Transfer Function of two Common Emitters
To determine this transfer function, many methods exist. I will use the fast analytical techniques or FACTs described in the book I wrote. The principle consists of individually determining the time constants associated with each energy-storing element. You do that to express the denominator which, is the present case, is defined as: \$D(s)=1+sb_1+s^2b_2\$. The numerator, where the zeroes lie, is determined using a generalized expression as follows: \$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{21}\tau_2\tau_{21}\$

\$H_0\$ is straightforward: open all the caps and the output is 0 V then \$H_0=0\$.

To determine the time constants, you go through two small sketches in which you determine the resistance driving the capacitor. You do that using inspection, without writing a line of algebra. For \$\tau_1\$ and \$\tau_2\$, you examine the below sketch:



Once you have the time constants on hand, you write \$b_1=\tau_1+\tau_2\$.

For \$\tau_{21}\$, capacitor \$C_2\$ is replaced by a short circuit and you determine the resistance driving \$C_1\$ in this mode:



Then you form \$b_2=\tau_2\tau_{21}\$. This is it, you have the denominator which is of second order and considering a low-\$Q\$ approximation, you can factor it as \$D(s)\approx \frac{1}{(1+\frac{1}{\omega_{p1}})(1+\frac{1}{\omega_{p2}})}\$ with \$\omega_{p1}=\frac{1}{b_1}\$ and \$\omega_{p2}=\frac{b_1}{b_2}\$.

To determine the zeroes, you will alternatively set the capacitors in their high-frequency state (a short) while one remains in its dc state (open) and calculate the gain in this mode. If you do that, you find \$H_1=H_2=0\$. So the term in \$s\$ in the numerator is null. Then, to determine the second-order zeroes, put all the caps in their high-frequency state (a short) and determine the gain \$H^{12}\$ of the below circuit:



In this case, \$H(s)=s^2H^{12}\tau_2\tau_{21}\$ reusing the time constants already determined. This is it, the transfer function is obtained and given in the below Mathcad sheet rearranged in a way to show the gain in the plateau region: remember, all of these techniques are for design-oriented analyses or D-OA, a term forged by the late Dr. Middlebrook.





The final transfer function involves two inverted poles which allow you to factor the plateau gain as the leading term since this is what is wanted. The terms could obviously be simplified if we ignore the biasing resistors but I wanted to show the complete expression for the sake of the exercise. As you could see, all was done by inspection, without a single line of algebra and this is the power of the FACTs.