solutions_final_section_001_f05

solutions_final_section_001_f05 - ECE 220, Section 001...

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ECE 220, Section 001 Final Exam Solutions 13 December 2005 Problem 1. (10 points) Consider the pulse, p ( t ), in Equation 1: p ( t ) = 1 , 0 t < 10 , 0 , otherwise. (1) Express the signal s ( t ) in Equation 2 in terms of p ( t ). s ( t ) = 0 , t < 0 , - 2 , 0 t < 5 , 1 , 5 t < 15 , 0 , t 15. (2) Solution The signal contains two pulses, one of width 5-0 = 5, and another of width 15-5 =10. The scaled signal s 1 ( t ) = p (2 t ) = 1 , 0 2 t < 10 , 0 , otherwise. = 1 , 0 t < 5 , 0 , otherwise, has a width of five. The signal s 2 ( t ) = p ( t - 5) = 1 , 0 t - 5 < 10 , 0 , otherwise. = 1 , 5 t < 15 , 0 , otherwise, has a width of 10. With these two signals we can write s ( t ) = - 2 s 1 ( t ) + s 2 ( t ), so finally, s ( t ) = - 2 p (2 t ) + p ( t - 5) 1
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Problem 2. (10 points) Consider the complex-valued function z ( t ) in Equation 3; t is a real number: z ( t ) = 1 + j + e - j π 16 · e - j 6 . (3) Determine the period, T , of this function. Clearly explain your answer. Guesses won’t count. Solution Let T denote the period of this function; T satisfies the property z ( t + T ) = z ( t ) , t ( -∞ , ) . (4) We have z ( t + T ) = 1 + j + e - j π 16 · e - j ( t + T ) π 6 = 1 + j + e - j π 16 · e - j 6 e - j 6 = z ( t ) e - j 6 (5) From equations 4 and 5, we see that T must be the smallest positive number that satisfies the property: e - j 6 = 1 . (6) Since 1 = e - j 2 π , we can write e - j 6 = e - j 2 π and thus 6 = 2 π So, finally, T = 12 2
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Problem 3. (10 points) Find all of the solutions to the equation z 3 - 27 j = 0 (7) where z is a complex number. Express the solutions in exponential format. Solution Express j in exponential format as j = e jπ/ 2 Using the fact that e j 2 πk = 1 for any integer k , rewrite equation 7 as z 3 = 27 j = 27 e jπ/ 2 = 27 e jπ/ 2 e j 2 πk From this last equation, we can write (after taking the third root): z = 27 1 / 3 e jπ/ 2 · 1 / 3 e j 2 πk · 1 / 3 = 3 e jπ/ 6 e j 2 πk/ 3 (8) Let now k = 0 , 1 , 2 in equation 8: z 1 = 3 e jπ/ 6 e j 2 π 0 / 3 = 3 e jπ/ 6 z 2 = 3 e jπ/ 6 e j 2 π 1 / 3 = 3 e j 5 π/ 6 z 3 = 3 e jπ/ 6 e j 2 π 2 / 3 = 3 e j 9 π/ 6 So, finally, z 1 = 3 e jπ/ 6 , z 2 = 3 e j 5 π/ 6 , z 3 = 3 e j 9 π/ 6 3
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Problem 4. (10 points) Consider the following set of equations: bx 1 + x 2 = 0 x 2 - x 3 = 1 x 1 + bx 3 = 0 b is a real-valued constant. Determine, if possible, all values for b that will make the system of equations have no solutions. Justify your answer. Solution Apply Gaussian elimination. Start with the array: B = b 1 0 0 0 1 - 1 1 1 0 b 0 Swap rows 1 and 3: B = 1 0 b 0 0 1 - 1 1 b 1 0 0 Multiply the first row times - b and add it to the third one, to make the B 31 element equal to 0. We get B =
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This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.

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solutions_final_section_001_f05 - ECE 220, Section 001...

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