Fourier Transforms of Special Functions http://www.google.com/search?hl=en&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=unit+step+fourier+transform&spell=1 主講者:虞台文 Content Introduction More on Impulse Function Fourier Transform Related to Impulse Function Fourier Transform of Some Special Functions Fourier Transform vs. Fourier Series Introduction Sufficient condition for the existence of a Fourier transform | f (t ) |dt That is, f(t) is absolutely integrable. However, the above condition is not the necessary one. Some Unabsolutely Integrable Functions Functions: cos t, sin t,… Unit Step Function: u(t). Sinusoidal Generalized – – Functions: Impulse Function (t); and Impulse Train. Fourier Transforms of Special Functions More on Impulse Function Dirac Delta Function 0 t 0 (t ) t 0 and (t )dt 1 Also called unit impulse function. 0 t Generalized Function The value of delta function can also be defined in the sense of generalized function: (t )(t )dt (0) (t): Test Function We shall never talk about the value of (t). Instead, we talk about the values of integrals involving (t). Properties of Unit Impulse Function (t t0 )(t )dt (t0 ) Pf) Write t as t + t0 (t t0 )(t )dt (t )(t t0 )dt (t0 ) Properties of Unit Impulse Function 1 (at)(t )dt | a | (0) Pf) Write t as t/a Consider a>0 (at)(t )dt 1 t (t ) dt a a 1 (0) |a| Consider a<0 (at)(t )dt 1 t (t ) dt a a 1 (0) |a| Properties of Unit Impulse Function f (t )(t ) f (0)(t ) Pf) [ f (t )(t )](t )dt (t )[ f (t )(t )]dt f (0)(0) f (0) (t )(t )dt [ f (0)(t )](t )dt Properties of Unit Impulse Function f (t )(t ) f (0)(t ) Pf) 1 (at) (t ) |a| 1 1 (at)(t )dt | a | (0) | a | (t )(t )dt 1 (t )(t )dt | a | Properties of Unit Impulse Function f (t )(t ) f (0)(t ) 1 (at) (t ) |a| t(t ) 0 (t ) (t ) Generalized Derivatives The derivative f’(t) of an arbitrary generalized function f(t) is defined by: f ' (t )(t )dt f (t )' (t )dt Show that this definition is consistent to the ordinary definition for the first derivative of a continuous function. f ' (t )(t )dt f (t )(t ) f (t )' (t )dt =0 Derivatives of the -Function ' (t )(t )dt (t )' (t )dt ' (0) d(t ) ' (t ) , dt d(t ) ' (0) dt t 0 (t )(t )dt (1) (0) ( n) n d (t ) ( n) (t ) , n dt n ( n) n d (t ) ( n) (0) n dt t 0 Product Rule [ f (t )(t )]' f ' (t )(t ) f (t )' (t ) Pf) [ f (t )(t )]' (t )dt [ f (t )(t )]' (t )dt (t )[ f (t )' (t )]dt (t ){[ f (t )(t )]' f ' (t )(t )}dt (t )[ f (t )(t )]'dt (t )[ f (t )'(t )]dt ' (t )[ f (t )(t )]dt (t )[ f (t )'(t )]dt [' (t ) f (t ) (t ) f ' (t )](t )dt Product Rule f (t )' (t ) f (0)' (t ) f ' (0)(t ) Pf) f (t )' (t ) [ f (t )(t )]' f (t )'(t ) [ f (0)(t )]' f (0)' (t ) f ' (0)(t ) Unit Step Function u(t) Define u(t )(t )dt (t )dt 0 u(t) 0 t 1 t 0 u (t ) 0 t 0 Derivative of the Unit Step Function Show that u' (t ) (t ) u ' (t )(t )dt u (t )' (t )dt ' (t )dt 0 [() (0)] (0) (t )(t )dt Derivative of the Unit Step Function (t) u(t) Derivative 0 t 0 t Fourier Transforms of Special Functions Fourier Transform Related to Impulse Function Fourier Transform for (t) (t ) 1 F F [(t )] (t )e jt dt e jt t 0 1 F(j) (t) F 0 t 1 0 Fourier Transform for (t) Show that 1 jt (t ) e d 2 1 j t 1 jt e d 1 e d (t ) F [1] 2 2 1 1 j t e d converges to The integration 2 in the sense of generalized function. (t ) Fourier Transform for (t) 1 Show that (t ) cos td 0 1 1 jt (cos t j sin t )d (t ) e d 2 2 1 j cos td sin td 2 2 1 Converges to (t) in the sense of cos td generalized function. 