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

analogcapacitorcircuit analysisfilterinductor

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%?

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: - 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: -

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\$$.