# Chapter 8 : Noise in Amplitude Modulation System¶

## Example 8.1 Page No : 436¶

In :
#Given frequency range fc =  1MHz to fc  =  1.0005Mhz
#Single side message bandwidth is fM
fM =  (1.0005 - 1)*10**6;
print 'Message bandwidth is ',fM,' Hz'
#The textbook contains a calculation error here. The calculated fM in the textbook is 500kHz instead of 5kHz, following which all the solutions are erroneous

#Given input signal strength Si =  1mW
#Let output signal strength be So
#So = Si/4
Si =  10.**(-3);
So =  Si/4;
print 'Signal output strength is ',So,' dB'

#Given Power Spectral Density n  =  10**-9 W/Hz
#Let output noise strength be No
n =  10.**-9;
No =  (n*fM)/4;
print 'Output Noise Strength is ',No,' dB'

#Let SNR at filter output be SNR
SNR =  So / No;
print 'Output SNR is ',SNR,' dB'

#By reduction of message signal Bandwidth the Output Noise strength changes
#Let the new output noise strength, bandwidth and SNR be be No_new, fM_new and SNR_new respectively
fM_new  =  75./100*fM;
No_new  =  n*fM_new/4;
SNR_new  =  So / No_new;

print 'Changed SNR is %.4f'%SNR_new,' dB'

Message bandwidth is  500.0  Hz
Signal output strength is  0.00025  dB
Output Noise Strength is  1.25e-07  dB
Output SNR is  2000.0  dB
Changed SNR is 2666.6667  dB


## Example 8.2 Page No : 436¶

In :
#Given frequency range fc - fm  =  0.995MHz to fc + fm  =  1.005Mhz
#Double side message bandwidth is fM
fM =  (1.005 - 0.995)*10**6 / 2;
print 'Message bandwidth is ',fM,' Hz'
#The textbook contains a calculation error here.
#The calculated fM in the textbook is 500kHz instead of 5kHz,
#Following which all the solutions obtained here are erroneous.

#Given input signal strength Si =  1mW
#Let output signal strength be So
#So = Si/2
Si =  10.**(-3);
So =  Si/2;
print 'Signal output strength is ',So,' dB'

#Given Power Spectral Density n  =  10**-9 W/Hz
#Let output noise strength be No
n =  10.**-9;
No =  (n*fM)/2;
print 'Output Noise Strength is ',No,' dB'

#Let SNR at filter output be SNR
SNR =  So / No;
print 'Output SNR of the DSB-SC wave is ',SNR,' dB'

#By reduction of message signal Bandwidth the Output Noise strength changes
#Let the new output noise strength, bandwidth and SNR be be No_new, fM_new and SNR_new respectively
fM_new  =  75./100*fM;
No_new  =  n*fM_new/4;
SNR_new  =  So / No_new;
print 'Changed SNR is %.4f'%SNR_new,' dB'

Message bandwidth is  5000.0  Hz
Signal output strength is  0.0005  dB
Output Noise Strength is  2.5e-06  dB
Output SNR of the DSB-SC wave is  200.0  dB
Changed SNR is 533.3333  dB


## Example 8.3 Page No : 446¶

In :
#Given bandwidth of signal is fM  =  4kHZ
fM  =  4.*10**3;
#Given power spectral density of white noise n  =  2*10**-9 W/Hz
n  =  2.*10**-9;
#Also given that minimum output SNR is 40dB
#Signal undergoes a loss of 30dB

#For SSB:
# Required minimum output SNR  =  Si_min_SSB / (n*fM)  =  40 dB  =  10**4
Si_min_SSB  =  (10.**4)*n*fM;
# Required minimum signal strength at transmitter output Si_tran  =  Si_min * 30 dB
Si_tran_SSB  =  Si_min_SSB * 10**3;
print 'Required minimum SSB signal strength at transmitter output is',Si_tran_SSB,' W'

#For DSB-SC:
# Required minimum output SNR  =  (Si_min_DSB/3) / (n*fM)  =  40 dB  =  10**4
Si_min_DSB  =  3*(10**4)*n*fM;
# Required minimum signal strength at transmitter output Si_tran  =  Si_min * 30 dB
Si_tran_DSB  =  Si_min_DSB * 10**3;
print 'Required minimum DSB signal strength at transmitter output is',Si_tran_DSB,' W'

Required minimum SSB signal strength at transmitter output is 80.0  W
Required minimum DSB signal strength at transmitter output is 240.0  W


## Example 8.4 Page No : 447¶

In :
#Given bandwidth of signal is fM  =  60 kHZ
fM  =  60.*10**3;

#Given power spectral density of white noise n  =  2*10**-6 W/Hz
n  =  2.*10**-6;

#Given time average of square of mssg signal P  =  0.1W
P  =  0.1;

#Noise power at input baseband range NM
NM  =  n * fM;

#Threshold occurs at carrier power Pc  =  2.9 * NM
Pc_Threshold  =  2.9 * NM;

#For carrier power Pc  =  10W, output SNR
Pc  =  10.;
SNRo  =  Pc * P / NM ;
print 'Output SNR is %.4f'%SNRo,' dB'

#Carrier power is reduced by 100 times making the new power Pc_new
Pc_new  =  Pc / 100;

#In the given solutions the NM value is 1.2W instead of 0.12W

Output SNR is 8.3333  dB