Chapter 4 Fourier transform Example 4.4 G)=11s 0t|>T T0 Tt F 2T sinc(oT) 0|t>T sin o t F→X(o) O≤O 丌t 0 Q>ac 11
11 Chapter 4 Fourier Transform –T1 0 T1 t 1 x(t) Example 4.4 ( ) = 0 t 1 t 1 1 T T x t ( ) 1 0 c c X j = F ( ) sin c t x t t = ( ) 1 1 ( ) 2T sinc T = 0 t 1 t 1 1 T T x t F
Chapter 4 Fourier transform Example 1 +1t>0 son 1() lim e a→0 sgnl 0t=0 -1t<0 timeout a→>0t sgn(t)<F 2 Jo Example 2 n()<>z6(a)+ JO 12
12 Chapter 4 Fourier Transform Example 1 ( ) = + = -1 0 0 0 1 0 sgn t t t t t −1 1 sgn(t) ( ) j t 2 sgn ⎯F → e u(t) at a − → + 0 lim e u( t) at a − − → + 0 lim Example 2 u t( ) ( ) F 1 j ⎯→ +
Chapter 4 Fourier transform Example3 x(t) (-) e Jo J@ot F 2rdo-@o
13 Chapter 4 Fourier Transform Example 3 ( ) j t x t e 0 = F ( ) ( ) 2 −0 j t x t e 0 = (t t − 0 ) F j t0 e ⎯→ −
Chapter 4 Fourier transform 54.2 The Fourier Transforms for Periodic Signals cosOpt F >cda+oo)+Ido-oo) J sin ot"):(、则j 6(a L Oo a x(t)=∑ X(o)=2z∑a1o(o-ka) 14
14 Chapter 4 Fourier Transform §4.2 The Fourier Transforms for Periodic Signals ( ) ( ) 0 0 F 0 cos t ⎯→ + + − −0 0 ( ) ( ) 0 0 F 0 sin ⎯→ − − + j j t − j −0 0 j ( ) jk t k x t a e 0 + − = ( ) ( ) 2 a k0 X j k k = − + =−
Chapter 4 Fourier transform Periodic square wave Ga ●●● Periodic impulses train 2I xGo ()=∑(-kr) ●●● k=-∞ -000020o15
15 Chapter 4 Fourier Transform T 2 -ω0 0 ω0 2ω0 X(j) −0 0 X(j) Periodic impulses train x(t) (t kT) k = − + =− Periodic square wave