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My question is: why do we choose cut-off frequency to satisfy that the magnitude of transfer function \$ H(\omega) = \frac{1}{\sqrt{2}} \$?

Therefore I am not asking how cut-off frequency is calculated as \$ \sqrt{\frac{R}{L}}\$, but I am asking why it is chosen that way.

I know I sound a bit confusing but it is because I am confused. Thanks in advance

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    \$\begingroup\$ To add more confusion: This is really only a convention. You can pick other magnitudes to call the cut-off point. -3 dB is pretty useful, though. \$\endgroup\$
    – pipe
    Nov 14, 2017 at 10:17
  • \$\begingroup\$ @pipe huh, I needed this answer to satisfy myself, then my next question is: In what way it is useful so that it is chosen as cut-off frequency? \$\endgroup\$ Nov 15, 2017 at 17:42
  • \$\begingroup\$ I don't think I can answer that better than how Bimpelrekkie did. Since there is never a hard limit in a physically realizable filter, it makes sense to put it where half the power is on one side and half on the other. \$\endgroup\$
    – pipe
    Nov 15, 2017 at 17:52

1 Answer 1

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H(w) =1/ (square root of 2) That sounds a bit confusing, we usually refer to the cutoff point as the -3 dB point.

That is the same though, -3 dB is half the power.

Let me explain: take your \$H(\omega) = \frac{1}{\sqrt2}\$

That means that at that \$\omega\$ the voltage is divided by \$\sqrt2\$, if this voltage is applied across a (load) resistor at the output of the filter then the current through that same load resistor will also be divided by \$\sqrt2\$.

What does that mean for the Power?

I means that the power is halved.

On a dB Power scale that means -3 dB

Using that "half of the power" as a reference point is useful because if we divide a wideband signal into a low frequency part and a high frequency part then we can do that using a low pass filter and a highpass filter with the same cutoff frequency. At that cutoff frequency half of the power ends up at the output of the lowpass filter and the other half ends up at the output of the highpass filter.

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  • \$\begingroup\$ So "why do we choose cut-off frequency to satisfy that the magnitude of transfer function H(w) =1/ (square root of 2)?" is answered with "I means that the power is halved.". And... why do we choose when the power has been halved? Maybe this isn't OP's question, but this will be his/hers next question. \$\endgroup\$ Nov 14, 2017 at 10:10
  • \$\begingroup\$ I added an explanation for that. \$\endgroup\$ Nov 14, 2017 at 10:11
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    \$\begingroup\$ And that deserves a +1 from me. \$\endgroup\$ Nov 14, 2017 at 10:12
  • \$\begingroup\$ Doesn't this assume unit gain at DC? Wouldn't $$H(\omega) = \frac{1}{\sqrt2} H (0)$$ be better? \$\endgroup\$ Nov 14, 2017 at 12:15
  • \$\begingroup\$ @RodrigodeAzevedo Only for a Low pass filter. Consider a high pass filter which will often completely block DC. More accurately the -3dB point is at a frequency relative to the passband. \$\endgroup\$ Nov 14, 2017 at 12:38

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