On Sat, 25 Apr 2015 16:49:25 -0400, robert bristow-johnson wrote:> you might think that this is a settled issue, especially before any > "digitization" of the transfer function. so, for the moment, let's > leave the z-plane outa it. > > originally, i always took the definition of "Q" and resonant frequency > "w0" (pronounced "omega-naught" not "dubya-zero" even though we all know > he's a zero) as coming from the classic RLC series circuit (with output > voltage taken over the R) having transfer function: > > H(s) = R / ( R + sL + 1/(sC) ) > > which if you re-arrange it a little, you get this BPF: > > H(s) = (1/Q)(s/w0) / ( (s/w0)^2 + (1/Q)(s/w0) + 1 ) > > > for w0 = 1/sqrt(LC) and Q = w0*L/R > > when s = j*w0 you get H(j*w0) = 1 which has greater magnitude than any > other real frequency. > > but this is not necessarily the frequency at which the circuit "rings" > at if whacked with an impulse. > > h(t) = w0 * exp(-(w0/2Q)*t) * cos( sqrt(1-(1/2Q)^2)*w0*t + phi ) > > phi = arctan(1/2Q) i think > > > the "ringing frequency" is sqrt(1-(1/2Q)^2)*w0 . if the Q is smaller > than 1/2, this thing doesn't even ring and you have two damped > exponentials in the impulse response rather than a damped sinusoid. > (instead of a pair of complex-conjugate poles, there is a pair of real > poles.) > > now, can we agree that the resonant frequency of this analog transfer > function is w0 which exists for any Q>0 ? even if this thing does not > ring? > > the reason i am asking is, reviewing some very old notes regarding Hal's > SVF *and* with the Lattice filters, i am wondering if the "frequency" > that the filters are tuned to are the ringing frequency (that exists > only for Q > 1/2) and not the resonant frequency (which still has > meaning for low-pass filters that don't ring at all). > > so which is it? (consider this a polling question.)The resonant frequency is where the phase=0 ;) More seriously (risking being called a moral relativist) why the need for a single definition? Isn't this one of those cases where we might usefully depend on circumstances/requirements?
Low pass filters in the time domain?
Started by ●April 20, 2015
Reply by ●April 26, 20152015-04-26
Reply by ●April 26, 20152015-04-26
On Sat, 25 Apr 2015 16:49:25 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:> H(s) = (1/Q)(s/w0) / ( (s/w0)^2 + (1/Q)(s/w0) + 1 ) > >for w0 = 1/sqrt(LC) and Q = w0*L/R > >when s = j*w0 you get H(j*w0) = 1 which has greater magnitude than any >other real frequency. > >but this is not necessarily the frequency at which the circuit "rings" >at if whacked with an impulse. > > h(t) = w0 * exp(-(w0/2Q)*t) * cos( sqrt(1-(1/2Q)^2)*w0*t + phi ) > > phi = arctan(1/2Q) i think > > >the "ringing frequency" is sqrt(1-(1/2Q)^2)*w0 .I seem to recall a control systems class (the controls folks are always interested in time-domain performance) wherein it was acknowledged that there is a difference between "f0" (= w0/2PI) and the "ringing frequency". It has to do with the effect of damping upon the rate of change of the signal, and the ringing frequency equals f0 when there is no damping (i.e., on the jw axis). In fact, and this is ancient knowledge so it may be a little dusty, I seem to recall that the actual ringing frequency corresponds to the imaginary part of the pole location on the s-plane -- that seems reasonable, because for Q<0.5 the poles lie on the real axis.>now, can we agree that the resonant frequency of this analog transfer >function is w0 which exists for any Q>0 ? even if this thing does not ring?I really prefer this definition for its mathematical consistency -- w0 is always the same regardless of the value of Q.
Reply by ●April 26, 20152015-04-26
On Sun, 26 Apr 2015 08:55:04 -0500, Greg Berchin wrote:> On Sat, 25 Apr 2015 16:49:25 -0400, robert bristow-johnson > <rbj@audioimagination.com> wrote: > >> H(s) = (1/Q)(s/w0) / ( (s/w0)^2 + (1/Q)(s/w0) + 1 ) >> >>for w0 = 1/sqrt(LC) and Q = w0*L/R >> >>when s = j*w0 you get H(j*w0) = 1 which has greater magnitude than any >>other real frequency. >> >>but this is not necessarily the frequency at which the circuit "rings" >>at if whacked with an impulse. >> >> h(t) = w0 * exp(-(w0/2Q)*t) * cos( sqrt(1-(1/2Q)^2)*w0*t + phi ) >> >> phi = arctan(1/2Q) i think >> >> >>the "ringing frequency" is sqrt(1-(1/2Q)^2)*w0 . > > I seem to recall a control systems class (the controls folks are always > interested in time-domain performance) wherein it was acknowledged that > there is a difference between "f0" (= w0/2PI) and the "ringing > frequency". It has to do with the effect of damping upon the rate of > change of the signal, and the ringing frequency equals f0 when there is > no damping (i.e., on the jw axis). In fact, and this is ancient > knowledge so it may be a little dusty, I seem to recall that the actual > ringing frequency corresponds to the imaginary part of the pole location > on the s-plane -- that seems reasonable, because for Q<0.5 the poles lie > on the real axis.w0 is called the "natural frequency" in that case. The older I get the less I care, because if Q > 10 or so then (a) unless all the components are passive you're likely to be in a world of hurt (or working on an old radio), and (b) the resonant frequency today is different from the resonant frequency yesterday and tomorrow both. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●April 26, 20152015-04-26
On Sat, 25 Apr 2015 16:49:25 -0400, robert bristow-johnson wrote:> you might think that this is a settled issue, especially before any > "digitization" of the transfer function. so, for the moment, let's > leave the z-plane outa it. > > originally, i always took the definition of "Q" and resonant frequency > "w0" (pronounced "omega-naught" not "dubya-zero" even though we all know > he's a zero) as coming from the classic RLC series circuit (with output > voltage taken over the R) having transfer function: > > H(s) = R / ( R + sL + 1/(sC) ) > > which if you re-arrange it a little, you get this BPF: > > H(s) = (1/Q)(s/w0) / ( (s/w0)^2 + (1/Q)(s/w0) + 1 ) > > > for w0 = 1/sqrt(LC) and Q = w0*L/R > > when s = j*w0 you get H(j*w0) = 1 which has greater magnitude than any > other real frequency. > > but this is not necessarily the frequency at which the circuit "rings" > at if whacked with an impulse. > > h(t) = w0 * exp(-(w0/2Q)*t) * cos( sqrt(1-(1/2Q)^2)*w0*t + phi ) > > phi = arctan(1/2Q) i think > > > the "ringing frequency" is sqrt(1-(1/2Q)^2)*w0 . if the Q is smaller > than 1/2, this thing doesn't even ring and you have two damped > exponentials in the impulse response rather than a damped sinusoid. > (instead of a pair of complex-conjugate poles, there is a pair of real > poles.) > > now, can we agree that the resonant frequency of this analog transfer > function is w0 which exists for any Q>0 ? even if this thing does not > ring? > > the reason i am asking is, reviewing some very old notes regarding Hal's > SVF *and* with the Lattice filters, i am wondering if the "frequency" > that the filters are tuned to are the ringing frequency (that exists > only for Q > 1/2) and not the resonant frequency (which still has > meaning for low-pass filters that don't ring at all). > > so which is it? (consider this a polling question.)I believe the ringing frequency is different from the frequency at maximum gain. Either one might get called the "resonant frequency". Check the ingredients list on the package before consuming the product. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
