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(a) Let x(n) = u(n) Find ya (n) by first convolving x(n) with h1 (n) and then convolving the result with h2 (n) ie ya(n) = x(n) * h1 (n) * h2 (n) (b) Again let x(n) = u(n) Find yb(n) by convolving x(n) with the result of the convolution of h1 (n) and h2 (n) ie yb(n) = x(n) * h1(n) * h2(n) Your results for parts (a) and (b) should be identical, illustrating. 24 c JFessler,May27,04,1310(studentversion) 212 Classication of discretetime signals The energy of a discretetime signal is dened as Ex 4= X1 n=1 jxnj2 The average power of a signal is dened as Px 4= lim N!1 1 2N 1 XN n= N jxnj2 If E is nite (E < 1) then xn is called an energy signal and P = 0 If E is innite, then P can be either nite or innite. ª v j O Ò i ë 4 u t y Ò i v W C u O C u O y ê é ¬ l ¾ ñ ¾ r M n q ê é ¬ ³ ß W \ O ¾ y W Ë u O ※ Ï I v r q z O z ¸ ü u J ² v á S ê é ¬ 2 " v U \ 2 " W Ë s.
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Nunx e. May 16, 15 · The answer to this (and your other similar question) is most likely found in the somewhat unusual notation used for the system In ordinary terms, if you just take your system as a map between input and output signal, then you're absolutely right, it would not be time invariant. EC52 2 The zTransform deﬁnition (cont) The transfer function H(z) = X∞ k=−∞ hkz−k Generally, let z = rejΩThen, H(rejΩ) = X∞ n=−∞ hnz−n e−jΩn Thus, H(z) is the DTFT of hnr−nThe inverse DTFT of H(rejΩ) must be hnr−n. ABB 3 615 # ¾ n © Relion® º ª f Ã 1YZA Ô r J } Ö 1YZAcn @ ª é ( ÖA © ò 11 ABB ò ï ¶.

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