數(shù)字電路第3章習(xí)題課

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1、單擊此處編輯母版標(biāo)題樣式,單擊此處編輯母版文本樣式,第二級(jí),第三級(jí),第四級(jí),第五級(jí),*,*,*,組合邏輯電路習(xí)題課,1,一、組合邏輯電路的基本概念,1.,定義,2.,結(jié)構(gòu)特點(diǎn),(1),電路由邏輯門構(gòu)成，不含記憶元件；,(2),輸入信號(hào)是單向傳輸?shù)模娐分胁缓答伝芈罚?3.,功能描述,真值表；表達(dá)式；卡諾圖；電路圖；波形圖,2,二、,SSI,構(gòu)成的組合邏輯電路的分析和設(shè)計(jì),1.,分析步驟,(1),從輸入端開始，逐級(jí)推導(dǎo)出函數(shù)表達(dá)式,;,(2),列真值表,(3),確定邏輯功能,2.,設(shè)計(jì)步驟,(1),列真值表；,(2),寫最簡(jiǎn)表達(dá)式；,(3),畫邏輯電路,3,三、,MSI,組合邏輯電路的工作原理

2、及應(yīng)用,1.,功能表、簡(jiǎn)化邏輯符號(hào),2.,典型應(yīng)用,(1),用二進(jìn)制譯碼器設(shè)計(jì)組合邏輯電路,(2),用數(shù)據(jù)選擇器設(shè)計(jì)組合邏輯電路,四、組合邏輯電路中的競(jìng)爭(zhēng)和冒險(xiǎn),1.,競(jìng)爭(zhēng)和冒險(xiǎn)的概念,(1)1,型冒險(xiǎn)和,0,型冒險(xiǎn)；,(2),邏輯冒險(xiǎn)和功能冒險(xiǎn)；,4,2.,邏輯冒險(xiǎn)、功能冒險(xiǎn)的檢查,3.,冒險(xiǎn)的消除方法,五、例題講解,5,例,1,：分析下圖電路的邏輯功能。,4,位加法器,4,位數(shù)值比較器,B,3,A,3,B,2,A,2,B,1,A,1,B,0,A,0,1,0,0,1,D,3,D,2,D,1,D,0,0,(AB),i,F,(AB),74,85,1,A,3,A,2,A,1,A,0,B,3,B,2

3、,B,1,B,0,S,3,S,2,S,1,S,0,Y,3,Y,2,Y,1,Y,0,CI,0,C,O,C,O,0,74LS283,0,6,解：邏輯真值表,分析：,當(dāng),D,3,D,0,9,時(shí)，,F,（,AB,）,0,，,Y,3,Y,0,等于,D,3,D,0,，即為十進(jìn)制數(shù)的,0,9,；,當(dāng),D,3,D,0,9,時(shí)，,F,（,AB,）,1,，則加法器將,D,3,D,0,加上,6,，,Y,3,Y,0,就等于調(diào)整后的十進(jìn)制數(shù)的,個(gè)位,，同時(shí),CO,1,表示十進(jìn)制數(shù)的,十位,。,結(jié)論：,此電路是將,4,位二進(jìn)制數(shù),D,3,D,0,轉(zhuǎn)化為十進(jìn)制數(shù)的,8421BCD,碼的電路。,7,例,2,：試用,4,位超前

4、進(jìn)位加法器,74LS283,構(gòu)成,4,位減法器。,解：設(shè)被減數(shù)為,A,3,A,2,A,1,A,0,，減數(shù)為,B,3,B,2,B,1,B,0,。由二進(jìn)制運(yùn)算法則可知，,A,3,A,2,A,1,A,0,減去,B,3,B,2,B,1,B,0,等于,A,3,A,2,A,1,A,0,加上,B,3,B,2,B,1,B,0,的補(bǔ)碼。而,補(bǔ)碼等于反碼加,1,。故,B,3,B,2,B,1,B,0,的補(bǔ)碼可以利用非門求,B,3,B,2,B,1,B,0,的反碼，利用低位進(jìn)位輸入端,CI,接,1,實(shí)現(xiàn),B,3,B,2,B,1,B,0,的反碼加,1,。,8,4.1,分析圖,P4.1,電路的邏輯功能,9,解：（,1,）推

5、導(dǎo)輸出表達(dá)式,Y,2,=X,2,；,Y,1,=X,1,X,2,；,Y,0,=(MY,1,+X,1,M),X,0,(2),列真值表,M X,2,X,1,X,0,Y,2,Y,1,Y,0,0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111,000,001,011,010,110,111,101,100,000,001,011,010,111,110,100,101,10,（,3,）邏輯功能：,當(dāng),M=0,時(shí)，實(shí)現(xiàn),3,位自然二進(jìn)制碼轉(zhuǎn)換成,3,位循環(huán)碼。,當(dāng),M=1,時(shí)，實(shí)現(xiàn),3,位循環(huán)碼轉(zhuǎn)換

