Ракетный блок. YZ-1. [Редактировать]

YZ1 (Yuanzheng-1 или Expedition-1) -верхняя ступень ракеты-носителя которая функционирует шесть с половиной часов. Ее отличительной особенностью является возможность многократного включения. В качестве топлива ступень использует несимметричный диметилгидразин как топливо и тетраоксид диазота как окислитель. Двигательная установка ступени выдает тягу 663 кгс. Удельный импульс 315 сек.

Отличительной особенность модификаций YZ-1A/2 является возможность многократного включения.

Дополнительные наименования

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Технические характеристики

#ХарактеристикаЗначение
1Тяга, кгс663
2Удельный импульс, с315

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Дата загрузки: 2017-04-09
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.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan... lines below (where y = x’) x y x+y 0 0 0 0 1 1 1 0 1 1 1 1 11 Is X+YZ = (X+Y)(X+Z)? X Y Z X+Y X+Z YZ X+YZ (X+Y)(X+Z) 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 12 DeMorgan’s Theorem • We can... simplifications x’y’ + xyz + x’y = x’(y’ + y) + xyz = x’•1 + xyz = x’ + xyz = (x’ + x)(x’ + yz) = 1 • (x’ + yz) = x’ + yz 1. 3. 5. 7. 9. 10. 12. 14. 16. x+0=x x+1=1 x+x=x x + x’ = 1 (x’)’ = x x+y=y+x x + (y + z) = (x + y) + z x(y + z) = xy.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan... second form 15 Another Example F= X’YZ+ X’YZ’+XZ = X’Y(Z+Z’)+XZ (14) = X’Y . 1 + XZ (7) = X’Y+ XZ...’ = x x + x’y = x + y 4. 5. 6. x(x + y) = x (x + y)(x + y’) = x x(x’ + y) = xy 17 Consensus Theorem • XY + X’Z + YZ = XY + X’Z or its dual (X+Y)(X’+Z)(Y+Z) = (X+Y)(X’+Z) • The... the theorem: XY + X’Z + YZ = XY + X’Z + YZ(X + X’) = XY + X’Z + XYZ + X’YZ = XY + XYZ + X’Z + X’YZ = XY(1 + Z) + X’Z(1 + Y) = XY... keep “pushing” the complements inwards f(x,y,z) = x(y’z’ + yz) f’(x,y,z) = ( x(y’z’ + yz) )’ = x’ + (y’z’ + yz)’ = x’ + (y’z’)’ (yz)’ = x’ + (y + z)(y’ + z’) • [ complement both sides ] [ because (xy... then complement each literal – If f(x,y,z) = x(y’z’ + yz)… – …the dual of f is x + (y’ + z’)(y + z)… – …then... must be a product of literals f(x,y,z) = y’ + x’yz’ + xz • The advantage is that..., such as f(x,y,z), has 23 = 8 minterms: x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz • Each... one combination of inputs: Minterm x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz Is... where the function output is 1. x y z f(x,y,z) f’(x,y,z) 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 0 0 1 0 0 0 0 0 1 1 0 1 f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’ = m0 + m1 + m2 + m3... the corresponding maxterm Mi Minterm x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz • Shorthand...



Дата загрузки: 2017-04-23
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... consumption in kWh/annum: - Value ‘YZ’: Calibri bold 20 pt, 100... 100 % – round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 15 pt, 100... corners: 3,5 mm, - Value ‘WXYZ’ or ‘YZ’: Calibri bold at least 20... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 15 pt, 100... corners: 3,5 mm, - Values ‘WXYZ’ or ‘YZ’: Calibri at least 13 pt... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 15 pt, 100... corners: 3,5 mm, - Values ‘WXYZ’ or ‘YZ’: Calibri at least 13 pt... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 45 pt, 100... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 20 pt, 100... 100 % – round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 45 pt, 100... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 20 pt, 100... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 45 pt, 100... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 20 pt, 100... 100 %, round corners: 3,5 mm, - Values ‘YZ’: Calibri at least 15 pt... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 20 pt, 100... 100 %, round corners: 3,5 mm, - Values ‘YZ’: Calibri at least 18 pt... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 37,5 pt, 100... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 20 pt, 100... 100 %, round corners: 3,5 mm, - Values ‘YZ’: Calibri at least 12 pt... 100 %, round corners: 3,5 mm, - Value ‘YZ’: Calibri bold 15 pt, 100...



