Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Please,please help me - Communication Systems - JEE Main

In an amplitude modulator circuit, the carrier wave is given by, C(t)=4sin(20000\pi t) while modulating signal is given by m(t)=2sin(2000\pi t). The values of the modulation index and lower sideband frequency are:


Option 1)  0.5 and  10  kHz

Option 2)  0.4 and  10 kHz

Option 3)  0.3 and 9 kHz

Option 4)  0.5 and  9 kHz

Answers (1)

\\\text{Modulation index is given by } \mathrm{m}=\frac{\mathrm{A}_{\mathrm{m}}}{\mathrm{A}_{\mathrm{c}}}=\frac{2}{4}=0.5\\ \text{(a) carrier wave frequency is given by } =2 \pi f_{c} \\=2 \times 10^{4} \pi\\ \mathrm{f}_{\mathrm{c}}=10 \mathrm{kHz}

\\\text{(b) modulation wave frequency }$\left(\mathrm{f}_{\mathrm{m}}\right)$ \\ $2 \pi f_{m}=2000 \pi$ \\ $\Rightarrow f_{m}=1 \mathrm{kHz}$ \\ $lower side band frequency $\Rightarrow f_{c}-f_{m}$ $\Rightarrow 10 \mathrm{kHz}-1 \mathrm{kHz}=9 \mathrm{kHz}$

 

Posted by

lovekush

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question