Test 2 with Answers | Computer Organization | CSCI 2150, Exams of Computer Architecture and Organization

Download Test 2 with Answers | Computer Organization | CSCI 2150 and more Exams Computer Architecture and Organization in PDF only on Docsity! Points missed: _____ Student's Name: _________________________________ Total score: _____ /100 points East Tennessee State University Department of Computer and Information Sciences CSCI 2150 (Tarnoff) – Computer Organization TEST 2 for Fall Semester, 2005 Section 001 Read this before starting! • The total possible score for this test is 100 points. • This test is closed book and closed notes. • All answers must be placed in space provided. Failure to do so may result in loss of points. • 1 point will be deducted per answer for missing or incorrect units when required. No assumptions will be made for hexadecimal versus decimal, so you should always include the base in your answer. • If you perform work on the back of a page in this test, indicate that you have done so in case the need arises for partial credit to be determined. • Calculators are not allowed. Use the tables below for any conversions you may need. Leaving numeric equations is fine too. Binary Hex Binary Hex Power of 2 Equals 0000 0 1000 8 23 8 0001 1 1001 9 24 16 0010 2 1010 A 25 32 0011 3 1011 B 26 64 0100 4 1100 C 27 128 0101 5 1101 D 28 256 0110 6 1110 E 29 512 0111 7 1111 F 210 1K 220 1M 230 1G “Fine print” Academic Misconduct: Section 5.7 "Academic Misconduct" of the East Tennessee State University Faculty Handbook, June 1, 2001: "Academic misconduct will be subject to disciplinary action. Any act of dishonesty in academic work constitutes academic misconduct. This includes plagiarism, the changing of falsifying of any academic documents or materials, cheating, and the giving or receiving of unauthorized aid in tests, examinations, or other assigned school work. Penalties for academic misconduct will vary with the seriousness of the offense and may include, but are not limited to: a grade of 'F' on the work in question, a grade of 'F' of the course, reprimand, probation, suspension, and expulsion. For a second academic offense the penalty is permanent expulsion." Short answers – 2 points each For the following four circuits, identify the value of the output Q from the following choices. Consider the D-latch a rising edge triggered latch. a.) 1 b.) 0 c.) Q0 (stored value of Q) d.) Q0 (inverse of stored value of Q) e.) undefined 5. The expression is not in proper Sum-of-Products format. What boolean algebra operation would you need to apply to correct this? a.) It's not a problem; illegal term drops out b.) Distributive Law c.) Use "F-O-I-L" d.) Take the inverse of the inverse e.) DeMorgan's Theorem f.) It can't be fixed 6. True or False: There exists exactly one pattern of rectangles for every pattern of 1's and 0's in a Karnaugh map. All it takes is finding one case where this isn't true. The following is one of those cases. (Note the red rectangle.) D S Q CLK R 0 1 1 0 1. Answer: ______ D S Q CLK R 1 1 0 ↑ 2. Answer: ______ D S Q CLK R 1 1 0 1 3. Answer: ______ D S Q CLK R 0 0 1 0 4. Answer: ______ _ ___ A⋅B⋅C + A⋅B⋅C + A⋅B a c b e 00 01 11 10 00 0 1 1 0 01 0 0 1 0 11 0 0 1 1 10 0 0 0 0 AB CD 00 01 11 10 00 0 1 1 0 01 0 0 1 0 11 0 0 1 1 10 0 0 0 0 AB CD 21. If a group of four rows or columns in a Karnaugh map is identified with two variables, it is numbered 00, 01, 11, 10 instead of 00, 01, 10, 11. Why? Because any horizontal or vertical movement between adjacent cells can only have 1 input variable change. Numbering 00, 01, 10, 11 has two variables changing when going from 01 to 10 and when wrapping around the table from 00 to 11. 00, 01, 11, 10 does not have this problem. 22. For the multiplexer/selector shown to the right, what is the output Y? Remember that a multiplexer is like a digitally controlled channel changer using the inputs S1 and S0, to select either D0, D1, D2, or D3 to be output to Y. The way it works is that the decimal equivalent of the bits at the inputs S1 and S0 identify the subscript of the D input being routed to Y. Since S1 and S0 equal 0 and 1 respectively, and since 012 = 110, then D1 is routed to Y. Since D1 equals 1, then the output at Y equals 1. 23. For the Karnaugh map to the right, identify the problems with each of the three rectangles shown. (2 points each) Medium answers – 4 points each 24. Convert the Sum-of_Products circuit shown below to NAND-NAND logic, i.e., a circuit that contains nothing but NAND gates (and possible inverters at inputs). 25. Complete the truth table to the right with the values for the following Sum-of-Products expression: )()()( CBACBACAX ⋅⋅+⋅⋅+⋅= A B C X 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 AB CD Rectangle 1 Rectangle 2 Rectangle 3 A B C D0 D1 D2 D3 Y S1 S0 0 1 1 0 0 1 Rectangle 1: Rectangle 2: Rectangle 3: This rectangle could be made larger by doubling it downward to include the whole column. This rectangle encloses a non-power of 2 number of cells, specifically 6. This rectangle contains a zero. A B C Remember that each product generates one's in the truth table for the case where 1⋅1⋅1 = 1. _ A·C _ A·B·C _ _ A·B·C 26. In the Karnaugh map to the right, draw the best pattern of rectangles you can. Do not derive the SOP expression. 27. Create a Karnaugh map from the truth table below. Do not worry about making the rectangles. A B C X 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 28. Show the D latch output waveform Q based on the inputs D, S , R , and clock indicated in the graph to the right. Assume the latch captures on the rising edge. (The figure below is just for a reference.) Longer answers – Points vary per problem 29. Make the state diagram that will output a ‘1’ when the sequence ‘010’ is detected in a serial stream of bits. For example, if the following binary stream is received: then 1’s will be output here. (Notice that the last zero of the second pattern is shared with the first zero of the third pattern.) At all other times, the system will output zeros. Label the input D. (8 points) 00 01 11 10 00 1 1 1 1 01 0 0 0 X 11 0 1 0 1 10 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 0 1 1 D Q Q Clock D Clock Q AB CD Connects with other half of rectangle on top of K-map. 0 1 00 1 1 01 0 1 11 1 0 10 1 1 AB C No digits 0 Initial state 1 digits 0 2 digits 0 3 digits 1 D = 0 D = 0 D = 1 D = 1 D = 0 D = 0 D = 1 D = 1 30. Derive the minimum SOP expression from the Karnaugh map below. (6 points) The final answer is: _ _ A·C + B·D 31. Create the next state truth table and the output truth table for the state diagram to the right. Use the variable names S1 and S0 to represent the most significant and least significant bits respectively of the binary number identifying the state. Label the output 'X'. (8 points) 00 01 11 10 00 1 0 0 1 01 0 0 0 0 11 0 0 1 1 10 1 0 1 1 AB CD 00 1 01 0 11 0 K=1 10 1 K=0 K=0 K=1 K=1 K=0 K=1 K=0 Rectangle 1 Rectangle 2 Rectangle 1 A B C D 1 1 1 1 1 1 1 0 1 0 1 1 1 0 1 0 B and D drop out. Since A & C are 1, neither is inverted. A·C Rectangle 2 A B C D 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 A and C drop out. B & D are both 0, so both are inverted. _ _ B·D Next State T.T. S1 S0 K S1' S0' 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 1 1 0 0 1 1 1 0 1 0 0 1 1 0 0 1 1 1 1 1 0 Output T.T. S1 S0 X 0 0 1 0 1 0 1 0 1 1 1 0