Electronic – Misunderstanding of calculating gain of amplifier

Gain is set for each \$ V_{in} \$.

\$ A_{Vin1} = -\dfrac{R_3}{R_1}\$.

\$ A_{Vin2} = -\dfrac{R_3}{R_2}\$.

and so on, \$ A_{Vin N} = -\dfrac{R_3}{R_N}\$.

If you were to analyze this circuit using superposition, you would find that for each input signal, it is just an inverting amplifier. The signals are amplified or attenuated individually. Therefore, gain is only a meaningful quantity with respect to each individual signal. Since gains are all independent of each other, summing them does not yield a meaningful result.

The gain of a true common-emitter (emitter at AC common or "ground") is approximately:

$$A_v = -g_mR_C||R_L = -g_m 7.5 \Omega$$

for your resistor values. For a (magnitude) gain of 7, the required transconductance is:

$$g_m = \frac{7}{7.5} = \frac{I_C}{V_T} \Rightarrow I_C \approx 23mA$$

But, according to your schematic:

$$I_C \approx 8mA$$

Back to the drawing board!

Also, do realize that the above calculation is for small-signal gain. Since you have the emitter at AC ground, this amplifier is quite non-linear for large signals and the distortion is quite evidently large in both transient simulations you provide.