Plotting frequency spectrum for a signal with 20MHz bandwidth and sampling frequency of 12.5GHz
Your fundamental limitation is the (comparatively) short window length of $M=400,000$ samples, at such a prohibitively high sampling frequency of $12.5$ GHz. This will roughly yield a few hundred kHz of effective FFT resolution per bin of the FFT spectral samples. 

Using the formula for the spectral resolution of various windows we can deduce the following resolution, for example, for a standard Hamming window as

$$\Delta w = 8\pi / M$$ where $M$ is the window's length, that is $M=400,000$, which yields $$\Delta w = 8\pi / 400,000 = 6.2832 \times 10^{-5}$$ radians.

This resolution is determined by the window length, and its frequency equivalent under the sampling rate of $f_s = 12.5 \times 10^9$ Hz is 

$$ \Delta f = \frac{\Delta w}{2\pi} f_s = \frac{6.2832 \times 10^{-5}}{2\pi} 12.5 \times 10^9 = 125,000 Hz$$ i.e., the effective frequency resolution of your window is $\Delta f = 125$ kHz. Now given your bandwdith of $B=20$ MHz, this equates to a number of $20 \times 10^6 / 125\times 10^3 = 160$ samples.

On the other hand the apparent count of the number of FFT samples (apparent resolution) inside the $20$ MHz region yields something about $k = 4\times 10^5 \times 12.5\times 10^9 / 20\times 10^6 = 640$ samples, which however does not carry more information than those inferred from $160$ subsamples. 

Hence a bandwidth of about $20$ MHz would yield about less than 200 true-samples of a spectral region inside $400,000$ many apparent-FFT samples. This is quite narrow. You would either use a much longer observation interval or use a much reduced sampling frequency as @MBaz commented.