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Unformatted text preview: ECE 301 Division 2 Final Exam Solutions, 12/17/2010, 79pm 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 closedbook 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 multiplechoice questions. • Each correct answer is worth two points, for a total of 40 points. • Please only turn in the answer sheets. 1 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 ,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 ,a 1 ,... . Question 1. x ( t ) = e j 2 πt . Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , , otherwise . 2. Representation (1) does not exist. 3. a k = 1 2 j , for k = 1 , 1 2 j , for k = 1 , , 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 squareintegrable 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 , , otherwise . Answer : 1. 2 Question 2. x ( t ) = sin ( 2 πt 10 ) . Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , , otherwise . 2. Representation (1) does not exist. 3. a k = 1 2 j , for k = 1 , 1 2 j , for k = 1 , , 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 squareintegrable 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 , , otherwise . Answer : 3. Question 3. x ( t ) = braceleftbigg 1 , for 10 n ≤ t ≤ 10 n + 2 , for all integer n , for 10 n + 2 < t < 10 n + 10 , for all integer n Select the correct statement: 1. a k = braceleftbigg 1 , for k = 10 , , otherwise ....
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This note was uploaded on 02/14/2012 for the course ECE 301 taught by Professor V."ragu"balakrishnan during the Fall '06 term at Purdue.
 Fall '06
 V."Ragu"Balakrishnan

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