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

Question

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?

Answer 1

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)\$):

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

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Using this on your transfer function, we get:

1. dB/decade: \begin{equation} \begin{split} 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(\lim_{\omega\to\infty}\sqrt{\frac{1+\left(\frac{\omega_0}{\omega}\right)^2}{100+\left(\frac{\omega_0}{\omega}\right)^2}}\right)\\ \\ &=20\log_{10}\left(\sqrt{\frac{1+0}{100+0}}\right)\\ \\ &=20\log_{10}\left(\frac{1}{\sqrt{100}}\right)\\ \\ &=20\log_{10}\left(\frac{1}{10}\right)\\ \\ &=-20\space\text{dB/decade} \end{split}\tag5 \end{equation}
2. dB/octave: \begin{equation} \begin{split} 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} \end{split}\tag6 \end{equation}

Answer 2

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.