Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
548 views
in General by (237k points)
closed by
In phase modulated signal, the frequency deviation is proportional to
1. Frequency only
2. Amplitude only
3. Both (1) and (2)
4. none of the above

1 Answer

0 votes
by (239k points)
selected by
 
Best answer
Correct Answer - Option 3 : Both (1) and (2)

The frequency deviation in a phase-modulated carrier is proportional to the amplitude and frequency of the modulating signal both.

Derivation:

A general expression for a phase-modulated wave is:

xPM (t) = A cos [2πfct + kpm(t)]

The instantaneous angle is given as:

ϕi(t) = 2πfct + kp m(t)

The instantaneous frequency (in Hz) will be obtained as:

\({f_i}\left( t \right) = \frac{1}{{2\pi }}\frac{{d{\phi _i}\left( t \right)}}{{dt}}\)

\({f_i}\left( t \right) = \frac{1}{{2\pi }}\frac{d}{{dt}}\left[ {2\pi {f_c}t + {k_p}m\left( t \right)} \right]\)

\( = \frac{1}{{2\pi }} \cdot 2\pi {f_c} + \frac{{{k_p}}}{{2\pi }} \cdot \frac{{d\;m\left( t \right)}}{{dt}}\)

\({f_i}\left( t \right) = {f_c} + \frac{{{k_p}}}{{2\pi }}\frac{{dm\left( t \right)}}{{dt}}\)

Frequency deviation is given by:

\(Δ f= \frac{{{k_p}}}{{2\pi }}\frac{{dm\left( t \right)}}{{dt}}\)

For a single tone modulation:

m(t) = Am sin ωmt

The frequency deviation becomes:

\(Δ f= \frac{{{k_p}}}{{2\pi }}\frac{{d(A_msin\omega _mt)}}{{dt}}\)

\(Δ f= \frac{{{k_pA_m\omega _mcos\omega _mt}}}{{2\pi }} \)

Δf ∝ Afm

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...