There is a relation between those two sampling frequencies. Let's assume you want to design a digital lowpass filter $h[n]$ from an analog prototype lowpass filter $h(t)$ by the method of impulse invariance.
Assume that the analog lowpass filter has a cutoff frequency of $f_c$ in Hz. Then you must use a filter sampling frequency, $f_{sf}$, that's at least twice greater than this $f_c$ to avoid aliasing in the resulting digital filter's spectrum.
After you've obtained the digital filter $h[n]$, its corresponding cutoff frequency in discrete-time frequency will be:
$$ w_c = 2\pi \frac{f_c}{f_{sf}} $$
Forex. with $f_c = 1$ kHz and $f_{sf} = 4$ kHz, then the cutoff frequency of $h[n]$ will be $w_c = \pi / 2$
When this filter is used within a DSP system where the input signal is sampled at $f_{ss}$ Hz, then the effective analog cutoff frequency of the digital filter will be
$$f_{ec} = \frac{\omega_c}{2\pi } f_{ss}$$
Hz. For the above example if you select a signal sampling frequency of $f_{ss} = 10$ kHz, then the effective analog cutoff frequency will be: $$f_{ec} = \frac{\pi/2}{2\pi } 10k = 2.5 \text{kHz}$$
Therefore there is a two stage process of finding the actual effective frequency. In order to avoid this two stage projection of the sampling frequencies on the actual frequencies, it's better to start by the specifications of the digital filter in the digital domain such as the required cutoff frequency $\omega_c$ and then project this into the analog filter prototype design, at an arbitrary filter sampling frequency $f_{sf}$. Then the resulting digital filter will always have the requested digital cutoff frequency $\omega_c$ and will behave based only on the signal sampling frequency $f_{ss}$ as expected.