0 Two Identities for (t) 1 jxy ( y ) e dx 2 1 ( y ) cos xydx 0 These two ordinary integrations themselves are meaningless. They converge to (t) in the sense of generalized function. Shifted Impulse Function (t t0 ) e F jt0 Use the fact F [ f (t t0 )] F ( j)e jt0 (t t0) |F(j)| F 0 t0 t 1 0 Fourier Transforms of Special Functions Fourier Transform of a Some Special Functions Fourier Transform of a Constant f (t ) A F ( j) A2() F F ( j) F [ A] Ae jt d 1 j ( ) t 2A e dt 2 2A() Fourier Transform of a Constant f (t ) A F ( j) A2() F F(j) 0 A2() F A t 0 Fourier Transform of Exponential Wave f (t ) e F [ f (t )e j0t j0t F ( j) 2( 0 ) F ] F[ j( 0 )] F [1] 2() F [e j0t ] 2( 0 ) Fourier Transforms of Sinusoidal Functions cos0t ( 0 ) ( 0 ) F sin 0t j( 0 ) j( 0 ) F F(j) (+0) (0) f(t)=cos0t t F 0 0 0 Fourier Transform of Unit Step Function Let F [u(t )] F ( j) F [u(t )] F ( j) u(t ) u(t ) 1 (exceptfor t 0) F [u(t ) u(t )] F [1] F [u(t )] F [u(t )] 2() F ( j) F ( j) 2() F(j)=? Can you guess it? Fourier Transform of Unit Step Function Guess F ( j) k() B() k F ( j) F ( j) k() k() B() B() 2k() B() B() 0 B() must be odd F ( j) F ( j) 2() Fourier Transform of Unit Step Function k Guess F ( j) k() B() u' (t ) (t ) F [u (t )] F ( j) 1 B() j F [u' (t )] F [(t )] 1 F [u ' (t )] jF ( j) j[() B()] j() jB() 0 Fourier Transform of Unit Step Function Guess F ( j) k() B() 1 u (t ) () j F k 1 B() j Fourier Transform of Unit Step Function |F(j)| f(t) F 1 t 0 1 u (t ) () j F () 0 Fourier Transforms of Special Functions Fourier Transform vs. Fourier Series Find the FT of a Periodic Function Sufficient condition --- existence of FT | f (t ) |dt Any periodic function does not satisfy this condition. How to find its FT (in the sense of general function)? Find the FT of a Periodic Function We can express a periodic function f(t) as: f (t ) c e n n jn0t , 2 0 T F ( j) F [ f (t )] F cn e jn0t cnF [e jn0t ] n n c 2( n ) n n 0 2 cn ( n0 ) n Find the FT of a Periodic Function We can express a periodic function f(t) as: f (t ) c e n jn0t n , 2 0 T F ( j) 2 cn ( n0 ) n The FT of a periodic function consists of a sequence of equidistant impulses located at the harmonic frequencies of the function. Example: Impulse Train 3T 2T T T (t ) 0 T 2T 3T t (t nT ) n Find the FT of the impulse train. Example: Impulse Train 3T 2T T 0 T 2T 3T t 1 jn0t Find the FT of the T (t ) (t nT ) T (t ) e impulse T train. n n cn 2 F [T (t )] ( n0 ) Example: T n 0 Impulse Train 3T 2T T 0 T 2T 3T t 1 jn0t Find the FT of the T (t ) (t nT ) T (t ) e impulse T train. n n cn 2 F [T (t )] ( n0 ) Example: T n 0 Impulse Train 3T 2T T 0 T 2T 3T 0 20 30 t F 2/T 30 20 0 0 Find Fourier Series Using Fourier Transform f(t) t T/2 f (t ) c e n jn0t n 1 cn T Fo ( j) f o (t )e jt T / 2 T /2 T / 2 f (t )e jn0t 1 cn Fo ( jn0 ) T T /2 T/2 f (t )e jt T/2 fo(t) t T/2 Sampling the Fourier Transform of fo(t) with period 2/T, we can find the Fourier Series of f (t). Find Fourier Series Using Fourier Transform f(t) t T/2 f (t ) c e n jn0t n 1 cn T Fo ( j) f o (t )e jt T / 2 T /2 T / 2 f (t )e jn0t 1 cn Fo ( jn0 ) T T /2 T/2 f (t )e jt T/2 fo(t) t T/2 Example: The Fourier Series of a Rectangular Wave f(t) 1 1 d f (t ) 0 jn0t c e n n fo(t) t t 0 Fo ( j) d /2 e jt dt d / 2 2 d 1 sin cn Fo ( jn0 ) 2 T 2 1 n0 d n0 d sin sin Tn0 2 n 2 F ( j) 2 cn ( n0 ) Example: n The Fourier Transform of a Rectangular Wave f(t) 1 d f (t ) t 0 jn0t c e n n F [f(t)]=? 2 n0 d F ( j) sin ( n0 ) 2 n n 1 cn Fo ( jn0 ) T 2 1 n0 d n0 d sin sin Tn0 2 n 2