2018年4月9日 星期一

計概A-04同位元檢查與布林代數-公職試題

【選擇題】

D01.


C02.布林函數F = x + yz'的真值表中共有幾種輸入組合可以使得F = 1 (A)3 (B)4 (C)5 (D)6[109地方四等資處]

x   y   z'   F

0   1   0   1

1   0   0   1

1   1   0   1

1   0   1   1

1   1   1   1

 

D03.

AB

C

m0

m1

m3

m2

題目中輸入0, 1, 2, 3m0, m1, m2, m3表示

按照卡諾圖,當AB00時,C沒有作用,而且m0輸出為1

m1m2輸出為Cm3輸出為0,故選(D)

00

01

11

10

0

1

 

 

 

1

1

1

 

1


C04.下列關於布林運算(Boolean operations)的敘述何者錯誤 (A)0 XOR 0 = 1 AND 0 (B)NOT (0 AND 1) = (0 OR 1) AND 1 (C)NOT (0 OR 1) = 1 XOR 0 (D)(1 OR 0) AND 1 = NOT (1 XOR 1)[109地方四等電子]

(A)0 XOR 0 = 1 AND 0 = 0

(B)NOT (0 AND 1) = (0 OR 1) AND 1 = 1

(C)NOT (0 OR 1) = NOT 1 = 0 1 XOR 0 = 1

(D)(1 OR 0) AND 1 = NOT (1 XOR 1) = 1

 

D05.已知函數F(A, B, C, D) = Σm(0, 4, 6, 7, 8, 10, 11, 12, 14, 15)下列何者為此函數F最大項之積(product of maxterms) (A)Σm(0, 4, 6, 7, 8, 10, 11, 12, 14, 15) (B)Σm(1, 2, 3, 5, 9, 13) (C)ΠM(0, 4, 6, 7, 8, 10, 11, 12, 14, 15) (D)ΠM(1, 2, 3, 5, 9, 13)[109地方四等電子]

布林代數的函數式,Sum of Product(SOP)Product of Sum(POS)兩者互補。

SOP = Σm(0, 4, 6, 7, 8, 10, 11, 12, 14, 15)POS = ΠM(1, 2, 3, 5, 9, 13)

 

D06.


D07.布林函數A + BC等於 (A)(A + B)C (B)AB + AC (C)AB + AB + BC (D)(A + B)(A + C)[109關務四等]

分配律A + BC = (A + B)(A + C)

 

D08.一個1位元比較器輸入為布林變數XY,輸出有FX < Y(X小於Y,表示X = 0Y = 1)FX > Y(X大於Y,表示X = 1Y = 0)FX = Y(X等於Y),下列敘述何者錯誤? (A)FX < Y = X'Y (B)FX > Y = XY' (C)FX = Y + FX > Y = X + Y' (D)FX = Y + FX < Y = X' + Y'[109關務]

(A)X < Y表示X = 0Y = 1X' = 1X'Y = 1

(B)X > Y代表示X = 1Y = 0Y' = 1XY' = 1

(C)X = YX > Y相加,表示不是X < Y~X' + ~Y = X + Y'

(D)X = YX < Y相加,表示不是X > Y~X + ~Y' = X' + Y

 

A09.若你的電腦系統採用偶同位(even parity),則下列數字何者會被視為有錯誤發生? (A)01110000 (B)01000001 (C)11010100 (D)11011110[109鐵路員級]

偶同位檢查:資料位元與同位元中,1的個數總和為偶數。(A)為奇同位檢查。

 

B10.下列何者是布林函數F(x, y, z) = (x + y)(yʹ + z)積之和(sum of products)表示法 (A)F(x, y, z) = xʹyʹ + yzʹ (B)F(x, y, z) = xyʹ + yz (C)F(x, y, z) = x + yzʹ (D)F(x, y, z) = xʹ + yz[109鐵路員級]

x

y

z

x + y

y' + z

(x + y) * (y' + z)

積之和

F

F

F

F

T

F

 

F

F

T

F

T

F

 

F

T

F

T

F

F

 

F

T

T

T

T

T

x'yz

T

F

F

T

T

T

xy'z'

T

F

T

T

T

T

xy'z

T

T

F

T

F

F

 

T

T

T

T

T

T

xyz

F(x, y, z) = x'yz + xy'z' + xy'z + xyz = (xy'z' + xy'z) + (x'yz + xyz) = xy'(z' + z) + yz(x' + x) = xy' + yz

 

C11.


C12.若布林函數F(A, B, C)=AB+A'C,以正規最小項和(Canonical Sum of Minterms)表示時,其結果為 (A)Σm(0, 1, 6, 7) (B)Σm(1, 2, 5, 6) (C)Σm(1, 3, 6, 7) (D)Σm(2, 3, 5, 6)[110地方四等資處]

AB + A'C = A'B'C + A'BC + ABC' + ABC = 001 + 011 + 110 + 111 = (1, 3, 6, 7)

 

D13.


