I have not used the mlab functions in matplotlib, but from what I can discern from your time plot, I see a pattern that repeats roughly every 15,000 samples, at 1.25 Gsps according to your code.
This means we should see a strong spectral component at $1.25e9/15000=83.3$ KHz.
The reason we do not see this, I suspect is the relatively short FFT length (or very high sampling rate compared to the spectral occupancy, or both!).
Let me explain: The FFT length is currently 256 samples. At 1.25GHz this is only $256/1.25e9 = 204.8$ ns.
The frequency resolution prior to using any additional window (meaning using a rectangular window) is $1/T$ where T is the length of the FFT in time (for further detail on that see:
What happens when N increases in N-point DFT )
So in your case the frequency resolution is greater than $1/204.8e-9$ = 4.88 MHz!! Any window will reduce sidelobes but at the expense of increasing the mainlobe, so the width of your mainlobe given the Hanning window used will cause the resolution to be even wider than 4.88 MHz. (But definitely good to include windowing, otherwise the additional sidelobes would be much higher). You can see this post for additional details on the windowing choices:Why would one use a Hann or Bartlett window? (From this we see the Hanning window is approximately twice as wide as the rectangular window, so your example would have approximately 10 MHz of resolution bandwidth per bin in the FFT)
Looking at your plot, it looks like we see exactly this, which is the FFT bin of your strong DC component, and this single FFT bin is approximately 10MHz wide in bandwidth! So that is burying any signals of interest that you want to see.
My recommendation is to increase the FFT length within what your processing will allow. And if that starts to get problematic then reduce the sampling rate to get further resolution (however there are challenges in doing that properly that I won't get into except to say that you must be sure higher frequency components are filtered out prior to resampling). Without requiring a sampling rate change, you would want to use an FFT length of at least $1.25e9/10e3$ = 125,000 to get 20 KHz of resolution BW while using the Hanning window. That's a reasonable length and I suspect you can go much higher than that for finer resolution.
Also as a side note I don't see where the impedance is set, so I guess it is assumed to be normalized to 1 ohm. Just mentioning in case you cared about knowing the actual power levels.