Communications Test 2

Result

Communications Test 2

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Weekly Quiz Competition

Question 1
1 / -0

An angle modulated signal is given as \(v\left( t \right) = 5\cos \left[ {10^62\pi t + 5\sin \left( {{{10}^3} \times 2\pi t} \right)} \right]\). The maximum phase deviation and frequency deviations are

A
10 rad and 5 kHz

B
10 rad and 10 kHz

C
5 rad and 10 kHz

D
5 rad and 5 kHz

Solution

v(t) = 5cos [2π × 10⁶t + 5 sin (2π × 10³t)]
Instantaneous phase = 2π × 10⁶t + 5 sin (2π × 10³t)
Maximum phase deviation = 5 rad
Differentiating instantaneous frequency
= 2π × 10⁶ + 5 × 2π × 10³ cos (2π × 10³t)
Maximum frequency deviation \( = \frac{{10\pi \times {{10}^3}}}{{2\pi }}\)
= 5 kHz
Question 2
1 / -0

A modulated signal is given
x(t) = 10 cos (2π × 10⁵t) + 6 cos (2π × 10³t)cos (2π × 10⁵t)
This signal is to be detected by a linear diode. The most suitable value of resistance if the capacitor is of 100 pF is

A
10 kΩ

B
1 kΩ

C
1 MΩ

D
100 MΩ

Solution

Concept:
For demodulation of Am using envelope detector the time constant of circuit is given by
\(\frac{1}{{{f_c}}} < {RC} < \frac{1}{w}\)
ω → Bandwidth of the message signal
Application
The message signal can be rewritten as:
10 (1 + 0.6 cos (2π × 10³t)) cos (2π × 10⁵ t)
here
f_c = 10⁵ Hz
f_m = 10³ Hz
\(\frac{1}{{{{10}^5}}} < RC < \frac{1}{{{{10}^3}}}\)
\(\frac{1}{{{{10}^5}C}} < R < \frac{1}{{{{10}^3}C}}\)
\(\frac{1}{{{{10}^5} \times 100 \times {{10}^{ - 12}}}} < R < \frac{1}{{\begin{array}{*{20}{c}} {{{10}^3} \times 100 \times {{10}^{ - 12}}} \end{array}}}\)
10⁵ < R < 10⁷
10 kΩ < R < 10 MΩ
Question 3
1 / -0

A FM is fed with a message signal m(t) = A_M cos(2π × 1.6 × 10³)t. The value of AM is gradually increased from 0 and it is found that at AM = 2V the carrier component goes to zero for the first time. Determine at what value of AM the carrier component will again go to zero ______.
Given J₀(2.4) = 0 J₀(5.52) = 0

A
4.5-4.7

B
4.54-4.71

C
4.57-4.76

D
4.59-4.72

Solution

The amplitude of carrier in WBFM is given by \(\frac{{{A_c}{J_0}\left( \beta \right)}}{2}\)
If amplitude of carrier goes to zero for the first time
⇒ J₀(β) = 0
⇒ β = 2.4
\(\beta = \frac{{{\rm{\Delta }}f}}{{{f_m}}} = 2.4\)
Δf = 2.4 × f_m = 2.4 × 1600
Δf = kf × A_m = 3.84 kHz
\(kf = \frac{{3.84}}{2} = 1.92\;kHz/V\)
For 2^nd time carrier to be zero
\(\frac{{{A_c}{J_0}\left( \beta \right)}}{2} = 0\)
\(\beta = 5.52 = \frac{{{\rm{\Delta }}f}}{{{f_m}}} = \frac{{kf \times A_m'}}{{{f_m}}}\)
\(A_m' = \frac{{5.52 \times {f_m}}}{{kf}}\)
\(= \frac{{5.52 \times 1.6}}{{1.92}}\)
= 4.6 V