6、成,3,位自然二進(jìn)制碼。,11,圖,P 4.2,4.2,分析圖,P4.2,電路的邏輯功能。,12,解：,(1),從輸入端開始，逐級(jí)推導(dǎo)出函數(shù)表達(dá)式,F,1,=ABC,F,2,=A(BC)+BC,=A BC+ABC+ABC+ABC,(2),列真值表,13,A B C,F,1,F,2,0 0 0,0 0,0 0 1,1 1,0 1 0,1 1,0 1 1,0 1,1 0 0,1 0,1 0 1,0 0,1 1 0,0 0,1 1 1,1 1,(3),確定邏輯功能,假設(shè)變量,A,、,B,、,C,和函數(shù),F,1,、,F,2,均,表示一位二進(jìn)制數(shù)，那么，由真值表可知，該電路實(shí)現(xiàn)了全減器的功能。,14,A

7、,、,B,、,C,、,F,1,、,F,2,分別表示被減數(shù)、減數(shù)、來(lái)自低位的借位、本位差、本位向高位的借位。,A,B,C,F,1,F,2,被減數(shù),減 數(shù),借 位,差,15,4.4,設(shè),ABCD,是一個(gè),8421BCD,碼，試用最少與非門設(shè)計(jì)一個(gè)能判斷該,8421BCD,碼是否大于等于,5,的電路，該數(shù)大于等于,5,，,F=1,；,否則為,0,。,解：,(1),列真值表,16,A B C D,F,0 0 0 0,0,0 0 0 1,0,0 0 1 0,0,0 0 1 1,0,0 1 0 0,0,0 1 0 1,1,0 1 1 0,1,0 1 1 1,1,A B C D,F,1 0 0 0,1,1

8、0 0 1,1,1 0 1 0,1 0 1 1,1 1 0 0,1 1 0 1,1 1 1 0,1 1 1 1,17,(3),畫邏輯電路，如下圖所示：,(2),寫最簡(jiǎn)表達(dá)式,AB,CD,1,1,10,11,1,1,1,01,00,10,11,01,00,F=A+BD+BC,=A BD BC,18,題,4.4,圖,19,4.10,電話室對(duì),3,種電話編碼控制，按緊急次序排列優(yōu)先權(quán)高低是：火警電話、急救電話、普通電話，分別編碼為,11,，,10,，,01,。試設(shè)計(jì)該編碼電路。,解：設(shè)火警為,A,，,急救為,B,，,普通為,C,，,列真值表為：,A B C,F,1,F,2,1,0,1,0 0,1,0

9、 0 0,20,A,BC,F,1,=A+B,A,BC,F,2,=,21,4.11,試將,2/4,譯碼器擴(kuò)展成,4/16,譯碼器,A,1,EN,Y,3,A,0,2/4,Y,2,譯碼器,Y,1,Y,0,EN,A,1,2/4(1),A,0,Y,0,Y,1,Y,2,Y,3,EN,A,1,2/4(2),A,0,Y,0,Y,1,Y,2,Y,3,EN,A,1,2/4(3),A,0,Y,0,Y,1,Y,2,Y,3,EN,A,1,2/4(4),A,0,Y,0,Y,1,Y,2,Y,3,A,3,A,2,A,1,A,0,Y,0,Y,1,Y,2,Y,3,Y,4,Y,5,Y,6,Y,7,Y,8,Y,9,Y,10,Y,11

10、,Y,12,Y,13,Y,14,Y,15,22,4.12,試用,74138,設(shè)計(jì)一個(gè)多輸出組合網(wǎng)絡(luò)，它的輸入是,4,位二進(jìn)制碼,ABCD,，,輸出為：,F,1,：,ABCD,是,4,的倍數(shù)。,F,2,：,ABCD,比,2,大。,F,3,：,ABCD,在,8,11,之間。,F,4,：,ABCD,不等于,0,。,23,解：由題意，各函數(shù)是,4,變量函數(shù)，故須將,74138,擴(kuò)展為,4-16,線譯碼器，讓,A,、,B,、,C,、,D,分別接,4-16,線譯碼器的地址端,A,3,、,A,2,、,A,1,、,A,0,，,可寫出各函數(shù)的表達(dá)式如下：,=m,0,m,4,m,8,m,12,=Y,0,Y,4,Y

11、,8,Y,12,24,=m,8,m,9,m,10,m,11,=m,0,m,1,m,2,=Y,0,Y,1,Y,2,=Y,8,Y,9,Y,10,Y,11,=Y,0,25,實(shí)現(xiàn)電路如下圖所示：,26,4.13,試將八選一,MUX,擴(kuò)展成三十二選一,MUX,。,EN,A,2,A,1,A,0,D,0,D,1,74151(1)Y,D,2,D,3,D,4,D,5,D,6,D,7,EN,A,2,A,1,A,0,D,0,D,1,74151(8)Y,D,2,D,3,D,4,D,5,D,6,D,7,1,A,2,Y,0,A,1,Y,1,A,0,Y,2,741,38,Y,3,E,1,Y,4,E,2A,Y,5,E,2B,

