On Fri, 11 Dec 2015 11:46:16 -0500, bitrex wrote:> Suppose we have the equation for a damped harmonic oscillator > driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the > damping parameter. > > Now suppose we allow A to be any _continuous_ function of x, i.e. > steps, delta functions, and their permutations are disallowed. > > Is there a way to determine which such function damps the system > out in the minimum time, _without_ allowing the function to overshoot > the final equilibrium point (i.e. no ringing)?Wait -- you're suggesting that rather than a linear system with A as a constant, you want to use some A(x)? I suspect that it will be easy to find A(x) for a specific step input, but that as soon as you change the size of the step you'll find that your magical A(x) no longer works. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com

# Oscillator mathematcal problem

Started by ●December 11, 2015

Reply by ●December 11, 20152015-12-11

Reply by ●December 11, 20152015-12-11

Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> Wrote in message:> On Fri, 11 Dec 2015 12:43:14 -0500 (EST), bitrex > <bitrex@de.lete.earthlink.net> wrote: > >>Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> Wrote in message: >>> On 12/11/2015 11:46 AM, bitrex wrote: >>>> >>>> Suppose we have the equation for a damped harmonic oscillator >>>> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the >>>> damping parameter. >>>> >>>> Now suppose we allow A to be any _continuous_ function of x, i.e. >>>> steps, delta functions, and their permutations are >>>> disallowed. >>>> >>>> Is there a way to determine which such function damps the system >>>> out in the minimum time, _without_ allowing the function to >>>> overshoot the final equilibrium point (i.e. no ringing)? >>>> >>> >>> Sure, that's a fairly straightforward calculus of variations problem. >>> >>> Cheers >>> >>> Phil Hobbs >>> >>> -- >>> Dr Philip C D Hobbs >>> Principal Consultant >>> ElectroOptical Innovations LLC >>> Optics, Electro-optics, Photonics, Analog Electronics >>> >>> 160 North State Road #203 >>> Briarcliff Manor NY 10510 >>> >>> hobbs at electrooptical dot net >>> http://electrooptical.net >>> >> >>Also IIRC non-conservative forces such as viscous damping require >> something funky to be added to the Lagrangian... > > K-Y Jelly ?>:-} > > ...Jim Thompson > --Nice 1 -- ----Android NewsGroup Reader---- http://usenet.sinaapp.com/

Reply by ●December 11, 20152015-12-11

On Friday, December 11, 2015 at 12:43:31 PM UTC-5, bitrex wrote:> Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> Wrote in message: > > On 12/11/2015 11:46 AM, bitrex wrote: > >> > >> Suppose we have the equation for a damped harmonic oscillator > >> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the > >> damping parameter. > >> > >> Now suppose we allow A to be any _continuous_ function of x, i.e. > >> steps, delta functions, and their permutations are > >> disallowed. > >> > >> Is there a way to determine which such function damps the system > >> out in the minimum time, _without_ allowing the function to > >> overshoot the final equilibrium point (i.e. no ringing)? > >> > > > > Sure, that's a fairly straightforward calculus of variations problem. > > > > Cheers > > > > Phil Hobbs > > > > -- > > Dr Philip C D Hobbs > > Principal Consultant > > ElectroOptical Innovations LLC > > Optics, Electro-optics, Photonics, Analog Electronics > > > > 160 North State Road #203 > > Briarcliff Manor NY 10510 > > > > hobbs at electrooptical dot net > > http://electrooptical.net > > > > Also IIRC non-conservative forces such as viscous damping require > something funky to be added to the Lagrangian... > -- > > > ----Android NewsGroup Reader---- > http://usenet.sinaapp.com/I think you want velocity damping (like you've got written.) Air drag goes as the velocity squared (over some region) and that gives a long tail. Friction damping (proportional to distance) will (in general) leave you with a non-zero amplitude when it stops.. x doesn't go all the way to zero. For the fastest I think you want it critically damped... more damping than that and you get a longer decay time. George H.

