ABSTRACT
Take the second-order term and break it into two terms in the form of the two Laplace transform identities given above.
! -~- s+! s2 + s + 1 - s (s + !i + (yi
Finally, the time response:
( ) _11 (J3 ) 1 _11 . (J3 ) y t = 1 - e 2 cos - · t - - e 2 sm - · t 2 J3 ' 2
There are several things to be learned from this example. First, it is a method that provides a way of obtaining the time response of systems containing complex conjugate roots. The method becomes more important when different inputs are combined and the standard step input is not the only one present. This leads to the second point to be made. The example used here falls into a very common class; one already examined as some length, the step response of a second-order system. If the goal of this section was to simply obtain the time response (without teaching a method applicable to more general systems), all we would have to calculate is the natural frequency and the damping ratio of the system and we would know the time response. Again, this is true in this case because the input is a step function and we can compare this response to a standard response and determine the generalized parameters.