Дата загрузки: 2017-04-08
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Дата загрузки: 2017-04-23
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...! E(%*4-%!A#! c.,*4! ZKHLI! ! ! Q;)4*/&! J! ! ! A+4/$+*;(6! Yf! ! ! B8=! "If! fbLI! ! 8/+(;! YZ! ! ! ! IJJb! ! ! ! IWKI! V"We! V"Wb! IWIb.../*.g&!+6+5/*%5%/.!&1))6.L!!Q/!$(&!(2!%2&/(66+,!5()(5%/.!47!fZK!B`!%2!%/&!:*%,!2+/'4*CF!47! '$%5$!YZ"!B`!%&!$.,*4!(2,!IJI!B`!%&!/$+*;(6L!!#(2S(2%(g&!;(X4*!$.,*4)4'+*!)6(2/&!(*+!G%,(/1!NIJb...



Дата загрузки: 2017-04-09
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...: a. xyz + (xyz)' Ans. a. x 0 0 0 0 1 1 1 1 b. x(yz'+x'y) b. y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 xyz 0 0 0 0 0 0 0 1 (xyz)' 1 1 1 1 1 1 1 0 xyz + (xyz)' 1 1 1 1 1 1 1 1 x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 yz' 0 0 1 0 0 0 1 0 x'y 0 0 1 1 0 0 0 0 (yz' + x'y) 0 0 1 1 0 0 1 0 x(yz '+ x'y) 0 0 0 0 0 0 1 0 ______________________________________________________________________________ Page 2 Last... for the following: x 0 0 0 0 1 1 1 1 a. xyz + x(yz)' + (xyz)' Ans. a. y z xyz x(yz)' (xyz)' 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 0 b. (x+y)(x+z)(x'+z) Sum 1 1 1 1 1 1 1 1 b. x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 x+y 0 0 1 1 1 1 1 1 x+z 0 1 0 1 1 1 1 1 x' + z 1 1 1 1 0 1 0 1 (x + y)(x + z)(x' + z) 1 1 1 1 0 1 0 1 ______________________________________________________________________________... the complement of F if F(x,y,z) = x(y' + z). Ans. F(x,y,z) = x(y' + z) F'(x,y,z) = (x(y' + z))' = x'+(y' + z)' = x' + yz' ______________________________________________________________________________ 4. Using DeMorgan's Law, write... of F if F(x,y,z) = xy + x'z + yz'. Ans. F(x,y,z) = xy + x'z + yz' F'(x,y,x) = (xy + x'z + yz')' = (xy)'(x'z)'(yz')' = (x' + y')(x + z')(y' + z) (not simplified) ______________________________________________________________________________... of F if F(w,x,y,z) = xyz' (y'z + x)' + (w'yz + x' ). Ans. F(w,x,y,z) = xyz'(y'z + x)'+(w'yz + x) F'(w,x,y,z) = (xyz'(y'z + x)'+(w'yz + x))' = (xyz')'+((y'z + x)')'(w'yz + x)' = (xyz')'+((y'z + x)(w'yz + x')') = x'+y'+z+((y'z + x)((w'yz)'x'') = x'+y'+z+((y'z + x)((w+y' + z')x)) ______________________________________________________________________________ 6. a. Use... step. a. x'yz + xz b. (x + y)'(x' + y')' c. (x'y''z)' Ans. a. x'yz + xz = x'yz + xz(1) Identity = x'yz + xz(y + y') Inverse = x'yz + xyz + xy'z Distributive and Commutative = x'yz... + xy'z) Associative = (x' + x)yz + (y + y')xz Distributive (two applications) = (1)(yz) + (1)xz Inverse = yz + xz Identity...' b. x'yz + xz c. wx + w(xy + yz') Ans. a. xy + x'y b. x'yz + xz = = = = = = = = = = = = x(y + y') Distributive x(1) Inverse x Identity x'yz + xz(1) x'yz + xz(y + y') x'yz + xzy + xzy' x'yz + (xzy + xzy) + xzy' (x'yz + xzy) + (xzy + xzy') (x'yz + xyz) + (xyz + xy'z) (x' + x)yz + xz(y + y') (1)yz + xz(1) yz + xz.... yz + xyz' + x' y' z = xy + x' z Ans. Using identities: yz + xyz' + x'y'z = = = = = Using truth tables: x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 yz 0 0 0 1 0 0 0 1 (x +x')yz + xyz' + x'y'z xyz + x'yz + xyz' + x'y'z (x'y'z + x'yz) + (xyz + xyz') x'z(y' + y) + xy (z + z') x'z + xy xyz' 0 0 0 0 0 0 1 0 x'y'z 0 1 0 0 0 0 0 0 yz + xyz... + x' z Ans. xy + x'z + yz = xy + x'z + (1)yz = xy + x'z + (x + x')yz = xy + x'z + xyz + x'yz = (xy + xyz) + (x'z + x'yz) = xy(1 + z)+x'z(1 + y) = xy... simplifies to: a = x b = x'yz + xy'z + xyz' + xyz = yz + xz + xy c = x'yz' + xy'z' + xyz' + xyz... 1 10 1 1 1 0 0 0 0 00 1 01 1 11 1 10 0 1 1 1 1 1 YZ X c. YZ X Ans. a. x'z + xz' b. x'z + x'y + xy'z' c. x + y' + z ______________________________________________________________________________ 2. Create... following functions: a. F(x,y,z) = x'y'z' + x'yz + x'yz' b. F(x,y,z) = x'y'z' + x'yz' + xy'z' + xyz' c. F(x,y,z) = y'z' + y'z + xyz' Ans. a. x'y'z' + x'yz + x'yz' Simplifies to: x'y + x'z' b. x'y'z' + x'yz' + xy'z' + xyz' Simplifies to: z' YZ X 0 00 1 01... 1 1 1 1 1 1 1 1 01 11 1 10 1 1 1 Ans. w'x' + wx + w'y + yz' + x'z' c. YZ WX 00 01 00 1 01...: a. F(w,x,y.z) = w'x'y'z' + w'x'yz' + w'xy'z + w'xyz + w'xyz' + wx'y'z' + wx'yz' b. F(w,x,y,z) = w'x'y'z' + w'x'y'z + wx'y'z + wx'yz' + wx'y'z' c. F(w,x,y,z) = y'z + wy' + w'xy + w'x'yz' + wx'yz' Ans. a. w'xz + w'xy + x'y'z' + x'yz b. x'y' + wx'z' YZ YZ... reduce to 1 term. X Ans. x'y'z+x'yz+xy'z+xyz = = = = YZ 00 0 0 01 1 11 1 10... defined by the following K-map: YZ YZ 00 01 11 10 00... F'. Ans. w 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 a. x 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 F 1 0 1 0 0 1 1 1 1 0 1 0 0 0 0 0 b. F = x'z' + w'xz + w'xy OR x'z' + w'xz + w'yz' c. To find F', circle the zeros...



