Before exploring the other methods, let's first find the Fourier series for y(t), shown below, which is a periodic triangular function with a period of T. y(t) will be useful for the Method(c) y(t) (t) ()+dk We will find the Fourier series for a(t) and from it, we will find the Fourier series for y() z(t)e-Jkwo dt swot T Ti(-ikwo) (En-c6)=7n(=k AwoL kuo1⊥a-jku0oT1 Kwoi)I Thus dk=eljko 2-2 cos(hwoTi) TTk2w2� � � Before exploring the other methods, let’s first find the Fourier series for y(t), shown below, which is a periodic triangular function with a period of T. y(t) will be useful for the Method(c): y(t) t 1 T −T1 T1 T · · · · · · − 2 2 z(t) = dy(t) dt 1 T1 · · · − T1 · · · T −T1 T t 2 2 −1 T1 F Let z(t) = dy(t) , z(t) �⇒ ek , and y(t) F = jk�0 ek dt �⇒ dk 1 We will find the Fourier series for z(t) and from it, we will find the Fourier series for y(t), as follows: ⎪ ⎪ T1 1 z(t) e−jk�0t dt = 1 ⎩⎪ 0 1 )e−jk�0t dt + (−1 )e−jk�0t ek = ( dt T T T −T1 T1 0 T1 1 1 � 1 � T T1 (−jk�0) e−jk�0t 0 − e−jk�0t T1 � 1 jk�0T1 + 1 − e−jk�0T1 � = |−T1 |0 = T T1 (−jk�0) 1 − e � −1 = −1 2 − (ejk�0T1 + e−jk�0T1 ) � = T T1jk�0 [2 − 2 cos(jk�0T1)] . T T1jk�0 d Thus, k = ek( 1 ) = 2 − 2 cos(k�0T1) . jk�0 T T1k2�2 0 6