I am trying to design an active filter, which I assume is an active high pass filter according to what I see from the Bode plot.
I can get the transfer function from this plot and then I need to design an active filter .
The transfer function given:
I need to design something like this:
When I made errors in my assumptions for specs. My choice words fit, there is no significant change in the 2nd order equation simplifying it to a 1st order equation. s^2 factors cancel out when the roots of the first order are only an octave apart. There is no significant change or benefit.
Hence @Antonio51 is 100% right.
Doing a sensitivity analysis with the derivative of H(s) proves that.
mY error: fmax is H(s)=0 dB with breakpoints 1 octave apart and 12dB space can never achieve 40 dB/dec or 12 dB/octave and almost 4dB /octave. where 3 dB/octave is used for Pink noise filters.
From the symmetry of this question, there is a simple passive approach.
The more mistakes you correct, the more careful you try to be in making fast assumptions. But then you don't have to be brilliant ("I arn't") just work harder to correct them and gain more experience. Most of my Analog/ SCADA experience was my 1st 5 years in Aerospace 40 years ago. NEVER ASSUME ANY DESIGN IMPLEMENTATION SPECIFICS before you identify the key measurable design specs. (remember there is a U between "ass and me" if I tell you to use this when you know there is a better way.
There must be some shunt attenuation for partial gains. This is also called a Lead-Lag attenuator and is common for PLL compensation loops as it adds +ve phase margin to the loop when used near unity gain. However, there the breakpoints are better selected one decade (10:1) apart and not just 1 octave apart (2:1) which is only 12 dB difference between a gain of 0.25 and 0.25. and over a much wider span 12 dB/octave.
Thus you never achieve 12 dB/octave because of the -3dB breakpoints typically select for the half-power bandwidth. So it is only 6dB/octave unless you raise the Q for more phase shift (which is not needed here) So you might assume maximally flat Butterworth or something else as long as you define that assumption not given.
While we're at it, since there is no gain, we might as well consider a passive filter with say 1k to 10k input impedance.
DC gain = 0.25 (~-12 dB = -20 log(4)) 2nd order ramp up at 10 rad/s HF gain = 0.5 (~-6 dB) 2nd order ramp to flat at 20 rad/s
2nd order equation implies THERE MUST BE TWO REACTIVE ELEMENTS minimum.
The Breakpoints and BW are defined by the "half-power point" ~ -3dB so each point reduces the slope from 2nd order 12 dB/octave to only 6 dB actual at each breakpoint -12,-6 dB and "appears like a 1st order filter" yes is a 2nd order filter. It isn't because the asymptotes are too close together and thus interact with each other. Inverting / Normal (optional) Lead-lag phase frequency response with 6 dB dirfference ramp up from asymptotes 10 to 20 rad/s This can be done with Resistor ratios or Capacitor ratios but in this case as a lead-lag filter or flat with a HP then LP in between.
All the boxes are checked. Design complete.
However , your Prof may expect you to do it the hard way the 1st time.
Opinion
My simple approach only used a few minutes to create with experience and >30 minutes to write this up. You will be able to do this later too.
With experience, Feinmann could do this in his head.
UPDATE: I made confusion with the OP function between omega and frequency:
As the function in the "picture" is not related to the specification itself.
I used the idea of @Tony Stewart EE75, thanks ...
Here is a simulation that can "answer" the OP question. Idea is ok, but values of components must be updated. Will do later.
It can also be "synthesized" with "one-order" function (numerator and denominator).
Update: corrected f -> omega. Transfer functions not calculated, only adjusted by hand with microcap v12 tools.
In my opinion, it is important to explain how to derive an equivalent transfer function from the starting expression. First off, it has to be shaped in a low-entropy form where gain, poles and zeros appear. The normalized polynomial for a second-order expression is \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$. From there, you must rearrange the numerator and the denominator to approach this format: divide all the numerator terms by 100 and all denominator terms by 400. This gives you a leading term, call it \$G_0\$, which is a dc gain, the value taken by the transfer function when \$s=0\$. It amounts to 0.25 or -12 dB in this case.
Once there, you can identify the resonant frequencies in the numerator and the denominator as well as their respective quality factors \$Q\$. In this example, the resonant frequency in the numerator is 10 rad/s while it is 20 rad/s in the denominator. In this particular example, a quality coefficient of 0.5 signals that the poles or the zeroes are coincident meaning that the numerator can be expressed as \$N(s)=(1+\frac{s}{\omega_z})^2\$ while the denominator is \$D(s)=(1+\frac{s}{\omega_p})^2\$. Finally, the complete transfer function becomes \$G(s)=G_0\frac{(1+\frac{s}{\omega_z})^2}{(1+\frac{s}{\omega_p})^2}\$. A few lines in Mathcad confirm this approach:
This is it for the poles-zeroes location and you now can concentrate on the practical realization of this filter.