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... последовательности символов: А АВ ABC ... AB..YZ на экран. Решение. Последовательность символов...



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... 7SHJLZ [OYV\NO ;LJOUVSVN` PZ [OL ÄYZ[ PU[LYKPZJPWSPUHY` JVUMLYLUJL VYNHUPZLK PU... HUK 4HZ[LYWSHU -VY [OL ÄYZ[ [PTL [OL V\[W\[Z of the model...



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.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan... lines below (where y = x’) x y x+y 0 0 0 0 1 1 1 0 1 1 1 1 11 Is X+YZ = (X+Y)(X+Z)? X Y Z X+Y X+Z YZ X+YZ (X+Y)(X+Z) 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 12 DeMorgan’s Theorem • We can... simplifications x’y’ + xyz + x’y = x’(y’ + y) + xyz = x’•1 + xyz = x’ + xyz = (x’ + x)(x’ + yz) = 1 • (x’ + yz) = x’ + yz 1. 3. 5. 7. 9. 10. 12. 14. 16. x+0=x x+1=1 x+x=x x + x’ = 1 (x’)’ = x x+y=y+x x + (y + z) = (x + y) + z x(y + z) = xy.... 15. 17. x x x x •1=x •0=0 •x=x • x’ = 0 xy = yx x(yz) = (xy)z x + yz = (x + y)(x + z) (xy)’ = x’ + y’ Commutative Associative Distributive DeMorgan... second form 15 Another Example F= X’YZ+ X’YZ’+XZ = X’Y(Z+Z’)+XZ (14) = X’Y . 1 + XZ (7) = X’Y+ XZ...’ = x x + x’y = x + y 4. 5. 6. x(x + y) = x (x + y)(x + y’) = x x(x’ + y) = xy 17 Consensus Theorem • XY + X’Z + YZ = XY + X’Z or its dual (X+Y)(X’+Z)(Y+Z) = (X+Y)(X’+Z) • The... the theorem: XY + X’Z + YZ = XY + X’Z + YZ(X + X’) = XY + X’Z + XYZ + X’YZ = XY + XYZ + X’Z + X’YZ = XY(1 + Z) + X’Z(1 + Y) = XY... keep “pushing” the complements inwards f(x,y,z) = x(y’z’ + yz) f’(x,y,z) = ( x(y’z’ + yz) )’ = x’ + (y’z’ + yz)’ = x’ + (y’z’)’ (yz)’ = x’ + (y + z)(y’ + z’) • [ complement both sides ] [ because (xy... then complement each literal – If f(x,y,z) = x(y’z’ + yz)… – …the dual of f is x + (y’ + z’)(y + z)… – …then... must be a product of literals f(x,y,z) = y’ + x’yz’ + xz • The advantage is that..., such as f(x,y,z), has 23 = 8 minterms: x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz • Each... one combination of inputs: Minterm x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz Is... where the function output is 1. x y z f(x,y,z) f’(x,y,z) 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 0 0 1 0 0 0 0 0 1 1 0 1 f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’ = m0 + m1 + m2 + m3... the corresponding maxterm Mi Minterm x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz • Shorthand...



