2
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So to obtain intermediate frequency Fif, why it is preferred to use Flo > fc

So we use this equation

Fif = Flo - Fc

Instead of

Fif = Fc - Flo

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  • \$\begingroup\$ Look up image rejection. \$\endgroup\$ May 12, 2017 at 21:19
  • \$\begingroup\$ I know about image response, it will happen in either cases, the question is why we choose the local oscillator frequency to be higher than carrier frequency not smaller, as both will result in the IF frequency ? \$\endgroup\$
    – Mostafa
    May 12, 2017 at 21:33
  • \$\begingroup\$ Sometimes LO is lower. \$\endgroup\$
    – user16324
    May 12, 2017 at 21:41
  • \$\begingroup\$ Depending upon the particular IF and input frequency you can force the image to a negative frequency with the LO above the carrier. Spectrum analyzers for example may put the IF at 3GHz even for receiving low frequency inputs. \$\endgroup\$ May 12, 2017 at 21:47
  • \$\begingroup\$ No you don't, look up image rejection as the man said. \$\endgroup\$
    – Andy aka
    May 12, 2017 at 22:54

1 Answer 1

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I don't think an "image rejection" answer is complete enough. There are other reasons.

For example, AM band case that ranges from \$530\:\textrm{kHz} \le f_{RF}\le 1700\:\textrm{kHz}\$, using \$f_{IF}=455\:\textrm{kHz}\$. Assuming upper LO injection, there will be in-band images and this fact puts pressure on having a very selectable and tunable 1st band-pass filter. (Lower LO injection will also have in-band images.) The reason here for using upper LO injection isn't as much about image rejection (the argument is moot) but instead about the required dynamic range of the LO. Upper LO injection: \$985\:\textrm{kHz} \le f_{LO}\le 2155\:\textrm{kHz}\$ for a dynamic range of only about a factor of 2 or so. Relatively easy to do. Lower LO injection: \$75\:\textrm{kHz} \le f_{LO}\le 1245\:\textrm{kHz}\$ for a dynamic range of a factor of 16 or so. Much, much harder to do well.

In this case, I think the argument for the upper LO injection is more about tuning the LO than about image rejection.

Now assume \$f_{IF}=10.7\:\textrm{MHz}\$. Lower LO injection: \$10.17\:\textrm{MHz} \ge f_{LO}\ge 9\:\textrm{MHz}\$. Upper LO injection: \$11.23\:\textrm{MHz} \le f_{LO}\le 12.4\:\textrm{MHz}\$. The worst case image frequency for lower LO injection is \$19.7\:\textrm{MHz}\$ and for upper LO injection is \$21.93\:\textrm{MHZ}\$. A relatively simple 2-pole low pass filter is kind of easy for 40 db per decade and given that it has to pass the entire AM band, should be about \$-40\:\textrm{dB}\$ at \$17\:\textrm{MHz}\$. This suggests about \$-42.5\:\textrm{dB}\$ for lower LO injection and about \$-44.4\:\textrm{dB}\$ for upper LO injection.

In this case, tuning the LO isn't the issue at all. Now it's more about image rejection.

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