A14.至少需要幾個2-input NOR閘,才能實現布林函數(X + Y)(X + Z) (A)3 (B)4 (C)5 (D)6[110身心四等]

迪摩根定律((X + Y)' + (X + Z)')' = (X + Y)(X + Z)

 

A15.欲傳輸一7位元的ASCII1010110,若採用奇同位元(Odd Parity Bit)檢查,則被傳輸的訊息為何? (A)11010110 (B)01010110 (C)10101111 (D)10101100[110身心四等]

奇同位元(Odd Parity Bit):資料位元與同位元中,1的個數總和為奇數。

 

B16.

F(0, 1, 0, 1) = 1110 + 011 + 001 + 000 = 0 + 0 + 0 + 0 = 0

 

D17.關於下列布林恆等式,何者錯誤? (A)(x + y)(x + z) = x + (yz)' (B)x(y + z) = xy + xz (C)x(x + y) = x (D)(x + y)(x' + y) = y[110國安五等]

(x + y)(x' + y) = xx' + xy' + x'y + yy' = 0 + xy' + x'y + 0 = xy' + x'y

 

D18.


A】19.


D20.在布林(Boolean)代數中,下列何者是DeMorgan's Law(+OR,.為AND'NOT) (A)X+YZ=(X+Y)(X+Z)X(Y+Z)=XY+XZ (B)X+Y=Y+XXY=YX (C)X+XY=XX(X+Y)=X (D)(X+Y)'=X'Y'(X∙Y)'=X'+Y'[110普考電子]

(A)分配律。(B)交換律。(C)消去律。(D)迪摩根定律。


C21.


D22.ABCD四個變數構成之函數,若由卡諾圖(Karnaugh Map)中可得到F = B'D' + B'C' + A'C'DF' = AB + CD + BD'。則下列何者代表函數F之和項積(product of sums) (A)B'D' + B'C' + A'C'D (B)AB + CD + BD' (C)(B + D)(B + C)(A + C + D') (D)(A' + B')(C' + D')(B' + D)[110關務四等]

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

F' = AB + CD + BD' = (A' + B')(C' + D')(B' + D)

 

A23.布林函數(B + C)(A + B + C)可化簡為 (A)B + C (B)A + B + C (C)A(B + C) (D)A + BC[110關務四等]

B + C = 0(B + C)(A + B + C) = 0

B + C = 1(B + C)(A + B + C) = 1

(B+C)(A+B+C) = B + C

 

B24.下列那一組位元資料不符合奇同位元檢查(odd-parity check) (A)00011111 (B)01010101 (C)01100111 (D)11000010[111地方四等資處]

奇同位元檢查,1的個數總和為奇數,(B)01010101有個1

 

D24.請用布林代數(Boolean Algebra)化簡AB+A(B+C)+B(B+C),其最簡結果為何? (A)B (B)0 (C)1 (D)B+AC[111身心五等]

AB+A(B+C)+B(B+C)=AB+AB+AC+B+BC=B(A+1+C)+AC=B+AC

 

B25.假設XY為布林變數,符號「*」、「+」、「~」、「⊕」分別代表ANDORNOTXOR(exclusive-OR)四種運算子。下列何者與函數XY等價? (A)X*Y+(~X)*(~Y) (B)X*(~Y)+(~X)*Y (C)(X+Y)*((~X)+(~Y)) (D)(X+(~Y))*((~X)+Y)[111地方四等電子]

  XTrueYTrue

(A)X*Y+(~X)*(~Y)

  T and T or (not T) and (not T)

  T or F and F

  T or F

  T

(B)X*(~Y)+(~X)*Y

  T and (not T) or (not T) and T

  T and F or F and T

  F or F

  F

(C)(X+Y)*((~X)+(~Y))

  (T or T) and ((not T) or (not T))

  T and (F or F)

  T and F

  F

(D)(X+(~Y))*((~X)+Y)

  (T or (not T)) and ((not T) or T)

  (T or F) and (F or T)

  T and T

  T

XY T xor T F,答案:(B)(C)


D26.請用布林代數(Boolean Algebra)化簡AB+A(B+C)+B(B+C),其最簡結果為何? (A)B (B)0 (C)1 (D)B+AC[111身心五等]

AB+A(B+C)+B(B+C)=AB+AB+AC+B+BC=B(A+1+C)+AC=B+AC

 

A27.



