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Tuto6_sol - 1 2 2 2 j X e dt t x d j F − − ⎯→ ←...

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EE3210 Tutorial 6 (Solution) 1. (Problem 5.21) a) Using the Fourier transform analysis equation, we have ω ω ω ω ω ω ω j j j j n n j n n j j e e e e e e n x e X 5 4 3 2 5 2 ] [ ) ( = −∞ = + + + = = = b) Similar to part (a), we use the Fourier transform analysis equation: ω ω ω ω ω ω ω ω j j j j n n j n n j n n n j j e e e e e e e n x e X = = = = = = −∞ = −∞ = 2 2 1 1 2 / 2 1 2 1 ] [ ) ( 1 1 2. (similar to Problem 4.6 in the Textbook) a) Using the time-reversal property, we have ) ( ) ( ω j X t x F ⎯→ . Using time-shifting property on this, we have ) ( )) 1 ( ( ) ( 1 ω ω j X e t x t x j F ⎯→ = . b) Using the time-scaling property, we have ) 3 ( 3 1 ) 3 ( ω j X t x F ⎯→ Using the time-shifting property on this, we have ) 3 ( 3 1 )) 2 ( 3 ( ) ( 2 2 ω ω j X e t x t x j F ⎯→ = c) Using the differentiation property, we have ) ( ) ( ω ω j X j dt t dx F ⎯→ Using this property again, we have ) ( ) ( 2 2 2 ω ω j X dt t x d F ⎯→
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