[time-nuts] Down-conversion to IF and sampling

[Attila Kinali]

On Sat, 23 Dec 2017 21:00:39 +0000
Stephan Sandenbergh <ssandenbergh at gmail.com> wrote:

> I guess I'm just worried that I might be missing something obvious here?
> And I know there is no better place to ask such a question other than on
> time-nuts :)

No, your derivation is correct. Though conventionally, it would be
written as: V_{IF}(t) = sin[(ω_{RF} - ω_{LO})t - Δφ_{RF-LO}(t)]
Hence, the sampled signal becomes:
V_{IF}[nT_s] = sin[(ω_{RF}*(nT_s) - ω_{LO}*(nT_s) - Δφ_{RF-LO}[nTS]]

In this notation, it is a bit more obvious what's going on. Assuming
both RF and LO frequency are constant, then the sampled voltage only
depends on the difference of the frequencies, and the initial phase offset
at time t = 0*T_s. Be aware that this only holds true if you either
use a low pass filter after the mixer or use complex down conversion.
In all other cases you have to account for the (ω_{RF} + ω_{LO}) component
as well.

Using x(t) = x_0 + y_0*t confuses things a bit, as this means
that you are modulating the phase with a frequency of y_0, which you
probably do not intend.

Any phase noise you have in the system, you can fold into φ_{RF-LO}(t).

Please note, the above has the implicit assumption, that:
ω_{IF} is < 0.5 * ω_{LO}, ie that the IF signal is in the first
Nyquist zone. Otherwise you have to treat the ADC as another mixer stage,
with it's own ω_{LO_{ADC}} and φ_{LO_{ADC}}.

As Tim Shoppa mentioned, you do not want to have a ratio with small integers
between the LO frequency and the sampling frequency, as any feedthrough of
the LO and its harmonics will lead to a DC offset and spurs. The amplitude
of both will depend on the exact phase relation between the LO frequency
and the sampling frequency, which is usually stable, but not time-nuts stable.


				Attila Kinali
-- 
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