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

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.

• dB measures level, not frequency. There isn't enough information to give an answer. – Ignacio Vazquez-Abrams Jul 8 '14 at 16:48
• Recall that dB is a ratio (Vout/Vin). – dext0rb Jul 8 '14 at 16:51
• @dext0rb So yo you mean I can replace the value of Vout/Vin with -3 – Lifestohack Jul 8 '14 at 16:53
• No, take a quick minute to review what a decibel is defined as. That is the crux of your problem, I think. – dext0rb Jul 8 '14 at 16:57
• -3dB translates to a Vout/Vin value. You have to do the algebra to separate the $\omega$ variable. – Kaz Jul 8 '14 at 18:42

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).

• Kinda spoonfed this one... – dext0rb Jul 8 '14 at 16:59
• Good. Note that it's not exactly -3dB, but close. – Spehro Pefhany Jul 8 '14 at 17:02
• @dext0rb I thought this is better than giving an answer $f=(2\pi RC)^{-1}$ directly. – nidhin Jul 8 '14 at 17:27
• @SpehroPefhany not exactly -3dB, but it probably is exactly what people usually mean when they say "-3dB", which is exactly "half" (even though "half" is more like -3.010299956639812 dB). – Phil Frost Jul 8 '14 at 18:16

$$\omega = 2 \pi f$$

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.

• Can I directly substitute the value of f as 3 and insert in that formula to get R*C – Lifestohack Jul 8 '14 at 16:50
• Were you told the -3 dB freuquency is 3 Hz? – The Photon Jul 8 '14 at 16:52
• @ThePhoton Nope. I have to choose the value of register and capacitor so that the circuit will be realized at -3db cut off frequency. – Lifestohack Jul 8 '14 at 17:20
• No matter what values of R and C you choose, your circuit will have a 3-dB cutoff frequency. The question is, what do you want that frequency to be? Different R and C values will give -3 dB at different frequencies, according to the formula given in these answers. – The Photon Jul 8 '14 at 17:21