What do you mean by -3db cut off frequency in low-pass-filters?

Question

I have a low pass filter as follows.

^{simulate this circuit – Schematic created using CircuitLab}

Using voltage divider I get

$$\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1+(\omega RC)^2}} $$

But now I got a problem. I have been given the property of low pass filter as -3db cut off frequency. How am I supposed to put the value of \$\omega\$ if no frequency is given but only -3db?

If It helps the reader to understand more the question: Actually I have been given two resistors with 1kΩ and 5kΩ and two capacitors with 2nF and 4nF and I have to choose only three of the component, to realize the circuit with -3db cut of frequency. So I came up with this schematic that could also be the possibility.

Answer 1

From the voltage divider rule,

$$\left|\frac{V_{out}}{V_{in}}\right| = \frac{1}{\sqrt{1+(wRC)^2}}$$ Expressing in dB, $$\left|\frac{V_{out}}{V_{in}}\right|_{dB} = 20\log\left(\frac{1}{\sqrt{1+(wRC)^2}}\right)$$

at \$w=1/RC,\$ $$\left|\frac{V_{out}}{V_{in}}\right|_{dB} = 20\log(\frac{1}{\sqrt2}) = -3.01\mathrm{dB}$$

So -3bB frequency is the frequency at which the the voltage gain of the filter falls to \$1/\sqrt2\$ times of the maximum value. For a simple RC low pass filter the -3dB frequency is given by, $$w_c = 2\pi f_c = \frac{1}{RC}$$ or, $$f_c = \frac{1}{2\pi RC}\tag1$$

So if you are asked to design a low pass filter with -3dB frequency = \$f_c\$, choose the value of R and C such that it satisfies equation (1).

Answer 2

$$ \omega = 2 \pi f $$

You can also get the required values for C and R for a -3dB point at a frequency \$f\$ by the formula:

$$ f = \frac{ 1 }{ 2\pi R C } $$

\$f\$ is your -3dB frequency.