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
FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+……+(FN– 1)/ G1G2G3 GN
F1, F2, F3 .. FN, G1,G2, G3…. GN 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
FN = 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 (Vn) and the signal bandwidth (B) are related as
a. Vn is directly proportional to bandwidth
b. Vn is directly proportional to √bandwidth
c. Vn is inversely proportional to absolute temperature
d. Vn is inversely proportional to bandwidth
ANSWER: (b) Vn is directly proportional to √bandwidth
9) Noise factor for a system is defined as the ratio of
a. Input noise power (Pni) to output noise power (Pno)
b. Output noise power (Pno) to input noise power (Pni)
c. Output noise power (Pno) to input signal power (Psi)
d. Output signal power (Pso) to input noise power (Pni)
ANSWER: (b) Output noise power (Pno) to input noise power (Pni)
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 (FN) is given by (F1, F2, F3 .. FN, G1, G2, G3….GN) are the noise factors and the gains of the amplifiers at different stages:
a. FN = F1 + F2/ G1 + F3/ G1G2+ ..+ FN/ G1G2G3GN
b. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ (G1+G2)+ ..+(FN – 1)/ (G1+G2+G3+…+GN)
c. FN = F1 + F2/ G1 + F3/ (G1+G2) +…+ FN/ (G1+G2+G3+…+GN)
d. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN
ANSWER: (d) FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN
12) For a two stage amplifier, first amplifier has Voltage gain = 20, Input Resistance Rin1=700Ω, equivalent Resistance Req1=1800Ω and Output Resistor Ro1 = 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 = Rin1 + Req1 = 700 + 1800 = 2500Ω
R2 = (Ro1 Rin2)/ (Ro1 + Rin2) + Req2 = 30 * 80/(30 + 80) + 12 = 40.92KΩ
R3 = Ro2 = 1.2MΩ
Equivalent input noise resistance of a two stage amplifier is given by
Req = 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 Vn and absolute temperature T are related as
a. Vn = 1/ √(4RKTB)
b. Vn = √(4RK)/ (TB)
c. Vn = √(4RKTB)
d. Vn = √(4KTB)/R
ANSWER: (c) Vn = √(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 Vn = √(4R KTB)
Where, K = 1.381×10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by individual resistors
Vn1 = √(4R1 KTB)
= √(4 * 10 * 103 * 1.381 * 10-23 * 3000 * 10 * 103)
= √16.572 * 10-12
= 4.071 * 10-6 V
Vn2 = √(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 Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in series is
Vn = √{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 Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in parallel is
Vn = √{4R KTB}
Here for resistors to be in parallel,
1/R = 1/R1 + 1/R2
= 1/10K + 1/30K
= 0.1333
R = 7.502KΩ
Vn = √{4 * 7.502 * 103 * 1.381×10-23 * 300 * 10 * 103}
= √124.323 * 10-14
= 11.15 * 10-7