[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.
[1989: 2 M] A carrier Accos ωc t is frequency modulation by a signal Em cos ωm t. The modulation index is mf. The expression for the resulting FM signal is.
a) Ac cos [ωc t + mf sin ωc t ]
b) Ac cos [ωc t + mf cos ωc t ]
c) Ac cos [ωc t + 2πmf sin ωc 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}XFM(t)Ac cos [ωc t + mf sin ωc t ]
So, option (a) is the correct answer.
[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.
[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.
[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 ,
fc = 4 GHz = 4000 MHz
fm = 2 MHz
The upper frequency is given by
fH = fc + fm = 4000 + 2 = 4000 MHz
The lower frequency is given by,
fL = fc – fm = 4000 – 2 = 3998 MHZ
f_{s}=\frac{2f_H}Kand 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 .
[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.
[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
XAM(t) = Ac(1 + mcosωmt )cos ωct
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.
[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,
fsi = fs + 2IF
fsi = 1200 + 2(455)
fsi = 2110 kHz
THerefore, 2110 kHz is the correct answer.
[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.
[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 + fm)
BW = 2(50 + 0.5)
BW = 101 kHz
[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