To characterize a signal it's a common use to determine bandwidth at -3dB from the peak (half power). Also, filter's cutoff frequency is pointed at -3dB. Why we use this rule?


3 Answers 3


As an addition to Tendero's answer, it is mathematically simpler to define the bandedges as the -3.01 dB or "half-power" frequencies. The simplest example is a first-order low-pass filter:

$$ H(s) = \frac{1}{1 + \frac{s}{\omega_0}} $$

The frequency response is obtained by substituting: $s \leftarrow j \omega$ . Then

$$ H(j \omega) = \frac{1}{1 + j \frac{\omega}{\omega_0}} $$

Then the power spectrum of the frequency response is the magnitude-square of the frequency response:

$$\begin{align} \Big| H(j \omega) \Big|^2 &= \left| \frac{1}{1 + j \frac{\omega}{\omega_0}} \right|^2 \\ \\ &= \frac{1}{1 + \left(\frac{\omega}{\omega_0} \right)^2} \\ \end{align}$$

You can see that when $\omega = 0$ (or DC), the power gain is $\big| H(j 0) \big|^2 = 1$ which is the peak gain.

And when $\omega = \omega_0$, the power gain is $\big| H(j \omega_0) \big|^2 = \frac12$ which is half the peak power gain. When $s$ is normalized by $\omega_0$ in the first-order low-pass filter, then $\omega_0$ is the half-power frequency.


It is a common value because at $-3 \ \mathrm{dB}$ the power of the signal is reduced to half its value. I'll show a brief example to make it clearer.

Suppose that you have a signal whose amplitude is $1 \ \mathrm{V}$. If we reduce its amplitude by $-3 \ \mathrm{dB}$, the new amplitude would be close to $ 0.707 \ \mathrm{V} $.

The power of a signal is proportional to the square of its amplitude. For simplicity, let's assume a resistance of $1 \ \Omega$ on which we are going to calculate the dissipated power. Therefore, originally we would have had a power of $$(1 \ \mathrm{V})^2\cdot\frac{1}{1 \ \Omega}=1 \ \mathrm{W}$$

But after the reduction, we have a power of

$$(0.707 \ \mathrm{V})^2\cdot\frac{1}{1 \ \Omega}=0.5 \ \mathrm{W}$$

The power has decreased to half its original value. That is why the $-3 \ \mathrm{dB}$ point is commonly used to characterize frequency-related properties.

  • 1
    $\begingroup$ Is half-power some sort of minimum guarantee for signal pass through the passband (in case of filter)? There are other choices but half-power is widely used? Why the design at half-power? $\endgroup$
    – jomegaA
    Feb 3, 2020 at 18:41
  • $\begingroup$ @jomegaA It actually depends on what your requirements for your filter are. There has to be some criteria to inform what the bandwidth is, and half-power seems an easy-to-remember choice. There is no actual reason behind it, it's just a convention. IMHO, it is a good choice for manufacturers because 1) it makes their products seem "better" (the bandwidth would be reduced if we chose, for instance, -1 dB as cutoff point) and 2) it would be unacceptable (to me) to say that a frequency such that the power is less than half its original value is inside the bandwidth. $\endgroup$
    – Tendero
    Feb 3, 2020 at 18:46

Another source of the 3(dB) rule-of-thumb is that is how far two targets/signals/sources must be separated in order for the human eye to distinguish between the two. For example, if you have two rectangle functions, the edge of one must be past the -3(dB) edge of the other for them not to look like one single rectangle function.


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