Introduction to Computer Systems COMP-273 Assignment #1 Due: September 24, 2009 on Web CT at 23:55

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1 Introduction to Computer Systems COMP-273 Assignment #1 Due: September 24, 2009 on Web CT at 23:55 The purpose of this first assignment is to develop the fundamental skills needed when constructing solutions to more advanced problems in circuits and assembler programming. Question 1: LOGISIM (2 points) For the circuit assignments and the mini-project you will be asked to build logical circuits using the LOGISIM simulator. This useful simulator not only helps you draw pretty circuit diagrams but it also executes them. You can actually watch them run to completion graphically. This will both help you and the TA validate the correctness of your circuits. All your circuits must be validated in this way for you to receive full marks. There is a link to the LOGISIM program on the course web page. You can get there through WEB CT in the following way: WEB CT Select COMP273 Select Course Information on the right hand side will be a link to the LOGISIM JAR file. Download that file and then double click it to run it. It is programmed in JAVA so it should just run when you double click the JAR file, assuming you have Java installed on your computer. Most computers come with Java already installed. This question wants you to do the following: 1. Download LOGISIM 2. Run LOGISIM's tutorial 1. Go to the help menu 2. Select tutorial (it will pop-up a 4 step Beginner's Tutorial ) 3. Construct the Beginner's Tutorial circuit 4. Save this circuit in a file called BEGINNER. 5. Upload this file to WEB CT as your answer. Question 2: Exploring Memory (5 points) In this question, I am asking you to draw a circuit diagram using the elementary registers and gates we covered in class. When making circuit diagrams wires must interconnect all the registers and gates. These connections allow the desired logic and functionality to occur at the physical level. Use LOGISIM to create the following circuit diagram: I would like you to design a partial 4-byte RAM memory system. Page 1 of 5

2 To make things simpler, assume that we are only creating the circuitry for retrieving bytes of information that are already in RAM. Do not consider how the data will get into your 4-byte RAM. You can optionally use LOGISIM's black box feature to help you simplify the diagram, but you must create the black box yourself do not use the pre-built ones from LOGISIM. You are only permitted to use one LOGISIM black box, the BIT. They can be found in the program at: Project Load Library Built In Libraries. BIT is called a FLIP FLOP and it can be found in the MEMORY library. You can use any FLIP FLOP model. (Actually, take some time and look around at all the pre-built stuff but don't get tempted to use any of them, except for the flip-flops) A black box is a mini circuit diagram. LOGISIM lets you create these. They help make your diagrams simpler to read. When you create a black box LOGISIM will only display a picture of a square on the screen with input and output wire/terminals protruding out from the box. The mini circuit is assumed to be contained in the square. The input/output terminals connect to the input/output wires in your mini circuit. As I said, black boxes are optional and do not effect your mark, but you may find it useful. Save the LOGISIM circuit diagram in a file called RAM. This is what you will upload to WEB CT. Your partial RAM will be composed of the following things: 4-bytes of memory (where each byte is addressable independently). Do not draw the circuitry for the bits. Use LOGISIM's black box that defines a single BIT. Group these bits together into your own byte construction. A 2-bit address register (that when loaded with an address will activate that byte in memory). Do not include circuitry for how this register receives its address. Only provide circuitry from that register to RAM. LOGISIM lets you input values directly into flip-flops. Do this for your address register when testing. An 8-bit data register (that will receive the byte from RAM addressed by the address register) A 1-bit status register (when set to 0 no data is sent to the data register but when set to 1 data is put into the data register) You do not need to assume a clock wire. The simulator will automatically do this for you. Your address register uses binary numbers to address RAM. This means that if the address register is loaded with the binary number zero then the first byte of RAM is addressed and its data exits to the data register if and only if the status register is set to 1. And so on until the binary number 3 is put into the address register to address the last byte of RAM. Page 2 of 5

