TUYN TP THI V BI GII MN X L TN HIU S(TRANG 1)Li ni u:Cc bn thn mn!Ph cng c hc mn x l tn hiu s ca thy ng Hng Lnh v th Ph bit c mn hc ny tng i kh hc. L thuyt nhiu nhng rtkh hiu v vy Ph c gng son li gii cho cc thi ca cc nm trc! Thit ngh, gip ngi khc l mt nim vui, t to ng lc Ph vit ra ti liu ny. Hy vng n s gip cc bn khng cn phi lo lng khi hc mn ny na! Mnh dm khng nh rng nu cc bn n theo nhng g m Ph vit y th kh nng im A mn ny s nm trong tm tay ca cc bn. Nu thy khng chu i mi cch ra thi th chc chn l s nm trong ti liu ny m thi! Cc bn yn tm nha! s dng mt cch hiu qu nhtti liu ny, mnh ngh cc bn nn ti v ri in ra giy cho d hc hn v sn khi no cng c th ly ra c c. Nh vy kin thc mi chc chn c.V nh l phi chia s v gii thiu cho bn b v mi ngi bit n ti liu ny vi nh! V hy lun c gng v im 10 nha cc bn! Cui cng, iu m mnh mun gi ti cc bn l khi cc bn s dng ti liu ny th mong cc bn tn trng ngi vit ra n nh! ng hin tng n cp thng hiu, bn quyn xy ra nha! Bi v c c ti liu ny, Ph bao nhiu l tm huyt vo y! Bn cnh , do thi gian khng cho php nn trong qu trnh gii c th c mt s sai st (rt t hoc khng c u) th mong cc bn thng cm cho Ph vi. Chc cc bn thnh cng!Mi kin ng gp v thc mc xin gi v a ch:Email: phumot@gmail.comHocin thoi: 01682577107Xin chn thnh cm n!Vinh, ngy 11 thng 11 nm 2010Bin son:Cung nh PhLi giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 2) 46 A-B-K TIN Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 3)BI GIICu 1: (5)a. Xc nh phng trnh quan h vo raTa c:H(Z)=141 2Z + 121 3Z H(Z) = ( ) ( )( ) ( )1 11 14 1 3 2 1 21 2 1 3Z ZZ Z + = 1 11 1 24 12 2 41 3 2 6Z ZZ Z Z + + H(Z) = 11 26 161 5 6ZZ Z +MH(Z) = ( )( )Y ZX Z = 11 26 161 5 6ZZ Z + Y(Z)(1-5Z-1+6Z-2) = 6X(Z)(6-16Z-1) Y(Z)-5Z-1Y(Z)+6Z-2Y(Z) = 6X(Z)-16Z-1X(Z) Y(Z) = 5Z-1Y(Z)-6Z-2Y(Z)+6X(Z)-16Z-1X(Z) y(n)=5y(n-1)-6y(n-2)+6x(n)-16x(n-1)Vy phng trnh quan h vo ra l:y(n)=5y(n-1)-6y(n-2)+6x(n)-16x(n-1)b. Xc nh p ng xungTa c: ROC1: |Z| X(Z) = 111 ZY(Z) =( ) ( ) ( )1 20, 25y 2 0, 25y 1 Z 0, 25Z Y Z + + +111 ZThay vo iu kin ban u, ta c:Y(Z) =( )1 20, 25.1 0, 25.0.Z 0, 25Z Y Z + + +111 Z Y(Z)(1-0,25Z-2) = 0,25+111 Z Y(Z)(1-0,25Z-2) = ( )110, 25 1 11ZZ + Y(Z)(1-0,25Z-2) = 110, 25 0, 25 11ZZ + = 111, 25 0, 251ZZ Y(Z) = ( ) ( )11 21, 25 0, 251 1 0, 25ZZ Z = ( ) ( ) ( )11 