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Re: explaining i/q

From: Fons Adriaensen
Subject: Re: explaining i/q
Date: Thu, 5 Nov 2020 13:37:34 +0100

On Wed, Nov 04, 2020 at 07:13:09PM -0600, David Hagood wrote:
> 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?

With complex samples at rate R, the frequency domain is periodic modulo R.
So DC and the R map to the same signal.

It's just a matter of interpretation. You could look at the complex signal
bandwidth as ranging from -0.5 R to +0.5 R, or as 0 to R, any in many other
> 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.

If the complex signal is created in the analog domain you can get all sorts
of problems at DC. And others as well if gains or phases are not exactly

But there are more robust ways to produce a complex signal given a real one.

A popular one takes a real signal at rate R, and maps 1/4 R to DC, producing
a complex signal at rate R/2 with a frequency range -1/4 R to +1/4 R.
Any DC offset in the input is translated to the lower edge at -1/4 R which
is usually out of the region of interest. Such a system will not have any
anomaly at DC. 


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