5
\$\begingroup\$

All radio receivers seem to use the heterodyne principle. I can see that it will be easier to work with a fixed frequency than demodulating from a wide range. Questions:

  • what's the difference between heterodyne and superheterodyne
  • what does a double-superheterodyne add to that
  • is there a reason why everyone uses the same intermediate frequency? is there something special about 10.7 Mhz which makes it better than 10 Mhz?
  • is the heterodyne principle also used in modulators?
\$\endgroup\$
  • 1
    \$\begingroup\$ @Frederico Russo: That's a lot of questions. Perhaps you should split them up \$\endgroup\$ – Brian Carlton Sep 6 '12 at 16:03
5
\$\begingroup\$

Superheterodyne and heterodyne are effectively the same. The "super-" merely indicates that the intermediate frequency is ultrasonic. See part way through the history section of Wikipedia's Superheterodyne receiver.

Double-superheterodyne is a radio system which uses two different frequency conversion stages. This is used in analog TV to recover the audio band or in very wide tuning range radios in order to improve image rejection. See the wikipedia article's Advanced Designs section.

Why 10.7MHz? In addition to producing the desired intermediate frequency signal, heterodyning also produces an image signal at twice the intermediate frequency. This is removed by the bandpass filter; however, image rejection is improved by increasing the intermediate frequency so that the image is better attenuated by the filter. At the time of FM radio development, ~10MHz was near the limits of what was possible.

The choice of 10.7 vs 10 was so that the local oscillator radiation would fall in between stations since the US FM radio frequency allocation places stations every 0.2MHz. (summarized from the radio board's Why 10.7MHz IF?)

Yes, the heterodyne principle may be used in modulation. Wikipedia Heterodyne page

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ The statement that 10MHz was the limit of what was possible doesn't make sense if they were able to transmit on 10 times that frequency at the same time. The 10 is dictated by the fact that it's half the width of the VHF band 88-108MHz, so that the first image is at least half the dial away, and also by an equation somewhere in the theory that I can never find when I look that gives THD as an inverse function of IF frequency. There are tuners that use 12.5MHz. \$\endgroup\$ – Marquis of Lorne Mar 19 '13 at 22:16

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