Communication miscellaneous Easy Questions and Answers | Page - 25

Communication miscellaneous

The plot for frequency deviation is—

1. A
1. B
1. C
1. D

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We know that, ωi – ωc = |kf m(t)|max
Δω = kf m(t)
So frequency deviation will be amplitude of m (t) multiplied by kf which is given in fig. (B).
Hence alternative (B) is the correct choice.

Correct Option: B

We know that, ωi – ωc = |kf m(t)|max
Δω = kf m(t)
So frequency deviation will be amplitude of m (t) multiplied by kf which is given in fig. (B).
Hence alternative (B) is the correct choice.

Let message signal m(t) = cos (4π 10³t) and carrier signal c(t) = 5 cos (2π 10⁶)t are used to generate a FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term, cos [2π(1008 × 10³t)t] in the FM signal would be—

1. 5J4 (3)
1. 5 J₈
  2
  (3)
1. 5 J₈
  2
  (4)
1. 5 J₄(6)

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Given that, m(t) = cos (2π × 10³t)
c(t) = 5 cos (2π × 10⁶t)
and
Δf = 3 (transmission bandwidth)
Δf = 3(2f_m) = 6f_m
Now, modulation index
β = Δf = 6f_m = 6
f_m f_m

The FM signal is represented in terms of Bessel function as
S_FM(t)=A_c ∑^∞ _{n = – ∞} Jn (β) cos (ω_c + nω_mt)
given that
S_FM (t) = [cos 2π (10008 × 10³t)] or
S_FM (t) = cos[(2π × 10⁶ + 8π × 10³)t]
= A_c ∑^∞_{n = – ∞} J_n (β) cos(ω_c + ω_m)t
n = 4
Thus coefficient, A_c J_n (β) = 5j₄ (6)

Correct Option: D

Given that, m(t) = cos (2π × 10³t)
c(t) = 5 cos (2π × 10⁶t)
and
Δf = 3 (transmission bandwidth)
Δf = 3(2f_m) = 6f_m
Now, modulation index
β = Δf = 6f_m = 6
f_m f_m

The FM signal is represented in terms of Bessel function as
S_FM(t)=A_c ∑^∞ _{n = – ∞} Jn (β) cos (ω_c + nω_mt)
given that
S_FM (t) = [cos 2π (10008 × 10³t)] or
S_FM (t) = cos[(2π × 10⁶ + 8π × 10³)t]
= A_c ∑^∞_{n = – ∞} J_n (β) cos(ω_c + ω_m)t
n = 4
Thus coefficient, A_c J_n (β) = 5j₄ (6)

A carrier wave of 1 MHz and amplitude 3V is frequency modulated by a sinusoidal modulating signal frequency of 500 Hz and peak amplitude of 1V. The frequency deviation is 1kHz. The peak level of the modulating waveform is changed to 2kHz. The expression for the new modulation waveform is—

1. cos [2π × 10⁶t + 2·5 cos (4π × 10³t)]
1. cos [2π × 10⁶t + 5 cos (4π × 10³t)]
1. 3 cos [2π × 10⁶t + 2·5 sin (4π × 10³t)]
1. 3 cos [2π × 10⁶t + 5 cos (4π × 10³t)]

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Since for FM modulated wave, frequency sensitivity constant Kf remains constant. So first we will calculate,
K_f = Δf = 1 × 10³ = 1kHz/V
A_m 1

Now for the second case,
A_m = 5V and f_m = 2 kHz
Δf = K_f × A_m = 1 × 5 = 5 kHz
Modulation index,
β = Δf = 5 kHz = 2.5
f_m 2 kHz

The FM signal say S(t),
s(t) = 3 cos [2π × 10⁶ + β sin (2π·2 × 10³t)]
= 3 cos [2π × 10⁶t + 2·5 sin (4π × 10³t)]

Correct Option: C

Since for FM modulated wave, frequency sensitivity constant Kf remains constant. So first we will calculate,
K_f = Δf = 1 × 10³ = 1kHz/V
A_m 1

