I would argue that the ADC has a 4th input in addition to the three cited by Fred: its clock. At least for some types of ADCs, jitter or phase noise on the clock can impact ADC measurements.
You say you have a 25 MHz oscillator but are running the ADC at 30 MHz, so you have some PLL involved in the generation of its clock. If that is not working well, its irregularity could be a source of conversion noise. Can you try changing the software configuration (even temporarily) to not use the PLL and just run off of a the input clock or divided down from it?
I believe some microcontrollers also have a mechanism for suspending most of the digital circuitry while taking an ADC reading in order to reduce noise. You might look into seeing if something like that is possible.
A system with a single node: some R and some C, will have the total integrated noise defined EXACTLY by sqrt (K*T/C).
Thus a 10pF cap, at 290 degree K, produces exactly 20 microVolts RMS, regardlss of the value of the resistor.
Also, a 100pF cap would produce 1/sqrt(10) less noise, or about 6 microVolts RMS, regardless of the value of resistor.
To measure 6uV RMS in 10GHz bandwidth, the system noise density must be lower than 6uS/sqrt(10GHZ) = 6uV/sqrt(10^10) = 6uV/100,000 = 0.06 nanoVolts per rootHertz.
Given a 60 ohm resistor produces 1nanoVolt/rtHz noise density, and we need 0.06nV, or 16X smaller than 1nV, the total front end Rnoise my be 60/(16*16) 60/256 or 0.25 ohms.
Achieving Rnoise of 0.25 ohms will not happen.
Typical Rnoise (the rbb') of bipolars is 10 ohms or 100 ohms.
I cannot speak for typical rbb' for silicon-germanium bipolars.
Best Answer
Yes, those equations are correct and the same.
In the first equation, you just replace \$x\$ with its expression following from \$y=\bar{y}+x\$.
For the second equation, you can do the following: The square of the rms value of y is equal to $$ \begin{aligned} rms(y)^2\ &= \frac{1}{N}\sum_{n=1}^N (y[n])^2\\ &= \frac{1}{N}\sum_{n=1}^N (\bar{y} + x[n])^2 \\ &= \frac{1}{N}\sum_{n=1}^N (\bar{y}^2 + 2 \bar{y} x[n] + (x[n])^2) \end{aligned} $$
The first term (\$\frac{1}{N}\sum\limits_{n=1}^N \bar{y}^2\$) is just equal to \$\bar{y}^2\$.
The second term (\$\frac{1}{N}\sum\limits_{n=1}^N 2 \bar{y} x[n]\$) is zero since you can take the constant factor \$2 \bar{y}\$ out of the sum and \$\bar{x}=\frac{1}{N}\sum\limits_{n=1}^N x[n]\$ is zero.
The third term (\$\frac{1}{N}\sum\limits_{n=1}^N (x[n])^2\$) is the square of the rms value of x.
Thus $$ rms(y)^2 = \bar{y}^2 + rms(x)^2 $$ and thus $$ rms(x) = \sqrt{rms(y)^2 - \bar{y}^2} $$
The example of Marcus Müller in the comments is still valid: \$\bar{y}=2\$, \$rms(x)=0\$, and thus \$rms(y)=2\$. Note that an rms value is not the same as a standard deviation if the signal (\$y\$ here) has a non-zero mean value.