This presentation covers types of noise in communication system, noise modelling, thermal noise, shot noise, experimental determination of noise figure, noise figure, friss formula with numerical.

1. Noise
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2. What is noise?
In electrical terms, it is an unwanted signal, which interfere with the desired signal.
Example:
 In radio receiver: ‘Hiss’ sound from speaker
In TV: ‘Snow’ or ‘confetti’(colored snow) display on the picture.
Drawback of Noise
o For a given transmitted power ,it limit the range of systems
o It affects the Sensitivity of receivers
o Sometimes Force reduction in bandwidth of the system
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3. Sources of Noise
External Noise : Noise created outside the receiver
 Internal Noise : Noise created inside the receiver
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Source of External Noise:
• Atmospheric Noise
• Extraterrestrial Noise
• Solar Noise
• Cosmic Noise
• Industrial Noise
Source of Internal Noise:
• Thermal Agitation Noise or Johnson Noise
• Shot Noise
• Transit Time Noise

4. . Atmospheric Noise or Static noise
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caused by lighting discharges in thunderstorms,
the electrical impulses are random in nature and spread over most of the RF spectrum
used in broadcasting.
Atmospheric Noise consists of false radio signals and distributed over wide range of
frequencies.
It is propagated over the earth and at any point on the ground atmospheric noise will be
received from the thunderstorms.
Since Field strength is inversely proportional to frequency , so this Noise will interfere
more with radio receiver then the television.
Atmospheric noise is less severe above about 30 MHz.

5. Extraterrestrial Noise or Space Noise
caused by radiation of RF noise by sun and distant stars.
Sun is a large body at a very high temperature over 6000 0C on the surface) and Radiates over a very
broad frequency spectrum which includes the frequencies we Use for communication.
Distant stars are also suns and have high temperatures, they radiate RF noise in the same manner as our
sun .
The noise received is called thermal (or black-body ) noise and is distributed uniformly over the entire
sky.
This noise is observable at frequencies in the range from about 8 MHz to 1.43 GHz
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6. Industrial Noise
This noise is effective in industrial and densely populated area
Source of industrial noise are
1.Automobile and aircraft ignition
2.Leakage from high voltage lines
3.Heavy electric machines
4.Fluorescent light ,etc
This noise is observable at frequencies in the range from 1MHz to 600 MHz
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7. Internal Noise
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Created by the active or passive devices present in the receiver
Distributed randomly over the entire radio spectrum
Impossible to treat instantaneous voltage basis ,in place of that it is easily describe
statistically.
Source of Internal Noise
1.Thermal,agitation,white or Johnson noise
2. Shot Noise
3.Transit-Time Noise

8. Thermal agitation or white or Johnson noise
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• Randomly distributed over the bandwidth
• Generated in resistance or the resistive component
• Generated due to the random motion of atoms and electrons inside the components
• Noise generated by the resistor is proportional to its absolute temperature and bandwidth
• Noise power output from a resistor,
Pn = KTf
Here K = Boltzmann constant = 1.38X10-23J/k, T= 0A,
B=bandwidth of interest
Question: If the resistor is operating at 27oc and the bandwidth of interest is 2 MHz,
then what is the maximum noise power output of a resistor ?
Ans 0.0828X10-12 Watts

9. Noise Voltage across resistor R
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Maximum power transfer will happen when R= Ri
Then noise power at the input of next stage is =
𝑉𝑛
2
4𝑅
= 𝑃𝑛 = 𝐾𝑇𝛿𝑓
i.e equivalent rms noise voltage, 𝑉𝑛 = 4𝐾𝑇𝛿𝑓 𝑅
Similarily, equivalent rms noise voltage, 𝐼 𝑛 = 4𝐾𝑇𝛿𝑓 𝐺
where conductance 𝐺 =
1
𝑅
R
R
noiseless
Vn
R
(noiseless)
Vn
RiVn/2 R R
noiseless
In
R
(noiseless)
Vn
RiVn/2 R R
noiseless
In

10. Noise Voltage across resistor R
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Question : An amplifier operating over the frequency range from 18 to 20 MHz has
a 10-Kiloohm input resistor. What is the rms noise voltage at the input to this
amplifier if the ambient temperature is 27oC?
Ans
K= 1.38 x 10-23J/0K, T= 27oC = 300oK, B= (20-18) MHz = 2 x 106Hz, R= 10x 103
Vn = (4KTf R)
= (4 x 1.38 x 10-23 x 300 x 2 x 106x 10x103 )
= (331.2x10-12)
= 18.2x10-6 V
=18.2 V

