Multiple Choice Questions and Answers on Noise

1)   For a three stage cascade amplifier, calculate the overall noise figure when each stage has a gain of 12 DB and noise figure of 8dB.

a. 12
b. 24
c. 13.55
d. 8

ANSWER:  (c) 13.55

Explanation: 
As the signal passes through various stages of an amplifier, the output has the original signal and some noise that gets amplified at different stages of amplifiers. So the final noise figure of the cascaded amplifier is obtained by

F_N = F₁ + (F₂ – 1)/ G₁ + (F₃ – 1)/ G₁G₂+……+(F_N– 1)/ G₁G₂G₃ G_N
F₁, F₂, F₃ .. F_N, G₁,G₂, G₃…. G_N are the noise figures and the gains respectively of the amplifiers at different stages.
F₁ = 12, F₂ = 12, F₃ = 12
G₁ = 8, G₂ = 8, G₃ = 8
F_N = 12 + (12- 1)/ 8+ (12- 1)/ 8 * 8
= 12 + 11/8 + 11/64
= 13.55

2)   The Hilbert transform of the signal sinω1t + sinω₂t is

a. sinω₁t + sinω₂t
b. cosω₁t + cosω₂t
c. sinω₂t + cosω₂t
d. sinω₁t + sinω₁t

ANSWER:  (b) cosω₁t + cosω₂t

Explanation: 
In Hilbert transform, the signal gets shifted by 900.

So the signal sinω₁t+ sinω₂t gets shifted by 900
sinω₁(t+900)+ sinω₂(t+900)
= cosω₁t+ cosω₂t

3)   The noise due to random behaviour of charge carriers is

a. Shot noise
b. Partition noise
c. Industrial noise
d. Flicker noise

ANSWER:  (a) Shot noise

4)   Transit time noise is

a. Low frequency noise
b. High frequency noise
c. Due to random behavior of carrier charges
d. Due to increase in reverse current in the device

ANSWER:  (b) High frequency noise

5)   Figure of merit γ is

a. Ratio of output signal to noise ratio to input signal to noise ratio
b. Ratio of input signal to noise ratio to output signal to noise ratio
c. Ratio of output signal to input signal to a system
d. Ratio of input signal to output signal to a system

ANSWER: (a) Ratio of output signal to noise ratio to input signal to noise ratio

6)   Signum function sgn(f), for f>0, f=0 and f<0, has the values:

a. -1 to +1
b. +1, 0, -1 respectively
c. -∞ to + ∞
d. 0 always

ANSWER: (b) +1, 0, -1 respectively

Explanation: 
The sgn(f) is a signum function that is defined in the frequency domain as
sgn(f) = 1, f> 0
= 0, f = 0
= -1, f< 0
Mathematically, the sign function or signum function is an odd mathematical function which extracts the sign of a real number and is often represented as sgn

7)   In Hilbert transform of a signal, the phase angles of all components of a given signal are shifted by

a. +/- π
b. +/- π/4
c. +/- π/2
d. Any angle from 00 to 3600

ANSWER:  (c) +/- π/2

8)   The noise voltage (V_n) and the signal bandwidth (B) are related as

a. V_n is directly proportional to bandwidth
b. V_n is directly proportional to √bandwidth
c. V_n is inversely proportional to absolute temperature
d. V_n is inversely proportional to bandwidth

ANSWER:  (b) V_n is directly proportional to √bandwidth

9)   Noise factor for a system is defined as the ratio of

a. Input noise power (P_ni) to output noise power (P_no)
b. Output noise power (P_no) to input noise power (P_ni)
c. Output noise power (P_no) to input signal power (P_si)
d. Output signal power (P_so) to input noise power (P_ni)

ANSWER:  (b) Output noise power (P_no) to input noise power (P_ni)

10)   Noise Factor(F) and Noise Figure(NF) are related as

a. NF = 10 log₁₀(F)
b. F = 10 log₁₀(NF)
c. NF = 10 (F)
d. F = 10 (NF)

ANSWER: (a)  NF = 10 log₁₀(F)

11)   The Noise Factor for cascaded amplifiers (F_N) is given by (F₁, F₂, F₃ .. F_N, G₁, G₂, G₃….G_N) are the noise factors and the gains of the amplifiers at different stages:

a. F_N = F₁ + F₂/ G₁ + F₃/ G₁G₂+ ..+ F_N/ G₁G₂G₃G_N
b. F_N = F₁ + (F₂ – 1)/ G₁ + (F₃ – 1)/ (G1+G2)+ ..+(F_N – 1)/ (G1+G2+G3+…+G_N)
c. F_N = F₁ + F₂/ G₁ + F₃/ (G1+G2) +…+ F_N/ (G1+G2+G3+…+G_N)
d. F_N = F₁ + (F₂ – 1)/ G₁ + (F₃ – 1)/ G₁G₂+…+(F_N – 1)/ G₁G₂G₃G_N

