1
\$\begingroup\$

I have a low pass filter as follows.

schematic

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.

\$\endgroup\$
  • \$\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
  • 1
    \$\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
  • 2
    \$\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
5
\$\begingroup\$

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).

\$\endgroup\$
  • 3
    \$\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
  • 2
    \$\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
2
\$\begingroup\$

$$ \omega = 2 \pi f $$

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.

\$\endgroup\$
  • \$\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
  • 1
    \$\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
  • 1
    \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.