11. Noise rms Voltage: resistance in series/ parallel
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12. Reactance and Equivalent noise bandwidth
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Inductance and capacitance do not generate noise
Let us assume that capacitance C1 generate noise power P1 while
capacitance C2 generate noise power P2
then according to thermal equilibrium, power generated must be
equal to power consumed.
As reactance do not consume power, so
P1 = P2 =0 i.e reactance do not generate noise, instead it limits
the noise power.
C1 C2
P2P1

13. Reactance and Equivalent noise bandwidth
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𝑉𝑏0 =
1
𝑗2𝜋𝑓𝐶
𝑅 +
1
𝑗2𝜋𝑓𝐶
𝑉𝑛 =
1
1 + 𝑗2𝜋𝑓𝐶𝑅
𝑉𝑛
Frequency response 𝐻 𝜔 =
𝑉𝑛0
𝑉𝑛
=
1
1+𝑗2𝜋𝑓𝐶𝑅
Output noise spectral density 𝑆 𝑛0 = 𝐻 𝑓 2
𝑆 𝑛 𝑓
As 𝑆 𝑛 𝑓 =KT
𝑆 𝑛0 𝑓 =
𝐾𝑇
1 + (2𝜋𝑓𝐶𝑅)2
Considering only positive frequencies, noise power at the output
𝑃𝑛0 =
0
∞
𝐾𝑇
1 + (2𝜋𝑓𝐶𝑅)2
𝑑𝑓 =
𝐾𝑇
2𝜋𝑅𝐶
𝑡𝑎𝑛−1
(2𝜋𝑓𝑅𝐶) 0
∞
=
𝐾𝑇
4𝑅𝐶
Effective bandwidth 𝐵 𝑒𝑓𝑓 =
1
4𝑅𝐶
Noise voltage 𝑉𝑛0 = 4𝐾𝑇𝑅𝐵𝑒𝑓𝑓 =
𝐾𝑇
𝐶

14. Reactance and Equivalent noise bandwidth:
Tuned circuit
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Resonant frequency 𝑓0 =
1
2𝜋 𝐿𝐶
Quality factor 𝑄 =
2𝜋𝑓0 𝐿
𝑅
=
1
2𝜋𝑓0 𝐶
=
𝑓0
∆𝑓
𝑉𝑛0
2
= 4𝑄2
𝑅𝐾𝑇∆𝑓= 4𝑅 𝐷 𝐾𝑇∆𝑓
Where dynamic resistance 𝑅 𝐷=𝑄2
𝑅
Noise power at the output
𝑃𝑛0 =
𝐾𝑇
4𝑅 𝐷 𝐶
= 𝐾𝑇𝐵𝑒𝑓𝑓
Thus effective bandwidth 𝐵 𝑒𝑓𝑓 =
1
4𝑅 𝐷 𝐶
3 dB bandwidth 𝐵3 𝑑𝐵 =
2
𝜋
𝐵𝑒𝑓𝑓

15. Example
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Ex. A 100 K resistance is connected in parallel with 1 pf capacitor at room temperature (27 0C). Calculate
effective bandwidth and equivalent noise voltage at the output.
Sol:
Effective bandwidth 𝐵 𝑒𝑓𝑓 =
1
4𝑅𝐶
=
1
4 𝑥 105 𝑥 10−12 = 2.5 𝑀𝐻𝑧
Noise voltage 𝑉𝑛0 = 4𝐾𝑇𝑅𝐵𝑒𝑓𝑓 =
𝐾𝑇
𝐶
=
1,38 𝑥 10−23 𝑥 300
10−12 = 64.34 𝜇𝑉

