finals02 - Sample Final Exam Covering Chapters 1-9...

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Sample Final Exam Covering Chapters 1-9 (finals02) 1 Sample Final Exam (finals00) Covering Chapters 1-9 of Fundamentals of Signals & Systems Problem 1 (20 marks) The unit step response of an LTI system was measured to be 3 () 2 s in ( ) () () 6 t st e t ut tut π =− + . (a) [10 marks] Find the transfer function Hs of the system. Specify its ROC. Sketch its pole-zero plot. Answer: 3 66 3 3 3 2 s ( 6 2( ) ( ) ( ) 2 31 22 ) ( ) ( ) 2 t jt t t t sSs s e t ee s e j je s e j s ππ −− −+  == +     =+  + = L L L L 3 33 2 11 3s i n c o s ( ) ( )( ) (3 ) 1 ( jt t jt tj t t t t tt j e e su t u t t u t j set et u t u t t u t s s s +− + + ++ L L 2 2 32 3 2 2 2 3) 1 [ ( 3 )1 ] (1 ) [ 2 34 ] ) [( 1] 1] (4 2 3) 4 (2 3 1) 2 3) 4 [( 1] [( s s s ss s s s s s −+ + + − + ROC: Re{ } 0 s >
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Sample Final Exam Covering Chapters 1-9 (finals02) 2 Poles are 1 2 3 0 3 3 p p j p j = =− + , Zeros are zeros of 2 0.21748 1.62331 ss +− : 1 2 3 1.3875, 1.1700 z z z = =∞ (b) [4 marks] Is the system causal? Is it stable? Justify your answers. Answer: System is causal: ROC is an open RHP and transfer function is rational . This system isn't stable as ROC doesn't include the imaginary axis (or because rightmost pole 0 has a nonnegative real part.) (c) [6 marks] Give the direct form realization (block diagram) of H s () . Answer: 2 32 (2 3 1) (4 2 3) 4 23 4 Hs s −+ = ++ Re{s} Im{s} -1.7 -1.39 1.17 -1 1 + - W s 1 s X s 1 s sW s sW s 2 4 Y s 1 s 3 0 4 23 1 423
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Sample Final Exam Covering Chapters 1-9 (finals02) 3 Problem 2 (20 marks) The following circuit has initial conditions on the capacitor v C () 0 and inductor i L 0 . (a) [4 marks] Transform the circuit using the unilateral Laplace transform. Answer: (b) [8 marks] Find the unilateral Laplace transform of vt . Answer: Let's use mesh analysis. For mesh 1: VI I I IV I sc s Cs s s vR s s s R s Rs v RCs s ( ) [ () ] ( ) ( ) () −−−− = =− + + + 11 00 01 1 1 2 2 1 1 For mesh 2: sR L ss L i s R RRL s s L R i L L 2 2 2 1 1 12 2 1 1 0 [( ) ( ) ]( )( ) ( II I −− + + = =+ + Substituting, we obtain I 2 1 1 2 1 1 2 2 2 1 1 1 1 2 1 2 2 1 1 0 10 1 0 ) ( [ ( )( )] ( ) ( ) ( ) ( ) ( ) [ ( ) ] () s R s v RCs R L R i R Cs R Cs R R Ls s R Cs s R Cv R Cs Li LRCs L RRC s R R s s RCv R L L + + + L N M O Q P −+ = + + + + = + + 1 0 Cs Li L ) Solving for I 2 s , we get C R 1 R 2 + - L + - s + - + - + + I 1 s V s V s s 1 Cs Ls I 2 s R 1 Li L 0 1 0 s v c R 2
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Sample Final Exam Covering Chapters 1-9 (finals02) 4 I V 2 1 1 2 12 1 2 11 1 2 1 2 10 0 () ( )( ) s RCs s LR Cs L R R C s R R RCs Li RCv LR Cs L R R C s R R sL c = ++ + +− −− And finally the output voltage is 22 1 2 2 1 2 2 1 2 2 2 1 1 2 1 2 2 2 1 2 ( 1 ) (0) ( ) {(1 ) [ ( ) ( )]} (0 c L L s L LR Cs s R Cs L si R CLsv ssL i Li LR Cs L R R C s R R LR Cs L R R C s R R LR Cs s R Cs L s L R Cs L L R R C s L R R i LR Cs L R R C s R R =−= + +−+ + + + =+ V VI V 1 2 2 1 2 2 2 1 2 1 2 1 211 2 1 2 0 ) [ ( ) ] (0) c c RCLsv LR Cs L R R C s R R LR Cs s LR R Cs L R R i R CLsv LR Cs L R R C s R R LR Cs L R R C s R R −+ + V (c) [8 marks]
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finals02 - Sample Final Exam Covering Chapters 1-9...

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