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

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Unformatted text preview: EE3210 Tutorial 3 (Solution) 1. Note that T = 2. Therefore, ∫ − = 2 ) ( 2 1 dt e t x a t jk k π . Now, 5 . 1 2 1 5 . 1 2 1 2 1 1 = − = ∫ ∫ 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 . 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 ) ( ω a) If x(t) is even, then x(t) = x(-t). Now i) k letting (by -k) i letting (by ) ( = = = = = − ∑ ∑ ∑ ∞ −∞ = − ∞ −∞ = − ∞ −∞ = − 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 ) ( * * * * = = = = = ∑ ∑ ∑ ∞ −∞ = − ∞ −∞ = − ∞ −∞ = − 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 * . 3. Note that the signal is periodic with fundamental period equal to 6. Note that the signal is periodic with fundamental period equal to 6....
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This note was uploaded on 11/22/2009 for the course ELECTRONIC EE3210 taught by Professor Sungchiwan during the Spring '09 term at École Normale Supérieure.

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

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