Chapter 4 Noise

Example 4.2.1,Pg.no 120

In [1]:
from math import sqrt
T=290.0
BW=1*10**6          #Noise bandwidth in hertz
k=1.38*10**-23      #Boltzman constant in J/K
R=50.0              #Determination of thermal noise power Pn
Pn=k*T*BW
#Pn=round(Pn,1)
print 'The value of thermal noise power is',Pn,'W'
#Determination of RMS noise voltage
En=sqrt(4*R*k*T*BW)
En=En*(10**6)
En=round(En,2)
print 'The value of RMS noise voltage is',En,'uV'
The value of thermal noise power is 4.002e-15 W
The value of RMS noise voltage is 0.89 uV

Example 4.2.2,Pg.no.122

In [2]:
from math import sqrt
R1=20000.0
R2=50000.0
k=1.38*10**-23              #Boltzman constant in J/K
T=290.0
BW=100*10**3       #Determination of thermal noise voltage for 20Kohm resistor
En1=sqrt(4*R1*k*T*BW)
En1=En1*(10**6)
En1=round(En1,2)
print 'a) i )The value of RMS noise voltage is',En1,'uV'
#Determination of thermal noise voltage for 50 kohm resistor
En2=En1*sqrt(R2/R1)
En2=En2*10**6
print '   ii)The value of RMS noise voltage is',En2,'uV'
#Determination of thermal noise voltage for 20K & 50k resistor in series
Rser=R1+R2           #Series combination of R1 & R2
En3=En1*sqrt(Rser/R1)
En3=En3*10**6
En3=round(En3,2)
print 'b)The value of RMS noise voltage is',En3,'uV'
#Determination of thermal noise voltage for 20K & 50k resistor in parellel
Rpar=(R1*R2)/(R1+R2)         #parallel combination of R1 & R2
En4=En1*sqrt(Rpar/R1)
En4=En4*10**6
En4=round(En4,2)
print 'c)The value of RMS noise voltage is',En4,'uV'
a) i )The value of RMS noise voltage is 5.66 uV
   ii)The value of RMS noise voltage is 8949245.77828 uV
b)The value of RMS noise voltage is 10588890.4 uV
c)The value of RMS noise voltage is 4783573.08 uV

Example 4.2.3,Pg.no.128

In [3]:
from math import sqrt,pi
f=120*10**6
c=25*10**-12         #capacitance of 12 pF
Q=30.0               #Q−factor of the ckt is 30
BW=10*10**3          #channel BW of the receiver is 10 KHz
k=1.38*10**-23       #Boltzman constant in J/K
T=290.0              #Room temp
#Determination of effective noise voltage Rd appearing at i /p at room temp
Rd=Q/(2*pi*f*c)
Rd=Rd/1000
Rd=round(Rd,2)
print 'The value of Rd is',Rd,'kohm'
Vn=sqrt(4*Rd*k*T*BW)
Vn=Vn*10**6
Vn=round(Vn,2)
print 'The value of effective noise voltage is',Vn,'uV'
The value of Rd is 1.59 kohm
The value of effective noise voltage is 0.02 uV

Example 4.3.1,Pg.no.131

In [4]:
from math import sqrt
Idc=10**-3
Bn=10**6           #Effective noise BW=1 MHz
q=1.6*10**-19      #Charge on electron in coulombs
#Determination of noise component current In in DC current of Idc=1 mA
In=sqrt(2*Idc*q*Bn)
In=In*10**9
In=round(In,2)
print 'The value of noise current In is',In,'nA'
The value of noise current In is 17.89 nA

Example 4.11.1,Pg.no.135

In [5]:
import math
from math import pi,sqrt
#An amplifier is given
Rn=300.0        #Equivalent noise resistance
Ieq=5*10**-6    #Equivalent noise current is 5 uA
Rs=150.0        #Amplifier fed from 150 ohm,10 uV rms sinusoidal source
Vs=10*10**-6
Bn=10*10**6     #Noise BW is 10 MHz
#Assume the following
kT=4*10**-21    #k is Boltzman constant in J/K & T is room temp
q=1.6*10**-19   #Charge on electron in coloumbs
#Determination of shot noise current
Ina=sqrt(2*q*Ieq*Bn)
Ina=Ina*(10**9)
print 'The value of shot noise current Ina is',Ina,'nA'
#Noise voltage developed by this across source resistance is
V=Ina*Rs
V=V*(10**6)
V=round(V,2)
print 'The value of noise voltage across Rs is',V,'uV'
#Noise voltage developed across Rn resistance is
Vna=sqrt(4*Rn*kT*Bn)*10**6
Vna=round(Vna,2)
print 'The value of noise voltage across Rn is',Vna,'uV'
#Determination of thermal noise voltage from source
Vns=sqrt(4*Rs*kT*Bn)*10**6
Vns=round(Vns,2)
print 'The value of thermal noise voltage at Rs is',Vns,'uV'
#Determination of total noise voltage at input
Vn=(((V)**2)+((Vna)**2)+((Vns)**2))**(1/2)
Vn=Vn*(10**6)
print 'The value of total noise voltage Vn is',Vn,'uV'
#Determination of signal to noise ratio in dB
SNR=20*(math.log10(Vs/Vn))
print 'The value of signal to noise ratio is',SNR,'dB'
The value of shot noise current Ina is 4.0 nA
The value of noise voltage across Rs is 600000000.0 uV
The value of noise voltage across Rn is 6.93 uV
The value of thermal noise voltage at Rs is 4.9 uV
The value of total noise voltage Vn is 1000000.0 uV
The value of signal to noise ratio is -220.0 dB