Reply by ●December 11, 20152015-12-11

Tim Wescott <seemywebsite@myfooter.really> Wrote in message:> On Fri, 11 Dec 2015 11:46:16 -0500, bitrex wrote: > >> Suppose we have the equation for a damped harmonic oscillator >> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the >> damping parameter. >> >> Now suppose we allow A to be any _continuous_ function of x, i.e. >> steps, delta functions, and their permutations are disallowed. >> >> Is there a way to determine which such function damps the system >> out in the minimum time, _without_ allowing the function to overshoot >> the final equilibrium point (i.e. no ringing)? > > Wait -- you're suggesting that rather than a linear system with A as a > constant, you want to use some A(x)? > > I suspect that it will be easy to find A(x) for a specific step input, > but that as soon as you change the size of the step you'll find that your > magical A(x) no longer works. > > -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com >Well, for example, if you use Wolfram Alpha to solve the critically damped harmonic oscillator equation x'' + 2x' + x = 0 for x(0) = 1, and the following equation with A(y) = sec(y), y'' + 2sec(y)y' + y = 0, y(0) it is clear from the phase portrait of x vs x' and y vs y' that the latter damps out quicker without overshoot. Essentially, if discontinus functions were allowed, the way you would get it to equilibrium is if you set up the equation right on the boundary of overshooting, and then slammed up an infinite potential "wall" when the oscillator hit the X axis. But you can't do that when only continuous functions are allowed, because no potential will grow rapidly enough to stop the oscillator from crossing the axis. So you have to have some kind of damping potential that is a function of x of the right variety that "engages" at the right point - too soon and the oscillator will damp much slower than the oscillator with linear damping, too late and it won't damp significantly faster. -- ----Android NewsGroup Reader---- http://usenet.sinaapp.com/

Reply by ●December 11, 20152015-12-11

On Friday, December 11, 2015 at 8:46:35 AM UTC-8, bitrex wrote:> Suppose we have the equation for a damped harmonic oscillator > driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the > damping parameter. > > Now suppose we allow A to be any _continuous_ function of x, i.e. > steps, delta functions, and their permutations are > disallowed. > > Is there a way to determine which such function damps the system > out in the minimum time, _without_ allowing the function to > overshoot the final equilibrium point (i.e. no ringing)?Well, you departed from physical reality once, in exciting the oscillator with a Heaviside 'function'. It is easy to imagine non-physical components (like the complex conjugate of the oscillator impedance) that rapidly remove energy, and the x(t) of the connection node for that 'load' would potentially be a good damping solution. The description 'damps the system out in the minimum time' , however, implies some measure of damping and of a time at which the damping is 'finished', neither of which is provided. So, do you minimize integral_0^infty (|x| *t )dt or integral_0^infty (x**2 *t )dt or t_decay such that x(t) .ge. 0 .and. x(t) .le. x(0)/100 whenever t > t_decay ? What mathematical definition of decay time do you intend?

Reply by ●December 11, 20152015-12-11

"Tim Wescott" wrote in message news:Uv6dnVJSLPQUj_bLnZ2dnUVZ5umdnZ2d@giganews.com...>Wait -- you're suggesting that rather than a linear system with A as a >constant, you want to use some A(x)? > >I suspect that it will be easy to find A(x) for a specific step input, >but that as soon as you change the size of the step you'll find that your >magical A(x) no longer works.For example, using a nonlinear capacitor in an R+C damper (like a too-low-voltage X7R), or a regenerative snubber circuit (meaning: a capacitor coupled into a full wave doubler rectifier, so that the energy is regenerated into a supply rail; [quasi- and] resonant snubbers can exhibit similar behavior), tends to result in a linear rather than exponential decay of the ringing envelope. The rationale being, the nonlinear behavior is able to remove a fixed amount of energy or delta V per cycle, rather than proportionally so. You have the gain of saving the energy (well, in the regenerative case anyway), but the loss of having an uglier transient. Tim -- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com

Reply by ●December 11, 20152015-12-11

whit3rd <whit3rd@gmail.com> Wrote in message:> On Friday, December 11, 2015 at 8:46:35 AM UTC-8, bitrex wrote: >> Suppose we have the equation for a damped harmonic oscillator >> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the >> damping parameter. >> >> Now suppose we allow A to be any _continuous_ function of x, i.e. >> steps, delta functions, and their permutations are >> disallowed. >> >> Is there a way to determine which such function damps the system >> out in the minimum time, _without_ allowing the function to >> overshoot the final equilibrium point (i.e. no ringing)? > > Well, you departed from physical reality once, in exciting the oscillator > with a Heaviside 'function'. > It is easy to imagine non-physical components (like the complex conjugate > of the oscillator impedance) that rapidly remove energy, > and the x(t) of the connection node for that 'load' would potentially > be a good damping solution. The description 'damps the system out in the > minimum time' , however, implies some measure of damping and of a time > at which the damping is 'finished', neither of which is provided. > So, do you minimize > integral_0^infty (|x| *t )dt > or > integral_0^infty (x**2 *t )dt > or > t_decay such that x(t) .ge. 0 .and. x(t) .le. x(0)/100 whenever t > t_decay ? > > What mathematical definition of decay time do you intend? >It might be wise to use the definition of "fall time" for overdamped systems, so 90% to 10% of final value? -- ----Android NewsGroup Reader---- http://usenet.sinaapp.com/

