Assuming the center frequency of the synthesizer is 2GHz with a spur at an offset of 200kHz, what is the offset frequency and amplitude of the spur when the signal goes through a divide-by-2? This divider is outside of the synthesizer.
I would like to see intuitive insight and math proof for it.
The offset frequency of the spur will keep at 200kHz. Just think a 2GHz carrier with a 200kHz frequency modulation. The modulation cycle time is 5ms no mater the signal is divided by 2 or not.
Modulation frequency is the same. So offset frequency of ther spur remains 200 kHz. Phase or frequency deviation is 2 times smaller. Therefore spur level is 6 dB lower.