ECE108HW%238sol_06

# ECE108HW%238sol_06 - ECE 108 HW#8 Due Solution 5.1(b n1 n 1...

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ECE 108 HW#8 Due 4/3/06 — Solution 5.1(b) () 11 1 1 1 22 2 nn n x nu n u n −− +  == +   Then the DTFT is 00 01 0 1 1 2 2 1 1 2 2 1 2 2 2 jj n j n j n j n n n jn j j n n j j X ee e e e eee e e e −+ ∞∞ Ω− =−∞ = =−∞ = = = −Ω =+ = + = + ∑∑ 1 2 1 42 3 3 4 5 54 c o s cos 4 j j j j e e e e = +− + 5.5 Then 3 3 4 4 2 2 4 4 33 3 2 2 2 3 sin 2 j xn e e d e n n π ππ =Ω = =− =

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2 Now if n was a continuous variable, then it would have zero values for 3 for 1,2, 42 nm m π  −= =   " Solving for n 3 4 f o r 1 , 2 , 8 m =− = " But n must be an integer value and these values are not integers. So only for do we get n =±∞ zero values for x ( n) . 5.21 (c) () 11 2 f o r 2 33 nn xn u n n = Then 2 22 1 1 1 1 3 3 3 1 39 3 n n j jn j j j n n jj j X ee e e e e e −∞ Ω− =−∞ = = = ΩΩ == = = = −− ∑∑ (g) sin cos 2 n n =+ Then sin cos 2 j X en n  =ℑ +ℑ   Using Table 5.2 1 2 1 2 j k k X ek k j kk ππ δω =−∞ =−∞ + + +
3 (j) () ( ) 1 1 3 n xn n  =−   Then () 1 0 11 00 33 1 1 3 3 nn jj n j n n n j j n n Xe n e n e ee n e n e −∞ Ω−

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ECE108HW%238sol_06 - ECE 108 HW#8 Due Solution 5.1(b n1 n 1...

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