If I have a sampled cosine wave signal with Fc = 80kHz and Fs = 200kHz (so the Nyquist region is between 100kHz and -100kHz) and I want to preform a decimation by the factor of 2.
Will my input signal be outside of the new Nyquist region (50kHz to -50kHz) and I will see only the Fc - Fs = -20kHz product of the signal or the input signal will be shifted to the new Nyquist region somehow ?
It may help to first read this existing post which provides a more intuitive understanding of aliasing in the A/D conversion process which equally applies to resampling: the A/D process is simply resampling from continuous time, while the OP is resampling from 200 kHz to 100 kHz: all spectral aliasing that occurs with sampling applies to resampling.
Decimating to the new sampling frequency $f_s= 100 $ kHz is resampling and will have the same result as directly sampling the 80 kHz sine wave at that sampling rate.
The result is identical to what you would get if you instead sampled a 20 kHz sine wave: everything in the first Nyquist zone (from -50 kHz to +50 kHz) repeats exactly at all higher Nyquist zones; so what is at $0 \pm f_s/2$ repeats the same at every $n \pm fs/2$, where $n$ is any integer. For a sine at 20 kHz sampled at 100 KHz, the resulting spectrum in the first Nyquist zone will have components at $\pm 20$ kHz (associated with $n=0$). Thus it will also have components at $100 \pm 20$ kHz, $200 \pm 20$ kHz, etc.
Similarly if we sampled 80 kHz directly with $f_s = 100$ kHz, it would have components at $\pm 80$ kHz (associated with $n=0$. When $n=1$ we get $100 \pm 80$ kHz, or 20 kHz and 180 kHz. If you work through each $n$ you see that this will also be all the same components as $nf_s \pm 20 $ kHz, the case when we sampled 20 KHz.
For those familiar with frequency translation by multiplying in the time domain (“mixing” for RF engineers), the following post provides further explanation:
