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Unformatted text preview: 1 and 2 are, ( ) y t will contain for sure the following terms: 1 1 2 1 2 ( ) ... ... 1 t y t e s s = + + = + + + L g ) If ( ) M U s s = then: ( ) ( ) 2 2 1 1 1 1 2 1 1 ( ) ( ) ( ) 1 1 K K K s MK s K K M Y s G s U s s s s s s + + + + = = = + + Proceeding as before, a partial fraction expansion of ( ) Y s would give: 3 1 2 2 ( ) 1 Y s s s s = + + + Therefore ( ) y t will contain the following terms: 1 3 1 2 1 2 3 2 ( ) 1 t y t t e s s s = + + = + ⋅ + + L Since 2 t ⋅ is not bounded, ( ) y t is not bounded for a step change. Note that the term 2 1 s is an integrating element....
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 Spring '09
 RafiqulGani
 partial fraction expansion, overall transfer function, K1 K1

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