Why are ratios relevant, but not the increments when defining center frequency of a band-pass filter?

Question

First of all, here is my directly related previous question: Why is center frequency of a bandpass filter is given by the geometric average of the two cutoff frequencies?

Olin Lathrop explained very well why the geometrical mean is taken when calcuating the center frequancy. Because ratios matter when defining the center frequency not the linear distance.

But now what I'm wondering the idea behind defining this center frequency by considering ratios. Why was that defined like that? Is that something to do with half power point? What is the benefit? Anyway to illustratively or mathematically explain the idea behind?

Answer 1

The physics works either way (ratios or increments). What you are really asking is why are ratios more convenient to us for mathematical manipulations of amplitudes and frequencies than increments.

First, this is how humans perceive frequencies and amplitudes. For example, the musical scale is all about ratios, with everything repeating each octave (factor of 2). You can take a song, scale all its frequencies by some amount, and it's still the same melody. However, if you were to add some amount to all the frequencies it wouldn't be the same melody at all anymore.

Alexander Graham Bell found long ago that humans perceive loudness on a logarithmic scale. A fixed factor of loudness change sounds more like a fixed increment to us. This is why he invented a logarithmic scale for measuring loudness, which is where we got dB (deci-Bels) from. A +6 dB change in loudness sounds like a increment to us, even though it's actually a ratio change (about a factor of 4 in the case of 6 dB). For example, going from 50 to 56 dBA, sounds roughly like the same increment to us as going from 60 to 66 dBA, even though the second is over 3x more when accounted for as a power increment.

Not only do humans think of amplitude and frequency logarithmically, but the math also falls out nicer that way. Both these reasons is why we commonly use Bode plots to show frequency responses. Note that a Bode plot is Log(amplitude) as a function of Log(frequency).

Some convenient things happen when you view filters this way. Plot the response of two simple R-C filters on a Bode plot and on a linear plot as comparison. Let's say the first is a low pass filter with a 1 kHz rolloff point, and the second at 10 kHz. On a Bode plot these will look identical except that one is shifted a decade in frequency from the other. On a linear plot these will be rather different, and much harder to see they are really the same thing at two different frequencies. Try it. No, really, actually plot these out to see what I mean.

Answer 2

By way of illustration, consider the bandpass filter below.

The frequency response function is: \$\small \frac{V2}{V1}= G(j\omega)=\large\frac{R}{R+j(\omega L-\frac{1}{\omega C})}\$

and the gain is: \$\small G(\omega)=\large \frac{R}{\sqrt{R^2+(\omega L-\frac{1}{\large \omega C})^2}}\$

The resonant frequency, \$\small \omega_r\$, is defined when \$\small \omega L=\large \frac{1}{\omega C}\$, giving \$\omega_r =\frac{1}{\sqrt{LC}}\$, and the gain at resonance is unity.

We can determine the bandwidth of this filter by calculating the two corner frequencies, say \$\small \omega_l\$ and \$\small \omega_u\$, where the gain is 3dB down from the gain at resonance, or in other words where the gain is \$\frac{1}{\sqrt{2}}\$.

From the gain equation, \$\small \omega_l\$ and \$\small \omega_u\$ must be defined by \$\small (\omega L-\large \frac{1}{\omega C})^2=\small R^2\$, and we can determine \$\small \omega_l\$ and \$\small \omega_u\$ by solving the quadratic equation: \$\small \omega^2 LC-\omega RC-1=0\$, noting that \$\omega\$ must be a positive value.

Thus: \$\omega_l=\frac{\sqrt{R^2C^2+4LC}-RC}{2LC}\$ and \$\omega_u=\frac{\sqrt{R^2C^2+4LC}+RC}{2LC}\$

Now, the geometric mean, \$\omega_c\$, of \$\omega_l\$ and \$\omega_u\$, is defined when: \$\large \frac{\omega_u}{\omega_c}=\frac{\omega_c}{\omega_l}\$, or \${\omega_c}^2=\omega_l \: \omega_u=\frac{1}{LC}={\omega_r}^2\$. Hence $$\large \omega_c=\omega_r$$

That is, the geometric mean of the corner frequencies is the resonant frequency.

Answer 3

I don`t think it is correct to say that the center frequency of a bandpass "was defined like that" (based on the mentioned ratios). This property (geometric mean value of the 3dB-cutoffs) is only the RESULT of the definition. And the definition of a bandpass center frequency (second order) is simply based on the magnitude maximum at f=fo. This maximum is reached when the denominator assumes its minimum. In this case, both real parts of the denominator cancel each other and we have H(jwo)=Ajwo/Bjwo=A/B.

This is a real ratio. We therefore define: At the center frequency f=fo the second-order bandpass has maximum gain (minimum damping) and zero phase shift.

This is the definition - all other properties are the result of this definition and can be used to make this definition descriptive and logical.

EDIT: The explanations above are based on the classical ("normal") form of a second-order bandpass with a denominator written as second-order polynominal in s : H(s)=As/[1+Bs+(s/wp)²].