{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

finals02

# finals02 - Sample Final Exam Covering Chapters...

This preview shows pages 1–4. Sign up to view the full content.

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 H z is given by the following causal difference equation initially at rest: 1 2 0 1 2 [ ] [ 1] [ 2] [ ] [ 1] [ 2] y n a y n a y n b x n b x n b x n + + = + + (a) [8 marks] Find the controllable canonical state-space realization of the filter ( ) d H z , 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 x t c ( ) y t c ( ) [ ] d y n ( ) d H z + + + + - - + + [ ] y n z 1 [ ] u n z 1 1 a 2 a 0 b 1 b 2 b

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Sample Final Exam Covering Chapters 10-17 (finals02) 2 N 1 1 2 2 1 2 [ 1] 0 1 [ ] 0 [ ] [ 1] [ ] 1 B A x n x n u n x n a a x n +     = +     +     ±²³²´ . The output equation: [ ] N 2 2 0 1 1 0 0 [ ] [ ] [ ] D C y n b a b b a b x n b u n = + ±²²²³²²²´ (b) [10 marks] The filter ( ) d H z is designed to approximate the unity-gain, second-order, continuous- time, causal LTI filter 2 2 ( ) , Re{ } 1 3 2 H s s s s = > − + + . Find the values of the parameters 1 2 0 1 2 { , , , , } a a b b b of ( ) d H z using the "c2d" transformation. Specify the ROC of ( ) d H z . Answer: The state equation for ( ) H s : N 1 1 2 2 ( ) ( ) 0 1 0 ( ) ( ) ( ) 2 3 1 B A x t x t u t x t x t   = +     µ µ ±²³²´ . The output equation for ( ) H s : [ ] ( ) 2 0 ( ) C y t x t = ±³´ Diagonalize to compute matrix exponential: 1 1 1 0 1 1 , 0 2 1 2 A V AV V = = = 1 1 2 2 2 2 2 2 2 1 2 1 1 2 1 0 1 2 1 1 0 1 1 0.9909 0.0861 2 2 1 2 0.1722 0.7326 2 2 2 3/ 2 1/ 2 2 1 0 T A T AT d T T T T T T T T T T T T T AT d e A e Ve V e e e e e e e e e e e e e e B A e I B = = = = = = + + = = 2 2 2 2 2 2 2 2 0 1 1 2 2 2 1 1 1 3/ 2 1/ 2 0.0045 2 2 1 0 0.0861 2 1 T T T T T T T T T T T T T T T T d e e e e e e e e e e e e e e e C C      + +   + + = = = + = Transfer function:
Sample Final Exam Covering Chapters 10-17 (finals02) 3 [ ] [ ] 1 1 ( ) ( ) 0.9909 0.0861 0.0045 2 0 0.1722 0.7326 0.0861 0.0045( 0.7326) (0.0861)(0.0861) 1 2 0 ( 0.1722)(0.0045) (0.0861)( 0.9909) ( 0.9909)( 0.7326) 0.0148 0.009( 0.732 d n d d d z C zI A B D z z z z z z z = + = + = + + = H 2 1 2 1 2 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 z z z z z z z z + + + = + + = + Hence, 0 1 2 1 2 0, 0.009, 0.0082, 1.7235, 0.7407 b b b a a = = = = − = (c) [5 marks] Sketch the pole-zero plot of ( ) d H z . Compute the gain of

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}