4
\$\begingroup\$

It is said that in frequency modulation the normal wave represents a '0' (False) and a more frequent wave represents a '1' (True).

  1. Is there any particular rule that the normal wave's frequency should continuously be the same? How actually is this "normal frequency" found out?
  2. Consider an example of a wave that has first 3 waves in a time period, then 2 waves and finally one wave. How would it frequency modulate it?
\$\endgroup\$
  • 7
    \$\begingroup\$ When modulating from a digital signal, this is referred to as "frequency-shift keying". "Frequency modulation" usually applies to analog signals, I believe. \$\endgroup\$ – endolith Nov 5 '10 at 21:40
  • \$\begingroup\$ But the word modulation,itself means the process of conversion of digital signals to analog signals.Applying this logic,it should be frequency modulation and the opposite should be frequency demodulation \$\endgroup\$ – user1799 Nov 28 '10 at 6:10
5
\$\begingroup\$

This is not a perfect explanation, but I hope it helps.

When something is frequency modulated there are 2 frequencies, I would not define one as "normal".

There is a frequency that is a digital 1, there is a frequency that is a digital 0.

So, when you are receiving, lets look at a conceptually simple detector. You have a bandpass for each frequency, and on the output of these, you have a rectifier which measures your relative power.

When you are receiving, you need to know the bitrate, or you need a shared clock (which requires a second connection to clock the data, or a second communication channel that cycles between its two frequencies to send the data). Often you read off of the two rectified signals as if you were reading a UART line.

Normally your data rate is significantly slower than your modulation frequency. Normally by a factor of infinity, which in engineering we can get by as approximating as 10.

Please let me know if there is something I can expand on to make this make more sense.

\$\endgroup\$
  • 2
    \$\begingroup\$ you just went over my head. \$\endgroup\$ – user1799 Nov 5 '10 at 17:57
  • 3
    \$\begingroup\$ Wouldn't this be FSK? I suppose that's what fahad was getting at, though. \$\endgroup\$ – Jesse Nov 5 '10 at 18:43
  • 2
    \$\begingroup\$ "which in engineering we can get by as approximating as 10." LOL! \$\endgroup\$ – akohlsmith Nov 8 '10 at 18:51
  • 1
    \$\begingroup\$ @andrew, You wish I was joking. \$\endgroup\$ – Kortuk Nov 8 '10 at 19:46
  • 1
    \$\begingroup\$ @Kortuk nope, as an engineer I know you're 100% correct. Approximation is the heart and soul of engineering. \$\endgroup\$ – akohlsmith Nov 9 '10 at 15:25
0
\$\begingroup\$

As for (1): before modulating a carrier, data are usually encoded - one reason is to guarantee that there are no long trains of zeros or ones. Other reason is to add some data redundancy, so that the receiver can decode original data even if some of them were demodulated incorrectly due to channel noise or interferers.

As for (2): the transition time is (should be) negligible comparing to the data rate. Transition times may in fact be further extended by some filtering (band-pass or low-pass) often required to remove interferences.

\$\endgroup\$

Your Answer

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