 dBm) – (-100.5573 dBm) =15.5573 dB 
b) 
Taking care of inter-conversions between dB and non dB values, 
 F= 10
(NF/10)
= 10
(6/10)
=  3.981 
Eq. system noise at IP: N1 = (F-1)No = (3.981-1)*(4.00 x 10
-18
 mW/Hz)= 1.1924*10
-17
 mW/Hz
Equivalent noise at IP: No+N1=4.00 *10
-18
 mW/Hz +1.1924*10
-17
 mW/Hz =1.5924 *10
-17
 mW/Hz
Equivalent noise at IP in dB:  
= 10*log10(No+N1) = -167.98 dBm 
=No,dB+NF =-174dBm+6dB= -168dBm 
c) Total:  
Nt =   No + Nmm + N1 =  4.00 x 10
-18
 + 3.981*10
-17
mW/Hz  + 1.1924*10
-17
  mW/Hz  =5.5734 x 10
(-17)
 
mW/Hz =  -162.

Text: Wireless Communication Andrea Molisch. Problem-1 Problem-2 Problem-3 Problem-4 Problem-5 Problem-6

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