Fall 2001 16.318-7 State-Space Transformations e State space representations are not unique because we have a lot of freedom in choosing the state vector Selection of the state is quite arbitrary, and not that important In fact, given one model, we can transform it to another model that is equivalent in terms of its input-output properties To see this, define Model 1 of G(s)as i(t)=A.c(t)+ Bu(t y t)= Ca(t)+ Du(t) . Now introduce the new state vector 2 related to the first state a through the transformation m=Tz Tis an invertible(similarity)transform matrix T-(AT+ Bu T-(ATz+ Bu (T-AT)z+T-Bu= Az+ Bu an y=Ca+Du=CTz+ Du=Cz+ Du ● So the new model is Az+B +D Are these going to give the same transfer function? They must if these really are equivalent modelsFall 2001 16.31 8–7 State-Space Transformations • State space representations are not unique because we have a lot of freedom in choosing the state vector. – Selection of the state is quite arbitrary, and not that important. • In fact, given one model, we can transform it to another model that is equivalent in terms of its input-output properties. • To see this, define Model 1 of G(s) as x˙(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) • Now introduce the new state vector z related to the first state x through the transformation x = T z – T is an invertible (similarity) transform matrix z˙ = T −1 x˙ = T −1 (Ax + Bu) = T −1 (AT z + Bu) = (T −1 AT)z + T −1 Bu = Az¯ + Bu¯ and y = Cx + Du = CTz + Du = Cz¯ + Du¯ • So the new model is z˙ = Az¯ + Bu¯ y = Cz¯ + Du¯ • Are these going to give the same transfer function? They must if these really are equivalent models