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hw2soln

Course: ECE 202, Spring 2012
School: Purdue
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Word Count: 192

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Spring ECE202: 2012 HW 2 Solutions page 1 Problem 5: a) = 15 = 15 = + = + | + | = 15 + + + + = + = 15 + + = 15 = + b) = = + , , , + + , = + +1 = , = + , =1 + = | 2 + = + + , = = = + = +1 = | | + = . . + + + +1 = , = = + + + = =0 + = + . + = + = , , . ECE202: Spring 2012 = = 2 = lim = =...

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Spring ECE202: 2012 HW 2 Solutions page 1 Problem 5: a) = 15 = 15 = + = + | + | = 15 + + + + = + = 15 + + = 15 = + b) = = + , , , + + , = + +1 = , = + , =1 + = | 2 + = + + , = = = + = +1 = | | + = . . + + + +1 = , = = + + + = =0 + = + . + = + = , , . ECE202: Spring 2012 = = 2 = lim = = + | HW 2 Solutions = 1 +2 | 12 =2 + + =2 =2 2 | = 2 +4 8 2 = 4 =2 +2 page 2 2 +4 8 = 2 +2 + + + 2 + +2 =1 = 1 1 1 2 +2 + MATLAB Code: 1 B=[2 -1 -2 4 -8]; % coefficients of the polynomial (numerator) A=[1 0 -4 0 0]; % coefficients of the polynomial (denominator) [R,P,K] = RESIDUE(B,A) % function which calculates the partial fraction expansion parameters R= 1.0000 % corresponds to (s-2) -1.0000 % corresponds to (s+2) -1.0000 % corresponds to (s) 2.0000 % corresponds to (s^2) P= 2.0000 -2.0000 0 0 % % % % corresponds corresponds corresponds corresponds to to to to (s-2) (s+2) (s) (s^2) K= 2 % corresponds to Lim F(s), as s->inf Therefore, the MATLAB code verifies our answers. ECE202: Spring 2012 HW 2 Solutions page 3 e 6 + 17 + 115 = + 5 + 33 + 29 = = = +1 +1 | 6 + 17 + 115 = + +1 + 4 + 29 +1 +2 + + 2 + 25 + +2 +2 + + 2 + 25 + 25 =4 To find A and B, we should generate two equations to get A and B (evaluate F(s=-2) and F(s=0) then you can calculate A,B). 4 2 5 4 + = + +1 + +2 2 + 25 +1 = =4 = 2, A=2, B=-5 +2 = 1, = 5, cos 5 = 2, 15 = 5, 1 2 +2 + 2 + 25 sin 5 5 + 2 + 25 = 4. Problem 6: a) = 120 , = = 1 2 = 140 + 1 4 = 240, + || + 10 = 140 : || = 240 + + || || + 12 + 8 24 4 + +2 +4 4 +2 = 24 2 sin 2 8 +2 : 2 + + 1 + 24 + cos 2 + 140 8 +4 +4 4 +4 +2 + 140 ECE202: Spring 2012 HW 2 Solutions page 4 b) = 90 , = = 3 5 = 12 1 5 = 30, + + 12 , + = 12 : + + +1 18 1 18 : + +2 1 +4 +1 18 = = 30 1 cos 3 + 12 = +4 18 1 sin 3 1 3 1 + sin 2 6 Problem 7: a) 12 , + 16 = = b) : 12 + 2 + 2 + 16 : = = = 12 + + 16 + 2 12 + 16 = 12 192 + 16 1 1 2 +1 + 3 +1 +4 2 + cos 2 3 ECE202: Spring 2012 : = : = = Now put Rule used = 0, = = + page 5 = + 12 192 + 16 = = = = : 0 HW 2 Solutions = 12 + 2 192 + 2 + 16 + .This shows that M(s)=H(s+a) = = 0 = + 12 +2 + 2 + 16 e) = First of all we find Inverse Laplace of = = +2 + = = = , + (by the frequency shift property) : 0 ECE202: Spring 2012 HW 2 Solutions : a) i) = = 1 + 1 + = = + = 4 +2 = + = 2 0 + 1 +0= + b) i) = = +1 + 1 1 ++ +2 2 +2 +1 + 2 page 6 ECE202: Spring 2012 = = HW 2 Solutions 1 + 1 = 1 + 1 1 1 1= + = 1 + + 1 = +1+ 1 + = + 0 page 7 + + 1 3 2 1 0
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Purdue - ECE - 202
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