High Pass Filter Design from Transfer Function
When s=0 and infinity the G(s)=1.

The transfer function at mean midpoint of sqrt(530*2220)=1070 inserted for s becomes ,

G(1030)=0.486

Thus a unity gain low pass then high pass parallel &amp; series T filter is needed or an active equivalent. 

Hypothesis

So the transfer function resembles a passive loudness switch that cuts the audio midrange 20log0.486=-6.3dB. My Bogan stereo tube amp during the ‘60’s and ‘70’s had the -15 dB preferred solution over the bass-boom box version used since then. 


  One reason for this type of response to cut the loudness in the midrange only, rather increase, may be due a phone call, so we cut the midrange only then restore to flat for full loudness. It seems only Bogan got it right. Much after Baxandall invented it, stereo designers used the loudness switch boost crappy speakers bass response below 100 Hz rather than cut the midrange to match our hearing response according to the well known Fletcher-Munson curves. developed before 1933.  So this transfer function is a simple approximation for these curves to listen at -6dB very slightly lower volumes.


The next clue is the passive Baxandall dual T filter fixed mid-range filter. Peter J.Baxandall invented this filter just before I was born in 1952.   The variable bass treble version is still in use today in old tuners.

Can you imagine an RCR//CRC filter to do this with the R ratios being 2200/520? 

If so , you may be as smart as Bax.  If not learn how to find his answer.  This is for your education, learning how to learn on your own without spoon-feeding.

That’s MY reason for not giving you the solution.