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.

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

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    \$\begingroup\$ Kinda spoonfed this one... \$\endgroup\$ – dext0rb Jul 8 '14 at 16:59
  • \$\begingroup\$ Good. Note that it's not exactly -3dB, but close. \$\endgroup\$ – Spehro Pefhany Jul 8 '14 at 17:02
  • \$\begingroup\$ @dext0rb I thought this is better than giving an answer \$f=(2\pi RC)^{-1}\$ directly. \$\endgroup\$ – nidhin Jul 8 '14 at 17:27
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    \$\begingroup\$ @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). \$\endgroup\$ – 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.

  • \$\begingroup\$ Can I directly substitute the value of f as 3 and insert in that formula to get R*C \$\endgroup\$ – Lifestohack Jul 8 '14 at 16:50
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    \$\begingroup\$ Were you told the -3 dB freuquency is 3 Hz? \$\endgroup\$ – The Photon Jul 8 '14 at 16:52
  • \$\begingroup\$ @ThePhoton Nope. I have to choose the value of register and capacitor so that the circuit will be realized at -3db cut off frequency. \$\endgroup\$ – Lifestohack Jul 8 '14 at 17:20
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    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Jul 8 '14 at 17:21

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