0
\$\begingroup\$

I have a problem understanding the circuit below where the LC oscillator is used to produce the negative cycle wave for the amplitude modulation.

I am having a difficulty in grasping the fact that the oscillator produces the negative cycle. Does the LC get a pulse from the half wave and then this pulse is converted to sine form?

enter image description here

Secondly, I have my carrier wave given to the signal but with the changing values of the LC my frequency should change accordingly. Should I choose a value of the oscillation equal to the resonant frequency?

\$\endgroup\$
5
  • 1
    \$\begingroup\$ The "LC" is not an oscillator; it's a tuned circuit (sometimes referred to as a tank). The value of the LC doesn't change and, accordingly, the frequency should not change either. The clue is in the phrase "amplitude modulation" and not frequency modulation. \$\endgroup\$
    – Andy aka
    Commented Oct 23, 2023 at 9:13
  • \$\begingroup\$ But how does it generate the nagitive cycle when the rectifier makes it positive \$\endgroup\$
    – kam1212
    Commented Oct 23, 2023 at 9:22
  • \$\begingroup\$ @Andyaka by. Changing the values I mean that changing the values of the LC changes the results and the output wave. What does the values depend on.? \$\endgroup\$
    – kam1212
    Commented Oct 23, 2023 at 9:29
  • \$\begingroup\$ There does seem to be some confusion, (and maybe a few things are being mixed up too). What is the given part of your question - the two frequencies or the two LC components? Did you calculate the resonant frequency of the existing LC components? Was it a frequency of interest? \$\endgroup\$
    – Nedd
    Commented Oct 23, 2023 at 11:19
  • \$\begingroup\$ @Nedd I meant that how do we select the two LC components value?? Also what's the concept behind using an LC tank here?? I am not getting it conceptually \$\endgroup\$
    – kam1212
    Commented Oct 23, 2023 at 11:43

1 Answer 1

2
\$\begingroup\$

AM modulation in the simplest way is done like this: enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.