3 Question 3: Assembler (4 points) A friend of yours has come to you for help. They attempted to write an assembler program to solve the following equation: x = (x + y) - 10 * y. They initialized x and y with the integer numbers 50 and -2 respectively. Assuming the assembler language they used has the following 32-bit instruction format: OPCODE DESTINATION ARG1 ARG2 Where: OPCODE is 4 bits, DESTINATION is an 8-bit address, ARG1 is an 8-bit address (optional argument) and ARG2 is either an 8-bit address or a 12-bit binary integer constant and the following language chart: INSTRUCTION MNEMONIC OPCODE SYNTAX Add with constant addi 0000 nnnnnnnn nnnnnnnn cccccccccccc Add with variables add 0001 nnnnnnnn nnnnnnnn nnnnnnnnxxxx Subtract with constant subi 0010 nnnnnnnn nnnnnnnn cccccccccccc Subtract with variables sub 0011 nnnnnnnn nnnnnnnn nnnnnnnnxxxx Mult with constant muli 0100 nnnnnnnn nnnnnnnn cccccccccccc Mult with variables mul 0101 nnnnnnnn nnnnnnnn nnnnnnnnxxxx Define a constant def 1000 nnnnnnnn xxxxxxxx cccccccccccc Where: n is a binary integer address number (represents address of variable) c is a binary 2's compliment capable integer number (c for constant) x that bit is not being used This is the code that does not work: ADDRESS BINARY Page 3 of 5

4 Do the following for this questions: 1. Convert your friend's assembler program from Binary to Mnemonics (for example add x y z is x = y + z, or addi x y 10 is x = y + 10 ). 2. State, in English or French, what he/she did wrong 3. Using Mnemonics rewrite the program correctly 4. Now convert your Mnemonic to binary (as your imaginary friend did for you, above) Question 4: Data Representation (4 marks) A standard computer modem requires specific code sequences that cause it to behave in the manner the programmer desires. Modems have three special registers. There is an 8- bit data register. This register contains the byte of data the modem either transmits or it contains the byte of data the modem received. Modems also have a command register. The command register is used to control the modem. Depending on the code placed in that register the modem performs an operation. The operation is executed by the modem immediately (ie. As soon as the command register receives a code it is executed). Finally the modem has a status register. This register is set by the modem itself. It contains information about the state of the modem. Programmers read the status register, write to the command register and either read or write to the data register depending on what they want to do at that moment. Assume the following: The status register is 8 bits long and records the following information: o Bit 1 = 1 if modem is connected to network, zero otherwise o Bit 2 = 1 if modem has timed-out, zero otherwise o Bit 3 = 1 if data was just received, zero otherwise o Bit 4 = 1 if transmitted data failed, zero otherwise o Bit 5-7 = binary number indicating the speed of the modem, zero if off o Bit 8 = 1 if an unknown error occurred, zero otherwise The 16-bit command register is assigned binary integer numbers: o Integer 1 (in binary) = hang up o Integer 2 (in binary) = connect o Integer 5 (in binary) = send data o Integer 6 (in binary) = wait for data o Integer 64, 1024, 2048 (in binary) = different modem speeds In binary, make an ordered list of what you would place in each register for each step as described below (Indicate the step number and which register or registers are being set, and in binary the contents of those registers. In the case of the status register also indicate the bit-arithmetic used to extract the information you want). Do the following: Page 4 of 5

5 1. Set the modem speed to Connect to network 3. Verify that you connected 4. Send the following message SOS 5. Check if the transmission failed or timed out 6. Wait for any one character reply 7. Hang up Note that when testing the status register more than one bit can be accessed. You cannot assume that the bit you are interested in is the only one set. Instead, you need to use boolean logic AND, OR, NOT, XOR bit operations to extract the bits you are interested in. Make sure to show how you did this. Question 5: Circuit Two (5 Marks) Assume you have an application that requires you to design a circuit that receives as input a 2-bit integer number (00, 01, 10, 11) and must output a 2-bit integer number that is one less but wraps around the number sequence. In other words: INPUT OUTPUT Notice that 1 goes to zero, 2 goes to 1, 3 goes to 2 and 0 raps around to 3. Draw and validate this circuit using LOGISIM. Save the file as CIRCUIT2. ASSIGNMENT SUBMISSION INSTRUCTIONS Submit your solution to Web CT before the due date 5% is removed for each day late up to 2 days at which time the submission box closes Please submit your assignment in electronic form. The circuit diagrams must be in LOGISIM while text can be in PDF, RTF or TXT file formats. If you have a large number of files (more than 5), to make things easier for the TA, ZIP all you files into one ZIP file. GRADING INSTRUCTIONS This assignment is worth a total of 20 points The breakdown is stated on each question Each question is graded by an even distribution of points. In other words, if your question is 50% correct you get 50% of the marks. Page 5 of 5

Basic Von Neumann Designs

COMP 273 Assignment #2 Part One Due: Monday October 12, 2009 at 23:55 on Web CT Mini-Project Due: Monday November 9, 2009 at 23:55 on Web CT Basic Von Neumann Designs This assignment is divided into two