Method (d): using the analysis equation In the process of evaluating the analysis equation, the following integral will save us a lot of derivation steps te dt a≠0 2+t) dt dt+te"-3kwot dt-2t dr 21 wo jkwo kawa two -jkwo(1) 3 hwo , 6 jho 2 1(1) kwo jho e-jkwo(1) k2w2 k2w 2 RwG. 1 e-jhuo 3jhw kwo k2w2 kwa kwa 1 3(k2 12, 2ekup2 3k2∞6 k2w2 (remember that e- o ko2 for T=3) -eika), which is the same answer found in previous methods k2w� � � � � � � � � � � � � � � � � � � • Method (d): using the analysis equation: In the process of evaluating the analysis equation, the following integral will save us a lot of derivation steps: ⎪ t 1 at teatdt = a − a e , for any a = 0 2 √ ⎪ 1 ⎪ 1 1 � x(t)e−jk�0t dt = x(t)e−jk�0t ak = dt T 3 −2 ⎩⎪ −� ⎪ 1 � 1 0 (2 − 2t)e−jk�0t = (2 + t)e dt −jk�0t dt + 3 −2 0 1 ⎩ ⎪ 1 ⎪ 1 ⎪ � 0 e−jk�0t dt + te−jk�0t dt − 2 te−jk�0t = 2 dr � 3 −2 −2 0 � � � � 0 � e−jk�0t �1 � �1 1 � 1 t � 1 t � = 2 � + � e−jk�0t 3 −jk�0 � k2�2 − jk�0 e−jk�0t � − 2 k2�2 − −2 0 −2 0 jk�0 0 1 −2 e−jk�0(1) − e−jk�0(−2)� 1 1 (−2) e−jk�0(−2) = + 3 jk�0 k2�0 2 − k2�2 − 0 jk�0 ⎩� � �� 1 (1) 1 e−jk�0(1) −2 k2�2 − jk�0 − k2�0 2 0 1 −2 e−jk�0 2 k2 1 �0 2 − k2 1 �0 2 jk�02 2 − jk�0 jk�02 = + ejk�02 + e e 3 jk�0 jk�0 2 2 2 −k2�0 2 e−jk�0 + e−jk�0 + jk�0 k2�2 0 1 −2 1 2 e−jk�0 2 e−jk�0 2 e = + −jk�0 + + 3 jk�0 − k2�0 2 jk�0 k2�2 k2�2 0 0 2 1 2 + ejk�02 jk�02 jk�02 e e jk�0 − k2�0 2 − jk�0 1 2 3 1 + ejk�02 = 3 −k2�0 2 e−jk�0 k2�0 2 − k2�0 2 1 � = 3 − 2e−jk�0 jk�02� 3k2�0 2 − e = 1 1 − ejk�02� (remember that e−jk�0 = ejk�02 for T = 3) k2�2 0 = 1 1 − ejk 4 3 � , which is the same answer found in previous methods. k2�2 0 = 9 1 − ejk 4 3 � . 4k2�2 8