final_exam_solutions

final_exam_solutions - ECE 301 Division 2 Final Exam...

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ECE 301 Division 2 Final Exam Solutions, 12/17/2010, 7-9pm in LILY 1105. Your ID will be checked during the exam. Please bring a No. 2 pencil to fill out the answer sheet. This is a closed-book exam. No calculators are allowed. Instructor: Ilya Pollak Course: ECE 301 Date: 12/17/2010 Test: Final sign the answer sheet fill in your last name and the first six letters of your first name fill in your student ID for the section, fill in “0002” You have TWO hours to complete 20 multiple-choice questions. Each correct answer is worth two points, for a total of 40 points. Please only turn in the answer sheets. 1
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For Questions 1–5 , let φ k ( t ) = e j 2 πkt 10 , for k = 0 , ± 1 , ± 2 , . . . , and for all real t . Let . . . , a 1 , a 0 , a 1 , . . . be complex numbers, and consider the following representation of x as a linear combination of φ k ’s: x ( t ) = summationdisplay k = −∞ a k φ k ( t ) . (1) The objective of Questions 1–5 is to decide whether this representation is possible for a given signal x , and if so, to find these numbers . . . , a 1 , a 0 , a 1 , . . . . Question 1. x ( t ) = e j 2 πt . Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , 0 , otherwise . 2. Representation (1) does not exist. 3. a k = 1 2 j , for k = 1 , - 1 2 j , for k = - 1 , 0 , otherwise . 4. a k = k 5. a k = braceleftBigg 1 5 for k = 0 e - jπk 5 πk sin ( π 5 k ) for k = ± 1 , ± 2 , . . . . 6. All the above statements are incorrect. Solution. Signals φ k together form a CT Fourier basis with period T = 10. Since x is periodic with period 1, then it is also periodic with period 10. Also, it is square-integrable over any interval of length 10. Hence, x can be represented as a linear combination of φ k ’s. Now, x ( t ) = e j 2 πt = e j 2 π · 10 · t 10 . By inspection, we get: a k = braceleftbigg 1 , for k = 10 , 0 , otherwise . Answer : 1. 2
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Question 2. x ( t ) = sin ( 2 πt 10 ) . Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , 0 , otherwise . 2. Representation (1) does not exist. 3. a k = 1 2 j , for k = 1 , - 1 2 j , for k = - 1 , 0 , otherwise . 4. a k = k 5. a k = braceleftBigg 1 5 for k = 0 e - jπk 5 πk sin ( π 5 k ) for k = ± 1 , ± 2 , . . . . 6. All the above statements are incorrect. Solution. Since x is a CT signal that is periodic with period 10, and since it is square-integrable over one period, it can be represented as a linear combination of φ k ’s. Now, x ( t ) = 1 2 j e j 2 πt 10 - 1 2 j e j 2 πt 10 . By inspection, we get: a k = 1 2 j , for k = 1 , - 1 2 j , for k = - 1 , 0 , otherwise . Answer : 3. Question 3. x ( t ) = braceleftbigg 1 , for 10 n t 10 n + 2 , for all integer n 0 , for 10 n + 2 < t < 10 n + 10 , for all integer n Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , 0 , otherwise . 2. Representation (1) does not exist. 3. a k = 1 2 j , for k = 1 , - 1 2 j , for k = - 1 , 0 , otherwise . 4. a k = k 5. a k = braceleftBigg 1 5 for k = 0 e - jπk 5 πk sin ( π 5 k ) for k = ± 1 , ± 2 , . . . . 6. All the above statements are incorrect.
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