Break frequencies for directly coupled transistors

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Homework Statement


The problem asks to determine the break frequency at the base of Q2 (which is fine I have no problem with that), but the whole assignment never asks for the break frequency at the COLLECTOR of Q1, then later asks to produce a bode plot of the transfer function. See the image I've linked (pardon the sloppy paint job).

directcoupledtransistorsACmodel.jpg


Homework Equations


f(break) = 1/(2*pi*Ceq*Rth)


The Attempt at a Solution



Well I thought since the capacitor equivalent at collector1 (Cmiller(output) + Cce) wasn't actually in parallel (due to the non-bypassed emitter resistors) with the capacitor equivalent at base2 (Cmiller(input) + Cbe), I would have to treat them separately and produce two different break frequencies. Is this correct?
 
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still nothing?? Well, I've consulted with my colleague, and he told me that my interpretation of the miller-effect capacitance was incorrect. He told me that they always goes to ground (can anyone verify this??)...and he also told me to ignore the capacitances Cbe and Cce (ignore? never! mwhahaha)...

So looking at this NEW image:

directcoupledtransistorsACmodelpt2.jpg


You can see now that the "break frequency due to the base of Q2" probably means the two miller capacitances added together since now they are truly in parallel. BUT, to be 100% correct, wouldn't I now need a break frequency due to Cmiller(combined), Cce, AND Cbe(3 total)...not to mention the capacitors due to the rest of the circuit.

If I've confused anyone..I'm talking about break frequencies on Bode Plots of the gain vs. frequency response where the slope changes(20dB/decade...40dB/decade..so on), or the -3dB point on the actual transfer function (at least the first break frequency on the midband gain).

Can anyone help me understand this?
 
Well after extensive research (PSpice and a graphing calculator), I've concluded that adding the Miller effect to the parasitic capacitances (treating them as "parallel") gives nearly the same transfer function results as treating them separately. I guess it's just another one of those electronic approximations that are far too common.