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Ho Chi Minh City University of Technology FACULTY OF COMPUTER SCIENCE & ENGINEERING Digital Systems Exercises Lab Group 7: Nguyễn Viết An Trần Quốc Thắng Vũ Ngọc Thuận MSSV: 2112741 MSSV: 2110551 MSSV: 2112394 Ho Chi Minh City, 7/2022 Digital Systems – Lab Problem Simplify the following expressions using Boolean Algebra a � = ��� + �� = � (�� + �) = � (� + �) = �� + �� b � = (� + �)(� + �) = � (� + �) + � (� + �) = �� + �� + �� + �� = + �� + �� + = �� + �� = Q⨁R c � = ��� + ��� + � = �� (� + �) + � = �� + � = �� + � = � + � d � = ���(� + � + �) = (� + � + �) � � � = �.� �.� + �.� �.� + �.� �.� = � �.� + � �.� + � �.� = � �.� e � = � � � + ��� + ��� + �� � + ��� = (� � � + ���) + (��� + ���) + ��� = � �(� + �) + ��(� + �) + ��� = � � + �� + ��� = � � + �� + ��� = �� + � (� + ��) = �� + � (� + �) = �� + �� + �� f � = (� + �)(� + �) + � + � + � = (� + �)(� + �) + � � � = (� + �)(� + �) + ��� = �(� + �) + �(� + �) + ��� = �� + � � + ��� = � � + � (� + ��) = � � + � (� + �) = � � + �� + �� g � = (� + �) + ��� + �� � + � � �� = � � + ��� + �� � + ���� = �(� + ��) + �� � + ���� = �(� + �) + �� � + ���� = �� + �� + �� � + ���� h � = ��(��) + ��� + � � � = ��(� + �) + ��� + � � � = ��(� + �) + ��� + � � � = ��� + ��� + ��� + � � � Problem Simplify the circuits shown in the figures below using Boolean Algebra a � = ��� + �� (��) = ��� + �� (� + �) = ��� + �� (� + �) = ��� + �� + ��� Digital Systems – Lab = ��� + ��(1 + �) = ��� + �� = �(�� + �) = �(� + �) = �� + �� b � = ��� ��� ��� = ��� + ��� + ��� = ��� + ��� + ��� = ��(� + �) + ��� = �� + ��� = �� + ��� = �(� + ��) = �(� + �) = �� + �� Problem Use a K-map to simplify (all possible cases) a �(�, �, �) = A A (�, �, �, �, �, �) B C B C BC BC 1 1 1 b �(�, �, �, �) = (�, �, �, �, �, �, ��, ��) CD C D CD CD 1 AB 1 1 AB 1 0 AB 0 0 AB c �(�, �, �, �) = Vậy �(�, �, �) = �� + � + �� Vậy �(�, �, �, �) = �� + �� + �� (�, �, �, �, ��, ��, ��, ��) Digital Systems – Lab CD C D CD CD 0 AB 1 AB 1 AB 0 AB d �(�, �, �, �) = (�, �, �, �, ��, ��, ��, ��, ��) CD C D CD CD 0 AB 0 AB 1 AB 1 1 AB e �(�, �, �, �) = CD C D CD CD 0 AB 1 1 AB 1 AB 1 1 f �(�, �, �, �) = Vậy �(�, �, �, �) = ��� + �� + �� + �� (�, �, �, �, �, �, ��, ��, ��, ��, ��, ��) BA BA BA BA 1 DC 1 DC 1 1 DC 1 DC Vậy �(�, �, �, �) = BCD + BCD + AD + AB (�, �, �, �, �, �, �, ��, ��, ��, ��, ��) AB Vậy �(�, �, �, �) = ��� + �� + �� � Vậy �(�, �, �, �) = �� + �� + �� + �� Digital Systems – Lab g �(�, �, �, �) = (�, �, �, �, �, �, ��, ��, ��, ��) BA BA BA BA 1 0 DC 1 DC 1 DC 0 DC h �(�, �, �, �) = (�, �, �, ��, ��) + BA BA BA BA x x x DC x 0 DC x x x DC x 0 DC Vậy �(�, �, �, �) = �� + �� + � � � + ��� �(�, �, �, �, ��, ��, ��) Vậy �(�, �, �, �) = � � + � � + �� Problem Use a K-map to simplify (all possible cases) a �(�, �, �, �) = �(�, �, �, �, �, �, ��, ��, ��) + CD C D CD CD 1 x x AB 1 AB x 1 AB 0 AB b �(�, �, �, �) = �(�, ��) Vậy �(�, �, �, �) = �� + �� + �� �(�, �, �, �, ��, ��, ��, ��) �(�, �, �, �) Digital Systems – Lab CD C D CD CD x 0 AB 0 x x AB 0 AB x 1 AB c �(�, �, �, �) = �(�, �, �, �, ��, ��) + CD C D CD CD 1 x AB x x AB x x AB 1 AB d �(�, �, �, �) = CD C D CD CD 1 AB x x AB x 1 AB 0 x e �(�, �, �, �) = �(�, �, �, �, ��, ��) + BA BA BA BA 1 0 DC x x DC x x DC x x x DC �(�, �, �, ��, ��) Vậy �(�, �, �, �) = ���� + ��� + ��� + ��� �(�, �, �, �, �, ��, ��) �(�, �, ��, ��) AB Vậy �(�, �, �, �) = �� + ��� Vậy �(�, �, �, �) = (� + � + �)(� + � + �) (� + � + �) �(�, �, �, �, ��, ��, ��) Vậy �(�, �, �, �) = (� + � + �)(� + �) Digital Systems – Lab f �(�, �, �, �, �) = �(�, �, ��, ��, ��, ��, ��, ��) + CBA CBA CBA CBA CBA 1 0 ED 0 ED 0 ED x ED g �(�, �, �, �) = AB CD x 0 x 1 x 0 �(�, �, �, �, �, �) C D CD CD CBA CBA CBA 0 AB x AB 1 1 AB 1 �(�, ��, ��, ��) Vậy �(�, �, �, �) = ���� + ��� + ���� + ���� Vậy �(�, �, �, �) = (� + � + �)(� + � + �) (� + � + �) Problem Design a circuit that produces a HIGH out only when all three inputs are the same level a Use a truth table and K map to produce the SOP solution - Sử dụng Truth table: A B C x = f(A,B,C) 0 0 0 0 1 0 1 1 0 1 1 Từ bảng thực trị, ta viết công thức đại số: x=A B C + A B C + A B C + A B C + A B C + A B C + A B C + A B C → x = A B C + ABC Digital Systems – Lab - Sử dụng K map: A A Biểu đồ Karnaugh gồm LOOP1 B C B C B C BC 0 0 → x = A B C + ABC b Use two-input XOR and other gates to find a solution Theo đề, x = A=B=C Xét:Y = A⨁ B = (khi A=B) (1) Z = B ⨁ C =1 (khi B=C) (2) A = B = C (1) (2) Ta có mạch sau: Problem The following function is in minimum sum of products form Implement it using only two-input NAND gates No gate may be used as a NOT gate � = � �� + �� � + ��� + ��� Ta có: � = � �� + �� � + ��� + ��� = �(� � + ��) + �(�� + ��) Problem Construct the following circuit using two-input NAND gates only: Digital Systems – Lab Mạch viết dạng logich: � = �� � + ��� + ��� = ��� + �� � + ��� + ��� = � � + �(�� + ��) = �(� + �� + ��) = �(� + ��) = � + (� + ��) = � + �(� + �) Problem A manufacturing plant needs to have a horn sound to signal quitting time The horn should be activated when either of the following conditions is met: a It’s after o’clock and all machines are shutdown b It’s Friday, the production run for the day is complete, and all machines are shutdown * Gọi: * Truth table: A = After o’clock B = It’s Friday C = the production run for the day is complete D = all machines are shutdown X = Horn, X = 1: activated, X = 0: not activated → X = ABCD + ABC D + ABCD + ABCD Digital Systems – Lab * Rút gọn K-map CD CD AB AB AB AB CD CD 0 1 1 0 0 Vậy � = ��� + ��� * Mô mạch: Problem Design the logic circuit with these three switches as inputs so that the alarm will be activated whenever either of the following condition exists: a The headlights are on while the ignition is off b The door is open while the ignition if on * Gọi: L = headlights *Truth table: I = ignition D I L Y D = door 0 Gía trị công tắc trạng thái ON trạng thái OFF X 0 1 1 1 0 1 X - Điều kiện a: L=1, I=0 D mạng gái trị - Điều kiện b: D=1, I=1 L mang bát kể gái trị 10 Digital Systems – Lab * Rút gọn K-map D D IL IL IL IL 0 Vậy � = �� + �� * Mô mạch: Problem 10 A BCD code is being transmitted to a remote receiver The bits are A3, A2, A1, and A0, with A3 as the MSB The receiver circuitry includes a BCD error detector circuit that examines the received code to see if it is a legal BCD code (i.e.,