Now for the second case,
A_m = 5V and f_m = 2 kHz
Δf = K_f × A_m = 1 × 5 = 5 kHz
Modulation index,
β = Δf = 5 kHz = 2.5
f_m 2 kHz

The FM signal say S(t),
s(t) = 3 cos [2π × 10⁶ + β sin (2π·2 × 10³t)]
= 3 cos [2π × 10⁶t + 2·5 sin (4π × 10³t)]

Given message signal
m(t) = sin (2000 πt), K_f = 100 kHz/V and K_p = 10 rad/V. The bandwidth of FM and PM signals are respectively—

1. 204 kHz, 44 kHz
1. 202 kHz, 22 kHz
1. 402 kHz, 42 kHz
1. None of these

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Given m(t) = sin (2000πt)
K_f = 100 kHz/V
K_p = 10 rad/V.
for FM: BW = 2 (Δf + f_m)
Δf = |K_f A_m|_max = 1000 × 1 = 100 kHz
f_m = 1000 Hz or 1 kHz
Now BW = 2(100 + 1) = 202 kHz
for PM:
Δf = |Kp d/dt m(t)| Hz
2π

Δf = 10 × 2000π = 10kHz
2π

BW = 2(Δf + f_m) = 2 (10 + 1) = 22 kHz
Hence alternative (B) is the correct choice.

Correct Option: B

Given m(t) = sin (2000πt)
K_f = 100 kHz/V
K_p = 10 rad/V.
for FM: BW = 2 (Δf + f_m)
Δf = |K_f A_m|_max = 1000 × 1 = 100 kHz
f_m = 1000 Hz or 1 kHz
Now BW = 2(100 + 1) = 202 kHz
for PM:
Δf = |Kp d/dt m(t)| Hz
2π

Δf = 10 × 2000π = 10kHz
2π

BW = 2(Δf + f_m) = 2 (10 + 1) = 22 kHz
Hence alternative (B) is the correct choice.

An angle-modulated signal is given by
s(t) = cos 2π (2 × 10⁶t + 30 sin 150t + 40 cos 150t)
The maximum frequency and phase deviations of s(t) are —

1. 10.5 kHz, 140π rad
1. 6 kHz, 80π rad
1. 10.5 kHz, 100π rad
1. 7.5 kHz,100π rad

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Given angle modulated signal
s(t) = cos 2π (2 × 10⁶t + 30 sin 150t + 40 cos 150t)
From equation
θ(t) = 2 × 10⁶t + 30 sin 150t + 40 cos 150t
dθ(t) = 2 × 10⁶ + 30 × 150 cos 150t – 40 × 150 sin 150t
dt

or
dθ(t) = 2 × 10⁶ + 30 + 150 √ 30² + 40²cos (150t + φ)
dt

where
φ = tan⁻¹ −40
30

or
dθ(t) = 2 × 10⁶ + 7500 cos (150t + φ)
dt

ω_i – ω_c = Δω = |Kf f(t)|_max
where,
dθ(t) = ω_i
dt

2 × 10⁶ = ω_c
called maximum frequency deviation.
Hence Δω = 7500 = 7·5 kHz and max. phase deviation is
Δφ = √30²+40² × 2π = 100π radian.

Correct Option: D

Given angle modulated signal
s(t) = cos 2π (2 × 10⁶t + 30 sin 150t + 40 cos 150t)
From equation
θ(t) = 2 × 10⁶t + 30 sin 150t + 40 cos 150t
dθ(t) = 2 × 10⁶ + 30 × 150 cos 150t – 40 × 150 sin 150t
dt

or
dθ(t) = 2 × 10⁶ + 30 + 150 √ 30² + 40²cos (150t + φ)
dt

where
φ = tan⁻¹ −40
30

or
dθ(t) = 2 × 10⁶ + 7500 cos (150t + φ)
dt

ω_i – ω_c = Δω = |Kf f(t)|_max
where,
dθ(t) = ω_i
dt

2 × 10⁶ = ω_c
called maximum frequency deviation.
Hence Δω = 7500 = 7·5 kHz and max. phase deviation is
Δφ = √30²+40² × 2π = 100π radian.