8 14-2 Exponential Fourier series f(t) 0 +a1 cos at+ a2 coS 2at+.+b1 sin at+b2 sin 2at+ 2 j0t⊥a-j0t 20t ot ot 20t-i20t e e e jot e e e +l +b1 +6 i2 jb +(+22)20+(+-)e J Jat )e ot +— Let +jbu) 2 f(D)=…+A_2e -j2 +a ve ot +ale Jo 20 t 0 +a e 2 ∑Ane"o n=一0
§14-2 Exponential Fourier series = + a t + a t ++ b t + b t + a f t cos cos2 sin sin2 2 ( ) 1 2 1 2 0 + − + − + + + + = + − − − − 2 2 2 2 2 2 2 1 2 2 2 1 2 0 j e e b j e e b e e a e e a a j t j t j t j t j t j t j t j t = + + + + + + − + − + − j t − j t j t j t e a jb e a a jb e a jb e a2 jb2 2 1 1 0 1 1 2 2 2 ) 2 2 ) ( 2 2 ( 2 ) 2 2 ) ( 2 2 ( ( ) 2 1 n n n Let A = a − jb ( ) 2 1 n n n A− = a + jb f (t) = + A−2 e − j2 t + A−1 e − j t + A0 + A1 e j t + A2 e j2 t + jn t n An e + =− =
To obtain the evaluation integral for the a coefficients we multiply f(t on both sides by e- norAnd integrate over the full period: f(t)=…+A-120(+1_1e-0+A0+1e/iot +abeJar t o J(De-jinot 2元 2兀 d(at) e no d(ar) 0 2 2兀 2元 十 enot d(ar)+ o oe-jnat d(at 0√,ejat- inat d(x2x 2元 20t 0 -jni d(ot)+ 2兀 Note no d(ot=2/ A, 0 In=2Jo /(oe no d(@r)=T-T/2/(Oe no dt (an-jb,), (am+jb,), 2 一 an=2R,[An b,=-2ImIA
To obtain the evaluation integral for the An coefficients, we multiply f(t) on both sides by And integrate over the full period: jn t e − f t e dt T A f t e d t T T jn t jn t n − − − = = / 2 / 2 2 0 ( ) 1 ( ) ( ) 2 1 * ( ), 2 1 ( ), 2 1 n n n n n n A n An A = a − jb A− = a + jb − = ( ) 2 0 1 A e e d t j t jn t − − + − + − ( ) + − ( ) + 2 0 2 2 2 0 1 A e e d t A e e d t j t jn t j t jn t n jn t jn t An e e d t A ( ) 2 2 0 + = − ( ) 2 0 0 A e d t jn t − + f (t) = + A−2 e − j2 t + A−1 e − j t + A0 + A1 e j t + A2 e j2 t + 2 [ ] 2 [ ] n e n n m An a = R A b = − I − 2 0 f (t)e d( t) jn t ( ) 2 0 2 2 A e e d t j t jn t − − = + −