5
\$\begingroup\$

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?

\$\endgroup\$
1
  • 3
    \$\begingroup\$ "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. \$\endgroup\$
    – DKNguyen
    Oct 31, 2021 at 19:39

2 Answers 2

7
\$\begingroup\$

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

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.

\$\endgroup\$
1
  • 3
    \$\begingroup\$ 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. \$\endgroup\$ Nov 1, 2021 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.