HW6F13_sol

# We get 1 2 xn n un c we notice that x ej 2 3 1 sin 2

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Unformatted text preview: ct 348 ω π 3 · e −j ω Problem 13.23. Invert the following DTFTs: (a). X (ejω ) = cos( π ω ). 2 (b). X (ejω ) = 1 . 1 1− 2 e−jω (c). X (ejω ) = cos( π ω ) · sin( π ω ). 3 3 (d). X (ejω ) = cos2 ( π ω ). 6 (e). X (ejω ) = sin( π ω ) · cos2 ( π ω ). 3 4 Solution. (a). Recall the Euler formula and the deﬁnition of the inverse of DTFT: π j πω/2 e 1 + e−jπω/2 j nω e dω 2 π −π 2 1 1 1 j (n+ π )ω π j (n− π )ω 2 2 = + πe πe −π 4π j (n + 2 ) j (n − 2 ) x(n) = sin π n + π 2 (n + π ) 2 = 1 2π = π 1 sinc π n + 2 2 + sin π n − π 2 (n − π ) 2 + sinc π n − π −π π 2 (b). We get 1 2 x(n) = n u(n) (c). We notice that X (ejω ) = 2π ω 3 1 sin 2 which leads to π j 2π ω/3 1 − e−j 2πω/3 j nω e e dω 4 π −π 2j 1 1 π 1 j (n+ 23 )ω π = − πe −π 8j π j (n + 23 ) j (n − x(n) = = 1 4j π = 1 4j π sin π n + 23 2π (n +...
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## This note was uploaded on 12/01/2013 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.

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