RE reduces and stabilises gain of the bottom transistor.In fact gm becomes the reciprocal of RE .Example RE = 1K ohm so gm is 1 milliamp per Volt which is the reciprocal of 1 Volt per milliamp .Now the LC tank has a high impedence at resonance which is dependant on Q .100K was my calculated tank impedence at resonance on a 455KHz If strip but your job is different.So the gm is multiplied by the tank impedence to get the voltage gain .Voltage gains of more than 1000 are possible .
Since gmb2 is usually ignored when the substrate connection is properly grounded, I'll ignore it and make thing slightly simpler. So gmb2 = 0
Then looking "up" into the source of M2 I see:
$$1/(gm_2)$$
Looking right from the drain of M1 I see:
$$Rp$$
The current coming out of the drain of M1 is:
$$Id_1 = gm_1Vin$$
Now this current splits between that 1/gm2 and Rp:
$$Id_2 = Id_1\frac{gm_2Rp}{1+gm_2Rp}$$
(Note that this uses the "current dividing into two resistances" formula. It is similar to the voltage divider formula but then for current.)
Then filling in Id1 that makes:
$$Id_2 = gm_1Vin\frac{gm_2Rp}{1+gm_2Rp}$$
You can now replace gm2 with (gm2 + gmb2) if you like, the end result remains the same.
Drain current Id2 does not split anymore after M2, all the current goes into the drain load resistor Rd. So the voltage gain Av is simply Id2 as derived above multiplied by the value of the load resistor Rd. With a negative sign as a rising Vin results in a falling Vout.
Best Answer
Using this small-signal diagram
We clearly see a voltage divider at the input.
Therefore $$\frac{V_{B1}}{V_S} = \frac{R_1||R_2||(\beta_1 +1)r_{e1}}{R_S + R_1||R_2||(\beta_1 +1)r_{e1}}$$
Next, let us find the first stage voltage gain (Q1). This stage is working as a CE ( common-emitter) amplifier.
$$V_{B1} = I_B\cdot(\beta_1 +1)r_{e1}$$
$$V_{C1} = - I_{C1} \cdot r_{e2} = I_B \cdot \beta_1 \cdot r_{e2}$$
$$\frac{V_{C1}}{V_{B1}} = -\frac{I_B \cdot \beta_1 \cdot r_{e2}}{I_B\cdot(\beta_1 +1)r_{e1}} = -\frac{r_{e2}}{r_{e1}} \cdot \frac{\beta_1}{\beta_1 +1} = -\alpha \frac{r_{e2}}{r_{e1}} $$
Now we can find the second stage gain (Q2) that works as a CB (common-base) amplifier.
$$V_O = - I_{C2} \cdot R_C||R_L =-\alpha_2 I_{E2} R_C||R_L = -I_{C1} R_C||R_L \frac{\beta_2}{\beta_2 +1} $$
$$V_{C1} = - I_{C1}\times r_{e2}$$
$$\frac{V_O}{V_{C1}} = \frac{-I_{C1} R_C||R_L \frac{\beta_2}{\beta_2 +1}}{- I_{C1}\times r_{e2}} = \frac{R_C||R_L}{r_{e2}} \times \frac{\beta_2}{\beta_2 +1}$$
And finally, the overall voltage gain \$\frac{V_O}{V_S}\$ is
$$\frac{V_O}{V_S} = - \frac{R_1||R_2||(\beta_1 +1)r_{e1}}{R_S + R_1||R_2||(\beta_1 +1)r_{e1}} \times\frac{r_{e2}}{r_{e1}}\times\frac{\beta_1}{\beta_1 +1}\times\frac{R_C||R_L}{r_{e2}} \times \frac{\beta_2}{\beta_2 +1} $$
Any additional questions?