Sarthaks Test
0 votes
77 views
in Electronics by (54.3k points)

Determine the percent modulation of an AM wave whose total power content is 2500W and whose sidebands each contain 300W.

1 Answer

+1 vote
by (53.4k points)
selected by
 
Best answer

PT = Pc + PLSB + PUSB

2500 = Pc + 300 + 300

Pc = 2500 - 600

Pc = 1900W

PT = Pc(1 + \(\frac{m_a^2}{2}\))

2500 = 1900[1 + \(\frac{m_a^2}{2}\)]

\(\frac{2500}{1900}\) = 1 + \(\frac{m_a^2}{2}\)

1.315 - 1 = \(\frac{m_a^2}{2}\)

\(m_a^2\) = 2(0.315) = 0.631

ma\(\sqrt{0.631}\)

ma = 0.794

= 0.794 x 100

= 79.4%

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

...