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... рядку записано STAX YZ+ ; A → M(YZ+). В умовних позначках читаємо: YZ+ – вміст рег... MOV R1,R; R → R1 LXI YZ,D16; D16→YZ MVI R,D8; D8 → R SHLD...) LHLD adr; M(adr) → L, M(adr+1) → H ++ PUSH YZ ; YZ++→M(SP-1); M(SP-2) LDA adr; M(adr...-2); SP-2→SP STAX YZ+; A → M(YZ+) ++ POP YZ ; M(SP), M(SP+1)→YZ++ LDAX YZ+; M(YZ+) → A POP PSW; M(SP...; Ā→A INR”’ R; R+1→R INX YZ; YZ+1→YZ STC”; 1→C DAA’; Десятична корекція DCR”’ R; R-1→R DCX YZ; YZ-1→YZ Арифметичні і логичн...’ R; A ∪ R→A ORI’ D8; A ∪ D8→A DAD” YZ; HL+YZ→HL XRA’ R; A ⊕ R→A XRI’ D8; A ⊕ D8... комірки пам’яти M(HL) YZ – вміст регистрової пари BC..., HL або регістру SP (YZ в мнемониці замінюється на B, D, H або SP); YZ+++ – вміст регистрової пари BC або DE (YZ+ в мнемониці замінюється на B або D); YZ – вміст регистрової пари BC, DE, HL або PSW (YZ++ в мнемониці замінюється на...



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... a GrZibner Basis. 1895 (Gr2) zh(yz)(A - z)-' --+ Ah(zy)(A - z)-' - h(zy) We have...)-' and (1 - yz)-' (G9-3) 2-'h(zy)(A-z)-1 + A-'z-'h(zy)+X-'h(yz)(A-z)-' ((34) (A - 2)-'h(yz)(A - z)-1 - (A - z)-'h(zy)(A - z)-1 (preNF) z, y, z-', y-', (1 - z)-', (1 - y)-', (1 zy)-' and (1 - yz)-'. + A-'(h(yz)(X - z)-1 - (A - z)-'h(zy... by substituting h(zy) for S and h(yz) for T in the Grobner basis...). 2-'h(zy)(X - z)-1 FURTHER COMMON SITUATION S 3 + z-'h(zy) = h(yz)z-'(A - z)-1 = h(yz)(A-'2-1 +A-'(A - z)-') = A-'h(yz)z-1 + A-'h(yz)(A - z)-1 Possibly, this set of rules... h(zy) with h(zy)" and h(yz) with h(yz)" for any positive integer PZ... rules for a particular h . (NF) I ,y, z-', y-', ( l - ~ ) - ' ,(l-y)-', (l-zy)-', (1 - yz)-', (1 - zy)-'/' and (1 - y ~ ) - ' / ~ . These objects were... use is as follows. z < y < 2-1 < y-' < (1 - zy)-1 < (1 - yz)-' . The set of relations of... defining relations of z-l, y-l, (1 - zy)-l and (1 - yz)-' ( E & through EB7 below). These relations...)-' - (1 - zy)-' 1 EB5 = yz (1 - yz)-' - (1 - yz)-' EBG = (1 - zy)-'zy - (1 - zy)-' 1 1 EB7 = (1 - yz)-'yz - ( 1 - yz)-' EBa = (1 - yz)-' z-' - y(1 - zy)-' - z-' EBg = (1 - zy)-' y-' - z (1 - yz)-' - y-' + + + + EBlo EBll EB12 EB13 1896 = z-' (1 - zy)-' - y (1 - ~ y ) 1- - z-' = y-' (1 - yz)-' - I (1 - YZ)-' - y-' = (1 - yt)-' y - y (1 - ZY)-' = (1 - zy)-' z - z (1 - yz)-' . we follow the algebraist...