MỤC LỤC 1 TÓM TẮT !" #$%&'!()*+!,&-'. '/'."0'.1 &1.2341'. 45'6788.8*9&":.;< 4-=.>44?&+1$ +34.34 ?&4 &'."0;@&A1; (B.3 CDCEF 4'6' '/6"0;@&A EG&HIJ 34?&A;'64 * * &"03H A1C%A3&H(;&.=( *KL0" I. GIỚI THIỆU 0&M;BNCEF'6OP EG&H .4'/%O?&Q@&4'/1 4.%&*"R'! '/ SB-.3!.A3F"RG&HI4 34&*.T'6E4'/3O3+U3Q I' VBW CF1I;(@ &B CF1 3+UAV5 0;(B.3 CDCEFA'6 .Q(@&ABA;'6F"#X;(1 .EA;'61'$'6 '6." :Y'6&;(, • 0;>Z4.!44& [4" • 0;>4O*.\ !2Y E3O%$Y];" • ^A;/A;1'_E 4'Q+ 4'/14" • ` 3S;4E$E2.EIa.3S& A434?&" 2 • 0;4'61$ '/.?&A" II. PHÁT BIỂU BÀI TOÁN 2.1. Phát biểu bài toán b;, c>&.,def4,g4'/h1g49 ijf4$\E'!" c>&,k9(hl" 2.2. Hướng giải quyết #$f'!+H4?&+3;, ^'!9;(*E419 +E4 &./+*O/ 4V *3+ H*"&$1A 34?&.437UO 4E4'Q&+; '64.434 3.@'!@;$$.! 394<&M.="cA+H 'a 3?&V;BmbnmCWbF5 ^'!9 o(8'aE4+E ?&EG&H3+U"#EG&H3+U .8 &3+ HE1'.H&. /'63&()!.!3+!" &$1A&M.=.437U 3+U@6434; '64. 3.@&M.=&E"#$&I3+U(@ & 4$$'a3+U'6o(8;'K#0 BKCC #C 0pF1 KL0 BKCC L&C 0pF1 KN0 KCCN.C C0pBF5 031C*9&.+HEG&H;C'! 9f1;*>4?&3+ UKL0" RG&H&M;o(8I3+UKL04 E1I3+UKL0.!-E4'6 3 >;1@.>"cEG&H&M;3.@ V>@qE3&M;C KL04E+?&4C9&, BgF 0$, n,, , > '6 E4E+?&414'/14 >KL0" n : VE4.4<3+U KL0" #.'6!E4'a9,."`! OS;. !;Q. OS ;. !;" III. THUẬT TOÁN 3.1. Phép biến đổi DFT rI3+UKL0f&H;:%s, +− − = − = ∑ ∑ = N yv M xu j M x N y eyxfvuF π 2 1 0 1 0 ),(),( (2) 0$,t≤&≤:ng1t≤.≤sng1uv 1− " 0$,C u% v%wu%"&BfF'a'a.!, ∑∑ − = − = +−+= 1 0 1 0 )(π2sin)(π2cos),(),( M x N y N vy M ux j N vy M ux yxfvuF BxF nLB&1.F 9, wr>'a9.! wr>4'a9.!" 4 rI3+UKL0'6, + − = − = ∑ ∑ = N yv M xu j M u N v evuF MN yxf π 2 1 0 1 0 ),( 1 ),( (4) 0$,t≤%≤:ng1t≤≤sng R:vs, 0ABfFQ, + − − = − = ∑ ∑ = N vyxu j N x N y eyxfvuF π 2 1 0 1 0 ),(),( t≤&≤sng1t≤.≤sngBflF 0AByFQ, + − = − = ∑ ∑ = N yvux j M u N v evuF N yxf π 2 1 0 1 0 2 ),( 1 ),( t≤%≤sng1t≤≤sngBylF 3.2. Các bước nhúng thủy ấn z&A;('6C3'!' &, b'!g,#4h.4iEE'!{%{ Bf4$\E'!F"E37437B&|& F}~yj b'!f,b+UKL0;4E4 '64f4 .4" b'!x,•(8.EKL0'a94. 4C9BgF'6EE4; >KL0" b'!y,b+UhKL0BKL0'6FE4&'6-3'!1 &'64E+?&414<;" ae3'!;*KL0, 5 6 Hình 1:#3'!;(*KL0 IV. KỸ THUẬT CÀI ĐẶT 0?&AY&H.=C 9BgF1>4A.434*&>&, n^434; '64>&4.!41 X4344/*41($ 4 !.4S"z&*9&.o; 7E4. '/HO > '6'&,Bt"{€1 t"•€F.Bt1t‚1t1g‚F" n`!qE4>AY.E&434 "s+&ƒ(\gYV;4E 7 4AE„()(A'6fo(8 .-$E84;.>.!4E$ ;"0&*E&=EUe1(?&A -E E$.91*(83 E4.!\vt"g" R3+UKL04144E4 S3+UKL0.hKL01&*.4 E$E'!+A>4%C%I\6"0?& A3+UKL0&H.'614 X1e/q 4&E3+UO&4'Q4 =H E4<"s+&ES1A. 1; '64E+?&4>.!4' 4;"s'6 1+&4E !1/ 13+UKL0.hKL0 =&1; '64EI(>' 4A2 *"0e14E{%{1 E'!46 …1 434/1; '6 4E+?&4.4;" V. KẾT LUẬN # *9&.+H'a4 ;*>KL01.E+?&4'6434*&>&Y, 4;&'6Z1]I.!414 A;'E 4'Q+4" b*.;*>KL01< +;B.[C9BgFF+* E4; '614;& '6-'a*E3O; V)&"`('!= E+?&4&'6, k>&., 8 TÀI LIỆU THAM KHẢO †g‡u&r":1R"ˆ"ˆE1:RE ‰Š‹ K#0K`3 CNCE0C?&CphCŒ" †f‡s&)0+^enŠ` '6e.=4(*I 3+UKN0E+6I3+UK#0.H4[& *Œ‰0RV.#1~‚B€F,~y‰‚t" †x‡id1 K"KC%C1‹C:"iE &‰Ši3C(( :& CNCECKL0K•mD(^LC?&C b(Œ †y‡ ‹ m" RC( ( b d n Š‹ 3& C DCEC?&C3C(C& (C&(( C?&CCŒ †€‡c00P‰Šb4sH%o …4Œ" †~‡s&)Ž&=^&10>z&K_‰Š:&H.=4 *K#0Œ" 9 . !;Q. OS ;. !;" III. THUẬT TOÁN 3.1. Phép biến đổi DFT rI3+UKL0f&H;:%s, +− − = − = ∑ ∑ = N yv M xu j M x N y eyxfvuF π 2 1 0 1 0 ),(),( . 0;4'61$ '/.?&A" II. PHÁT BIỂU BÀI TOÁN 2.1. Phát biểu bài toán b;, c>&.,def4,g4'/h1g49 ijf4$E'!" c>&,k9(hl" 2.2 ∑ = N yvux j M u N v evuF N yxf π 2 1 0 1 0 2 ),( 1 ),( t≤%≤sng1t≤≤sngBylF 3.2. Các bước nhúng thủy ấn z&A;('6C3'!' &, b'!g,#4h.4iEE'!{%{ Bf4$E'!F"E37437B&|& F}~yj b'!f,b+UKL0;4E4