Electronic – Significance of cut-off frequency in a low-pass filter

cutoff frequencylow passtransfer function

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

Best Answer

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.