Digital Downconversion of Signal
For a centre frequency of $F_C$ for a modulated signal $v(t)$.
$$
r(t) = v_I(t) \sqrt{2} \cos 2\pi F_C t - v_Q(t) \sqrt{2}\sin 2\pi F_C t
$$

Assuming a direct RF sampling at a rate of $F_S=1/T_S$ before downconversion, we plug $t=nT_S$ above.
$$
r(nT_S) = v_I(nT_S) \sqrt{2} \cos 2\pi F_C nT_S - v_Q(nT_S) \sqrt{2}\sin 2\pi F_C nT_S
$$

The expression $F_C nT_S$ above becomes $\frac{F_C}{F_S}n$ to yield
$$
r(nT_S) = v_I(nT_S) \sqrt{2} \cos 2\pi \frac{F_C
}{F_S} n - v_Q(nT_S) \sqrt{2}\sin 2\pi \frac{F_C
}{F_S} n
$$

Perhaps this is the ratio of the centre frequency to the sample rate you are asking about.

EDIT: Downconversion can be embedded in a downsampling operation as follows. In a conventional operation, we will have an LO downconverting the signal to baseband, filter it and then downsample it according to the symbol rate, as shown below.



Here, the filter is operating at a higher rate which is unnecessary when we have to throw $M-1$ out of every $M$ samples anyway. This is a hint for what we need to do.

First, we interchange the operations of downconvesion and filtering so our filter needs to be implemented in passband now. If the lowpass filter impulse response is $h[n]$, the passband filter response is 
$$ g[n] = h[n] e^{j2\pi \frac{F_C}{F_S} n}$$

The interchange of downconversion and filtering is shown below.



Next, we slide the downsampling operation past the LO whose frequency gets multiplied by $M$, i.e., $F_C/F_S \cdot M$. We assume that
$$F_C = \frac{F_S}{M}$$ 
then our LO frequency after sliding it past the downsampler becomes
$$\frac{F_C}{F_S}M = \frac{F_S}{M}\frac{M}{F_S}=1$$
which implies that no downconversion is actually required!



However, remember that filter is still operating it at a higher rate. So we slide the downsampler past the filter now and implement it as a polyphase filter. The original spectrum is shown below.



This operation changes the spectrum such that all spectral replicas from $-F_S/2$ to $F_S/2$ end up at baseband overlapped on each other. The interesting part is that at the input of the polyphase arms, they possess a unique phase profile across the whole $360^{\circ}$ around the frequency IQ spectrum. A rough sketch is below.



This is only true in an exact manner for the sinusoidal signals, so the rest of the task for a signal with bandwidth is performed by the filter. When implemented as a polyphase filter, this filter operates with the delays matching those of the spectral replicas shown above. A final summation at the end cancels everything out leaving the desired spectrum.

This also implies that if the desired spectrum is at $F_S/M$ and we want to employ the initial lowpass filter $h[n]$ instead of the bandpass filter $g[n]$, a corresponding rotation is required after each polyphase arm to line everything up before the final cancelation.