I am having difficulty calculating the inverse Laplace transform of the following transfer function: H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) where: eta = damping ratio Gdc = DC gain w = undamped natural frequency What steps do you use? I am having difficulty getting the transfer function into a form that will allow me to take advantage of Laplace Transform tables for easy inversion. J

# transfer function

Started by ●September 4, 2007

Reply by ●September 4, 20072007-09-04

On Sep 4, 7:48 pm, Jennifer Williams <jjenniferwwilli...@gmail.com> wrote:> I am having difficulty calculating the inverse Laplace transform of > the following transfer function: > > H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) > > where: > eta = damping ratio > Gdc = DC gain > w = undamped natural frequency > > What steps do you use? I am having difficulty getting the transfer > function into a form that will allow me to take advantage of Laplace > Transform tables for easy inversion. > > JA method that will work, but is tedious, is good old partial fraction expansion. Factor the denominator to find your poles, then do a partial fraction expansion to find the coefficients that correspond to the time-domain term associated with each pole. I think this method is usually taught when Laplace transforms are examined. Jason

Reply by ●September 4, 20072007-09-04

On Sep 4, 8:11 pm, cincy...@gmail.com wrote:> On Sep 4, 7:48 pm, Jennifer Williams <jjenniferwwilli...@gmail.com> > wrote: > > > I am having difficulty calculating the inverse Laplace transform of > > the following transfer function: > > > H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) > > > where: > > eta = damping ratio > > Gdc = DC gain > > w = undamped natural frequency > > > What steps do you use? I am having difficulty getting the transfer > > function into a form that will allow me to take advantage of Laplace > > Transform tables for easy inversion. > > > J > > A method that will work, but is tedious, is good old partial fraction > expansion. Factor the denominator to find your poles, then do a > partial fraction expansion to find the coefficients that correspond to > the time-domain term associated with each pole. I think this method is > usually taught when Laplace transforms are examined. > > JasonYes. For some worked out examples, see Wikipedia: http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms John

Reply by ●September 5, 20072007-09-05

On Sep 5, 11:48 am, Jennifer Williams <jjenniferwwilli...@gmail.com> wrote:> I am having difficulty calculating the inverse Laplace transform of > the following transfer function: > > H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) > > where: > eta = damping ratio > Gdc = DC gain > w = undamped natural frequency > > What steps do you use? I am having difficulty getting the transfer > function into a form that will allow me to take advantage of Laplace > Transform tables for easy inversion. > > JA pic would be nice Jenny. We don't get many females on this NG of nerd do wells.

Reply by ●September 5, 20072007-09-05

On Tue, 04 Sep 2007 23:48:43 +0000, Jennifer Williams wrote:> I am having difficulty calculating the inverse Laplace transform of > the following transfer function: > > H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) > > where: > eta = damping ratio > Gdc = DC gain > w = undamped natural frequency > > What steps do you use? I am having difficulty getting the transfer > function into a form that will allow me to take advantage of Laplace > Transform tables for easy inversion. > > JIf eta < 1 then that's the Laplace transform of a damped sinusoid. Look in your tables for the two transforms of damped sinusoids (there'll be one for e^(this * t) * cos (that * t), and another one for e^(this * t) * sin(that * t)). Find the values of this and that to make it match your expression, then find the linear combination of the damped sin and the damped cosine. You'll spend some time with a machete hacking through the underbrush in mathemagic land, but you'll find an answer if you're diligent. For hacking through _real_ underbrush with a real machete, you should keep a whetstone in your pocket. For this sort of hacking I suggest beer... -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

Reply by ●September 5, 20072007-09-05

gay.highlander@yahoo.co.uk wrote:> A pic would be nice Jenny. We don't get many females on this NG of > nerd do wells. >...and now you know why. -- Jim Thomas Principal Applications Engineer Bittware, Inc jthomas@bittware.com http://www.bittware.com (603) 226-0404 x536 People think it must be fun to be a super genius, but they don't realize how hard it is to put up with all the idiots in the world - Calvin

Reply by ●September 11, 20072007-09-11

On Sep 4, 9:49 pm, Tim Wescott <t...@seemywebsite.com> wrote:> On Tue, 04 Sep 2007 23:48:43 +0000, Jennifer Williams wrote: > > I am having difficulty calculating the inverse Laplace transform of > > the followingtransferfunction: > > > H(s) = (Gdc * w^2) / (s^2 + 2*eta*w*s + w^2) > > > where: > > eta = damping ratio > > Gdc = DC gain > > w = undamped natural frequency > > > What steps do you use? I am having difficulty getting thetransfer > >functioninto a form that will allow me to take advantage of Laplace > > Transform tables for easy inversion. > > > J > > If eta < 1 then that's the Laplace transform of a damped sinusoid. Look > in your tables for the two transforms of damped sinusoids (there'll be one > for e^(this * t) * cos (that * t), and another one for e^(this * t) * > sin(that * t)). Find the values of this and that to make it match your > expression, then find the linear combination of the damped sin and the > damped cosine.Thanks for the responses. I now understand. As for a pic... <img src="http://www.swarthmore.edu/NatSci/echeeve1/Class/e12/Lecture/ L17_StateSpace/html/MatlabForSS_04.png">> > You'll spend some time with a machete hacking through the underbrush in > mathemagic land, but you'll find an answer if you're diligent. For > hacking through _real_ underbrush with a real machete, you should keep a > whetstone in your pocket. For this sort of hacking I suggest beer... > > -- > Tim Wescott > Control systems and communications consultinghttp://www.wescottdesign.com > > Need to learn how to apply control theory in your embedded system? > "Applied Control Theory for Embedded Systems" by Tim Wescott > Elsevier/Newnes,http://www.wescottdesign.com/actfes/actfes.html

Reply by ●September 11, 20072007-09-11

Jennifer Williams wrote: ...> <img src="http://www.swarthmore.edu/NatSci/echeeve1/Class/e12/Lecture/L17_StateSpace/html/MatlabForSS_04.png">Note the slight increase in slope that happens right near the origin. That indicates second order, but barely. some coefficients are evidently quite small. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