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# 請問以下邏輯設計問題 ? ( 原文的 )

註:「'」為「Bar」

1.+Prove the identity of each of the following Boolean equations,using algebraic manipulation:

(a)AB+BC'D'+A'BC+C'D=B+C'D

(b)WY+W'YZ'+WXZ+W'XY'=WY+W'XZ'+X'YZ'+XY'Z

(c)AC'+A'B+B'C+D'=(A'+B'+C'+D')(A+B+C+D')

2.Using DeMorgan's theorem,express the function

F=A'BC+B'C'+AB'

(a)with only OR and complement operations

(b)with only AND and complement operations

3.Draw the logic diagram for the following Boolean expressions.The diagram should correspond exactly to the equation.Assume that the complements of the inputs are not available.

(a)WX'Y'+W'Z+YZ

(b)A(BD'+B'D)+D(BC+B'C')

(c)WY'(X+Z)+X'Z(W+Y)+WX'(Y+Z)

4.*find the minterms of the following expressions by first plotting each expression on a map:

(a)XY+XZ+X'YZ

(b)XZ+W'XY'+WXY+W'YZ+WY'Z

(c)B'D'+ABD+A'BC

5.Optimize the following functions into (1) sum-of-products and (2) product-of-sums forms:

(a)F(A,B,C,D)=Sigma m(2,3,5,7,8,10,12,13)

(b)F(W,X,Y,Z)=IIM(2,10,13)

6.Use extraction to find a shared,minimum gate input count,multiple-level implementation for the pair of functions given using AND and OR gates and inverters.

(a)F(A,B,C,D)=Sigma m(0,5,11,14,15),d(A,B,C,D)=Sigma m(10)

(b)G(A,B,C,D)=Sigma m(2,7,10,11,14),d(A,B,C,D)=Sigma m(15)

7.(a)Connect the outputs of three 3-state buffers together,and add additional logic to implement the function

F=A'BC+ABD+AB'D'

Assume that C,D,and D' are data inputs to the buffers and A and B pass through logic that generates the enable inputs.

(b)Is your design in part(a) free of three-state output conflicts?If not,change the design ig necessary to be free of such conflicts.

請給我中文的過程和解答 , 我知道要翻譯有點困難 , 如果不行的話過幾天我會試著翻譯成中文再 PO 上來 , 謝謝

### 2 Answers

- Anonymous1 decade agoFavorite Answer
1. 用代數處理方式來證明下列布林代數等式成立

(a)AB+BC'D'+A'BC+C'D=(AB+A'BC)+(BC'D'+C'D)

=B(A+A'C)+C'(D+BD')=B(A+C)+C'(B+D)

=AB+(BC+BC')+C'D=(AB+B)+C'D=B(A+1)+C'D=B+C'D-->distributive

2007-12-19 21:35:05 補充：

(b) WY+W'XZ'+X'YZ'+XY'Z

= WY+W'XZ'+X'YZ'+XY'Z+XYZ'+WXZ+W'YZ'+W’XY'

=WY+W'YZ'+WXZ+W'XY'

2007-12-19 21:43:18 補充：

WYconsensusW'XZ'=XYZ'

WYconsensusXY'Z=WXZ

W'XZ'consensusX'YZ'=W'YZ'

W'XZ'consensusXY'Z=W'XY'

= WY+W'XZ'+X'YZ'+XY'Z+XYZ'+WXZ+W'YZ'+W'XY'

2007-12-19 21:46:55 補充：

X'YZ'+XYZ'=YZ'

YZ'consensusW'XY'=W'XZ'

WYconsensusW'YZ'=YZ'

WXZconsensusW'XY'=XY'Z

WYconsensusW'YZ'=YZ'

=WY+W'YZ'+WXZ+W'XY'

2007-12-19 21:51:43 補充：

004:increase four literals by consensus.

005:decrease four literals by consensus.

I don't have enough time to solve all questions, if you could add the question's time, I will do my best on solving them. I'm too busy to solve them at once.

2007-12-20 21:02:32 補充：

圖片參考：http://img263.imageshack.us/img263/9153/98782786uq...

Sorry, when I saw the question again, it is too late. I don't have enough time to solve else quesiotns, several days later, I will add.(if my sentence is wrong, you can modify them.)

2007-12-20 21:05:13 補充：

1. (a)(b)昨天已答於意見欄，請自行參閱，本來想全部解完，但是我時間有限也很忙，當我再看到問題時已經不夠時間了，剩下的我之後還會補充，造成麻煩實在抱歉

2007-12-20 21:25:47 補充：

5. (a)(1)A'B'CD'+A'B'CD+A'BC'D+A'BCD+AB'C'D'+AB'CD'+ABC'D'+ABC'D

(2)(A+B+C+D)(A+B+C+D')(A+B'+C+D)(A+B'+C'+D)(A+B'+C'+D)(A+B'+C+D)(A'+B'+C'+D)(A'+B'+C'+D')

2007-12-20 21:31:07 補充：

(b)(1)Σ(0,1,3,4,5,6,7,8,9,11,12,14,15)

=A'B'C'D'+A'B'C'D+A'B'CD+A'BC'D'+A'BC'D+A'BCD'+A'BCD+AB'C'D'+AB'C'D+AB'CD+ABC'D'+ABCD'+ABCD

(2)(A+B+C'+D)(A'+B+C'+D)(A'+B'+C+D')

Source(s): Me