16. Example
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Ex. A 10 K resistance is connected in RLC circuit of pass band 10 KHz, tuned at 1 MHz frequency. Calculate
the dynamic resistance of the circuit, equivalent noise voltage at the output across the capacitor. Also
calculate effective bandwidth at room temperature (27 0C). Capacitance used is 10 pf.
Sol:
Quality factor 𝑄 =
𝑓0
∆𝑓
=
1 𝑀𝐻𝑧
10 𝐾𝐻𝑧
= 100
dynamic resistance 𝑅 𝐷=𝑄2
𝑅 = 1002
𝑥 10 𝐾Ω = 100 𝑀Ω
Noise voltage 𝑉𝑛0 = 4𝐾𝑇∆𝑓𝑅 𝐷 = 4 𝑥 1,38 𝑥 10−23 𝑥 300 𝑥 104 𝑥 10 𝑥106 = 128.68 𝜇𝑉
Thus effective bandwidth 𝐵 𝑒𝑓𝑓 =
1
4𝑅 𝐷 𝐶
=
1
4 𝑥10 𝑥106 𝑥10 𝑥10−12 = 250 𝐻𝑧
3 dB bandwidth 𝐵3 𝑑𝐵 =
2
𝜋
𝐵𝑒𝑓𝑓 =
2
𝜋
𝑥 250 = 159.15 𝐻𝑧

17. Addition of noise due to amplifier in cascade
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The output resistance is R3, the noise voltage 𝑉𝑛3 = 4𝐾𝑇𝛿𝑓𝑅3
The noise voltage at the output will be same if R3 is assumed to be noiseless and a equivalent noisy resistance 𝑅3
′
is placed at the input of second stage, such that 𝑉𝑛3
′
=
𝑉𝑛3
𝐴2
=
4𝐾𝑇𝛿𝑓𝑅3
𝐴2
= 4𝐾𝑇𝛿𝑓𝑅3
′
i.e 𝑅3
′
=
𝑅3
𝐴2
2
Now total noise voltage at the input of second amplifier will be generated by equivalent resistance 𝑅 𝑒𝑞
′
such that
𝑅 𝑒𝑞
′
= 𝑅2 + 𝑅3
′
Similarly equivalent noise voltage at the output of first amplifier can be replaced by a noise voltage generated
𝑅2
′
at the input of first amplifier such that 𝑅2
′
=
𝑅 𝑒𝑞
′
𝐴1
2 =
𝑅2
𝐴1
2 +
𝑅3
𝐴1
2 𝐴2
2
The total noise voltage at the input will be generated by
equivalent resistance such that 𝑅 𝑒𝑞 = 𝑅1 + 𝑅2
′
= 𝑅1 +
𝑅2
𝐴1
2 +
𝑅3
𝐴1
2 𝐴2
2
Equivalent noise voltage at the input
( assuming all resistances as noiseless)
𝑉𝑛 = 𝐾𝑇 𝛿𝑓 𝑅 𝑒𝑞
R1
Vn1
R2
Vn2
R3
Vn3
Amplifier
gain A1
Amplifier
gain A2

18. Addition of noise due to amplifier in cascade
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• It is important to note that high resistance at the output R3 have noise effect equal to a low resistance
𝑅3
𝐴1
2 𝐴2
2
• Equivalent noise resistance Req is fictitious resistance, which generates total equivalent noise power at the input
of the device at room temperature while device is assumed to be noiseless. Req is fictitious resistance and does
not contribute to the input impedance of the device and is used for noise calculations only.
Ex. Two amplifiers of voltage gain 10 and 15 are cascaded and applied to a load resistance of 1 M. A 100 
resistance is applied to input of a first stage, while 200  resistance is connected to the input of second stage.
Calculate equivalent noise resistance at the input. At room temperature (27 0C) calculate equivalent noise voltage
at the input for measuring bandwidth of 100 KHz. What is equivalent noise resistance of load resistance at the
input
Sol:
Equivalent noise resistance Req = R1+
𝑅2
𝐴1
2+
𝑅3
𝐴1
2 𝐴2
2 =100+
200
102+
1𝑥 106
102 𝑥 152= 100+2 +
400
9
= 146.44 
Equivalent noise voltage 𝑉𝑛 = 𝐾𝑇 𝛿𝑓 𝑅 𝑒𝑞 = 1.38 𝑥 10−23 𝑥 300 𝑥 105 𝑥 146.44 = 492.45 nV
Equivalent load resistance at the input =
1𝑥 106
102 𝑥 152 = 44.44 

19. Shot noise
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• Shot noise is present in all active devices
• Caused due to random arrival of charge carriers.
• For diode shot noise is given as rms noise current
𝑖 𝑛 = 2𝑒𝑖 𝑝 𝛿𝑓
Where 𝑖 𝑝 is diode current in amperes, 𝑒 is charge of electron, 𝛿𝑓 is bandwidth of
measurement
• Equivalent noise resistance of shot noise is generally quoted in manufacturer
datasheet

