NAMA : ROLLY EGA SUGANDA NIM : J1D111025 LAPORAN SEMENTARA PERCOBAAN 4 PSD NO LISTING PROGRAM HASIL 4.1 Transformasi – Z 4.1 .2 B = [4, 5, 6]; zero1 = (-B(2)+sqrt(B(2)^2- 4*B(1)*B(3)))/(2*B(1)) zero2 = (-B(2)-sqrt(B(2)^2- 4*B(1)*B(3)))/(2*B(1)) A=[1, -2, -3]; pole1 = (-A(2)+sqrt(A(2)^2- 4*A(1)*A(3)))/(2*A(1)) pole2 = (-A(2)-sqrt(A(2)^2- 4*A(1)*A(3)))/(2*A(1)) 4.2 Pengenalan Diagram Pole- Zero 4.2 .1 zplane(B,A); -1 -0.5 0 0.5 1 1.5 2 2.5 3 -1.5 -1 -0.5 0 0.5 1 1.5 R ealP art Im aginary P art 4.3 I Invers Transformasi - Z
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NAMA : ROLLY EGA SUGANDA
NIM : J1D111025
LAPORAN SEMENTARA PERCOBAAN 4 PSD
NO LISTING PROGRAM HASIL4.1 Transformasi – Z4.1.2
B = [4, 5, 6];zero1 = (-B(2)+sqrt(B(2)^2-4*B(1)*B(3)))/(2*B(1))zero2 = (-B(2)-sqrt(B(2)^2-4*B(1)*B(3)))/(2*B(1))
B=[4,5,6];zero1=(-B(2)+sqrt(B(2)^2-4*B(1)*B(3)))/(2*B(1))zero2=(-B(2)+sqrt(B(2)^2-4*B(1)*B(3)))/(2*B(1))%Untuk mencari letak pole :A=[1,-2,-3];pole1=(-A(2)+sqrt(A(2)^2-4*A(1)*A(3)))/(2*A(1))pole2=(-A(2)+sqrt(A(2)^2-4*A(1)*A(3)))/(2*A(1))zplane(B,A);[R,P,K]=residuez(B,A)