B28.下列那一個布林運算式與(F=A xor B)完全等效? (A)F=((not A) or B) and (A or (not B)) (B)F=((not A) and B) or (A and (not B)) (C)F=(A and B) or ((not A) and (not B)) (D)F=(A or B) and ((not A) and (not B))[111身心四等]

(B)F = A xor B = A'B + AB' = ((not A) and B) or (A and (not B))

 

C29.10110110按位元進行NOT的運算,結果為下列何者? (A)00110110 (B)00110111 (C)01001001 (D)01101101[111身心四等]

NOT(10110110) = 01001001

 

B】30.簡化布林代數式(P+Q'+R')(P+Q'+R)(P+Q+R')的結果是 (A)(P'Q+R') (B)(P+Q'R') (C)(P'Q+R) (D)(PQ+R)[111普考資處]

(P+Q'+R')•(P+Q'+R)•(P+Q+R')

=P+(Q'+R')•(Q'+R)•(P+Q+R')

=P+Q'•(P+Q+R')

=P+Q'•(Q+R')

=P+Q'R'

 

C】31.布林函數XY+X'Z+YZ可化簡為 (A)XY+YZ (B)X'Z+YZ (C)XY+X'Z (D)X'Y+YZ[111普考資處]

 XY+X'Z+YZ

=XY+X'Z+YZ(X+X')

=XY(1+Z)+X'Z(1+Y)

=XY+X'Z

 

D32.下列何者可以簡化為布林代數式X+YZ (A)XY+YZ (B)(X+Y)Z (C)X'Y+XY'Z (D)(X+Y)(Y+Z)[111關務四等]

 

B33.


 C34.若一布林(Boolean)代數式XYYZX'ZYZ',可化簡為下列何者? (A)XYX'Z (B)XYYZ (C)YX'Z (D)ZX'Z[111鐵路員級]

XY+YZ+X'Z+YZ'=Y(Z+Z')+XY+X'Z=Y+XY+X'Z=Y(1+X)+X'Z=Y+X'Z


C35.


B36.


D37.布林函數F = XY' + X'Y + Z與下列那一個函數相等? (A)F(X, Y, Z) = Σm(0, 1, 2, 5, 6) (B)F(X, Y, Z) = Σm(0, 3, 5, 6, 7) (C)F(X, Y, Z) = Σm(1, 2, 3, 4, 5, 6) (D)F(X, Y, Z) = Σm(1, 2, 3, 4, 5, 7)[112地方四等資處]

 

B38.關於同位元檢查(parity check),下列何者正確? (A)若傳輸內容編碼為1011001,則偶同位檢查的同位位元為1 (B)只可用於偵測單一位元錯誤 (C)可用於修正位元錯誤 (D)傳送端與接收端無需事先協議採用之同位元檢查。[112地方四等資處]

(A)偶同位檢查的同位位元應為0

(C)同位元檢查只能偵測是否錯誤,不能修正錯誤。

(D)傳送端與接收端必須事先協議採用之同位元檢查。

 

B39.下列布林表示式(Boolean expressions),何者錯誤? (A)X'Y=XY' (B)XY=0,則XY=X'+Y (C)(X'Y)'=(XY')' (D)Y1=Y'[112身心五]

'」、「*」、「+」、「~」、「⊕」分別代表NOTANDORNOTXOR

X=1Y=0X And Y=0

(A)X'Y=XY'

Not 1 Xor 0 = 1 Xor Not 0

0 Xor 0 = 1 Xor 1

0 = 0

(B)XY=X'+Y

1 Xor 0 = Not 1 Or 0

1 Xor 0 = 0 Or 0

1 = 0 (錯誤)

(C)(X'Y)'=(XY')'

Not(Not 1 Xor 0) = Not(1 Xor Not 0)

Not(0 Xor 0) = Not(1 Xor 1)

Not 0 = Not 0

1 = 1

(D)Y1=Y'

0 Xor 1 = Not 0

1 = 1

 

B40.


C41.XYZ分別為4bit資料:100101011010,先將XYXOR運算,再與ZAND運算之結果,下列何者正確? (A)1001(2進位) (B)0021(3進位) (C)0020(4進位) (D)0011(5進位)[112國安五等資處]

(1001 XOR 0101) AND 1010 1100 AND 1010 1000(2進位) 0020(4進位)

 

D42.若有10101110資料透過網路傳送,以CRC為錯誤檢查方法,運算多項式為x4+x2+x+1,則CRC碼為下列何者? (A)1001 (B)1010 (C)0110 (D)0111[112國安五等資處]

除數:x4+x2+x+1 = 10111 (x3的係數為0) 除數有5位,k = 5

被除數:10101110後接(k-1)個零 101011100000

餘數 = 111CRC = 0111

 

A43.有關布林代數恆等式,下列何者錯誤? (A)(x+y)'=x'+y' (B)x+x'=1 (C)x(y+z)=xy+xz (D)x+yz=(x+y)(x+z)[112普考資處]

(A)(x+y)' = x'y'

(D)x+yz = x'(y'+z') = x'y'+x'z' = (x+y)(x+z)

 

C44.1011011010100111按位元(bit-wise)進行AND的運算,結果為下列何者? (A)00010001 (B)01011101 (C)10100110 (D)10110111[112普考電子]

1011 0110 AND 1010 0111 = 1010 0110

 

C45.0111011010001101按位元(bit-wise)進行XOR的運算,結果為下列何者? (A)00000000 (B)00000100 (C)11111011 (D)11111111[112鐵路員級]

0111 0110 XOR 1000 1101 = 1111 1011

 



 


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