《电路》（英文版）15-8 The transfer function(or network function)

8 15-8 The transfer function(or network function) H(S) We define transfer function H(s)as a ratio of the Laplace transform of system output (or response) vo(s) to the Laplace transform of the input (or forcing function) vi(s)when all initial conditions are zero. then H(S) or Vo(s)=H(svi(s) (s) when Vi(s)=l Vo(s)=H(s) and V(s)=1 U()=(t) The inverse function corresponding to the transfer function H(s)is the unit-impulse response of the circuit. h(t)=L [H(s)=L o(s=0(

§15-8 The transfer function(or network function) H(s) We define transfer function H(s) as a ratio of the Laplace transform of system output (or response) Vo (s) to the Laplace transform of the input (or forcing function) Vi (s) when all initial conditions are zero, then ( ) ( ) ( ) V s V s H s i O = or V (s) H(s)V (s) o = i when V (s) 1 V (s) H(s) i = o = and V (s) 1 (t) (t) i = i =  The inverse function corresponding to the transfer function H(s) is the unit-impulse response of the circuit. ( ) [ ( )] [ ( )] ( ) 1 1 h t L H s L V s t = = O =o − −

H(s) usually is a ratio of two polynomes containing s, by means of partial-fraction-expansion obtain H(S) N(S) D(s) S-P CB(=C1∑k|=∑ s-P i=1 815-9 The complex-frequency plane H(S) N(S) (S-x1)(S-z2)…(s-zmn) D(S) 0 (S-p1)(S-p2)…(S-Pn) Ho real number zeros P,p2… Pn poles

H(s) usually is a ratio of two polynomes containing s, by means of partial-fraction-expansion obtain = − = = n i i i s p k D s N s H s 1 ( ) ( ) ( )   = = − − = − = = n i p t i n i i i i k e s p k h t L H s L 1 1 1 1 ( ) [ ( )] [ ] §15-9 The complex-frequency plane ( )( )...( ) ( )( )...( ) ( ) ( ) ( ) 1 2 1 2 0 n m s p s p s p s z s z s z H D s N s H s − − − − − − = = H0 real number z1, z2, … zm zeros p1, p2, … pn poles

The complex frequency s=o+ja q q (s- plane)

The complex frequency s =  + j  j  j 0 (s − plane)

Example 2: Find: Z(=? 1+j5× s plane J 1+j5 j5× a) (b) (a) The pole-zero constellation of some impedance Z(s) (b)a portion of the rubber-sheet model of the magnitude of z(s) s+2 Z(s)=k s+2 (s+1-j5)(s+1+j5) or Z(s)=k s2+2s+26 J:z(0)=1,∴1=k→k=13 Z(s)=13-+2 s2+2s+26

Example 2: Find: Z(s)=? ( 1 5)( 1 5) 2 ( ) s j s j s Z s k + − + + + = 2 26 2 ( ) 2 + + + = s s s or Z s k 2 26 2 ( ) 13 2 + + +  = s s s 13 Z s 26 2 If : Z(0) = 1, 1 = k  k = (a) (b) (a) The pole-zero constellation of some impedance Z(s); (b) A portion of the rubber-sheet model of the magnitude of Z(s). j  − 2 −1+ j5 −1− j5 s plane j  −1+ j5