Example 4.12.1,Pg.no.136

In [6]:
import math
SNR1=60.0
l=3.0           #Determination of output signal to noise ratio
SNR=(SNR1) -10*math.log10(l)
SNR=round(SNR,2)
print 'The value of output signal to noise ratio is',SNR,'dB'
The value of output signal to noise ratio is 55.23 dB

Example 4.12.2,Pg.no.137

In [14]:
import math
SNRdB1=60.0      #SNR is 60 dB for Ist link
SNRdB2=60.0      #SNR is 60 dB for IInd link
SNRdB3=40.0      #SNR is 40 dB for IIIrd link
#Determination of power in watt
snr1=10**(-SNRdB1/10)
snr2=10**(-SNRdB2/10)
snr3=10**(-SNRdB3/10)
#Determination of overall SNR
SNR=snr3
#Determination of total SNR in dB
SNRdB=10*(-math.log10(SNR))
print 'The value of output signal to noise ratio is',SNRdB,'dB'
The value of output signal to noise ratio is 40.0 dB

Example 4.13.1,Pg.no.139

In [15]:
import math
SNRin=35.0 #SNR at i /p of amplifier
F=7.0      #Noise figure of an amplifier
#Determination of output SNR
SNRout=SNRin-F
print 'The value of output signal to noise ratio is',SNRout,'dB'
The value of output signal to noise ratio is 28.0 dB

Example 4.14.1,Pg.no.140

In [16]:
import math
f=13.0  #Noise figure of an amplifier
Bn=1*10**6
kT=4*10**-21   #k is Boltzman constant in J/K & T is room temp
F=10**(f/10)
#Determination of equivalent amplifier input noise
Pna=(F-1)*kT*Bn*10**12
Pna=round(Pna,2)
print 'The value of input noise is',Pna,'pW'
The value of input noise is 0.08 pW

Example 4.15.1,Pg.no.141

In [17]:
import math
f1=9.0     #Noise fig for amplifier
f2=20.0    #Noise fig for mixer
g=15.0     #power gain
#Converting dB in power ratio
F1=10**(f1/10)
F2=10**(f2/10)
G=10**(g/10)
#Determination of overall noise fig . reffered at i / p
F=F1+(F2-1)/G  #converting in dB
FdB=10*math.log10(F)
FdB=round(FdB,2)
print 'The overall noise fig is',FdB,'dB'
The overall noise fig is 10.44 dB

Example 4.17.1,Pg.no.143

In [18]:
import math
F=6.0      #Noise fig .=6 dB
#Determination of noise factor
Fn=10**(6/10)
print 'The value of noise factor is',Fn
The value of noise factor is 1

Example 4.18.1,Pg.no.144

In [19]:
import math
f=12.0 
Tm=290.0    #Room temp value
T=90.0
g=50.0      #calculating power ratio
F=10**(f/10)
G=10**(g/10)
#Determination of equivalent noise at room temp
Tem=(F-1)*Tm
Tem=round(Tem,2)
print 'The value of equivalent noise at room temp is',Tem,'k'
#Determination of equivalent noise at 90 k temp
Te=T+(Tem/G)
Te=round(Te,2)
print 'The value of equivalent noise at noise temp=90 is',Te,'K'
The value of equivalent noise at room temp is 4306.19 k
The value of equivalent noise at noise temp=90 is 90.04 K

Example 4.19.1,Pg.no.146

In [20]:
import math
enr=14.0
To=290.0          #Room temp in K
y=9.0             #Y−factor is 9 dB
#converting dB in power ratio
ENR=10**(enr/10)
Y=10**(y/10)      #From def of ENR the hot temp is
Th=To*(ENR+1)
Th=round(Th,2)
print 'The value of hot temp Th is',Th,'k'
#Determination of equivalent noise temp
Te=(Th-(Y*To))/(Y-1)
Te=round(Te,2)
print 'The value of equivalent noise temp Te is',Te,'K'
The value of hot temp Th is 7574.47 k
The value of equivalent noise temp Te is 759.14 K