Re: explaining i/q

[David Hagood]

On 11/4/20 6:26 PM, david vanhorn wrote:
> "Twice the bandwidth" but that doesn't account for the 0 Hz "hole" 
> where the incoming signal is exactly at the sampling rate.
> Or am I missing something?
What "hole" are you referring to? There is the "zero bang", which occurs 
because most systems aren't perfect and have a DC component to the LOs 
and mixing system, but a "perfect" system with zero DC in the LOs will 
resolve a DC signal. The imperfection of the implementation doesn't mean 
the concept is imperfect.

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