1 11, 25 0, 251 1 0, 5 1 0, 5ZZ Z Z + = ( )11AZ+( ) ( )1 11 0, 5 1 0, 5B CZ Z + + A( )11 0, 5Z( )11 0, 5Z++B( ) ( ) ( ) ( )1 1 1 11 1 0, 5 1 1 0, 5 Z Z C Z Z + + = A-0,25Z-2+B-0,5BZ-1-0,5BZ-2+C-1,5CZ-1+0,5CZ-2 = 1,25-0,25Z-11, 250, 5 1, 5 0, 250, 25 0, 5 0, 5 0A B CB CA B C+ + ' + =>4338724ABC 'Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 9) Y(Z) = ( )143 1 Z-( )138 1 0, 5Z+( )1724 1 0, 5Z+ y(n) = 4( )3U n-3(0, 5) ( )8nU n+7( 0, 5) ( )24nU n Vy:y(n) = 4( )3U n-3(0, 5) ( )8nU n+7( 0, 5) ( )24nU n HT 45B-K-A TinCu 1: (2 im)Mt h thng tuyn tnh bt bin c:( ) ( ) ( 1) ( 2) 2 ( 3) x n n n n n + ( ) ( ) ( 1) ( 2) ( 5) h n n n n n + + + Hy xc nh v v tn hiu ra y(n) ca h thng.Cu 2: (3 im)Mt b lc IIR c p ng xung c cho nh sau:20.4 0,1( )2(0.5) 2nkhinh nkhin '(ch : khin tc l khi n nha!)Hy xc nh phng trnh quan h vo ra ca b lc trn.Cu 3: (5 im)Mt h thng tuyn tnh bt bin c hm truyn t c cho nh sau:1 11 2( )1 0.5 1 0.3H ZZ Z + a) Xc nh phng trnh quan h vo rab) T hm truyn t, hy xc nh tt c cc p ng xung ca h thng.c) T phng trnh quan h tm c, hy v s thc hin h thng di dng chun tc 2 ri thc hin thut ton trong my tnh cho s ny.d) Trn c s ca hm truyn t, v s thc hin h thng di dng cc h thng ghp ni song song v xy dng thut ton trong my tnh cho s ny.e) Trn c s ca hm truyn t, hy v s thc hin di dng cc h thng ghp ni tip vi nhau v xy dng thut ton tnh cho s ny.BI GIILi giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 10)Cu 1:Ta c:30( ) ( ) ( )ky n x k h n k ( ) (0) ( ) (1) ( 1) (2) ( 2) (3) ( 3) y n x h n x h n x h n x h n + + + ( ) ( ) ( 1) ( 2) 2 ( 3) y n h n h n h n h n + M:( ) ( ) ( 1) ( 2) ( 5) h n n n n n + + + ( 1) ( 1) ( 2) ( 3) ( 6) h n n n n n + + + ( 2) ( 2) ( 3) ( 4) ( 7) h n n n n n + + + ( 3) ( 3) ( 4) ( 5) ( 8) h n n n n n + + + => ( ) ( ) 2 ( 1) ( 2) 2 ( 3) 3 ( 4) ( 5) ( 6) ( 7) 2 ( 8) y n n n n n n n n n n + + + V ? Qu d dng (cc bn t v ly nha, bn no khng th v c th ni vi Ph mt cu nha!)Cu 2:Ta c:0( ) ( ) ( )ky n h k x n k y(n) = h(0)x(n)+h(1)x(n-1)+h(2)x(n-2)+h(3)x(n-3)+y(n) = 0,4x(n)+0,4x(n-1)+2x(n-2)+2(0,5)x(n-3)+y(n-1) = 0,4x(n-1)+0,4x(n-2)+2x(n-3)+2(0,5)x(n-4)+ (1)Nhn (1) vi 0,5 ta c:0,5y(n-1) = 0,2x(n-1)+0,2x(n-2)+2.0,5x(n-3)+2(0,5)2x(n-4)+ y(n)-0,5y(n-1) = 0,4x(n)+0,2x(n-1)+1,8x(n-2) y(n) = 