THe following equation has a polynomial input 0125 xdoubledo

THe following equation has a polynomial input:

0.125 (xdoubledot) + 0.75(xdot) + x = y(t) = (-27/800)t^3 + (270/800)t^2

use simulink to plot x(t) and y(t) on the same graph. The inital condtions are zero.

Solution

Ans-

.Thus, we do not distinguish between polynomials of degree less or equal to three and 2by 2 matrices.With respect to the above equivalence arguments, we should probably clarify the notionof composition of two maps and an inverse map.ALGEBRA OF LIENAR TRANSFORMATIONS, THE SPACE HOM(V,W)Definition: (Composition of Linear Maps)Suppose we have L :V ® W and T :W ® U . Then the composition is defined as:(T o L)(v) = T (L(v))Example:Let us find out what happens when we compose two reflectionsLet L æ x ö = æ x ö be the reflection about x axis and T æ x ö = æ -x ö be the reflection about ç y ÷ ç -y ÷ ç y ÷ ç y ÷ è ø è ø è ø è øthe y axis. Then:(L o T )(u) = L æ æ x ö ö = L æ -x ö = æ -x ö çT ç y ÷ ÷ ç y ÷ ç -y ÷ è è ø ø è ø è øThe matrix representations are:[ L] = æ 1 0ö ç 0 -1ø÷ è[T ] = æ -1 0ö ç 0 ÷ è 1 ø[ L oT ] = æ -1 0ö = æ cosp - sinp ö ç 0 -1÷ø ç sin p ÷ è è cosp øNotice that the composition corresponds to the rotation by q = p .Notice also that the matrix for the composition is the product of the matrices for eachoperator:

two linear maps is the product of theindividual matrix representations.[L oT ] = [L]×[T ]Proof :L (T (v)) = [L]×T (v) = [L]×[T ]×vThe above proof works for operators, but it could easily be adjusted for more generaltransformations.Example: (Verification Example)Consider the following operators on P2 :( )L at2 + bt + c = 2at + bT (at2 + bt + c) = 2at2 + btThen we have:(L oT )(at2 + bt + c) = L(2at2 + bt) = 2(2a)t + b = 4at + bWe have the following matrix representation with respect to the standard ordered basis:(Please, verify) æ0 0 0ö ç ÷[ L] = ççè 2 0 0 ÷ø÷ 0 1 0 æ2 0 0ö ç ÷[T ] = ç 0 1 0 ÷ çè 0 0 0÷ø æ2 0 0ö ç ÷[ L oT ] = çèç 0 1 0 ÷ø÷ 0 0 0And notice that [L oT ] = [L]×[T ]As an exercise, do the same for T o L and notice that operator composition is notcommutative (which is not a surprise at this point)