12、Y,6,Y,7,1,0,0,A,5,A,4,A,3,A,2,A,1,A,0,D,0,D,1,D,7,D,56,D,57,D,63,Y,0,Y,7,Y,27,4.14,試用,74151,實(shí)現(xiàn)下列函數(shù)：,解：,(1),函數(shù)有,4,個(gè)輸入變量，而,74151,的地址端只有,3,個(gè)，即,A,2,、,A,1,、,A,0,，,故須對(duì)函數(shù)的卡諾圖進(jìn)行降維，即降為,3,維。,28,10,11,1,1,01,1,1,00,10,11,01,00,AB,CD,0,0,0,0,1,D,D,D,D,0,10,11,01,00,A,BC,D,6,D,7,D,5,D,4,1,D,2,D,3,D,1,D,0,0,10,11

13、,01,00,A,2,A,1,A,0,D,0,=D,3,=D,D,1,=D,2,=D,D,4,=D,5,=D,6,=D,7,=0,令,A=A,2,、,B=A,1,、,C=A,0,則：,29,相應(yīng)的電路圖如下所示：,30,(4),函數(shù)有,4,個(gè)輸入變量，而,74151,的地址端只有,3,個(gè)，即,A,2,、,A,1,、,A,0,，,故須對(duì)函數(shù)的卡諾圖進(jìn)行降維，即降為,3,維。,10,1,1,1,11,01,1,1,00,10,11,01,00,AB,CD,D,6,D,7,D,5,D,4,1,D,2,D,3,D,1,D,0,0,10,11,01,00,A,2,A,1,A,0,1,D,0,0,1,0,

14、0,D,D,0,10,11,01,00,A,BC,31,D,0,=D,7,=D,D,1,=D,D,2,=D,3,=D,4,=D,5,=0。,D,6,=1,相應(yīng)的電路圖,如右圖所示：,令,A=A,2,、,B=A,1,、,C=A,0,則：,32,4.15,用,74153,實(shí)現(xiàn)下列函數(shù)：,解：,(1),函數(shù)有,4,個(gè)輸入變量，而,74153,的地址端只有,2,個(gè)，即,A,1,、,A,0,，,故須對(duì)函數(shù)的卡諾圖進(jìn)行降維，即降為,2,維。,33,10,1,11,1,1,01,1,1,00,10,11,01,00,AB,CD,0,CD,0,CD,0,CD,1,1,A,B,D,2,D,0,0,D,1,0,D

15、,3,1,1,A,1,A,0,0,D,0,0,1,D,D,D,D,0,10,11,01,00,A,BC,34,D,0,=CD,D,1,=CD,D,2,=0,D,3,=CD,令,A=A,1,、,B=A,0,，,則：,相應(yīng)的電路圖如下圖所示：,35,4.16,試在圖,4.2.31,的基礎(chǔ)上增加一片,7485,，構(gòu)成,25,位數(shù)據(jù)比較器。,=,A,3,A,2,A,1,A,0,B,3,B,2,B,1,B,0,(A,B),i,(A=,B),i,7485(1),(AB,F,A=B,F,AB,F,A=B,F,A,B),i,(A=,B),i,7485(2),(AB,F,A=B,F,A,B),i,(A=,B),

16、i,7485(3),(AB,F,A=B,F,A,B),i,(A=,B),i,7485(4),(AB,F,A=B,F,A,B),i,(A=,B),i,7485(6),(AB,F,A=B,F,A,B),i,(A=,B),i,7485(5),(AB,F,A=B,F,A1001,時(shí)，須加,0110,修正項(xiàng)進(jìn)行調(diào)整，計(jì)算結(jié)果為,C,4,C,3,C,2,C,1,C,0,。,A,3,A,2,A,1,CO,A,0,S,3,CI 74283 S,2,B,3,S,1,B,2,S,0,B,1,B,0,A,3,A,2,A,1,A,0,B,3,B,2,B,1,B,0,1,A,3,A,2,A,1,CO,A,0,S,3,CI 74283 S,2,B,3,S,1,B,2,S,0,B,1,B,0,S,3,S,2,S,1,S,0,0,0,C,4,C,3,C,2,C,1,C,0,1,C,4,37,4.18,用,74283,將,8421BCD,碼轉(zhuǎn)換為余,3BCD,碼。,解：由于同一個(gè)十進(jìn)制數(shù)碼的余,3BCD,碼比相應(yīng)的,8421BCD,碼大,3,，故用一片,74283,既可以實(shí)現(xiàn)，電路圖如下所示：,38,4.19,用,74