Laplace Example 2.12 2001/01/29 2002/10/24 Elizabeth Chan lizychan@rice.edu Thanos Antoulas aca@rice.edu John Paul Slavinsky jps@alumni.rice.edu Elizabeth Chan lizychan@rice.edu Thanos Antoulas aca@rice.edu John Paul Slavinsky jps@alumni.rice.edu laplace homogeneous forced Example of a Laplace system

RLC circuit RLC circuit

y 0- -1 t1 y 0- 2 t2 y 0- -4 Find the step response for the system above, when ut is the input and yt is the output (i.e. find yt for u t step t ). Y s Admittance U s Admittance 1 Impedance 1 1 s 1 s 1 1 2 s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2 Y s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2 U s With our previous definition of q d ⅆ t y t p d ⅆ t u t we can define the Laplace domain equivalents of q and p as: q s s 3 2 s 2 s 2 p s 2 s 3 3 s 2 4 s 2 When we multiply qs times Ys, we have to remember to include terms relating to the initial conditions of yt. We normally think of the Laplace transform of tyt as sYs . However, in reality, the general transform is as follows: ℒ tn y t s n Y s s n 1 y 0- s n 2 t1 y 0- … s tn2 y 0- tn1 y 0- Therefore, using the initial conditions stated above, we can find the Laplace transforms of the first three derivatives of yt. ℒ t3 y t s 3 Y s s 2 2 s 4 ℒ t2 y t s 2 Y s s 2 ℒ t1 y t s Y s 1 We can now get a complete s-domain equation relating the output to the input by taking the Laplace transform of q d ⅆ t y t p d ⅆ t u t . The transform of the right-hand side of this equation is simple as the initial conditions of yt do not come into play here. The result is just the product of ps and the transform of the step function 1s. The left-hand side is somewhat more complicated because we have to make certain that the initial conditions are accounted for. To accomplish this, we take a linear combination of Laplace transform of the third derivative of y(t), Laplace transform of the second derivative of y(t), and Laplace transform of the first derivative of y(t) according to the polynomial qs. That is to say, we use the coefficients of the s terms in qs to determine how to combine these three equations. We take 1 of Laplace transform of the third derivative of y(t) plus 2 of Laplace transform of the second derivative of y(t) plus 1 of Laplace transform of the first derivative of y(t) plus 2. When we sum these components, collect the Ys terms, and set it equal to the right-hand side, we have: s 3 2 s 2 s 2 Y s s 2 1 2 s 3 3 s 2 4 s 2 1 s Rearranging, we can find the solution to Ys: Y s 2 s 3 3 s 2 4 s 2 s 3 2 s 2 s 2 1 s s 2 1 s 3 2 s 2 s 2 This solution can be looked at in two parts. The first term on the right-hand side is the particular (or forced) solution. You can see how it depends on ps and us. The second term is the homogeneous (or natural) solution. The numerator of this term describes how the initial conditions of the system affect the solution (recall that s 2 1 was the part of the result of the linear combination of Laplace transform of the third derivative of y(t), Laplace transform of the second derivative of y(t), Laplace transform of the first derivative of y(t)). The denominator of the second term is the qs polynomial; it serves to describe the system in general.