ANSWER: (d) F_N = F₁ + (F₂ – 1)/ G₁ + (F₃ – 1)/ G₁G₂+…+(F_N – 1)/ G₁G₂G₃G_N

12)   For a two stage amplifier, first amplifier has Voltage gain = 20, Input Resistance R_in1=700Ω, equivalent Resistance R_eq1=1800Ω and Output Resistor R_o1 = 30KΩ. The corresponding values of second amplifier are : 25, 80 KΩ, 12 KΩ, 1.2 MΩ respectively. What is the value of equivalent input noise resistance of the given two stage amplifier?

a. 2609.1Ω
b. 2607.1Ω
c. 107.1Ω
d. 2107.1Ω

ANSWER:  (b) 2607.1Ω

Explanation: 
R₁ = R_in1 + R_eq1 = 700 + 1800 = 2500Ω
R₂ = (R_o1 R_in2)/ (R_o1 + R_in2) + R_eq2 = 30 * 80/(30 + 80) + 12 = 40.92KΩ
R₃ = R_o2 = 1.2MΩ
Equivalent input noise resistance of a two stage amplifier is given by
R_eq = R₁ + R₂/ A²₁ + R₃/ (A²₁ A²₂)
= 2500 + 40.92 * 10³/(20)² + 1.2 * 10⁶/(20)²(25)²
= 2607.1Ω

13)   The noise temperature at a resistor depends upon

a. Resistance value
b. Noise power
c. Both a and b
d. None of the above

ANSWER: (b)  Noise power

14)   Noise voltage V_n and absolute temperature T are related as

a. V_n = 1/ √(4RKTB)
b. V_n = √(4RK)/ (TB)
c. V_n = √(4RKTB)
d. V_n = √(4KTB)/R

ANSWER:  (c) V_n = √(4RKTB)

15)   Notch filter is a

a. Band pass filter
b. Band stop filter
c. Low pass filter
d. High pass filter

ANSWER: (b)  Band stop filter

16)   Noise is added to a signal in a communication system

a. At the receiving end
b. At transmitting antenna
c. In the channel
d. During regeneration of the information

ANSWER:  (c) In the channel

17)   Noise power at the resistor is affected by the value of the resistor as

a. Directly proportional to the value of the resistor
b. Inversely proportional to the value of the resistor
c. Unaffected by the value of the resistor
d. Becomes half as the resistance value is doubled

ANSWER:  (c) Unaffected by the value of the resistor

18)   Low frequency noise is

a. Transit time noise
b. Flicker noise
c. Shot noise
d. None of the above

ANSWER:  (b) Flicker noise

19)   Hilbert transform may be used in

a. Generation of SSB signals
b. Representation of band pass signals
c. Designing of minimum phase type filters
d. All of the above

ANSWER:  (d) All of the above

20)   At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 20 KΩ when the bandwidth is 10 KHz.

a. 4.071 * 10^-6 V, 5.757 * 10^-6 V
b. 6.08 * 10^-6 V, 15.77 * 10^-6 V
c. 16.66 * 10^-6 V, 2.356 * 10^-6 V
d. 1.66 * 10^-6 V, 0.23 * 10^-6 V

ANSWER:  (a) 4.071 * 10^-6 V, 5.757 * 10^-6 V

Explanation: 
Noise voltage V_n = √(4R KTB)
Where, K = 1.381×10^-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power P_n is delivered.
T noise temperature
R is the resistance
Noise voltage by individual resistors
V_n1 = √(4R1 KTB)
= √(4 * 10 * 10³ * 1.381 * 10^-23 * 3000 * 10 * 10³)
= √16.572 * 10^-12
= 4.071 * 10^-6 V
V_n2 = √(4R₂ KTB)
= √(4 * 20 * 10³ * 1.381 * 10^-23 * 3000 * 10 * 10³)
= √33.144 * 10^-12
= 5.757 * 10^-6 V

21)   At a room temperature of 293K, calculate the thermal noise generated by two resistors of 20KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in series.

a. 300.66 * 10^-7
b. 284.48 * 10^-7
c. 684.51 * 10^-15
d. 106.22 * 10^-7

ANSWER:  (b) 284.48 * 10^-7

Explanation: 
Noise voltage V_n = √(4R KTB)
Where, K = 1.381 × 10^-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power P_n is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in series is
V_n = √{4(R₁ + R₂) KTB}
= √{4(20 * 10³ + 30 * 10³) * 1.381 × 10^-23 * 293 * 10 * 10³ }
= 284.48 * 10^-7

22)   At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in parallel.

a. 30.15 * 10^-3
b. 8.23 * 10^-23
c. 11.15 * 10^-7
d. 26.85 * 10^-7

ANSWER: (c)  11.15 * 10^-7

Explanation: 
Noise voltage V_n = √(4R KTB)
Where, K = 1.381 × 10^-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power P_n is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in parallel is
V_n = √{4R KTB}
Here for resistors to be in parallel,
1/R = 1/R₁ + 1/R₂
= 1/10K + 1/30K
= 0.1333
R = 7.502KΩ
V_n = √{4 * 7.502 * 10³ * 1.381×10^-23 * 300 * 10 * 10³}
= √124.323 * 10^-14
= 11.15 * 10^-7