Unformatted text preview: he presence of t he t erm u (t) in every element of"e At . This is t he case because
t he limits of t he convolution integral r un from 0 t o t [Eq. (13.56)J. Thus (_~e4t + ~e9t)u(t) * ~u(t)
eAt * B f(t) =
[ ¥( _ e 4t + e 9t )u(t) * ~u(t)
lse 4t u(t) =[ i s(e 4t 9t
 e )u(t) * u (t) (~e4t  ~e9' ) U(t) * u (t) * u (t) + ~e9tu(t) * U(t)]
_~e4tu(t) * u (t) + ~e9tu(t) * u (t) 1 = C [eAtx(O) + e AtB * f(t)J + D f(t) (13.62b) Now recall t hat t he c onvolution o f f (t) w ith t he u nit i mpulse 8(t) y ields f (t). L et u s
define a j x j d iagonal m atrix 6(t) s uch t hat a ll i ts d iagonal t erms a re u nit i mpulse
functions. I t is t hen o bvious t hat 6(t) * f (t) = f (t) a nd E q. ( 13.62b) c an b e e xpressed a s + e AtB * f(t)J + D 6(t) * f (t)
= C eAtx(O) + [ CeAtB + D6(t)J * f (t) y (t) = C [eAtx(O) W ith t he n otation ~(t) for eAt, E q. ( 13.63b) m ay b e e xpressed a s ( 13.63a)
(13.63b) ;
810 13 S tateSpace Analysis + [Ct/>(t)B + m (t)] * f(t) Ct/>(t)x(O) y et) = ~ .. (13.63c) 811 T he r eader c an verify t hat t he t ransferfunction m atrix H (s) i n Eq. (13.42) is t he L aplace
t ransform o f t he u nitimpulse response m atrix h (t) in Eq. (13.66) . • J v z erostate r esponse z eroinput r esponse T he z erostate response; t hat is, t he response when x(O) = 0 , is
y et) = [Ct/>(t)B
= h (t) 13.4 Linear Transformation O f S tate Vector 1 3.4 + DS(t)] * ret) (13.64a) * ret) (13.64b) w here
h (t) = Ct/>(t)B + DS(t) (13.65) T he m atrix h (t) is a k x j m atrix known as t he i mpulse r esponse m atrix. T he
r eason for t his d esignation is obvious. T he i jth e lement of h (t) is h ;j(t), which
r epresents t he z erostate response Yi w hen t he i nput f j(t) = I i(t) a nd w hen all o ther
i nputs ( and all t he i nitial conditions) are zero. I t c an also b e seen from Eq. (13.39)
a nd (13.64b) t hat l inear Transformation o f S tate Vectors I n Sec. 13.1 we saw t hat t he s tate o f a system c an b e specified in several ways.
T he s ets o f all possible s tate variables m ust b e r elatedin o ther words, if we are
given one s et o f s tate variables, we should be able t o r elate i t t o a ny o ther s et. We
are p articularly i nterested in a linear t ype o f relationship. L et X l, X 2, . .. , X n a nd
W I, W 2, . .. , W n b e two different sets of s tate v ariables specifying t he s ame s ystem.
L et t hese s ets b e r elated by linear equations as
W I = P llXI + P l2 X 2 + . .. + P lnXn
(13.67a) L:[h(t)] = H (s)
• W n = P nlXI E xample 1 3.8 F or t he s ystem d escribed by Eqs. (13.40a) a nd ( 13.40b), d etermine eAt using Eq.
(13.59b):
t/>(t) = eAt = C 1+(s) :} T his p roblem was solved earlier w ith f requencydomain techniques. F rom E q. (13.41),
we have t/>(t) = £ 1 [ ~ ( 8+1)(8+2)
2 ( 8+1)(8+2)
2 = £ 1 1 8 +1 [ 8 11 8+2 + 8 !2 2t 2e  t  e
= [ _ 2e t + 2 e 2t (8+1~(8+2) ] Wn ( 8+1)(8+2) 'v'
w 1 1] 8 +1  1
0 +1 8+2 o 2t] a nd et e
_ e t + 2 e 2t P l2 P In Xl P 21 P 22 P 2n X2 (13.67b)
. ..................
P nl P n2 P nn
v p Xn ' 'v'
x w...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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