33 p 303 17 now to test for this condition note that

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Unformatted text preview: = S (jβ )P (j (� − β ))dβ 2ω −� �� 2 1 = ω [α (� − 2 − β ) + α (� + 2 − β )]dβ 2ω −� 1 + β 2 � � �� 1 1 = + recalling that g (t)α (t − t0 )dt = g (to ) 1 + (� − 2)2 1 + (� + 2)2 −� Let’s double check our answer, using the analysis equation, and considering that the cosine function can b e expressed as a sum of complex exponentials. 13 X (j� ) = = = = = = = � � � x(t) e −j� dt = −� 0 � � −� � e−|t| cos 2t e−j� dt � e−t cos 2t e−j� dt e cos 2t e dt + 0 −� � � � � �0 �� 1 −j 2t −j� 1 −j 2t −j� t 1 j 2t −t 1 j 2t e e+e e e dt + e dt e+e 2 2 2 2 −� 0 � � 1 0 t(1+j 2−j�) 1 � t(−1+j 2−j�) t(1−j 2−j� ) e +e dt + e + et(−1−j 2−j�) dt 2 −� 20 � � � � 1 1 −1 1 1 −1 + + + 2 1 + j2 − j� 1 − j2 − j� 2 −1 + j 2 − j � −1 − j 2 − j � � � 1 1 1 1 1 + + + 2 1 + j (2 − � ) 1 − j (2 − � ) 1 − j (2 + � ) 1 + j (2 + � ) 1 1 + , which is t...
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This document was uploaded on 02/09/2014.

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