Homework LCS2 solution PDF

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Solution Guide for the Home Work 2, Linear Control System...

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Homework 2 Solutions

1 a Cayley-Hamilton’s Theorem states that An + an−1 An−1 + . . . + a1 A + a0 I = 0 ⇒ − a0 A−1 = An−1 + an−1 An−2 + . . . + a1 I. In other words, if A−1 exists, we have A−1 = −

1 (An−1 + an1 An−2 + . . . + a1 I). a0

b The characteristic polynomial is det(λI − A) = λ2 − (a + d)λ + ad − bc ⇒ a1 = −a − d, a0 = ad − bc, and hence

1 1 (A + a1 I) = ad − bc a0



et 6 0 = P −1 6 4 0 0

3 0 0 7 7 P, te4t 5 e4t

A−1 = −

d −c

−b a



.

2 From the book, we know that

eAt

2

0 e2t 0 0

0 0 e4t 0

where P is given in the homework set.

3 This is a linear, time-varying system. However, we note that   cos(t) cos(τ ) + tτ cos(t)τ + t cos(t) A(t)A(τ ) = = A(τ )A(t) t cos(τ ) + τ cos(t) tτ + cos(t) cos(τ ) and as such (t, t0 ) = e Φ Moreover

Z

Rt

t0

t

cos(s)ds = sin(t0 ) − sin(t), t0

1

A(s)ds

Z

.

t

sds = t0

1 2 (t − t02). 2

As such

2

x(t) = e

4

sin(t0 ) − sin(t) 1 2 (t − t02 ) 2

1 2 (t 2

− t02 ) sin(t0 ) − sin(t)

3 5

x0 ,

which is as far as this expression can be simplified.

4 First we note that the A matrix is already a Jordan block, i.e.,   −δ e δe−δ Aδ ˆ . A=e = 0 e−δ Moreover, Bˆ =

Z

0

δ

eA(δ−s) Bds =

Z δ 0

e−(δ−s) 0



ds =



1 − e−δ 0



.

5 When plotting and comparing the two solutions, we see (a) that they are indeed quite similar even for large δ. In fact, we need to push δ close to 1 to get any real difference between the exact and the approximate solutions, and (b) as δ grows, so does the difference between the exact and the approximate solutions. Moral of the story: Pick δ small and approximate :-)

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