10 To find ck, the Fourier series coefficients of v(t), let's take the period between -1 and 2, which contains two impulses Note that we can also take the period between -2 and 1, but we have to be careful not include the impulses at both -2 and 1. In other words, we can take the period between 2+0 and 1+d or the period between -2-d and 1-0 7/(-koat=1(/(-35()+35(-1)e-ud) [-6(t)+6(t-1) e-gkwot Now to find ak, we just need to use the integration property two times (from the Integration property, Table 3.1, O &W, p. 206) Kuo 2(e-jkwo_1) (1 e-k), which is the same answer found in Method(a)� � � � � −5 −4 −3 −2 −1 0 1 2 3 4 t v(t) −3 · · · · · · 3 3 To find ck, the Fourier series coefficients of v(t), let’s take the period between -1 and 2, which contains two impulses. Note that we can also take the period between -2 and 1, but we have to be careful not include the impulses at both -2 and 1. In other words, we can take the period between −2 + � and 1 + � or the period between −2 − � and 1 − �. ⎪ ck = 1 v(t)e−jk�0t dt = 1 �⎪ 2 [−3�(t) + 3�(t − 1)] e−jk�0t dt T T 3 −1 ⎪ 2 = [−�(t) + �(t − 1)] e−jk�0t dt −1 = −e−jk�0(0) + e−jk�0(1) = e−jk�0 − 1. Now to find ak, we just need to use the integration property two times: 1 1 ak = ck (from the Integration property, Table 3.1, O &W, p.206 ) jk�0 jk�0 = 1 e−jk�0 � (jk�0)2 − 1 = 1 1 − e−jk�0 � k2�2 0 1 = 1 − e− 3 jk 2� , which is the same answer found in Method(a). k2�2 0 5