Reply by ●December 11, 20152015-12-11

On Friday, December 11, 2015 at 3:27:31 PM UTC-5, bitrex wrote:> Tim Wescott <seemywebsite@myfooter.really> Wrote in message: > > On Fri, 11 Dec 2015 11:46:16 -0500, bitrex wrote: > > > >> Suppose we have the equation for a damped harmonic oscillator > >> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the > >> damping parameter. > >> > >> Now suppose we allow A to be any _continuous_ function of x, i.e. > >> steps, delta functions, and their permutations are disallowed. > >> > >> Is there a way to determine which such function damps the system > >> out in the minimum time, _without_ allowing the function to overshoot > >> the final equilibrium point (i.e. no ringing)? > > > > Wait -- you're suggesting that rather than a linear system with A as a > > constant, you want to use some A(x)? > > > > I suspect that it will be easy to find A(x) for a specific step input, > > but that as soon as you change the size of the step you'll find that your > > magical A(x) no longer works. > > > > -- > > > > Tim Wescott > > Wescott Design Services > > http://www.wescottdesign.com > > > > Well, for example, if you use Wolfram Alpha to solve the > critically damped harmonic oscillator equation x'' + 2x' + x = 0 > for x(0) = 1, and the following equation with A(y) = sec(y), y'' > + 2sec(y)y' + y = 0, y(0) it is clear from the phase portrait of > x vs x' and y vs y' that the latter damps out quicker without > overshoot. > > > Essentially, if discontinus functions were allowed, the way you > would get it to equilibrium is if you set up the equation right > on the boundary of overshooting, and then slammed up an infinite > potential "wall" when the oscillator hit the X axis. > > But you can't do that when only continuous functions are allowed, > because no potential will grow rapidly enough to stop the > oscillator from crossing the axis. So you have to have some kind > of damping potential that is a function of x of the right variety > that "engages" at the right point - too soon and the oscillator > will damp much slower than the oscillator with linear damping, > too late and it won't damp significantly faster.Ahh, I kinda lost you in there. For me math is used to approximate reality.. you shouldn't get tied to tight to any one model. If you've got feedback and control (and smarts), then you can slam on the brakes as much as they'll go... and then let off easy as you come to your stopping point, like braking your car at a stop sign. I don't think I'd use a damped oscillator as a model for that. What are you doing, or thinking of doing? George H.> > -- > > > ----Android NewsGroup Reader---- > http://usenet.sinaapp.com/

Reply by ●December 11, 20152015-12-11

bitrex wrote:> > Suppose we have the equation for a damped harmonic oscillator > driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the > damping parameter. > > Now suppose we allow A to be any _continuous_ function of x, i.e. > steps, delta functions, and their permutations are > disallowed. > > Is there a way to determine which such function damps the system > out in the minimum time, _without_ allowing the function to > overshoot the final equilibrium point (i.e. no ringing)? >Time one "hit" anti-phase and proper amplitude,and the oscillations are dead.

Reply by ●December 11, 20152015-12-11

John Larkin wrote:> On Fri, 11 Dec 2015 12:08:47 -0500, Phil Hobbs > <pcdhSpamMeSenseless@electrooptical.net> wrote: > >> On 12/11/2015 11:46 AM, bitrex wrote: >>> >>> Suppose we have the equation for a damped harmonic oscillator >>> driven by a step input, e.g. x'' + Ax' + x = H(t), where A is the >>> damping parameter. >>> >>> Now suppose we allow A to be any _continuous_ function of x, i.e. >>> steps, delta functions, and their permutations are >>> disallowed. >>> >>> Is there a way to determine which such function damps the system >>> out in the minimum time, _without_ allowing the function to >>> overshoot the final equilibrium point (i.e. no ringing)? >>> >> >> Sure, that's a fairly straightforward calculus of variations problem. >> >> Cheers >> >> Phil Hobbs > > Given enough amplitude, can't you kill it dead in an arbitrarily small > time? > > >Abso-tuve-ally.