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ID:33681387
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页数:29页
时间:2019-02-28
《数字电路习题疑难解答》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、习题1.4在图P1-4中,若输入均为宽1ms、幅度+5V的矩形脉冲vI(t),试分别画出RC电路中偏压VR为±5V时的输出波形vO(t),设电路时间常数τ=RC=100μs。习题1.8将下列二进制数转换成等值的十进制数:⑴.1;⑵11001.01;⑶10.0101;⑷0.11001;⑸100.001。解答:⑴[42.5]DEC⑵[25.25]DEC⑶[2.3125]DEC⑷[0.78125]DEC⑸[4.125]DEC习题1.11试将十进制数139写成:⑴8421BCD;⑵自然二进制码;⑶八进码;⑷16进码;⑸余3BCD码;⑹5421BCD码。解答:⑴[]842
2、1BCD⑵[]BIN⑶[213]OCT⑷[8B]HEX⑸[]EX3-BCD⑹[]5421BCD习题1.13将下列十进制数转换成:⑴[87]DEC=___________BIN=____________OCT=____________HEX⑵[94]DEC=___________BIN=____________OCT=____________HEX⑶[108]DEC=___________BIN=____________OCT=____________HEX⑷[175]DEC=___________BIN=____________OCT=____________H
3、EX解答:⑴[87]DEC=BIN=127OCT=57HEX⑵[94]DEC=BIN=136OCT=5EHEX⑶[108]DEC=BIN=154OCT=6CHEX⑷[175]DEC=BIN=257OCT=AFHEX习题1.14将下列十六进制数转换成:⑴[C5]HEX=___________BIN=____________OCT=____________DEC⑵[F4]HEX=___________BIN=____________OCT=____________DEC⑶[AB]HEX=___________BIN=____________OCT=_________
4、___DEC⑷[97]HEX=___________BIN=____________OCT=____________DEC⑸[3C8]HEX=___________BIN=____________OCT=____________DEC解答:⑴[C5]HEX=BIN=305OCT=197DEC⑵[F4]HEX=BIN=364OCT=244DEC⑶[AB]HEX=BIN=253OCT=171DEC⑷[97]HEX=BIN=227OCT=151DEC⑸[3C8]HEX=BIN=1710OCT=968DEC习题1.15将下列8421BCD数转换成:⑴[10010001]
5、8421=_____________DEC=_____________EX-3⑵[01010010]8421=_____________DEC=_____________EX-3⑶[10000111]8421=_____________5421=_____________2421⑷[01101001]8421=_____________5421=_____________EX-3解答:⑴[10010001]8421=91DEC=EX-3⑵[01010010]8421=52DEC=EX-3⑶[10000111]8421=5421=2421⑷[01101001]84
6、21=5421=EX-3习题1.17将下列数转换成:⑴[]原→[]反→[]补⑵[]原→[]反→[]补⑶[-78]DEC→[]反→[]补⑷[+187]DEC→[]反→[]补解答:⑴[]原→[]反→[]补⑵[]原→[]反→[]补⑶[-78]DEC→[]反→[]补⑷[+187]DEC→[]反→[]补习题2.4将下列各函数写成最小项之和表达式:⑴F(A,B,C)=AC′+B′C+(A′+B+C(A+C′))′⑵F(X,Y,Z)=(X′+YZ)(Y′+XZ′)+XYZ⑶F(A,B,C,D)=AB′C+BC′D+CD′A+A′BC′⑷F(W,X,Y,Z)=WX′(Y+Z′)
7、+(W′+X)Y′Z+(X′Y)′解答:⑴F(A,B,C)=AC′+B′C+(A′+B+C(A+C′))′=AC′+B′C+AB′C′(A+C′)=AC′+B′C+AB′C′=A(B+B′)C′+(A+A′)B′C+AB′C′=∑m(1,4,5,6)⑵F(X,Y,Z)=(X′+YZ)(Y′+XZ′)+XYZ=X′Y′+XYZ=X′Y′(Z+Z′)+XYZ=∑m(0,1,7)⑶F(A,B,C,D)=AB′C+BC′D+CD′A+A′BC′=AB′C(D′+D)+(A′+A)BC′D+A(B′+B)CD′+A′BC′(D′+D)=∑m(4,5,10,11,13,14
8、)⑷F(W,X,Y,Z)
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