0,5y(n-1)+0,4x(n)+0,2x(n-1)+1,8x(n-2)Vy phng trnh quan h vo ra ca h thng l:y(n) = 0,5y(n-1)+0,4x(n)+0,2x(n-1)+1,8x(n-2)Cu 3:a. Xc nh quan h vo ra:Ta c:1 11 2( )1 0.5 1 0.3H ZZ Z + = ( )( ) ( )1 11 11 0.3 2 1 0.51 0.5 1 0.3Z ZZ Z + = 11 23 1.31 0.8 0.15ZZ Z +M:Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 11)H(Z) = ( )( )Y ZX Z = 11 23 1.31 0.8 0.15ZZ Z + Y(Z)(1 21 0.8 0.15 Z Z + ) = X(Z)(13 1.3Z ) Y(Z)-0.8Z-1Y(Z)+0.15Z-2Y(Z) = 3X(Z)-1.3Z-1X(Z) Y(Z) = 0.8Z-1Y(Z)-0.15Z-2Y(Z)+3X(Z)-1.3Z-1X(Z) y(n) = 0.8y(n-1)-0.15y(n-2)+3x(n)-1.3x(n-1)Vy, phng trnh quan h vo ra l:y(n) = 0.8y(n-1)-0.15y(n-2)+3x(n)-1.3x(n-1)b. Xc nh p ng xungTa c: ROC1: |Z|0.50.4ZZ 'H(Z) = 30+11 222 291 0.9Z 0.2ZZ + = 30+1 11 0.5 1 0.4A BZ Z + 290.4 0.5 22A BA B+ ' 75104AB ' Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107S1 S2 S3TUYN TP THI V BI GII MN X L TN HIU S(TRANG 17) H(Z) = 30+1 175 1041 0.5 1 0.4 Z Z Ta c cc p ng xung l:+) ROC1: |Z|11( )113X ZZ Y(Z) = 111 1 1( 1) ( )12 213y Z Y ZZ + +Thay y(-1) = 1 vo: Y(Z) = 111 1 1( )12 213Z Y ZZ+ +111 1 1( ) 112 213Y Z ZZ _ + ,1111 11 11 2 3( ) 11213ZY Z ZZ _ + _ , ,113 12 6113ZZ Y(Z) = 11 13 12 61 11 13 2ZZ Z _ _

, ,Ta c:Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 23)Y(Z) = 1 11 11 12 3A BZ Z + Y(Z) = 1 11 11 11 13 21 11 12 3A Z B ZZ Z _ _ + , , _ _

, ,11 13 12 61 11 13 2ZZ Z _ _

, ,3213 2 6A BA B+ ' =>722AB ' Vy Y(Z) = 1 17221 11 12 3Z Z Do|Z|>13 cho nn:+) Nu 1 1| |3 2Z < 137 1 1( ) ( ) 2 ( )2 2 3n ny n U n U n _ _ , ,HT 48 Tn chLi giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 24)BI GIICu 3: (5 im)Li giiv bin son bi Cung nh Ph, lp 48k tin, trng i hc VinhEmail: phumot@gmail.comhotline: 01682577107TUYN TP THI V BI GII MN X L TN HIU S(TRANG 25)a) Ta c:1 11 2( )1 0.6 1 0.3H ZZ Z + = ( )( ) ( )1 11 11 0.3 2 1 0.61 0.6 1 0.3Z ZZ Z + = 11 23 1.51 0.9 0.18ZZ Z +M:H(Z) = ( )( )Y ZX Z = 11 23 1.51 0.9 0.18ZZ Z +=>( ) ( )1 2 1( ) 1 0.9 0.18 ( ) 3 1.5 Y Z Z Z X Z Z + => Y(Z)-0.9Z-1Y(Z)+0.18Z-2Y(Z) = 3X(Z)-1.5Z-1X(Z)=> Y(Z) = 0.9Z-1Y(Z)-0.18Z-2Y(Z)+3X(Z)-1.5Z-1X(Z)=>y(n) = 0.9y(n-1)-0.18y(n-2)+3x(n)-1.5x(n-1)Vy, phng trnh quan h vo ra ca h thng trn l:y(n) = 0.9y(n-1)-0.18y(n-2)+3x(n)-1.5x(n-1)b) Xc nh p ng xungTa c: ROC1: |Z|