C.6: CHT LNG IU KHIN C.6: CHT LNG IU KHIN H THNG IU KHIN S H THNG IU KHIN S 6.1. SAI LCH TNH • nh ngha: Sai lch gia đi lng đu vào và đi lng đu ra trng thái xác lp. 6.2. Kiu (loi) hàm truyn đt •Kiu (loi) hàm truyn đt bng s lng đim cc bng 1. 10 1 () 1 A zA Gz z + = − …kiu “1” 10 2 () A zA Gz z + = …kiu “0” ()( ) 10 3 () 10.5 Az A Gz zz + = −− …kiu “1” 10 3 32 () 2.5 2 0.5 Az A Gz zzz + = −+− ()( ) 10 2 10.5 Az A zz + = −− …kiu “2” 6.3. H thng có mt vòng kín G h (z) (-) X(z) Y(z)E(z) x(kT) e(kT) y(kT) lim ( ) t k s ekT →∞ = 1 1 lim ( ) z z E z z → − = 1 1() lim 1() z h z Xz z Gz → − =⋅ + nh ngha các hng s •Hng s bc thang 1 lim ( ) bt h z K Gz → = •Hng s bc mt () 1 1 lim 1 ( ) bm h z K zGz T → =− •Hng s bc hai () 2 2 1 1 lim 1 ( ) bh h z K zGz T → =− Tín hiu đu vào () 1 z Xz z ρ ⇒= − • Tín hiu đu vào là hàm bc thang: () .1() x kT kT ρ = 11 1() 1 lim lim 1() 1()1 tbt zz hh zXz z z ss zGz zGzz ρ →→ − − == ⋅ = ⋅ ⋅ + +− 1 1 lim 1()1lim() bt z hh z s Gz Gz ρ ρ → → == ++ 1 bt bt s K ρ = + Tín hiu đu vào () 2 () 1 zT Xz z ρ ⇒= − • Tín hiu đu vào là hàm t l bc mt vi thi gian: () .() x kT kT ρ = () 2 11 1() 1 lim lim 1() 1() 1 tbm zz hh zXz z zT ss zGz zGz z ρ →→ − − == ⋅ = ⋅ ⋅ ++ − 1 1 lim 11 1 ( 1) ( 1) ( ) lim( 1) ( ) bm z hh z s zzGz zGz TT T ρ ρ → → == −+ − − bm bm s K ρ = Tín hiu đu vào () 2 3 (1) () 2 1 zz T Xz z ρ + ⇒= − 2 () .() 2 xkT kT ρ = • Tín hiu đu vào là hàm t l bc hai vi thi gian: () 2 3 11 1() 11 (1) lim lim 1() 1()2 1 tbh zz hh zXz z zzT ss zGz zGz z ρ →→ −− + == ⋅ = ⋅ ⋅⋅ ++ − 1 2 22 2 22 1 (1) lim 1 11 lim( 1) ( ) 2(1) (1)() bh z h h z z s zGz zzGz T TT ρ ρ → → + == ⎡⎤ − −+ − ⎢⎥ ⎣⎦ bh bh s K ρ = Hàm truyn đt G h (z) ()() () 12 () ( ) ; 1; 1,2, , hi n Mz Gz z i n zz zz zz =∀≠= − − ⋅⋅⋅ − •G h (z) kiu “0”: ()() () ()() () 11 12 12 () lim ( ) lim (1) 11 1 bt h zz n bt n Mz KGz zz zz zz M Kconst zz z →→ == −−⋅⋅⋅− == −−⋅⋅⋅− 1 bt bt s const K ρ == +