I'm experimenting a little with feedbackloops (delwrite~ and vd~) with the vd~ driven by and oscillator, which is good fun. The vd~ output is (with a *~) feedbacked into the delwrite~.

When i put an lop~ in the feedback chain, the sound sounds well... lowpassed. But when i put a hip~ pd nearly instantly goes into overdrive and the peakmeters peak about 400 à 500. Could someone explain how this comes?

Thanks a lot,

D.

if the gain exceeds 1.0 at *any* frequency, then the feedback will explode eventually - there's a number of ways to fix it inside the loop (ie, just before delwrite~), including:

reducing gain [*~ 0.9]
"hard" clipping [clip~ -1 1]
"soft" clipping [expr~ tanh($v1)]
dynamic range compressor/limiter (see pd-list for some abstractions posted recently)

Ok, i understand that, but that still does not answer how it comes that a [lop~] reduces and a [hip~] not with the feedback in both cases set to 99.9% (the hip~ blows up the audiosignal even at lower feedback settings). It seems like the [hip~] is not only reducing certain (lower) frequencies but also attenuating other (high) frequencies. Is this correct behaviour?

have a look at the 'H' section of pd's browser help...H11-shelving filter, H12-peaking filter..etc

you can take out the [rpole] [rzero] bits used in those examples and put in whatever filter you want.

if you do [hip~ 20] , the output is pretty flat, except for a tiny jump at the end near 20000hz.

but if you do [hip~ 5000], then the higher frequencies are nearly doubled.

i'm not sure what the mathematical relationship would be between frequency and attenuation, but with a bit of trial and error, you could probably find a way to scale the output so hip~ never outputs above amplitude 1

Ok, thanx a lot hardy-one... But i always tought that [hip~] and [lop~] are non-resonant filters. Is this then normal behaviour (but that's just a little theoretical question which needn't be answered for my practical needs of course)?