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difx:channelisation [2009/10/26 11:06]
adamdeller
difx:channelisation [2009/10/26 11:07]
adamdeller
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 Basically, when you channelise B MHz of bandwidth into N channels you get actually N+1 channels, each spanning B/N MHz of bandwidth, and thus totalling B*(N+1)/N MHz of bandwidth. ​ How can that be?  Its because of the aliasing inherent to a discrete Fourier transform. ​ Each of the edge channels is actually centred on the very edge of the band, and includes half a channel'​s worth of aliased power. Basically, when you channelise B MHz of bandwidth into N channels you get actually N+1 channels, each spanning B/N MHz of bandwidth, and thus totalling B*(N+1)/N MHz of bandwidth. ​ How can that be?  Its because of the aliasing inherent to a discrete Fourier transform. ​ Each of the edge channels is actually centred on the very edge of the band, and includes half a channel'​s worth of aliased power.
  
-[[bandwidth.png]]+{{bandwidth.png}}
  
 For the DC channel (the one centred on the LO frequency, so the lowest channel for upper sideband data and the highest channel for lower sideband data) this aliasing is not too bad.  It leads to a little decorrelation,​ but at least the right fringe rate has been applied to the aliased power. For the DC channel (the one centred on the LO frequency, so the lowest channel for upper sideband data and the highest channel for lower sideband data) this aliasing is not too bad.  It leads to a little decorrelation,​ but at least the right fringe rate has been applied to the aliased power.
difx/channelisation.txt · Last modified: 2015/10/21 10:08 (external edit)