Computer Science Question

Computer Science

Question Description

The exam is on PST 01/28, starts at 9 am and ends on PST 01/29 at 9 am, so there is 24 hours period. You can choose any 1 hour you like, so the exam time limit is 1 hour.

I will post the practice exam, hope will help.

Exam topic will cover CMOS transistors, logic states, boolean algebra, K-maps, Multilevel logic, Mux, etc. I will post them toghther.

Computer Science Question
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First Name: Last Name: PID: Version A Left Neighbor Right Neighbor Seat No. CSE 140 Midterm 1 Prof. Tajana Simunic Rosing Winter 2018 ● ● ● ● ● ● ● ● ● ● Do not start the exam until you are told. The examination will be graded out of 100, with 10 extra bonus points. Write your name and PID at the top of every page. Do not separate the pages. Turn off and put away all your electronics. This is a closed-book, closed-notes, no calculator exam. You may only refer to one 8 ½ x 11” page of your handwritten notes. Do not look at anyone else’s exam. Do not talk to anyone but an exam proctor during the exam. If you have a question, raise your hand and an exam proctor will come to you. You have 80 minutes to finish the exam. When the time is finished, you must stop. Write your answers in the space provided. To get the most partial credit, clearly and neatly show all steps of your work. Problem 1 10 points Problem 2 15 points Problem 3 20 points Problem 4 20 points Problem 5 25 points Problem 6 20 points Total points 100 points + 10 bonus 1 First Name: Last Name: PID: 1. Suppose we have a printing device, with 3 sensors (A, B, C): A = detects if the device needs ink (1) / does not need ink (0) B = detects if the device needs paper (1) / does not need paper (0) C = detects if the device is jammed (1) / is not jammed (0) This printing device can raise the following alarms: Alarm 1 = When the device needs ink, needs paper, and is not jammed Alarm 2 = When the device needs paper only and it is jammed Alarm 3 = When the device needs ink only and it is jammed Alarm 4 = When the device needs both ink and paper, and it is also jammed a. An engineer puts together a boolean equation that outputs a logic 1 when one or more alarms are raised. Simplify this equation using Boolean algebra. Show all your steps. [5 pts] F(A,B,C) = (BC' + ABC + A'BC).(A'B' + AC + B'C) ​ Solution: F(A,B,C) = (BC' + ABC + A'BC).(A'B' + AC + B'C) = (BC' + ABC + A'BC).(A'B' + AC) = (BC' + (A + A') BC).(A'B' + AC) = (BC' + 1.BC).(A'B' + AC) = (BC' + BC).(A'B' + AC) = (B (C' + C)).(A'B' + AC) = B.1.(A'B' + AC) = B.(A'B' + AC) = BA'B' + ABC = 0.A' + ABC = ABC b. What type of alarm would be raised by the printing device when F(A,B,C)=1 in part a? [5 pts] Solution: Alarm 4 2 First Name: Last Name: PID: 2. a. Draw the logic gate implementation of Y(A, B, C) using a minimum number of NAND gates and inverters only. [10 pts] Solution: Y = (((A+B)(AC))’B)’ = (((A+B)’ + (AC)’)B)’ = ((A’B’ + (AC)’)B)’ = (A’B’B + (AC)’B)’ = ((AC)’B)’ b. How many CMOS transistors does your circuit in part (b) have? [5 pts] Number of transistors is: ​_ 4*2 = 8__________________________________ 3 First Name: Last Name: PID: 3. ​ ​The output Y(A,B,C,D) = 1 for any of the following conditions: (a) A = 1 and B = 0 (b) B = 0 and C = 1 (c) A = 0, B = 1, C = 0 and D = 1 The circuit designed will never get the input combination B = 1, C = 0 and D = 0. All other input combinations produce output Y(A,B,C,D) = 0. (i) Fill out the K-Map for the circuit. [ 5 pts] CD \ AB 00 01 11 10 00 0 X X 1 01 0 1 0 1 11 1 0 0 1 10 1 0 0 1 (ii) List ALL the prime implicates. [5 pts] (B’+C’), (A’+B’), (A+B+C), (B’+D), (A+C+D) _________________________________________________________________________ (iii) List the essential prime implicates. [5 pts] (B’+C’), (A’+B’), (A+B+C) _________________________________________________________________________ (iv) Minimum POS implementation is: [5 pts] Y= (B’+C’)(A’+B’)(A+B+C) _________________________________________________________________________ 4 First Name: Last Name: PID: 4. Given Y (A, B, C, D) = AC’D+A’D’ + BC + CD’. a) Fill out the K-Map table for the above function. [4 pts] CD \ AB 00 01 11 10 00 1 1 0 0 01 0 0 1 1 11 0 1 1 0 10 1 1 1 1 b) Define the inputs X0, X1, X2 and X3 (in the simplified form) to the following 4:1 multiplexer implementation of the function Y(A,B,C,D). Note that C is the MSB of the mux selection control lines, while D is the LSB. [8 pts] X0 A’ X1 A X2 1 X3 B 5 First Name: Last Name: PID: c) Define the inputs X0, X1, X2 and X3 (in the simplified form) of the following 2:1 multiplexer implementation of the function Y (A,B,C,D). [8 pts] X0 A’ X1 A X2 B X3 1 6 First Name: Last Name: PID: 5. Consider the following ALU. [25 pts] Box A​ (Q in the input lines can be either A or B.) 7 First Name: Last Name: PID: Define the operation that ALU performs for each combination of inputs X, Y, and Z. Inputs X Y Z Operation 0 0 0 S=A-B 0 0 1 S=A+B 0 1 0 S = B complement (B with all the bits flipped) 0 1 1 S = A complement (A with all the bits flipped) 1 0 0 S = B << 2 1 0 1 S = A >> 2 (logical shift) 1 1 0 S = A or B 1 1 1 S = A and B 8 First Name: Last Name: PID: 6. Consider the following circuit. Use the truth table given below to identify the mystery gate (marked MG on diagram). Show the steps or method used to reach your answer. [20 pts] A B C Y 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 Mystery Gate is ______​XNOR​_________________________________ 9 First Name: Last Name: PID: 10 ...
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