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5.1
a)
xDP(t) = hS(t) 2hS(t-tw) + hS(t-2tw) h xDP (s) = (1 - 2e-tws + e-2tws) s Response of a first-order process, K h -t s -2t s Y (s) = (1 - 2e w + e w ) s + 1 s or Y(s) = (1 - 2e-tw s + e-2tw s) 1 + 2 s s + 1 Kh Kh 1 = = Kh 2 = s + 1 s =0 s
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4.1
a) b) c) d)
iii iii v v
4.2
a) b) c)
5 10 10 s (10 s + 1) From the Final Value Theorem, y(t) = 10 when t Y (s) = y(t) = 10(1-e-t/10) , then y(10) = 6.32 = 63.2% of the final value.
d) e)
5 (1 - e - s ) (10s + 1) s From the Final Value Theor
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6.1
a)
By using MATLAB, the poles and zeros are: Zeros: (-1 +1i) , (-1 -1i) Poles: -4.3446 (-1.0834 +0.5853i) (-1.0834 0.5853i) (+0.7557 +0.5830i) (+0.7557 -0.5830i) These results are shown in Fig E6.1
Figure S6.1. Poles and zeros of G(s) plott