# How is the slope of the frequency response of an analog active filter defined?

I have seen different types of filters, and different slope values at -20 db, -40, etc.

The literature says that the order define the slope, but how? How can it be mathematically justified?

For example, I have a transfer function with a simple real pole $$\H(s)=\frac{1}{1+\frac{s}{\omega_{0}}}\$$ using $$\j\omega\$$ instead of $$\s\$$ $$\H(j\omega)=\frac{1}{1+j\frac{\omega}{\omega_{0}}}\$$ then the magnitude is:

$$|H(j\omega)|=\left|\frac{1}{1+j\frac{\omega}{\omega_{0}}}\right|=\frac{1}{\sqrt{1^{2}+(\frac{\omega}{\omega_{0}})^{2}}}$$

$$H(j\omega)|_{dB}=20log_{10}(\frac{1}{\sqrt{1+(\frac{\omega}{\omega_{0}})^{2}}})$$

Next, I understand there are three cases:$$\\omega<<\omega_{0}\$$ , $$\\omega>>\omega_{0}\$$ and $$\\omega=\omega_{0}\$$. Lets say, working when the frequency is the same as the cutoff frequency, the magnitude and angle are

$$|H(j\omega_{0})|_{dB}=20log_{10}(\frac{1}{\sqrt{1+(\frac{\omega}{\omega_{0}})^{2}}})=$$

$$20log_{10}(\frac{1}{\sqrt{2}})\approx-3\:dB$$

$$\angle H(j\omega)=-arctan(1)=-\deg{45}=-\frac{\pi}{4}\,rad$$

This results in a piecewise linear asymptotic Bode plot for magnitude from 0 dB until the cutoff frequency and then drops at 20 dB per decade (the famous -20 dB/decade.)

How are these modified to -40 or -60 dB?

This is the case for a simple real pole. How does it work with a real zero,a pole/zero at the origin,complex conjugate zero/poles, etc?

How does it work in the other two cases of the simple real pole?

• "How are these modified to -40 or -60 dB?" They are modified by that 20dB/decade you just mentioned. You might be misunderstanding what that is. It's not that the slope is changed to -20dB/decade. The slope is changed by -20dB/decade. That +/-20dB is the contribution of the pole or zero, therefore it is cumulative. Every time you pass a pole or zero as you sweep the frequency from lower to higher frequencies. Notice they are all multiples of 20. Oct 31, 2021 at 19:39

Well, generally we have two things that we look at:

1. dB/decade: $$\lim_{\omega\to\infty}\left(20\log_{10}\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|-20\log_{10}\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|\right)\tag1$$
2. dB/octave: $$\lim_{\omega\to\infty}\left(20\log_{10}\left|\underline{\mathscr{H}}\left(2\omega\text{j}\right)\right|-20\log_{10}\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|\right)\tag2$$

Using logarithm rules, we can write:

$$20\log_{10}\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|-20\log_{10}\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|=20\log_{10}\left(\frac{\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|}{\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|}\right)\tag3$$

Using the fact that the logarithm ($$\\log(x)\$$) is a continuous function (for $$\x>0\$$), we can write (and using $$\(3)\$$):

$$\lim_{\omega\to\infty}\left(20\log_{10}\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|-20\log_{10}\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|\right)=20\log_{10}\left(\lim_{\omega\to\infty}\frac{\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|}{\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|}\right)\tag4$$

Using this on your transfer function, we get:

1. dB/decade: $$20\log_{10}\left(\lim_{\omega\to\infty}\frac{\left|\underline{\mathscr{H}}\left(10\omega\text{j}\right)\right|}{\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|}\right)=20\log_{10}\left(\lim_{\omega\to\infty}\sqrt{\frac{\omega^2+\omega_0^2}{100\omega^2+\omega_0^2}}\right)=$$ $$20\log_{10}\left(\frac{1}{10}\right)=-20\space\text{dB/decade}\tag5$$
2. dB/octave: $$20\log_{10}\left(\lim_{\omega\to\infty}\frac{\left|\underline{\mathscr{H}}\left(2\omega\text{j}\right)\right|}{\left|\underline{\mathscr{H}}\left(\omega\text{j}\right)\right|}\right)=20\log_{10}\left(\lim_{\omega\to\infty}\sqrt{\frac{\omega^2+\omega_0^2}{4\omega^2+\omega_0^2}}\right)=$$ $$20\log_{10}\left(\frac{1}{2}\right)=-20\log_{10}\left(2\right)\approx-6.0206\space\text{dB/octave}\tag6$$

If you're talking about "typical" transfer functions (those that express the behavior of a linear ordinary differential equation), then they take the form

$$H(s) = \frac{s^m + b_{m-1}s^{m-1} + \cdots + b_0}{s^n + a_{n-1}s^{n-1} + \cdots + a_0}$$

where I'm using $$\s = j\omega\$$, to save on typing and on trying to keep track of minus signs. Note that the magnitude of $$\s\$$ is exactly equal to the magnitude of $$\\omega\$$.

The whole idea of the terminal slope of the transfer function is that when $$\s \gg 1\$$, then eventually the only thing that matters is the ratio $$\\frac{s^m}{s^n}\$$. At high enough frequencies, the transfer function acts as if it is just $$\\frac{1}{s^{n-m}}\$$.

This is where your slope comes from -- if $$\n-m\$$ is one, then you have 20dB/decade; if it's two, then you have 40dB/decade, etc.

• Maybe it would help OP if you'd say that s=j*2*pi*f, so 1/s becomes equivalent with 1/f, or 1/x, mathematically. Then it's easy to see that a decade backward, or foreward, the amplitude is more, or less, by a factor of 10. And, since the gain formula is 20*log10(10)=20*1=20, or 20*log10(0.1)=20*(-1)=-20. For f^2 it's 20*log10(100)=20*2=40, etc. Nov 1, 2021 at 15:39