11 p 288 18 by inspecting the values of the signals

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Unformatted text preview: (� − �0 ) + α (� + �0 )] j X (j� ) = j [α (� + 1) − α (� − 1)] − 3 [α (� − ω ) + α (� + ω )] 1ω 3 = [α (� − 1) − α (� + 1)] − ω [α (� − ω ) + α (� + ω )] ωj ω 1 3 � x(t) = sin t − cos ω t. ω ω (b) X (j� ) = 2 sin(2� − ω /2) X (j� ) = 2 sin(2� − ω /2) = 2 � 1 −j (2�− � ) 1 j (2�− � ) 2− 2 e e 2j 2j � 1 j 2� −j � 1 −j 2� j � e e 2− e e2 j j 1 j 2� 1 = e (−j ) − e−j 2� (j ) = −ej 2� − e−j 2� j j � x(t) = −α (t + 2) − α (t − 2) , from table 4.2 (O & W, p.3.29) . = Alternatively we can express X (j� ) as: X (j� ) = 2 sin(2� − ω /2) = −2 cos(2� ) Using the duality property (O & W, Section 4.3.6, p.309) : 1 1 F α (t − to ) + α (t + t0 ) �� cos � to 2 2 � � 1 1 � x(t) = −2 α (t − 2) + α (t + 2) = −α (t − 2) − α (t + 2). 2 2 16 Problem 7 Answer the questions asked in O&W 4.24 (a) for each of the following signals:...
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This document was uploaded on 02/09/2014.

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