# Ps4sol

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Unformatted text preview: the frequency response: �� �� −j�t e−4|t| e−j�t dt h(t)e dt = H (j� ) = −� −� �0 �� −4(−t) −j�t = e e dt + e−4(t) e−j�t dt −� 0 �0 �� = e(4−j�)t dt + e(−4−j�t)t dt −� 0 �0 � � �� 1 1 (4−j� )t � (−4−j� )t � + = e e � � 4 − j� −4 − j � −� 0 1 1 = (1 − 0) + (0 − 1) (remember that e−�+ja = 0 for any real a) 4 − j� −4 − j � 1 1 4 + j � + 4 − j � = + = 4 − j� 4 + j� 16 + � 2 8 H (j� ) = . 16 + � 2 3 Next, we ﬁnd the Fourier series coeﬃcients of x(t), ak : ak � � 1 12 −jk�0 t = x(t) e dt = [2α (t) − α (t − 1)] e−jk�0 t dt T T 3 −1 ⎩ 1 � −jk�0 (0) 2e − e−jk�0 (1) = 3 2 1 −jk�0 , for allk. = −e 3 3 Now we are ready to ﬁnd bk , the Fourier series representation of the output y (t): bk = ak H (jk0 ) (O & W, Section 3.8, p.226, and speciﬁcally eq.(3.124) ) � � �� 8 2 1 −jk�0 −e = 3 3 16 + (k�0 )2 �� 2� 8 2 − e−jk 3 = � ⎩ , for all k. 3 16 + k 2� 2 3 4 Pr...
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## This document was uploaded on 02/09/2014.

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