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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap04

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap04

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Chapter 4: Signal Representation using Fourier Series Problem 4.1 (a) Using Definition 4.4, the CT function x 1 ( t ) can be represented as x 1 ( t ) = c 1 φ 1 ( t ) + c 2 φ 2 ( t ) + c 3 φ 3 ( t ) with the coefficients c n , for n = 1,2, and 3, given by 0 ) ( ) ( ) ( ) ( 2 1 0 2 1 0 2 1 1 1 2 1 1 = + = + = φ = AT AT Adt dt A dt t t x c T T T T T T T T , ( ) , 0 ) 1 ( ) 1 ( ) ( ) 1 )( ( ) 1 )( ( ) ( ) ( 2 2 2 2 2 1 2 / 2 1 2 / 0 2 1 0 2 / 2 2 1 2 / 2 1 2 1 2 1 2 = + = + + φ + = φ = AT AT AT AT T T T T T T T T T T T T T T dt A dt A dt t A dt A dt t t x c and A AT AT dt A dt A dt t t x c T T T T T T T T = = + = φ = ) ( ) 1 ( ) 1 )( ( ) ( ) ( 2 1 0 2 1 0 2 1 3 1 2 1 3 . In other words, x 1 ( t ) = A φ 3 ( t ), which can also be proved by inspection. (b) By inspection, x 2 ( t ) = A φ 2 ( t ), which can also be proven by evaluating the coefficients c 1 = 0, c 2 = A , and c 3 = 0. (c) Using Definition 4.4, the CT function x 3 ( t ) can be represented as x 3 ( t ) = c 1 φ 1 ( t ) + c 2 φ 2 ( t ) + c 3 φ 3 ( t ) with the coefficients c n , for n = 1,2, and 3, given by 2 2 2 2 1 2 / 2 1 2 / 2 1 1 3 2 1 1 ) ( ) 1 )( ( ) 1 )( ( ) ( ) ( A AT AT T T T T T T T T T T dt A dt A dt t t x c = + = + = φ = , 2 2 2 2 1 2 / 2 1 2 / 2 1 2 3 2 1 2 ) ( ) 1 )( ( ) 1 )( ( ) ( ) ( A AT AT T T T T T T T T T T dt A dt A dt t t x c = = + = φ = , and 0 ) ( ) 1 )( ( ) 1 )( ( ) ( ) ( 2 1 0 2 1 2 / 2 1 3 3 2 1 3 = = + = φ = AT AT dt A dt A dt t t x c T T T T T T T T T . In other words, x 3 ( t ) = 0.5 A ( φ 1 ( t ) φ 2 ( t )), which can also be proved by inspection. Problem 4.2 Computing the integral ( ) = φ φ dt Ke e dt t t t t 4 2 2 1 1 ) ( ) ( Since the function inside the integral is even with respect to t , therefore,

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124 Chapter 4 ( ) 3 0 6 0 2 0 4 2 2 1 1 2 2 1 2 ) ( ) ( K t t t t dt e K dt e dt Ke e dt t t = = = φ φ For the functions to be orthogonal, . 3 0 1 3 = = K K Problem 4.3 The following derivation shows that the individual functions { P n ( x ), n = 0, 1, 2, 3} have nonzero finite energy. We use the notation P m , n to represent the integral P m , n = 1 1 ( ) ( ) m n P x P x dx . Computing the integrals [ ] 1 1 0,0 1 1 1.1 2 P dx x = = = , 1 1 2 3 1 2 1,1 3 3 1 1 P x dx x = = = , ( ) 1 1 1 2 2 4 2 5 3 9 9 1 1 1 1 2 2,2 4 4 4 5 2 5 5 1 1 1 (3 1) 9 6 1 2 ( 2 1) P x dx x x dx x x x = = + = + = + = , and ( ) 1 1 6 4 2 7 5 3 25 25 1 1 1 2 3,3 4 4 7 2 7 7 1 1 25 30 9 6 3 ( 6 3) P x x x dx x x x = + = + = + = , which shows that the functions P n ( x ) have nonzero finite energy. To show that the functions are orthogonal with respect to each other, we determine the integrals N 1 0,1 1 0 odd P x dx = = = , 1 1 2 3 1 1 0,2 2 2 1 1 (3 1) 0 P x dx x x = = = , 1 3 1 0,3 2 1 (5 3 ) 0 odd P x x dx = = = ±²³²´ , 1 3 1 1,2 2 1 (3 ) 0 odd P x x dx = = = ±³´ , 1 1 4 2 5 3 1 1 1,3 2 2 1 1 (5 3 ) 0 P x x dx x x = = = , and 1 5 3 1 2,3 4 1 15 14 3 0 odd P x x x dx = = + = ±²²³²²´ .
Solutions 125 Problem 4.4 The following derivation shows that the individual functions { T n ( x ), n

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Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap04

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