Communication miscellaneous
- The bandwidth required for amplitude modulation is—
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Refer synopsis.
Correct Option: B
Refer synopsis.
- Find the image rejection in a superheterodyne receiver with fIF = 455 kHz when tuned to MHz signal. Given Q. of the preselector and RF amplifier tuned circuit each is 65—
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We know that image rejection ratio, when quality factor is represented by
Image rejection ratio = √1 + ρ2Q2
where,ρ
=fimage − fc fc fimage
Given, fimage = fc + 2fIF
= 30 × 106 + 2 × 455 × 103
= 30.91 × 106ρ
=30.91 × 106 − 30 × 106 30 × 106 30.91 × 106
Image rejection ratio = √1 + (0.6)2 × 652
Image rejection ratio = √1 + (0.6)2 × 652
= √15.21
= 3.9Correct Option: A
We know that image rejection ratio, when quality factor is represented by
Image rejection ratio = √1 + ρ2Q2
where,ρ
=fimage − fc fc fimage
Given, fimage = fc + 2fIF
= 30 × 106 + 2 × 455 × 103
= 30.91 × 106ρ
=30.91 × 106 − 30 × 106 30 × 106 30.91 × 106
Image rejection ratio = √1 + (0.6)2 × 652
Image rejection ratio = √1 + (0.6)2 × 652
= √15.21
= 3.9
- A sinusoidal 10 cos ωct, ωc = 106 rad/sec is amplitude modulated using another sinusoidal Am cos ωmt, ωm = 103 with 100% modulation, then the expression for AM modulated wave is expressed as—
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AM modulated wave is expressed as
δ(t) = Ac[1 + μ cos ωmt] cos ωct
given,
Ac = 10μ = Am = Am = 1 Ac 10
(∴ 100% modulation)
or
Am = 10
ωm = 103 rad/sec
ωc = 106 rad/sec
Hence, s(t) = 10[1 + 1.cos 103t] cos 106tCorrect Option: D
AM modulated wave is expressed as
δ(t) = Ac[1 + μ cos ωmt] cos ωct
given,
Ac = 10μ = Am = Am = 1 Ac 10
(∴ 100% modulation)
or
Am = 10
ωm = 103 rad/sec
ωc = 106 rad/sec
Hence, s(t) = 10[1 + 1.cos 103t] cos 106t
- The sum of two signals e1 = 3 sin (4π × 103t) and e1= 5 sin(4π × 256t) is sampled at 1024 Hz. The sampled signal is passed through a low pass filter with cut off at 2048 Hz. The output of the filter will contain components at—
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Given that
e1= 3 sin(4π × 103t), fm1 = 2 × 103 Hz
e2 = 5 sin(4 π × 256t), fm2 = 256 Hz
sampling frequency, fs = 1024 Hz
cut-off frequency of LPF, ωc = 2048 Hz
Since for proper reconstruction of signal fs ≥ 2fm
For fm1, fs less than 2fm1 so it will pass through the LPF,
while for fm2, fs > 2fm2 [i.e. 1024 > (2 × 256)]
So the output of the filter will contain 256 Hz component only.Correct Option: C
Given that
e1= 3 sin(4π × 103t), fm1 = 2 × 103 Hz
e2 = 5 sin(4 π × 256t), fm2 = 256 Hz
sampling frequency, fs = 1024 Hz
cut-off frequency of LPF, ωc = 2048 Hz
Since for proper reconstruction of signal fs ≥ 2fm
For fm1, fs less than 2fm1 so it will pass through the LPF,
while for fm2, fs > 2fm2 [i.e. 1024 > (2 × 256)]
So the output of the filter will contain 256 Hz component only.
- In a broadcast transmitter, the RF output is represented as
e(t) = 50 [1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
What are the sidebands of the signals in radians?
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The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
Let 5000 = ωm1 or ωm1 = 0·005 MHz
9000 = ωm2 or ωm2 = 0·009 MHz
6 × 106 = ωc1 or ωc1 = 6 MHz
e(t) = 50[cos ωc1 t + 0·89 cos ωm1t. cos ωc1t + 0·30 sin ωm2t]
Sidebands of the signals are at
ωc1 ± ωm1 and ωc1 ± ωm2, frequency
i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
6·005 × 106, 5·995 × 106, 6·009 × 106, 5·991 × 106
Hence alternative (B) is the correct choice.Correct Option: B
The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
Let 5000 = ωm1 or ωm1 = 0·005 MHz
9000 = ωm2 or ωm2 = 0·009 MHz
6 × 106 = ωc1 or ωc1 = 6 MHz
e(t) = 50[cos ωc1 t + 0·89 cos ωm1t. cos ωc1t + 0·30 sin ωm2t]
Sidebands of the signals are at
ωc1 ± ωm1 and ωc1 ± ωm2, frequency
i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
6·005 × 106, 5·995 × 106, 6·009 × 106, 5·991 × 106
Hence alternative (B) is the correct choice.
