lecture 15 - 1 1 Summary of Lecture #15 9/27/2010 •...

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Unformatted text preview: 1 1 Summary of Lecture #15 9/27/2010 • Stable, unstable and metastable systems Examples • Classification of responses Complete response = ZIR+ZSR = TSR + SSR = NTR + FCR • Principle of superposition See me before 10/01 if your Test#1 score is less than 30. 2 • All poles are in the open left half plane ⇒ stable system or circuit • First order poles on j ϖ-axis ⇒ metastable system or circuit • 2nd and higher order poles on j ϖ-axis ⇒ unstable system or circuit • Any pole in open right half plane ⇒ unstable system or circuit Left half plane ( LHP ) 4 Ex. 2 Cs 1 ) s ( I ) s ( V ) s ( H in out = = Single pole at origin Metastable system i in (t) But if i in (t) = 16 u(t) V. Then in out 2 out 16 16 1 16 I (s) A, V (s) V s s s s v (t) 16 u(t) V, t = = ⋅ = = unbounded output Take C =1 F and i in (t) = 16 δ (t) A . Then in out out 1 16 I (s) 16 A, V (s) 16 , V s s v (t) 16u(t) V, = = ⋅ = = bounded output 5 Ex. 3 Double pole at origin Unstable system 2 2 1 in out s 1 C C A ) s ( I ) s ( V ) s ( H = = Take i in (t) = 16 δ (t) A , then 1 1 out out 2 1 2 1 2 16A 16A 1 V (s) V, v (t) u(t) V C C s C C t = = Take i in (t) = 16 u(t) A, then 1 1 out out 3 1 2 2 1 2 16A 8A 1 V (s) V, v (t) u(t) V C t C s C C = = Unbounded output 6 ) LC / 1 ( s 1 LC 1 ) s ( V ) s ( V ) s ( H 2 in out + = = Simple complex poles on j ϖ-axis Metastable system Ex. 4 Take L = ½ H, C= ½ F, v in (t)= 4 u(t) V, then out 2 2 out 16 4 s V (s) 4 s(s 4) s s 4 v (t) 4[1...
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lecture 15 - 1 1 Summary of Lecture #15 9/27/2010 •...

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