Second_Example_LTI_RP_additional

Second_Example_LTI_RP_additional - When computing the ACF...

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Unformatted text preview: When computing the ACF of Y (t), RY [k ] = RX [k ] ⋆ h[k ] ⋆ h[−k ] = (5 δ [k ] ⋆ h[k ]) ⋆ h[−k ] ∞ = 5 h[−k ] ⋆ h[k ] = 5 n=−∞ h[−n] h[k − n] 1 2 k =5 =5 1 2 ∞ −k u[−k ] ⋆ 1 2 ∞ −n u[k ] 1 2 k −n u[−n] 1 2 −2n u[k − n] n=−∞ =5 =5 the following were used: 1 2 1 2 k u[−n] u[k − n] n=−∞ k ∞ 4n u[−n] u[k − n] n=−∞ 1 a −b = ab xa xb = xab x2b = (x2 )b and L U +1 b −b , b=1 1−b n b= U − L + 1, b = 1 U n=L 1 ...
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This note was uploaded on 01/26/2011 for the course EEE 554 taught by Professor Duman during the Spring '10 term at ASU.

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