HW3Soln

# HW3Soln - ECE 301 Homework 3 Solutions Landis Human Purdue...

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ECE 301: Homework 3 Solutions Landis Huffman Purdue University, Summer 2009 Due July 9, 2009 in class 1. Chapter 3 Book problems 3.3, 3.8, 3.22(b), 3.30, 3.37(a) 3.3 Solution T = 6, w 0 = 2 π/ 6 = π/ 3. The signal can be written as a sum of complex exponentials using Euler’s formula x ( t ) = 2 + 1 2 ( e j 2 w 0 t + e - j 2 w 0 t ) + 4 1 2 j ( e j 5 w 0 t e - j 5 w 0 t ) = 2 + 1 2 e j 2 w 0 t + 1 2 e - j 2 w 0 t 2 je j 5 w 0 t + 2 je - j 5 w 0 t The non-zero Fourier series coefficients of x ( t ) are: a 0 = 2 , a 2 = a - 2 = 1 2 , a 5 = a * - 5 = 2 j. 3.8 Solution Since x ( t ) is real and odd (clue 1), its Fourier series coefficients a k are purely imaginary and odd (See Table 3.1). Therefore, a k = a - k and a 0 = 0. Also, since it is given that a k = 0 for | k | > 1, the only unknown Fourier series coefficients are a 1 and a - 1 . Using Parseval’s relation, 1 T integraldisplay <T > | x ( t ) | 2 dt = summationdisplay k = -∞ | a k | 2 , for the given signal we have 1 2 integraldisplay 2 0 | x ( t ) | 2 dt = 1 summationdisplay k = - 1 | a k | 2 . 1

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Using the information given in clue 4 along with the above equation, | a 1 | 2 + | a - 1 | 2 = 1 2 | a 1 | 2 = 1 . Therefore, a 1 = a - 1 = 1 2 j or a 1 = a - 1 = 1 2 j . The two possible signals which satisfy the given information are: x 1 ( t ) = 1 2 j e jπt 1 2 j e - jπt = 2 sin( πt ) . and x 2 ( t ) = 1 2 j e jπt + 1 2 j e - jπt = 2 sin( πt ) . 3.22(b) Solution x ( t ) periodic with period 2 and x ( t ) = e - t for 1 < t < 1 a k = 1 2 integraldisplay 1 - 1 e - t e - jkπt dt = 1 2 integraldisplay 1 - 1 e ( - 1 - jkπ ) t dt = 1 2 bracketleftbigg 1 1 jkπ e ( - 1 - jkπ ) t bracketrightbigg 1 - 1 = 1 2(1 + jkπ ) [ e (1+ jkπ ) e ( - 1 - jkπ ) ] = 1 2(1 + jkπ ) [ e ( 1) k e - 1 ( 1) k ] = ( 1) k 2(1 + jkπ ) [ e e - 1 ] . 3.30 Solution (a) N = 6 and w 0 = 2 π/ 6 = π/ 3. The signal x [ n ] can be written as a sum of complex exponentials using Euler’s formula. x [ n ] = 1 + 1 2 e - jw 0 n + 1 2 e jw 0 n = e j 0 + 1 2 e jw 0 n + 1 2 e j 5 w 0 n . 2
The Fourier series coefficients of x [ n ] are periodic with period 6, i.e., a k +6 = a k .

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