Analog Communication Test Questions - Set 8

ANSWER: 50

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

ANSWER: Both a and b

Explanation:
The bandwidth of the FM signal depends upon the frequency sensitivity k_f. When k_f 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.

ANSWER: Both a and b

Explanation:
The bandwidth of the FM signal depends upon the frequency sensitivity k_f. When k_f 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.

ANSWER: 170 KHz

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

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.

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.

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

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

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

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.

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.

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.

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.

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).

ANSWER: V_n is directly proportional to √bandwidth

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



R_L = R, for maximum power transfer from V_n to load resistor R_L
In the above figure, according to voltage divider method,
V = V_n/2
P_n = V²/ R
= (V_n/2)²/ R
= V_n²/ 4R
V_n² = 4R P_n
V_n = √(4R P_n)
Noise power P_n = 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
Therefore, V_n = √(4R KTB) or V_n is directly proportional to √bandwidth.

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

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

The input noise power is P_ni = 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., F_n = S_ni/ S_no

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 (P_no) to input noise power (P_ni):
F = P_no/ P_ni, at standard room temperature 290K
Noise Factor when expressed in decibels, it is called noise figure.
i.e. NF = 10 log₁₀(F)

ANSWER: F_N = F₁ + (F₂ - 1)/ G₁ + (F₃ - 1)/ G₁G₂+...+(F_N - 1)/ G₁G₂G₃G_N

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
F_N = F₁ + (F₂ - 1)/ G₁ + (F₃ - 1)/ G₁G₂+...+(F_N - 1)/ G₁G₂G₃G_N

ANSWER: 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

ANSWER: Noise power

Explanation:
The noise temperature T_n of any white noise source is defined by
T_n = P_n/KB
Where P_n 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 P_n is delivered.
The equivalent noise temperature T_n in terms of noise figure may be defined as
T_n = T₀ (F-1)
where F is the noise figure
T₀ = 170C
= 290K

ANSWER: V_n = √(4RKTB)

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



R_L = R, for maximum power transfer from V_n to load resistor R_L
In the above figure, according to voltage divider method,
V = V_n/2
P_n = V²/ R
= (V_n/2)²/ R
= V_n²/ 4R
V_n² = 4R P_n
V_n = √(4R P_n)
Noise power P_n = 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
Therefore, V_n = √(4R KTB)