Analog Communication Test Questions - Set 8

1)   A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.

a. 100
b. 50
c. 70
d. 90
Answer  Explanation  Related Ques

ANSWER: 50

Explanation:
Carrier frequency fc = 100MHz
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 500 KHz
Modulation index of FM signal is given by
mf = Δf/fm
= 500 * 103/ 10 * 103
= 50


2)   Narrow band FM has the characteristics:

a. The frequency sensitivity kf is small
b. Bandwidth is narrow
c. Both a and b
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
The bandwidth of the FM signal depends upon the frequency sensitivity kf. When kf is small, the bandwidth of the FM signal becomes narrow and this is known as Narrow Band FM signal. The bandwidth of a narrow band FM signal is almost same as that of an AM signal.


3)   Wide band FM has the characteristics:

a. The frequency sensitivity kf is large
b. Bandwidth is wide
c. Both a and b
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
The bandwidth of the FM signal depends upon the frequency sensitivity kf. When kf is large, the bandwidth of the FM signal becomes wide and this is known as Wide Band FM signal. A large number of sidebands are produced in a FM signal. The bandwidth of a wide band FM signal is very large as compared to that of an AM signal.


4)   Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz.

a. 170 KHz
b. 200 KHz
c. 100 KHz
d. 1000 KHz
Answer  Explanation  Related Ques

ANSWER: 170 KHz

Explanation:
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 75 KHz
According to Carson s rule, BW = 2(Δf + fm)
= 2 (75 + 10)
= 170 KHz


5)   FM signal is better than AM signal because

a. Less immune to noise
b. Less adjacent channel interference
c. Amplitude limiters are used to avoid amplitude variations
d. All of the above
Answer  Explanation  Related Ques

ANSWER: All of the above

Explanation:
FM signal is better than AM signal because FM signals are less immune to noise. Guard bands are provided for less adjacent channel interference so it is easy to be recovered. Amplitude limiters are used to avoid amplitude variations that are caused while transmission due to noise.


6)   FM is disadvantageous over AM signal because

a. much wider channel bandwidth is required
b. FM systems are more complex and costlier
c. Adjacent channel interference is more
d. Both a and b
Answer  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
The guard bands are provided to prevent the interference between adjacent channels in FM signals. Guard bands of 25 KHz are allowed on the either sides so the channel width becomes 2 (75+ 25) = 200KHz where 75KHz is the maximum permissible frequency deviation allowed for commercial FM broadcast. So a much wider channel width is required for FM transmission. FM systems are more complex and therefore costlier than AM transmitters and receivers.


7)   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  Explanation  Related Ques

ANSWER: 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


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

a. sinω1t + sinω2t
b. cosω1t + cosω2t
c. sinω2t + cosω2t
d. sinω1t + sinω1t
Answer  Explanation  Related Ques

ANSWER: 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


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

a. Shot noise
b. Partition noise
c. Industrial noise
d. Flicker noise
Answer  Explanation  Related Ques

ANSWER: Shot noise

Explanation:
Shot noise is caused due to random behaviour of charge carriers. Shot noise is generated due to random emission of electrons from cathodes in electron tubes. In semiconductor devices, shot noise is generated due to random generation and recombination of electron-hole pairs.


10)   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  Explanation  Related Ques

ANSWER: High frequency noise

Explanation:
Transit time noise is the noise caused due to increase in conductance with increase in frequency. This causes the increase in power spectral density of the signal.


11)   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  Explanation  Related Ques

ANSWER: Ratio of output signal to noise ratio to input signal to noise ratio

Explanation:
The figure of merit γ is the ratio of output signal to noise ratio to input signal to noise ratio of a receiver system.


12)   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  Explanation  Related Ques

ANSWER: +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.


13)   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  Explanation  Related Ques

ANSWER: +/- π/2

Explanation:
The Hilbert Transform g(t) of the signal g(t) is defined by
g(t) = (1/π)(τ)/(t - τ)d(τ), -∞ to + ∞
Hilbert transform may be obtained by first taking the Fourier Transform of the signal g(t), multiplying it by jsgn(f), then taking the inverse Fourier Transform and therefore obtaining g (t). jsgn(f) is j for the positive frequency f, and therefore the Hilbert Transform shifts the signal by -900 for a linear system whose input is g(t) and output obtained is g(t).


14)   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  Explanation  Related Ques

ANSWER: Vn is directly proportional to √bandwidth

Explanation:
Maximum Noise power is given by Pn = V2/ R

Relation between Noise voltage and bandwidth

RL = R, for maximum power transfer from Vn to load resistor RL
In the above figure, according to voltage divider method,
V = Vn/2
Pn = V2/ R
= (Vn/2)2/ R
= Vn2/ 4R
Vn2 = 4R Pn
Vn = √(4R Pn)
Noise power Pn = 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
Therefore, Vn = √(4R KTB) or Vn is directly proportional to √bandwidth.


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

a. Input noise power (Pno) to output noise power (Pni)
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  Explanation  Related Ques

ANSWER: Output noise power (Pno) to input noise power (Pni)

Explanation:
The noise factor of a system is defined as the ratio of output noise power (Pno) to input noise power (Pni):
Fn = Pno/ Pni, at standard room temperature 290K
= Pno/GKBT0

The input noise power is Pni = GKBT0,
G is the gain of the amplifier
K = 1.38 X 10-23 J/°K, Boltzmann's constant
B is the bandwidth of the network in hertz (Hz)
To = 290°K
Noise factor may also be defined as the ratio of Signal to Noise ratio at the output to Signal to Noise ratio at the input.
i.e., Fn = Sni/ Sno


16)   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  Explanation  Related Ques

ANSWER: NF = 10 log10(F)

Explanation:
Noise figure (NF) and noise factor (F) signify the degradation of the signal-to-noise ratio (SNR) for any system due to components in a radio frequency (RF) signal . They give the performance of the system.The noise factor of a system is defined as the ratio of output noise power (Pno) to input noise power (Pni):
F = Pno/ Pni, at standard room temperature 290K
Noise Factor when expressed in decibels, it is called noise figure.
i.e. NF = 10 log10(F)


17)   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  Explanation  Related Ques

ANSWER: FN = F1 + (F2 - 1)/ G1 + (F3 - 1)/ G1G2+...+(FN - 1)/ G1G2G3GN

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 factor of the cascaded amplifier is obtained by
FN = F1 + (F2 - 1)/ G1 + (F3 - 1)/ G1G2+...+(FN - 1)/ G1G2G3GN


18)   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  Explanation  Related Ques

ANSWER: 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


19)   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  Explanation  Related Ques

ANSWER: Noise power

Explanation:
The noise temperature Tn of any white noise source is defined by
Tn = Pn/KB
Where Pn is the noise power
K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
The equivalent noise temperature Tn in terms of noise figure may be defined as
Tn = T0 (F-1)
where F is the noise figure
T0 = 170C
= 290K


20)   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  Explanation  Related Ques

ANSWER: Vn = √(4RKTB)

Explanation:
In the following figure the maximum Noise power is given by Pn = V2/ R

Noise voltage

RL = R, for maximum power transfer from Vn to load resistor RL
In the above figure, according to voltage divider method,
V = Vn/2
Pn = V2/ R
= (Vn/2)2/ R
= Vn2/ 4R
Vn2 = 4R Pn
Vn = √(4R Pn)
Noise power Pn = 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
Therefore, Vn = √(4R KTB)