To demonstrate the difference, here is the basic form of a differential amplifier which makes up the input stage for an opamp:
Notice there are two signals input at each side. SIG and SIG_INV are a 1kHz differential input (SIG is 180° shifted in phase from SIN_INV), and SIG_COM is a 9kHz common mode input (same signal at each side referenced to ground, i.e. 0° phase difference)
These signals are both at a 10mV (20mV pk-pk) level.
Now lets have a look at the simulation:
We can see the input (referenced to ground) is the mix of both signals, but the output is only the 1kHz differential signal at roughly a gain of 100. The differential amplifier has rejected almost all of the 9kHz common mode signal.
To see exactly how much of the 9kHz signal gets through to the output, here is the simulation again with only the 9kHz signal present:
Now we can see the output is roughly 10mV pk-pk (+/-5mV), so there is a gain of 0.5. We can now calculate the CMRR as we know the differential gain is 100 and the common mode is 0.5, so 100/0.5 = 200 = 46dB.
This is not a very good ratio, but it's the most basic form of differential amplifier. An typical opamp will improve greatly on this figure by for example, using a current source instead of the common tail resistor (R3) (also other things too).
For interest's sake, I just replaced R3 with an ideal current source and this reduces the common mode output to 324uV pk-pk (for 20mV pk-pk in) so the common mode gain is 0.0162 and thus the CMRR is improved to 20 * log10 (100 / 0.0162) = ~75.8dB. A high quality opamp might reach 120dB or more.
Calculating CMRR from component values
In the differential amplifier above, we can calculate both differential gain and common mode gain pretty easily. Here are the formulas with a brief explanation:
The differential gain is:
Gdiff = Rc / (2 * (Re + re)) where Re is the emitter resistors value and re is the intrinsic emitter resistance, given by ~25mA / Ic.
So, for our circuit above, we get:
re = 25mA / 100uA = 250Ω
Gdiff = 75k / (2 * (100Ω + 250Ω)) = 107, which agrees with our simulation.
The common mode gain is given by:
Gcm = -Rc / ((2*Rtail) + Re + re) - the minus sign means the output is inverted (180° shift) Rtail is R3 in the schematic above (the differnetial pair is sometimes referred to as the "long tailed pair", so this is the "tail" resistor)
So, we get:
Gcm = -75kΩ / (2*75kΩ) + 100 Ω 250Ω) = ~-0.5, which again agrees with our simulation.
The CMRR can either be calculated using the above results, or can be calculated directly using:
20 * log10(Rtail / (Re + re)) = 20 * log10(75kΩ / (100 + 250)) = 46.6dB, which again agrees with what can can see in the simulation.
From the above formula, we can see that the ratio between the tail resistor and emitter resistor is the main factor controlling the CMRR, so using a high impedance current source improves things dramatically.
The above equations don't take everything into account (you will need to do some further reading for the more subtle effects), but get you close enough for most applications.
The trifilar transformer you have drawn will couple all three of its windings together. In the ideal transformer case, that means the voltage on each set of windings will be identical.
It will therefore effectively parallel all the signals together. The fact that you have drawn load impedances of Z on the end of each winding means that the transformer will not actually short your two noise sources together, which without the Zs could have made them malfunction. The average noise source voltage, ie half the sum, will be added to your signal.
Nevertheless, effectively connecting your two noise sources in parallel like this is a rather 'dirty' way to do it.
You can design a three input combiner, that isolates each signal from the other. Alternatively cascade a pair of two-input combiners. The design equations for these things are readily available, or you can buy the finished part, see MiniCircuits for instance.
If the sources are low impedance, then it would be better to cascade a pair of simple transformers, adding one noise source in each. The transformers will put the noise voltage in series with the signal voltage, so simply adding them. As the two noise series are then in series, they won't load each other.
Best Answer
Use this circuit (and equation) instead:
Source: https://www.analog.com/media/en/training-seminars/tutorials/MT-042.pdf
It should be easy to come up with resistors in spice that are matched (in real life not so much and so you should use the other circuit in the application note).
This has worked for me, if it doesn't work the model may not be that realistic.
Another note: While the diagram shown in the OP has no ground on the 18V supplies, I hope that there was a ground between the supplies in the spice netlist. Otherwise spice would have a very difficult time finding the DC operating point and simulating, as al the voltages of that portion of the circuit would be floating, and the simulation wouldn't function very well. Supplies need to be referenced from ground, which I hope was done.