BME343_fall2009_HW2_ANSWER

BME343_fall2009_HW2_ANSWER - Page 1 BME 343: Biomedical...

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Unformatted text preview: Page 1 BME 343: Biomedical Engineering Signal and Systems Analysis Homework 2 Do the following problems from the textbook. The point value of each problem is given below and 1 bonus points will be given for following the rules. Problem 2.2 -7 2.2 -8 2.3 -4 2.4 -3 2.4 -5 2.4 -8 2.4 -14 2.4 -24 2.4 -32 2.4 -35 2.4 -39 2.6 -6 2.7 -2 Point Value 6 6 6 7 8 8 8 10 10 7 11 6 6 2.2-7 The characteristic polynomial is 2 ( 1)( 5 6). + + + The characteristic equation is 2 ( 1)( 5 6) + + + = or ( 1)( 2)( 3) 0. + + + = The characteristic roots are -1, -2 and -3. The characteristic modes are t e- , 2 t e- and 3 t e- . Therefore, 2 3 1 2 3 ( ) t t t y t c e c e c e--- = + + and ' 2 3 1 2 3 ( ) 2 3 t t t y t c e c e c e--- = --- ' 2 3 1 2 3 ( ) 4 9 t t t y t c e c e c e--- = + + Setting t=0, and substituting initial conditions yields 1 2 3 1 2 3 1 2 3 2 2 3 1 4 9 5 c c c c c c c c c + + =- -- = - + + = 1 2 3 6, 7, 3 c c c = = - = Therefore, 2 3 ( ) 6 7 3 t t t y t e e e--- =- + Page 2 BME 343: Biomedical Engineering Signal and Systems Analysis 2.2-8 The zero-input response for a LTIC system is given as ( ) 2 3. t y t e- = + Since two modes are visible, the system must have, at least, the characteristic roots 1 = and 2 1 = - . (a) No, it is not possible for the systems characteristic equation to be 1 + = since the required mode at = is missing. (b) Yes, it is possible for the systems characteristic equation to be 2 3( ) + = since this equation has the two required roots 1 = and 2 1 = - . (c) Yes, it is possible for the systems characteristic equation to be 2 ( 1) + = . This equation supports a general zero-input response of 1 2 3 ( ) t t y t c c e c te-- = + + . By letting 1 3 c = , 2 2 c = , and 3 c = , the observed zero-input response is possible. 2.3-4 The characteristic equation is 2 2 6 9 ( 3) 0. + + = + = Therefore, 3 1 2 ( ) ( ) t n y t c c t e- = + and ' 3 1 2 2 ( ) [ 3( ) ] t n y t c c t c e- = - + + Setting t = 0, and substituting ' (0) 0, (0) 1 n n y y = = , we obtain 1 1 2 3 1 c c c =- + = 1 2 0, 1 c c = = 3 ( ) t n y t te- = Hence, ' 3 ( ) ( ) [ ( ) ( )] ( ) 0 ( ) [(2 9) ( )] ( ) [2 ( ) 9 ( )] ( ) (2 3 ) ( ) n n n n t h t b t P D y t u t t D y t u t y t y t u t t e u t...
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This note was uploaded on 09/28/2009 for the course BME 343 taught by Professor Emelianov during the Fall '09 term at University of Texas at Austin.

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BME343_fall2009_HW2_ANSWER - Page 1 BME 343: Biomedical...

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