Transfer function of a basic RC circuit

Question

I'm having a hard time getting my head around finding the transfer function.

I have a circuit that looks like this:

 o----------R-------C-----------o
                         |   
                         |
                         C
                         |
                         |
 o------------------------------o

The solutions in my book say the answer is $$\frac{(RC)^{-1}}{s + 2(RC)^{-1}}$$ where s = jω. How did they get this?

Help will be much appreciated.

Answer 1

At the risk of doing your homework for you,

Start with the voltage divider rule

$$\frac{V_o}{V_i}=\frac{Z_C}{R+Z_C+Z_C}$$

where \$Z_C\$ is the impedance associated with a capacitor with value C.

Now substitute

$$\frac{V_o}{V_i}=\frac{1/sC}{R+2/sC}$$

Now multiply by \$\frac{sC}{sC}\$

$$\frac{V_o}{V_i}=\frac{1}{sRC+2}$$

Now divide both the numerator and denominator by \$RC\$ to isolate \$s\$.

$$\frac{V_o}{V_i}=\frac{(RC)^{-1}}{s+2(RC)^{-1}}$$

Answer 2

use the Laplace/frequency analysis

\begin{gather} Z_{R} = R \\ Z_{C} = \frac{1}{sC} \\ \end{gather}

So than you use the normal circuit law you know. This circuit is a Voltage divider(https://en.wikipedia.org/wiki/Voltage_divider) . So:

\begin{gather} V_{out} = V_{in}\frac{Z_{C}}{Z_{C}+Z_{C}+Z_{R}} \\ V_{out} = V_{in}\frac{Z_{C}}{2Z_{C}+Z_{R}} \\ V_{out} = V_{in}\frac{\frac{1}{sC}}{2\frac{1}{sC}+R } \\ V_{out} = V_{in}\frac{1}{2+sCR } \\ V_{out} = V_{in}\frac{\frac{1}{RC}}{2\frac{1}{RC}+s } \\ \frac{V_{out}}{V_{in}} = \frac{\frac{1}{RC}}{2\frac{1}{RC}+s } \\ \end{gather}

Regards MathieuL.