Question

AQ, BS, C-→ Pand D→ R
7. In an amplitude modulator circuit, the
carrier wave is given by
C(t)= 4 sin(20000 it) while modulating
signal is given by, m(t)=2 sin (2000 met). The
values of modulation index and lower side
band frequency are
[JEE Main 2019, 12 April Shift-1
(a) 0.5 and 10 kHz (b) 0.4 and 10 kHz
(c) 0.3 and 9 kHz (d) 0.5 and 9 kHz​

Answer 1

SoluTion :-

$\sf {Modulation\ index\ is\ given\ by}\\\\\sf {m=\frac{A_m}{A_c}=\frac{2}{4}=0.5 }\\\\\\\sf {and\ (a)\ carrier\ wave\ frequency\ is\ given\ by}\\\\\sf {2 \pi f_c = 2 \times 10^4 \pi}\\\\\sf {f_c = 10kHz}\\\\\\\sf { (b)\ modulation\ wave\ frequency\ (f_m)}\\\\\sf {2 \pi f_m=2000 \pi}\\\\\sf {\Rightarrow f_m = 1kHz}\\\\\sf {lower\ side\ band\ frequency \Rightarrow f_c - f_m}\\\\\sf {10kHz-1kHz=9kHz}$

Answer 2

Answer:

Hey, u angry?

Explanation:

Please say