Understanding the Gain and Bode Plot of a Low Pass Filter Circuit with C1 > C2

[Fat Dapper Cat]

The question wants to know the gain of the this circuit, as well as a bode plot for the frequency response, and what type of filter this is when C1 is much greater than C2. Could someone double check my answers and assist in the bode plot?

My attempt:

For the Gain:

If C1 is much greater than C2 I said the filter would be a low pass filter.

I'm not sure how to proceed with the Bode Plot.

 

[Henryk]

Hi, I'm afraid your answer is not correct.
The simplest way to analyze the circuit is to note that $C_1$ and $C_2$ form a capacitive voltage divider connected to $V_{IN}$. The divider is equivalent of $C_1$ and $C_2$ in parallel connected to the voltage source $V_{eq} = V_{IN} \frac {C_1} {C_1 + C_2}$. Then connect the resistor between the output of the two capacitors and the ground. What you have is a high pass filter with the corner frequency given by $\omega _c = R *(C_1+C_2)$
If $C_1 \gg C_2$, then you can simply ignore the second capacitor. The equivalent circuits is showing you why you can do that.
I'm attaching a drawing of the equivalent circuit.

 

[Tom.G]

Won't there be a pole due to R1C2?

 

[NascentOxygen]

My attempt:

Please show your working.

 

[Tom.G]

Oops! You're right. No pole there. I just worked it out.