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

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%?

• 70.7% of amplitude = 50% the power. Why 50% power? Just a standard convention. Helps simplify some math like 1/(6.28*RC) – DKNguyen May 14 '20 at 17:34
• $\frac{1}{\sqrt 2}=0.707$ – Chu May 14 '20 at 18:20
• because that's the definition of cutoff frequency (for that type of filter anyway). Should your question be 'why do we choose to define cutoff frequency at the 70.7% point?' – Neil_UK May 15 '20 at 4:23
• see my answer here: electronics.stackexchange.com/questions/503713/… . – Sredni Vashtar Jun 8 '20 at 14:56

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\$$.

• What will happen if I take 50% instead of 70% ? Why we want 50% power reduction? Why not 75% ? – Jack Danniels May 14 '20 at 17:01
• The balance is between signals near the cutoff point. In the passband, the signal will be distorted near the cutoff frequency, and in the transition band the power of the signal drops below that half power (-3dB) point. Choosing 50% for example will give more signal distortion without as much power decrease. TL;DR; it's a pretty good tradeoff and optimization at 70.7% – nate May 14 '20 at 17:06
• It's not just convention; ti has useful properties too. Swap R and C : you now have a high pass filter. Pick any other convention and the two filters have different cutoff frequencies. – Brian Drummond May 14 '20 at 17:40
• For your single-pole RC filter fc = 1/($\omega RC$), any other number would result in ugly constants. – Spehro Pefhany May 14 '20 at 18:32
• "a 50% power reduction is equivalent to a voltage reduction of √0.50=0.70710678 or 70.7% approximately." - Nope. a 50% power reduction is equivalent to a voltage reduction to 70.7% of the original, or a voltage reduction of 29.3%. – abligh May 16 '20 at 8:04

The big advantage of the 50% power point is its symmetry : if you interchange the R and C in the classic low pass filter in Andy's answer, you get a high pass filter with the same cutoff frequency.

If you chose any other point (such as 50% of voltage, 25% of power) you would still get a high pass filter ... but you would have to re-calculate the cutoff frequency because it would be different for the same component values.

Only the 50% power point (-3dB gain, 3dB attenuation, 0.707 voltage) gives you the same cutoff frequency for equivalent filters with the same components.

With hindsight it's pretty obvious that passing 50% of the power is equivalent to stopping 50% of the power, and that holds for no other ratio.

This is exploited in some filter design texts by focussing in depth on one form of filter (generally the low pass filter) and describing the other filters in outline. You can design a high pass filter by following the low pass filter design process, then following a simple process to derive the equivalent HP (or bandpass or bandstop) filter you really wanted.

Well - it is not a simple task to explain WHY a certain definition was agreed upon. Of course, such a definition for the end of a passband should "make sense". But what does this mean?

• One possible explanation is - as mentioned in the existing contributions - based on power considerations.

• As another explanation the PHASE SHIFT could be used. Because such a first-order circuit allows a maximum phase shift of 90 deg (lowpass: Very large frequencies, highpass: Very low frequencies), it makes sense to define the frequency with 45 deg phase shift as end of the passband (cut-off). At this frequency the real and the imaginary part of the denumerator are equal. This is, of course, identical with the amplitude-based 3dB definition.

• It is to be noted that for higher-order filters there are other definitions - depending on the specific transfer function (Butterworth: 3 db cutoff, Chebyshev: application oriented). For the BESSEL-Thomson response, the cut-off is, in some applications, even defined in the time domain (based on group delay).

• Another reason for defining the inverse of the time constant RC as a passband end is the following: Using this definition, the 1st-oder filter parameters fit very well into the system of higher filter orders. The denominator of the first order transfer function is D(s)=1+sRC. The zero of this function gives the pole of the transfer function: sp=-1/RC in the complex frequency plane (in this simple case: On the negative real axis). The magnitude of this negative-real pole is identical to the so called "pole freqency".

• This is a good reason to define this pole frequency wp as the cut-off frequency wp=wc=1/RC. Why? Because also for all 2nd-oder filters (and for cascades of 2nd-order blocks), it is the pole frequency (wp=|sp|) which plays the primary role in the process of designing higher-order filter structures.

• Examples: For 2nd-order BUTTERWORTH responses we also have wp=wc (with a 3dB cut-off) and for all 2nd-order bandpass functions we also have center frequency identical to pole frequency (wo=wp).

Because 70.7% of output voltage means the half of the initial power:

$$P = \frac{V^2}{R_L}\\ P'=\frac{(0.707\ V)^2}{R} = 0.5\ \frac{V^2}{R_L} = 0.5\ P$$

If you ask "Why the half of the initial power", it has no explanation (at least, I don't know). Maybe about audio, or maybe another thing.

All we know is, the point where the initial power is reduced to its 50% is considered the cut-off point.

The cutoff is at 50% power, which is - 3dB. This is by convention.

For voltage, we calculate the Vo/Vi ratio in dB as 20 log(10) Vo/Vi.

Plugging - 3 into this equation for Vo/Vi yields 0.707 which is also sin45 and cos45.

The cut-off frequency is being defined as the point where there is a -3 dB roll-off. The value -3 dB is equal to 70.7%. The conversions between decibel-space and linear-space are shown below. $$10^{-3 \ \textrm{dB}/20 \ \textrm{dB} } = 0.707$$

$$20 \log_{10}\left(0.707\right) \ \textrm{dB}= -3 \ \textrm{dB}$$ Some complementary filters will have a cross over at -3 dB. Others will have a cross over at a different value. The value of -3 dB is arbitrary. However, it's intuitive for some specific cases.

According to my teachings, 3dB is the smallest change in sound pressure level detectable by the human ear. In the days when many measurements and standards were related to human parameters, it seems logical to adopt 3dB as a standard. Couple that with the happy 50% power coincidence and that standard translates well to electrical practices.

• 3dB as the smallest detectable change is a bit of a myth. A 1dB change in SPL at some frequencies relative to fixed SPL at other frequencies makes a clear difference to the sound of a loudspeaker : the BBC final adusted their studio monitors (like the LS5/9) by ear in 0.5 dB steps to match a reference speaker. – Brian Drummond May 20 '20 at 6:42