Next, we find the Fourier series coefficients of r(t),ak dt [26(t)-6(t-1) dt 3(2c-0 jkwo (1) 2 e-jkwo. for allk Now we are ready to find bk, the Fourier series representation of the output y(t) bk akHGkwo)(o& w, Section 3.8, p 226, and specifically eq (3.124)) 6+(ko)2 3 3/16+(k22, for all k� � � � � � Next, we find the Fourier series coefficients of x(t), ak: 1 ak = x(t) e−jk�0t dt = 1 � 2 [2α(t) − α(t − 1)] e−jk�0t dt T T 3 −1 = 1 2e−jk�0(0) − e−jk�0(1)⎩ 3 2 1 = − e−jk�0, for allk. 3 3 Now we are ready to find bk, the Fourier series representation of the output y(t): bk = akH(jk�0) (O & W, Section 3.8, p.226, and specifically eq.(3.124) ) 2 1 8 = − e−jk�0 3 3 16 + (k�0)2 � � 8 2 − e−jk 2 3 � = 3 16 + � k 2� ⎩2 , for all k. 3 4