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2. Noise and Frequency Spectrum
 AM communications system
 Noise
 Signal to Noise ratio
 Bandwidth
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Example in AM communications system
Information = low freq. sine
Carrier = high freq. sine
AM modulator
Transmitter
Propagate in the air medium
= propagation power loss
High power transmitted
Low power received
Amplified AM wave
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Information detected
from AM wave
Amplified output of
information signal
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Noise
Noise effect in receivers
Both Ps and Pn
amplified +
Ps Sin1

Pn Nin1
amplifier noise
Sout1 Sin1
1


Nout1 Nin1 NR1
Ps
Pn, en
Both Ps and Pn
amplified +
amplifier noise
Sin2 Sout1 Sin1
1



Nin2 Nout1 Nin1 NR1
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Sout 2 Sin2
1


Nout 2 Nin2 NR2
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Noise power , noise voltage , short noise
Noise power
Pn  kTf
Noise voltage
en  4kTf  R
Pn= Thermal Noise power(W)
k= Boltzmann’s constant(1.38x10-23J/K)
T=Resistance temperature (degree Kelvin)=300 at room temperature of 27°Celsius
f=Bandwidth of the system(Hz)
Pn dBm  10logkTΔTdBW  10logkTΔT  30dBm


 10log 1.38  10  23 273  27   10logBW  (if BW  Δf)
 230  26.17  30dBm  10logBW  174dBm  10logBW
Determine the noise voltage produced by a 1 MW resistor at room temperature of 27 C
1 MHz bandwidth



over a

en  4  1.38  10  23 273  27   1 10 6  1656  10 17  0.129μ volt
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Noise power , noise voltage , short noise (cont’d)
•Transistor Noise: the noise discussed before does not include transistor noise called “short
noise”.
•It is due to the discrete-particle nature of the current carriers in all forms of semiconductors.
These current carriers even in dc conditions, are not moving in an exactly continuous flow since
the distance they travel varies due to random paths of motion.
•The short noise sounds like a shower of lead shot falling on a metal surface. It adds to the
thermal noise.
•Equation of the short noise in a diode is: in  2qIdc Δf
where in  short noise (rms) amps
q  electron charge (1.6  1019 C)
Idc  dc current (amp)
(a) Determine the short noise current of a diode with a forward bias of 1mA over a 100 kHz bandwidth
(b) Determine the diode’s equivalent noise voltage
(c) If the diode is in circuit with 500W series resistance, calculate the total output noise voltage at 27 C
(a) in  2qIdc Δf
 2  1.6  10-19  1  103  105
 0.0056μA
26mV
 26Ω
1mA
 en  in  26  0.147 μV(rms)
(b) diode junc resistance  Rj 
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Signal to Noise ratio (S/N)
S Psignal

N Pnoise
(in ratio )
 S   10 log Psignal  10 log S
 
Pnoise
N
 N dB
Noise Ratio (NR) & Noise Figure (NF)
NR 
Si / Ni  (in ratio )
So / No 
NF  NR dB  10 log NR
10 log
Si / Ni   10 logS / N   10 logS / N   NF  NF
o
o
i
i
i
o
So / No 
Si / Ni = (S/N)I = 1000
So / No = (S/N)o = 631
(S/N)I = 10log1000=30dB
(S/N)o = 10log631=28dB
NR = ? NF = ?
NR = (S/N)i /(S/N)o = 1000/631=1.58
NF = 10log 1000 - 10log631 = 30dB-28dB = 2dB
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Noise figure of BJT used at different frequencies
NF = 2dB if 2N4957 is used at frequencies between 0.1MHz to 100MHz
Otherwise BJT will be very noisy and (S/N)o will be very low if used at other frequencies
Reactance Noise Effects
In noise calculatio ns, f and BW(bandwidth) are related by :
f  BW for resistive circuits

f  BW for L, C circuits
2
Therefore L, C circuits are more noisy
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Noise in amplifiers in cascade
PG1
PG2
PGN
NR1
NR2
NRN
NR = ? NF = ?
NR  NR1 
NRN  1
NR4  1
NR2  1 NR3  1



PG1
PG1  PG2 PG1  PG2  PG3
PG1  PG2  PG3    PG(N1)
NR = overall Noise Ratio , PG = Power Gain
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A three-stage amplifier system has a 3 dB bandwidth of 200kHz determined by LC
circuit at it’s input, operating at 22 C. The first stage has a power gain of14dB and a NF
= 3dB. The second and third stages each has a power gain of20dB and NF=8dB.
Input noise of the first stage is generated by 10kW resistor. The output is connected to
a 300W load. Calculate
(a) Noise voltage and the power at the input and output of this system if the amplifiers
are noiseless.
(b) Overall NF for the system
(c) The actual output noise voltage and noise power
Noiseless amplifiers
NF = 0dB
NR = 1
(S/N)in/(S/N)out= 1
(a )
For LC circuits,
Δf 
π
BW
2
PG1=14dB
π
  200kHz
2
PG2=20dB
PG3=20dB
PG =54dB
=log-1(54/10)=1054/10
= 2.51x105
 3.14  105 Hz
Pn (in)  kTΔT



 1.38  10  23 273  22  3.14  10 5
 1.28  10 15 W


en (in)  4  1.28  10 15  R



Pn (out)  Pn (in)  PG  1.28  10 15 W  2.51 10 5  3.22  10 10 W
en2 en2

 3.22  10 10  en  0.311mV
R 300
 4  1.28  10 15  10 4  7.15μ.
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Calculation including Amplifier Noise
PG1=14dB
=log-1(14/10)=25.1
NF1= 3dB
NR1= log-1(3/10)= 2
PG2= PG3 =20dB
=log-1(20/10)=100
NF3= 8dB
NR3= log-1(8/10)= 6.31
NF2= 8dB
NR2= log-1(8/10)= 6.31
(b) NR  NR1 
NR2  1 NR3  1
6.31  1
6.31  1

2

 2.212
PG1
PG1  PG2
25.1
25.1  100
NF  10log2.212  3.45dB
Nout
1
(c) NR  2.212  (S/N)in  Sin  Nout  1  Nout 

(S/N)out Nin Sout PG Nin
2.51  105 1.28  1015
 Nout  2.51  105  1.28  1015  2.212  7.11  1010 W
en2
 7.11  1010 W  en  300  7.11  1010  0.462mV
300
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Bandwidth
Frequency spectrum of information signals
1kHz square wave signal
(digital signals)
Frequency spectrum of a
1kHz square wave signal
(digital signals)
Bandwidth is the amount of important harmonics (whose amplitude is
at least one tenth of the fundamental amplitude) in the frequency
spectrum of information signals
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Harmonic content & BW of different signal waveforms
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Harmonic content and Bandwidth of sawtooth signal
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Effect of filter BW for a sawtooth information signal
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