finals02

# finals02 - Sample Final Exam Covering Chapters 10-17...

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Sample Final Exam Covering Chapters 10-17 (finals02) 1 Sample Final Exam (finals02) Covering Chapters 10-17 of Fundamentals of Signals & Systems Problem 1 (30 marks) Consider the system depicted below used for discrete-time processing of continuous-time signals. The sampling period is 100 milliseconds ( 0.1 T = s). The discrete-time filter () d Hz is given by the following causal difference equation initially at rest: 12 0 [] [ 1 ] [ 2 ] [ 1 ] [ 2 ] yn ayn ayn bxn bxn bxn +− = (a) [8 marks] Find the controllable canonical state-space realization of the filter d , i.e., sketch the block diagram and give the state-space equations. Answer: The state equation: CT/DT Τ DT/CT Τ d x n xt c yt c d yn d H z + - z 1 un z 1 1 a 2 a 0 b 1 b 2 b

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Sample Final Exam Covering Chapters 10-17 (finals02) 2 N 11 22 1 2 [1 ] 0 1 [ ] 0 [] ] [ ] 1 B A xn un a a xn +  =+  +−  ±²³²´ . The output equation: [ ] N 011 0 0 D C yn b ab b ab xn bun =− + ±²²²³²²²´ (b) [10 marks] The filter () d Hz is designed to approximate the unity-gain, second-order, continuous- time, causal LTI filter 2 2 ,R e {} 1 32 Hs s ss => ++ . Find the values of the parameters 12012 {, , ,,} aabbb of d using the "c2d" transformation. Specify the ROC of d . Answer: The state equation for : N 01 0 23 1 B A xt ut −− µ µ ±²³ ²´ . The output equation for : [ ] 2 0 () C yt = ±³´ Diagonalize to compute matrix exponential: 1 1 10 1 1 , 02 1 2 AVA V V == = 1 1 2 2 2 1 2 2 1 0 12 0 1 1 0.9909 0.0861 1 2 0.1722 0.7326 2 3/2 1/2 2 T AT AT d T TT T T T T T T T T AT d e Ae V e V e ee e BA e I B − − = = −+ + = 2 2 2 2 0 1 1 2 1 3/ 2 1/ 2 0.0045 1 0 0.0861 21 TTTT d e CC + + = = Transfer function:
Sample Final Exam Covering Chapters 10-17 (finals02) 3 [] 1 1 () ( ) 0.9909 0.0861 0.0045 20 0.1722 0.7326 0.0861 0.0045( 0.7326) (0.0861)(0.0861) 1 ( 0.1722)(0.0045) (0.0861)( 0.9909) ( 0.9909)( 0.7326) 0.0148 0.009( 0.732 dn d d d zC z IABD z z z z zz z =− + −−  =   −+ = + = H 2 12 6) 2(0.0861)(0.0861) ( 0.9909)( 0.7326) 0.0148 0.009 0.0082 1.7235 0.7407 0.009 0.0082 1 1.7235 0.7407 z + + + = + = Hence, 01 2 1 2 0, 0.009, 0.0082, 1.7235, 0.7407 bb b a a == = = = (c) [5 marks] Sketch the pole-zero plot of d Hz . Compute the gain of d at the highest frequency. Answer : 2 0.009 0.0082 1.7235 0.7407 0.009( 0.915) , 0.9051 ( 0.8184)( 0.9051) d z z z + = + => . 1 1 Re{ z } Im{ z -1 0.905 0.818 0.915

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Sample Final Exam Covering Chapters 10-17 (finals02) 4 Gain at highest frequency: 0.009 0.0082 ( 1) 0.00023 1 1.7235 0.7407 d H −+ −= = ++ (d) [7 marks] Suppose that the continuous-time signal to be filtered is given by: ( ) cos(6 ) cos(4 ) c x ttt ππ =− . Sketch the spectra of the continuous-time and discrete-time versions of the input signal () c X j ω and j d X e . Sketch the spectra j d Ye of the discrete- time output signal and c Yj of the continuous-time output signal c yt . Finally, give an expression for the output signal c . Answer: We need to compute 0.4 0.4 2.7521 0.8 0.4 0.6 0.6 2.3717 1.2 0.6 0.009 0.0082 ( ) 0.0106 0.0044 0.0115 1.7235 0.7407 0.009 0.0082 ( ) 0.0032 0.0031 0.0045 1.7235 0.7407 j j j d jj j j j d e He j e ee e j e π + == + = + + = .
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## This note was uploaded on 09/16/2011 for the course ECSE 361 taught by Professor Franciscodgaliana during the Winter '09 term at McGill.

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finals02 - Sample Final Exam Covering Chapters 10-17...

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