220Final - ECE 220, Section 001 Final Exam Solutions 13...

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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 5 , 1 , 5 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 10 , 0 , otherwise. = ± 1 , 0 5 , 0 , otherwise, has a width of Fve. The signal s 2 ( t p ( t - 5) = ± 1 , 0 t - 5 < 10 , 0 , otherwise. = ± 1 , 5 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 Fnally, 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 ² 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 satisFes the property z ( t + T z ( t ) , t ( -∞ , ) . (4) We have z ( t + T ± j + e - j π 16 ² e - j ( t + T ) π 6 = ± j + e - j π 16 ² e - j 6 e - j 6 = z ( t ) e - j 6 (5) ±rom equations 4 and 5, we see that T must be the smallest positive number that satisFes 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, Fnally, 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 2 = 27 e 2 e j 2 From this last equation, we can write (after taking the third root): z = 27 1 / 3 ± e 2 ² 1 / 3 ± e j 2 ² 1 / 3 =3 e 6 e j 2 πk/ 3 (8) Let now k =0 , 1 , 2 in equation 8: z 1 e 6 e j 2 π 0 / 3 e 6 z 2 e 6 e j 2 π 1 / 3 e j 5 π/ 6 z 3 e 6 e j 2 π 2 / 3 e j 9 6 So, ±nally, z 1 e 6 ,z 2 e j 5 6 3 e j 9 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 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.
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220Final - ECE 220, Section 001 Final Exam Solutions 13...

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