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PracticeProblemsMidterm_solu

# 0 s 0 1 1 1 1 g s h s d s s 1 g0h 0d0 since

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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 short-cut 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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