20. Signal to noise ratio (SNR)
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• Noise power alone is not sufficient for analysing noise performance of any system
• 𝑆𝑁𝑅 =
𝑆𝑖𝑔𝑛𝑎𝑙 𝑃𝑜𝑤𝑒𝑟
𝑁𝑜𝑖𝑠𝑒 𝑝𝑜𝑤𝑒𝑟
=
𝑆
𝑁
• 𝑆𝑁𝑅 𝑑𝐵 = 10𝑙𝑜𝑔10(𝑆𝑁𝑅)
• SNR is unit less quantity and is usually measured in dB
Ex. A 10 mV rms voltage is applied to 10 K noisy resistor at 27 0C. Calculate SNRdB of
the resistor of measuring bandwidth is 10 MHz.
Sol:
Signal power 𝑆 =
𝑉𝑠
2
𝑅
=
100 𝑥 10−6
104 = 100 𝑛𝑊
Noise power 𝑁 = 𝐾𝑇𝛿𝑓 = 1,38 𝑥 10−23 𝑥 300 𝑥 107 = 0.0414 pW
SNR=241545.89
Now 𝑆𝑁𝑅 𝑑𝐵 = 10𝑙𝑜𝑔10 𝑆𝑁𝑅 = 53.83 𝑑𝐵

21. Noise figure
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• Noise figure provides a measure of noise performance of receivers and amplifiers
• 𝐹 =
𝑆𝑁𝑅 𝑖
𝑆𝑁𝑅 𝑜
=
𝑆 𝑖
𝑁 𝑖
𝑆 𝑜
𝑁 𝑜
=
𝑆 𝑖
𝑆 𝑜
𝑥
𝑁 𝑜
𝑁 𝑖
=
1
𝐴2 𝑥
𝑁 𝑜
𝑁 𝑖
Where
A is voltage gain of amplifier,
Si signal power at the input,
So signal power at the output,
Ni noise power at the input,
No noise power at the input,
• A practical receiver adds noise power. So its noise figure is always greater than 1
• For ideal recover (which does nor add noise) noise figure is 1

22. Calculation of Noise figure from equivalent
resistance
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• Receiver is fed from antenna of resistance Ra. It feeds signal Vs and
generate noise Vn1.
• Receiver has input impedance Ri , Voltage gain A and output is
taken at load RL.
• Equivalent noise resistance of amplifier is Req.
• Let us define 𝑅 𝑒𝑞
′
such that 𝑅 𝑒𝑞
′
= 𝑅 𝑒𝑞 − 𝑅𝑖
Signal input to amplifier 𝑉𝑖 =
𝑅 𝑖
𝑅 𝑎+𝑅 𝑖
𝑉𝑠
Signal power at input of amplifier 𝑆𝑖 =
𝑉𝑖
2
𝑅 𝑖
= 𝑉𝑠
2 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)2
Signal power at output of amplifier 𝑆 𝑜 =
𝑉𝑜
2
𝑅 𝐿
=
𝐴2 𝑉𝑖
2
𝑅 𝐿
= 𝑉𝑠
2 𝐴2 𝑅𝑖
2
(𝑅 𝑎+𝑅 𝑖)2 𝑅 𝐿
Ra
Vn1
Ri RL
Vs
VoVi
Antenna Amplifier with voltage gain A

23. Calculation of Noise figure from equivalent resistance
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Noise at the input is due to antenna resistance and internal resistance of amplifier which are in parallel
Noise power at the input of amplifier 𝑁𝑖 =
𝑉𝑛1
2
𝑅 𝑖
=
4𝐾𝑇𝛿𝑓
𝑅 𝑎 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)
𝑅 𝑖
=
4𝐾𝑇𝛿𝑓𝑅 𝑎
(𝑅 𝑎+𝑅 𝑖)
Output noise power is sum of noise generated at input and added by amplifier itself i.e output noise
power will be due to equivalent noise resistance at input 𝑅 = 𝑅 𝑒𝑞
′
+
𝑅 𝑎 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)
Noise voltage at the output 𝑉𝑛𝑜 = 𝐴24𝐾𝑇𝛿𝑓(𝑅 𝑒𝑞
′
+
𝑅 𝑎 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)
)
Output noise power 𝑁𝑜 =
𝐴2(4𝐾𝑇𝛿𝑓𝑅)
𝑅 𝐿
Noise figure 𝐹 =
𝑆𝑁𝑅 𝑖
𝑆𝑁𝑅 𝑜
=
𝑆𝑖
𝑆 𝑜
𝑥
𝑁 𝑜
𝑁 𝑖
=
𝑉𝑠
2 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)2
𝑉𝑠
2 𝐴2 𝑅 𝑖
2
(𝑅 𝑎+𝑅 𝑖)2 𝑅 𝐿
𝑥
𝐴24𝐾𝑇𝛿𝑓 𝑅 𝑒𝑞
′ +
𝑅 𝑎 𝑅 𝑖
(𝑅 𝑎+𝑅 𝑖)
𝑅 𝐿
4𝐾𝑇𝛿𝑓𝑅 𝑎
(𝑅 𝑎+𝑅 𝑖)
= 1 + 𝑅 𝑒𝑞
′ 𝑅 𝑎+𝑅 𝑖
𝑅 𝑎 𝑅 𝑖
Under matched conditions (𝑅 𝑎 = 𝑅𝑖) , 𝐹 = 1 +
2𝑅 𝑒𝑞
′
𝑅 𝑎

