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# Tuto3_sol - EE3210 Tutorial 3(Solution 1 Note that T = 2...

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EE3210 Tutorial 3 (Solution) 1. Note that T = 2. Therefore, = 2 0 ) ( 2 1 dt e t x a t jk k π . Now, 0 5 . 1 2 1 5 . 1 2 1 2 1 1 0 0 = = dt dt a , and for k 0, [ ] π π π π jk t jk t jk k e j k dt e dt e a = = 1 2 3 5 . 1 2 1 5 . 1 2 1 2 1 1 0 . If you want to write it in polar form, then for k 0, = 2 sin 3 ) 2 / ( π π π k e k a jk k . 2. Note that . −∞ = = k t jk k e a t x 0 ) ( ω a) If x(t) is even, then x(t) = x(-t). Now i) k letting (by -k) i letting (by ) ( 0 0 0 = = = = = −∞ = −∞ = −∞ = k t jk k i t ji i k t jk k e a e a e a t x ω ω ω Hence, a k = a -k . b) If x(t) is real, then x(t) = x * (t). Now i) k letting (by -k) i letting (by ) ( 0 0 0 * * * * = = = = = −∞ = −∞ = −∞ = k t jk k i t ji i k t jk k e a e a e a t x ω ω ω Hence, a k = a -k * .

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