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
Method(c): By dissecting the signal into simpler components Here, we will dissect (t)into ni(t)and 2(t)which we know their Fourier Series(using the result of dk above and the time-shifting property) r(t)=a1(t)+r2(t), and let c1(t) bk and a2(t) ak= bk +Ck=(2) 9=30 2-2cos(ku0(1) (3)(1)k26 32(2+) Although this result looks different from those found in the previous methods, further simplification will show that they are identical a=2-208ko+ck)=3k4a(2-2csk(2+c) 32(、ckEo)(2+ek0 3k2(4+2ek0-2ek0-k2-2e~k0-e) 1 (4-ck02-2e-1k0-1) 3k26 (3 k2(1-c0 (remember that e-3kuwo ekwo2 for T=3) which is the same answer found in previous methods
� � � • Method (c): By dissecting the signal into simpler components: Here, we will dissect x(t) into x1(t) and x2(t) which we know their Fourier Series (using the result of dk above and the time-shifting property). x1(t) 2 −4 −3 −2 −1 0 1 2 3 4 t x2(t) −4 −3 −2 −1 0 1 2 3 4 t 1 x(t) = x1(t) + x2(t) , and let x1(t) F and x2(t) F �⇒ bk �⇒ ck � ak = bk + ck = (2) 2 − 2 cos(k�0(1)) + (1)2 − 2 cos(k�0(1)) e−jk�0(−1) (3)(1)k2�2 (3)(1)k2�2 0 0 = 2 − 2 cos k�0 (2 + ejk�0 ). 3k2�0 2 Although this result looks different from those found in the previous methods, further simplification will show that they are identical: 1 ak = 2 − 2 cos k�0 (2 + ejk�0 ) = 3k2�2 (2 − 2 cos k�0)(2 + ejk�0 ) 3k2�2 0 0 1 = 3k2�0 2 (2 − ejk�0 − e−jk�0)(2 + ejk�0 ) 1 � 4 + 2ejk�0 − 2ejk�0 − ejk�02 − 2e−jk�0 0 � = 3k2�2 − e 0 1 � 1 � = jk�02 − 2e−jk�0 � = jk�02 − 2e−jk�0 � − 1 3k2�0 2 4 − e 3k2�0 2 3 − 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 7