Analog Communication Set-1

[1987: 2 M] In a superheterodyne AM receiver, the image channel selectively is determined by

a) The preselector and RF stages

b) The preselector, RF and IF stages

c) The IF stages

d) all the stages

Ans: (a) The preselector and RF stages

Explanation:

By placing many no. of tank circuits, the image frequency rejection can be improved. Before IF stage it becomes impossible to remove it from the wanted signal.

Therefore, option (a) is the correct answer.

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[1989: 2 M] A carrier A_ccos ω_c t is frequency modulation by a signal E_m cos ω_m t. The modulation index is m_f. The expression for the resulting FM signal is.

a) A_c cos [ω_c t + m_f sin ω_c t ]

b) A_c cos [ω_c t + m_f cos ω_c t ]

c) A_c cos [ω_c t + 2πm_f sin ω_c t ]

d) A_c\cos\left[\omega_ct+\frac{2\pi m_fE_m}{\omega_m}\cos\omega_mt\right]

Ans: (a)

Explanation:

The expression for FM signal is given by,

X_{FM}(t)=A_c\cos\left[\omega_ct+k_f\int m(t)dt\right] X_{FM}(t)=A_c\cos\left[\omega_ct+k_f\int E_m\cos\omega_mtdt\right] X_{FM}(t)=A_c\cos\left[\omega_ct+\frac{k_fE_m}{\omega_m}\sin\omega_mt\right]

Modulation index

m_1=\frac{k_fE_m}{\omega_m}

X_FM(t)A_c cos [ω_c t + m_f sin ω_c t ]

So, option (a) is the correct answer.

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[1989: 2 M] Which of the following schemes suffer(s) from the threshold effect ?

a) AM detection using envelope detection

b) AM detection using synchronous detection

c) FM detection using a discriminator

d) SSB detection with synchronous detection

Ans: (c)

Explanation:

Correct option is (d) Because from the threshold effect , the Fm detection using a discriminator suffers from the threshold effect.

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[1989: 2 M] A signal x(t) = 2 cos (π. 10⁴ t) volts is applied to an FM modulation with the sensitivity constant of 10 kHz/volt. Then the modulation index of the FM wave is

a) 4

b) 2

c) 4/π

d) 2/π

Ans: (a)

Explanation: We know that , Modulation index is given by,

Modulation index = M_f\;=\frac{k_fA_m}{\omega_m}=\frac{\triangle\omega}{\omega_m}

M_f=\frac{(2\pi\times10\times10^3)\times(2)}{\pi\times10^4}

= 4

So, option (a) is the correct answer.

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[1990: 2 M] A 4 GHz carrier is DSB-SC modulated by a lowpass message signal with maximum frequency of 2 MHz. The resultant signal is to be ideally sampled. The minimum frequency of the sampling impulse train should be

a) 4 MHz

b) 8 MHz

c) 8 GHz

d) 8.004 GHz

Ans: (a)

Explanation: It is given that ,

f_c = 4 GHz = 4000 MHz

f_m = 2 MHz

The upper frequency is given by

f_H = f_c + f_m = 4000 + 2 = 4000 MHz

The lower frequency is given by,

f_L = f_c – f_m = 4000 – 2 = 3998 MHZ

f_{s}=\frac{2f_H}K

and K=\frac{f_H}{f_H-f_L}

=\frac{4002MHz}{4MHz}

=1000.5 ≃ 1000

f_{s}=\frac{2\times4002MHz}{1000}

= 8.004 MHz ≃ 8 MHz

So, (b) is the correct option .

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[1990: 2 M] In commercial TV transmission in India, picture and speech signals are modulated respectively as

(Picture) (Speech)

a) VSB and VSB

b) VSB and SSB

c) VSB and FM

d) FM and VSB

Ans: (c)

Explanation:

In iNdia, the speech or audio signal is modulated using FM modulation and the picture signal is modulated using VSB modulation. So, option (c) is the correct answer.

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[1993: 2 M] Which of the following demodulator(s) can be used for demodulating the signal x (t) = 5(1 + 2 cos 200πt) cos 20000 πt

a) Envelope demodulator

b) square-law demodulator

c) synchronous demodulator

d) None of the above

Ans: (c)

Explanation:

It is Given that

x(t) = 5(1 + 2 cos200πt)cos20000πt …..(i)

The standard equation for AM signal is

X_AM(t) = A_c(1 + mcosω_mt )cos ω_ct

Comapring Eq.(1) and (ii) , We have m = 2

As the modulation index is more than 1 here so it is the case of over modulation. When the modulation index of AM wave is more than 1 (over modulation) then the detection is possible only with synchronous modulation only. Such signals can not be detected with envelope detector.

So, option (c) is the correct answer.

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[1993: 2 M] A superheterodyne radio receiver with an intermediate frequency of 455 kHz is tuned to a station operating at 1200 kHz. The associated image frequency is _____kHz.

Explanation: image frequency is given by,

f_si = f_s + 2IF

f_si = 1200 + 2(455)

f_si = 2110 kHz

THerefore, 2110 kHz is the correct answer.

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[1994: 1 M] v(t) = 5[cos(10⁶ πt) – sin(10³ πt) × sin(10⁶ πt )] represents

a) DSB suppressed carrier signal

b) AM signal

c) SSB upper sideband signal

d) Narrow band FM signal

Ans: (d)

Explanation:

v(t) = 5 cos (10⁶ πt) – \frac52\cos(10^6-10^3)\pi t+\frac52(10^6+10^3)\pi t

Carrier and upper side-band are in phase and lower side band is out of phase with carrier and upper side-band.

the given signal is narrow band FM signal.

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[1994: 1 M] A 10 MHz carrier is frequency modulation by a sinusoidal signal of 500 Hz, the maximum frequency deviation being 50 kHz. The bandwidth required, as given by the Carson’s rule is _______.

Explanation: We know that Varson’s rule states that the bandwidth required to transmit an angle modulated wave is twice the sum of the epak frequency deviation and highest modulating signal frequency.

Therefore, By Carson’s rule

BW = 2(△f + f_m)

BW = 2(50 + 0.5)

BW = 101 kHz

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[1994: 2 M] Match List-I with List-II and select the correct answer using the code given below the Lists:

List-I List-II

A) SSB 1. Envelope detector

B) AM 2. Integrated and dump

c) BPSK 3. Hilbert transform

4. Ratio detector

5. PLL

Codes:

A B C

a) 3 1 2

b) 3 2 1

c) 2 1 3

d) 1 2 3

Explanation: THe correct matches are as follows.

SSB ⟶ Hilbert transform

AM ⟶ Envelope detector

BPSk ⟶ Integrate and dump

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