24. Calculation of Noise figure from equivalent resistance
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Ex. Two receivers has input impedance and equivalent noise resistance as (50  and 100 ) and (250
 and 100 ) respectively. Calculate the noise figure of both receivers if it is fed by an antenna of 50 
impedance.
Sol.
noise figure 𝐹 = 1 + 𝑅 𝑒𝑞
′ 𝑅 𝑎+𝑅 𝑖
𝑅 𝑎 𝑅 𝑖
For first receiver , noise figure 𝐹1 = 1 + 100𝑥
50+50
50 𝑥 50
= 5
For second receiver , noise figure 𝐹2 = 1 + 100𝑥
250+50
250 𝑥 50
= 3.4
• Noise contribution at input is same in both amplifiers as their noise equivalent resistance are same.
• Input impedance of Amplifier 2 is higher than Amplifier 1, it appears more noise will be generated
at input of amplifier 2.
• From the noise figure it is evident that noise performance of amplifier 2 is better than amplifier 1.

25. Amplifier Input noise in terms of noise figure
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𝐹 =
𝑆𝑁𝑅 𝑖
𝑆𝑁𝑅 𝑜
=
𝑆𝑖
𝑆 𝑜
𝑥
𝑁 𝑜
𝑁 𝑖
=
1
𝐴2 𝑥
𝑁 𝑜
𝑁 𝑖
𝑁𝑜 = 𝐴2
𝐹𝑁𝑖
𝑤ℎ𝑒𝑟𝑒 𝑁𝑖 = 𝐾𝑇𝛿𝑓
Total noise power at the output, 𝑁𝑜 = 𝐴2
𝐹𝐾𝑇𝛿𝑓
Total Noise power at the input =
𝑁 𝑜
𝐴2 = 𝐹𝐾𝑇𝛿𝑓
Contribution of amplifier in noise power at the input (so that amplifier can be assumed noiseless)
𝑁𝑎𝑖 = 𝐹 − 1 𝐾𝑇𝛿𝑓

26. Amplifiers in cascade
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 26
Noise power fed from the source at the input𝑁𝑖 = 𝐾𝑇𝛿𝑓
Equivalent noise power contributed by amplifier 2 at its input (assuming amplifier 2 noiseless)
𝑁𝑓2 = 𝐹2 − 1 𝐾𝑇𝛿𝑓
Total noise power at the input of second amplifier is sum of noise contributed by second
amplifier and noise at the out put of first amplifier,
𝑁𝑓 = 𝐴1
2
𝐹1 𝐾𝑇𝛿𝑓 + 𝐹2 − 1 𝐾𝑇𝛿𝑓
Total Noise power at the input of first amplifier
𝑁𝑓
𝐴1
2 = 𝐹1 𝐾𝑇𝛿𝑓 +
𝐹2 − 1 𝐾𝑇𝛿𝑓
𝐴1
2 = 𝐹𝐾𝑇𝛿𝑓
If F is the overall noise figure
i.e. 𝐹 = 𝐹1 +
𝐹2−1
𝐴1
2
In general, Friss equation
𝐹 = 𝐹1 +
𝐹2−1
𝐴1
2 +
𝐹3−1
𝐴1
2 𝐴2
2 + ------
Amplifier
gain A2
Noise
figure F2
Amplifier
gain A1
Noise
figure F1
Rs
Vn
Source
Nf
NoNi

