1
$\begingroup$

I cannot understand while the AM modulations have sidebands which are of a frequency higher than the carrier. If for example we are modulating a 4kHz signal to a carrier of 1MHz how we end up with a 1.004MHz if the 4kHz is in the amplitude not in the frequency. In the signal we still have the 1MHz which is the highest frequency.

$\endgroup$
  • $\begingroup$ Three AM outputs will be carrier, (carrier - modulation), (carrier + modulation). 1mhz, .991mhz, 1.004mhz $\endgroup$ – Optionparty Mar 10 '18 at 16:00
  • $\begingroup$ Did you at least google form the many resources doing the math? $\endgroup$ – Juancho Mar 10 '18 at 18:28
1
$\begingroup$

The answer is that the two frequencies are mixing to produce the sum and difference of the two. Note that this is simply the frequency component. So in your example, not only is 1.004 MHz produced but so is 0.996 MHz. Collectively these are called the sidebands. Depending on the desired transmission mode, one of these resulting frequencies may be filtered out. For example, if the desired transmission mode was LSB (lower side band), the original 1.0 MHz carrier and the 1.004 MHz signal would be filtered out leaving only the 0.996 MHz signal - the lower sideband.

In AM (and SSB and DSB), the amplitude of the frequencies resulting from the mixing is controlled by the amplitude of the audio modulating signal. So in the LSB example above, the amplitude of the 0.996 frequency is following the amplitude of the 4 kHz audio modulating signal.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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