Hi everyone I need help with how to transform T(s)=(omega^2)/((s(s+a))+(omega^2)). Because I cant seem to find a similar equation in the z transform table. Thank you

# z transform

Started by ●May 4, 2011

Reply by ●May 4, 20112011-05-04

On 05/04/2011 05:08 PM, saras3083 wrote:> Hi everyone > > I need help with how to transform T(s)=(omega^2)/((s(s+a))+(omega^2)). > Because I cant seem to find a similar equation in the z transform table.That is an expression in the Laplace domain, which describes a signal or a system's behavior in continuous time. The z domain describes signals and system behavior in discrete time. You cannot exactly express a continuous-time signal or a continuous-time system with sampled-time signals or system descriptions. Thus, there is no exact* transform between the Laplace domain and the z domain. I assume that you're showing a system transfer function. There _are_ ways to make sampled-time systems that approximate the behavior of a specified continuous-time system. There are exact ways to model how the behavior of a continuous-time system appears as 'seen' from a sampled time system. Look up the "bilinear transform" for one of the many flavors of the former -- I'm not sure what the name is of the latter. * Well, sorta kinda no exact transform -- if you know enough sampled-time signal processing theory to argue with me, you don't need to read any of the above. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html

Reply by ●May 11, 20112011-05-11

>I need help with how to transform T(s)=(omega^2)/((s(s+a))+(omega^2)). >Because I cant seem to find a similar equation in the z transform table.You can use the bilinear transform by setting s = 2/T * (1 - z^-1)/(1 + z^-1) then cranking to get the numerator and denominator into powers of z^-1. T is your sample period, of course. Mark