Electronic – Why is the output of a filter considered 70.7% of input while determining the cutoff frequency

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I have seen in many books that while finding the cutoff frequency of a filter (from the transfer function) the output is considered 70.7% of the input.

Why 70.7%?

Why not 50% or 20%?

Best Answer

Why 70.7%? Why not 50% or 20%?

When a voltage drops to 70.7%, the effective power it can produce into a resistive load is halved.

So, the important thing to note is that a 50% power reduction is equivalent to the voltage reducing to \$\sqrt{0.50} = 0.70710678\$ or 70.7% approximately.

Why 50% power and 70.71% voltage in a simple RC filter?

If you take a simple RC low pass filter like this: -

enter image description here

You'll find that the cut-off frequency for the filter, \$F_C\$ is when: -

$$R = |X_C|$$

You'll also find that the output voltage will be at 70.71% compared to the input voltage. This is because of Pythagoras and the impedance triangle: -

enter image description here

So, using Pythagoras, when \$R = |X_C|\$, the net input impedance \$ = \sqrt{R^2 + R^2} = \sqrt2\cdot R\$.

This means that the current into the RC filter is reduced by \$\sqrt2\$ compared to the current if \$V_{IN}\$ was applied to either of R or \$X_C\$. This of course means that the voltage amplitude at the output is reduced by \$\sqrt2\$. It also follows that the phase shift between output and input is 45°.

This is what we get for a simple RC filter (low-pass or high-pass) when we have equal magnitudes for R and \$X_C\$.