27. Amplifiers in cascade
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 27
Ex. Three amplifiers of voltage gain 10,20,40 and noise figure 2, 5, 20 are cascaded. Find overall
noise figure. What is contribution of amplifier in input noise if measuring bandwidth is 10 KHz
(room temperature is 300 0K), Calculate the noise figure if sequence of cascading is reversed.
Sol:
Noise figure 𝐹 = 𝐹1 +
𝐹2−1
𝐴1
2 +
𝐹3−1
𝐴1
2 𝐴2
2 = 2 +
5−1
100
+
20−1
100 𝑥 400
=2.040475=3.097 dB
Equivalent noise power contributed by amplifiers to its input (assuming amplifiers noiseless)
𝑁𝑖𝑎 = 𝐹 − 1 𝐾𝑇𝛿𝑓 = 2.040475 𝑥 1.38 𝑥 10−23 𝑥 300 𝑥 104 = 4.30 𝑥 10−17 𝑊
Noise figure for reversed order cascading
Noise figure 𝐹 = = 20 +
5−1
1600
+
2−1
1600 𝑥 400
=20.01

28. Experimental determination of Noise figure
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 28
R
Vn
In
Equivalent circuit for noise
source generating shot noise
Noise
generating
source
Amplifier
gain A
Noise
figure F
AC
Wattmeter
Shot noise power 𝐼 𝑛
2 = 2𝑒𝐼 𝑝 𝛿𝑓
Noise generating diode is applied to a resistance usually matched to input of amplifier
Then as per maximum power transfer theorem, noise power at the input of amplifier
𝑁𝑖 =
𝐼 𝑛
2
2
𝑅 =
𝐼 𝑛
2 𝑅
4
=
𝑒𝐼 𝑝 𝛿𝑓R
2
Noise power at the output due to amplifier itself, 𝑁𝑜𝑎 = 𝐴2
𝐹𝐾𝑇𝛿𝑓

29. Experimental determination of Noise figure
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 29
Total noise power at the output 𝑁𝑜 = 𝑁𝑜𝑎 + 𝐴2 𝑁𝑖
Let 𝑁𝑜 = 𝑄𝑁𝑜𝑎 where Q is a constant
Then 𝑄𝑁𝑜𝑎 = 𝑁𝑜𝑎 + 𝐴2 𝑁𝑖
Or (𝑄 − 1)𝑁𝑜𝑎 = 𝐴2 𝑁𝑖
Or (𝑄 − 1)𝐴2 𝐹𝐾𝑇𝛿𝑓 = 𝐴2 𝑒𝐼 𝑝 𝛿𝑓 𝑅
2
or 𝐹 =
𝑒𝐼 𝑝 𝑅
2𝐾𝑇(𝑄−1)
When 𝐼 𝑝 = 0 the source does not generate noise and in the output noise is due to
amplifier itself. Then 𝐼 𝑝 is increased such that output noise power doubles i.e. Q=2
Let T= 3000C
So 𝐹 =
𝑒𝐼 𝑝 𝑅
2𝐾𝑇
=
1.6 𝑥 10−19 𝑥 𝐼 𝑝 𝑅
2 𝑥 1.38 𝑥 10−23 𝑥 300
= 19.32 𝐼 𝑝 𝑅

30. Equivalent noise temperature
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 30
Noise associated with the amplifier at the input 𝐹 − 1 𝐾𝑇𝛿𝑓 can be
represented by equivalent noise temperature such that
𝐹 − 1 𝐾𝑇𝛿𝑓 = 𝐾𝑇𝑒 𝛿𝑓
i.e. 𝑇𝑒 = 𝐹 − 1 𝑇
Noise equivalent temperature is an alternative of noise figure
In general, Friss equation
𝑇𝑒 = 𝑇𝑒1 +
𝑇𝑒2
𝐴1
2 +
𝑇𝑒3
𝐴1
2 𝐴2
2+ ------

31. White noise
31-08-2016 IEC 503 ANALOG COMMUNICATION SYSTEM BY DR N R KIDWAI, INTEGRAL UNIVERSITY 31
Idealized form of noise which is independent of frequency
Power spectral density
𝑆 𝑤 𝑓 =
𝜂
2
where 𝜂=KT
𝜂
2
𝑅 𝑛 𝜏
𝜏
Auto correlation of white noisef
𝜂
2
𝑆 𝑤 𝑓
Power spectral density function of white noise