Search any question & find its solution
Question:
Answered & Verified by Expert
A signal of frequency \(10 \mathrm{kHz}\) and peak voltage \(10 \mathrm{~V}\) is used to amplitude modulate a carrier of frequency \(1 \mathrm{MHz}\) and peak voltage \(20 \mathrm{~V}\). The side-band frequencies in \(\mathrm{kHz}\) are
Options:
Solution:
2989 Upvotes
Verified Answer
The correct answer is:
1010,990
Given, frequency of signal, \(f_s=10 \mathrm{kHz}\)
peak voltage, \(V_p=10 \mathrm{~V}\)
carrier frequency, \(f_c=1 \mathrm{mHz}\)
and peak voltage, \(V_c=20 \mathrm{~V}\)
\(\therefore\) The side band frequencies are given as
\(f=f_c \pm f_s\)
Putting the given values, we get
\(f=1 \times 10^6 \pm 10 \times 10^3 \mathrm{~Hz}=(1000 \pm 10) 10^3 \mathrm{~Hz}\)
So, the side band frequencies in \(\mathrm{kHz}\) is \(1010 \mathrm{kHz}\) and \(990 \mathrm{kHz}\).
peak voltage, \(V_p=10 \mathrm{~V}\)
carrier frequency, \(f_c=1 \mathrm{mHz}\)
and peak voltage, \(V_c=20 \mathrm{~V}\)
\(\therefore\) The side band frequencies are given as
\(f=f_c \pm f_s\)
Putting the given values, we get
\(f=1 \times 10^6 \pm 10 \times 10^3 \mathrm{~Hz}=(1000 \pm 10) 10^3 \mathrm{~Hz}\)
So, the side band frequencies in \(\mathrm{kHz}\) is \(1010 \mathrm{kHz}\) and \(990 \mathrm{kHz}\).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.