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 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: 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 ) ( ) ( j X t x F ⎯→ . Using time-shifting property on this, we have ) ( )) 1 ( ( ) ( 1 j X e t x t x j F ⎯→ = . b) Using the time-scaling property, we have ) 3 ( 3 1 ) 3 ( j X t x F ⎯→ Using the time-shifting property on this, we have ) 3 ( 3 1 )) 2 ( 3 ( ) ( 2 2 j X e t x t x j F ⎯→ = c) Using the differentiation property, we have ) ( ) ( ωω j X j dt t dx F ⎯→ Using this property again, we have ) ( ) ( 2 2 2 j X dt t x d F ⎯→ Using the time-shifting property on this, we have
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Unformatted text preview: ) ( ) 1 ( 2 2 2 j X e dt t x d j F − − ⎯→ ← − 3. (Problem 5.29a) By direct calculation or by transform table, we know that ω j j e e H − − = 2 1 1 1 ) ( . We have j j e e X − − = 4 3 1 1 ) ( , and j j j j j e e e H e X e Y − − − × − = = 2 1 1 1 4 3 1 1 ) ( ) ( ) ( . Expanding it into partial fractions, we have j j j e e e Y − − − − − = 2 1 1 2 4 3 1 3 ) ( . Taking the inverse Fourier transform, we obtain ] [ 2 1 2 ] [ 4 3 3 ] [ n u n u n y n n ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ....
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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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Tuto6_sol - ) ( ) 1 ( 2 2 2 j X e dt t x d j F − −...

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