Exam Solutions 81-90

# Exam Solutions 81-90 - And therefore K = bracketleftbig 0 1...

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Unformatted text preview: And therefore K = bracketleftbig 0 1 bracketrightbig 1 3 bracketleftbigg 3- 1- 1 bracketrightbiggbracketleftbigg 3 9 12 bracketrightbigg = bracketleftbig- 3- 4 bracketrightbig Check: ( det [ sI- F + GK ]) = det parenleftbigg sI- bracketleftbigg 2 4- 3- 4 bracketrightbiggparenrightbigg = s 2 + 2 s + 4 2. Calculate the estimator for the system. The poles of the dynamics of the estimation error should be chosen in λ 1 =- 2 , λ 2 =- 4 . Due to the asked error dynamics, the estimator characteristic polynomial should be: α e ( s ) = ( s + 2)( s + 4) = s 2 + 6 s + 8 The observer gains can be simply directly calculated with Ackermann’s estimator formula. L = α e ( F ) O- 1 bracketleftbigg 1 bracketrightbigg We compute: O = bracketleftbigg H HF bracketrightbigg = bracketleftbigg 1- 3- 4 bracketrightbigg ⇒ O- 1 = 1 3 bracketleftbigg- 4- 1 3 bracketrightbigg and α e ( F ) = F 2 +6 F +8 I = bracketleftbigg- 1- 3- 4 bracketrightbiggbracketleftbigg- 1- 3- 4 bracketrightbigg +6 bracketleftbigg- 1- 3- 4 bracketrightbigg +8 bracketleftbigg 1 0 0 1 bracketrightbigg = bracketleftbigg 3- 3 0 bracketrightbigg And therefore L = bracketleftbigg 3- 3 0 bracketrightbigg 1 3 bracketleftbigg- 4- 1 3 bracketrightbiggbracketleftbigg 1 bracketrightbigg = bracketleftbigg- 1 1 bracketrightbigg Check: ( det [ sI- F + LH ]) = det parenleftbigg sI- bracketleftbigg- 1 1- 3- 5 bracketrightbiggparenrightbigg = s 2 + 6 s + 8 Question 5 The system G ( s ) has a transfer function: G ( s ) = s + τ ( s + 2) where the parameter τ is not exactly known, but bounded by 1 ≤ τ ≤ 2 This system can be described as a nominal system G ( s ) = ( s + 1) ( s + 2) with multiplicative uncertainty l ( s ) . 1. Describe the unstructured multiplicative uncertainty l ( s ) as a function of the uncertain τ . G ( s ) = s + τ ( s + 2) G ( s ) = ( s + 1) ( s + 2) From G ( s ) = G ( s )(1 + l ( s )) it follows l ( s ) = G ( s ) /G ( s )- 1 = s + τ ( s + 2) ( s + 2) ( s + 1)- 1 = τ- 1 ( s + 1) 2. Show that | l ( jω ) | ≤ 1 for all frequencies. Find maximum max ω | l ( jω ) | = max ω τ- 1 √ 1 + ω 2 = τ- 1 √ 1 + 0 2 = τ- 1 For all 1 ≤ τ ≤ 2 we find | l ( jω ) | ≤ 1 3. Prove that the controller D ( s ) = 1 s + 1 robustly stabilizes the system. To prove that the controller D ( s ) robustly stabilizes the system, we have to prove that | T- 1 ( jω ) | > | l ( jω ) | for all ω Compute T ( s ) for G ( s ) = ( s +1) ( s +2) and D ( s ) = 1 s +1 : T ( s ) = G D 1 + G D = ( s +1) ( s +2) 1 s +1 1 + (...
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Exam Solutions 81-90 - And therefore K = bracketleftbig 0 1...

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