Unformatted text preview: he transfer function from w to x is
X
1
=
.
W
1 + G( s ) H ( s ) D ( s )
By the ﬁnal value theorem, the steady state value of x(t) is
x(1) = lim sX (s) = lim s ·
s! 0 s! 0 1
1
1
·=
.
1 + G( s ) H ( s ) D ( s ) s
1 + G(0)H (0)D(0) Since D(0) = 1, write D(s) = Do (s)/s with Do (s) = a + bs. Then
s
0
0
=
= = 0.
s! 0 s + G ( s ) H ( s ) D o ( s )
0 + G(0)H (0)Do (0)
a x(1) = lim sX (s) = lim
s! 0 Similarly,
u(1) = lim sU (s) = lim s · G( s ) H ( s ) D o ( s )
G( s ) H ( s ) D ( s ) 1
· = lim
=
1 + G ( s ) H ( s ) D ( s ) s s! 0 s + G ( s ) H ( s ) D o ( s ) y (1) = lim sY (s) = lim s · G( s )
1
sG(s)
· = lim
= 0.
1 + G ( s ) H ( s ) D ( s ) s s! 0 s + G ( s ) H ( s ) D o ( s ) s! 0 s! 0 s! 0 s! 0 1 Hence the answer is B, C A shortcut to the answer goes like this. Note that D(s) has an integrator.
Hence, if the input to D(s) is nonzero in the steady state, then u would diverge to inﬁnity, violating the
stability assumption. So the input to D(s), which is y (1), must be zero. Since y (1) = 0, the static
gain relationship gives x(1) = G(0)y (1) = 0. The basic relation x = w + u then gives u(1) = 1
since w(1) = 1 fo...
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 Winter '09
 IDAN
 Steady State

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