E ejn h ej refer to o w section 32 p183 1 jn e 2 1 y

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Unformatted text preview: (−1)�0 ⎩ −1 1 + 2 e−j (−1)� −1 1 + 2(−1) = = b−1 = a−1 H e 1 1 1 −j (−1) � 4 2j 1 − 4 e 2j 1 − 1 ( �2 + j �2 ) 4 = 0.1247 − j 0.5806 � j (2)�0 ⎩ 1 + 2 e−j (2)� 1 + 2(1) = (−1) = (−1) b2 = a 2 H e 1 −j (2) � 4 1 − 1 (−j ) 1− 4e 4 = −2.8235 + j 0.7059 ⎩ � 1 + 2 e−j (−2)� 1 + 2(1) = (−1) b−2 = a−2 H ej (−2)�0 = (−1) 1 −j (−2) � 4 1 − 1 (j ) 1− 4e 4 = −2.8235 − j 0.7059 b3,4,5 = 0. As a double-check for our answer, notice that b−k = b� which indicates that the output y [n] k is real. We expect this b ecause the input is real and because the LTI system applies a real operation, i.e. the diﬀerence equation. We also expect this from a mathematical point of 9 view b ecause the input is real, i.e. a−k = a� , and H (e−j� ) = H � (ej� ), from the calculated k expression. 10 Problem 4 Specify the frequency response of a discrete-time LTI system so that if the input is x[n] = 2 + cos(ωn) − sin(ωn/2) + 2 cos(ωn/4 + ω /4) then the output is y [n] = 4 − 2 sin(ωn) + 2 cos(ωn/4). It will b e straight forward to determine the frequency response of the system, once we expand the expressions for x[n] and y [n] into their complex exponential components: x[n] = 2 + cos(ωn) − sin(ωn/2) + 2 cos(ωn/4 + ω /4) � �� � � � 1 j�n 1 −j�n 1 j�n/2 1 −j�n/2 1 j (�n/4+�/4) 1 −j (�n/4+�/4) − e +2 e − e +e = 2+ e +e 2 2 2j 2j 2 2...
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