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How does Identification Friend or Foe (IFF) work? Are radar images of friendly planes recorded, or do friendly planes have transponders on them?

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IFF systems use a transponder/interrogator combination to allow aircraft to issue challenges, receive responses, and do the friend or foe test. BAE Systems has a transponder product page with a little info, and Wikipedia (yeah yeah) has a reasonably thorough description and history of IFF.

According to NATO STANAG 4579 (abstract):

This NATO BRID is a millimetric wave question and answer system working in the Ka band. It comprises two main components, an interrogator, which allows a platform to question another platform with an encrypted message, and a transponder, which decodes the message and replies with an encrypted answer [emphasis mine]. Transponders will be fitted to all designated platforms, while interrogators will be fitted to only those designated platforms with an offensive or reconnaissance capability.

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    $\begingroup$ I assume these systems would be enabled in warzones. Wouldn't this give away the aircraft's position? $\endgroup$ Feb 23, 2014 at 1:59
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    $\begingroup$ That's one of the reasons it's a cryptographic handshake. If the query doesn't decode properly, the transponder doesn't answer at all. Of course this means that the interrogator lights up for a moment, but there are ways to manage that... and realistically, if you're close enough that IFF matters, you're already on radar. $\endgroup$
    – keshlam
    Feb 23, 2014 at 6:41
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    $\begingroup$ @keshlam However, if it does decode properly (i.e. is a friendly request), the transponder would light up as well, potentially giving away both their positions, wouldn't it? In this case they only need to be "close enough" to each other, not the enemy. $\endgroup$
    – falstro
    Feb 24, 2014 at 7:34
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    $\begingroup$ Call it the indeterminacy principle. A single ping giving current position doesn't say where it will be three seconds later, and that's presuming the ping can be recognized (I would assume it's spread-spectrum) and triangulated upon at all. $\endgroup$
    – keshlam
    Feb 24, 2014 at 16:27
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    $\begingroup$ It looks like STONAG link is broken. $\endgroup$
    – Gluttton
    Dec 21, 2014 at 13:47

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