# How to have an attenuation different from 40/20 dB per decade with a basic filter

I would like to do a theoritical filter with different attenuation according to the frequency. How can I do for having an attenuation different from 20 dB per decades. If for example I want to have an attenuation equal to 8.23 dB per decade ? How can I do ?

Have a nice day :)

• You can approximate it with a series of shelving networks, like a staircase. First you need to know the frequency span of interest and the accuracy required. Jun 7, 2021 at 14:43
• There is also fractional order filters. But currently they can ve realised using the method mentioned in above comments.
– AJN
Jun 7, 2021 at 14:46
• Use a graphic equalizer. Jun 7, 2021 at 15:23
• just to be sure: this is about implementing a complicated analog filter, not about a digital one, right? Because in the analog case, as user_1818839 correctly points out, realization 100% depends on the frequency range of interest. A filter with GHz bandwidths is implemented fundamentally different than one for audio, is implemented fundamentally different than for narrowband RF. Jun 7, 2021 at 16:30
• Thank you for all your replies :)
– Jess
Jun 8, 2021 at 6:23

Expression below will give a slope of +/- 20$$\{\gamma}\$$ dB/decade, between $$\{\omega}_0\$$ and $$\{\omega}_f\$$. N is number of pole-zero pairs.
$$\displaystyle\frac{\left(1+\frac{s}{\omega_0}\right)^\gamma}{\left(1+\frac{s}{\omega_f}\right)^\gamma} \space\space\space\space \approx \space\space\space\space \displaystyle\prod_{k=1}^{N} \frac { \left( 1+s\frac{ {q} ^ {(k-1) } }{ \omega_0 } \right) } { \left( 1+s\frac{ {q} ^ {-(k-1)} }{ \omega_f } \right) } \space\space\space\space\space\space\space\space\space\space q= \left( \frac{\omega_f}{\omega_0} \right) ^{ \frac{(N-\gamma)}{N(N-1)} }$$
I think it is equivalent to what you would get with the Oustaloup approximation for $$\s^\gamma\$$, which is described here for example.