JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Semiconductors

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The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulate by a 7.7 kHz sine wave of 5 V amplitude are

\frac{a}{10} V

and

\frac{b}{10} V

respectively. Then the value of

\frac{a}{b}

is ______.

Answer: 1

Solution

The modulation index is

μ = \frac{A _{m}}{A _{c}} = \frac{5}{15} = \frac{1}{3}

. An amplitude modulated wave splits half the modulated amplitude equally between the two side bands, each of amplitude

\frac{μ A _{c}}{2} = \frac{1}{3} \times \frac{15}{2} = 2.5 V = \frac{25}{10} V

The upper and lower side bands are always equal in amplitude, so

a = b = 25

and

\frac{a}{b} = 1

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer

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