Chemistry Hour Exam 2

Transcription

1 Chemistry Hour Exam 2 Fall 2003 Department of Chemistry Michigan State University East Lansing, MI Name Student Number Question Points Score Total 105 Answer any 7 of the 9 questions. The number of points earned on this exam will be added to your total points for determination of the final course grade. Notice that you may earn 5 bonus points above the advertised 100 points for this exam. Do your work on these pages. Use the backs of the pages for extra space if necessary. The exam is closed book. Calculators are allowed. A number of possibly useful tables are included at the end of the exam. Thursday, December 28,

2 Question 1 Data Acquisition Systems Question 1 Data Acquisition Systems V offset R offset FS03Exam2_DAC.cdr e 3 s 3 R 3 e 2 s 2 R 2 e in e 1 s 1 R 1 e 0 s 0 R 0 R f Switch Control e out b 3 b 2 b 1 b 0 Figure 1-4 Bit DAC The following are true for the 4 bit DAC shown in Figure 1. 1*R 0 = 2*R 1 = 4*R 2 = 8*R 3 = R = 16*R f R offset = R f V offset = 6 e in = e 0 = e 1 = e 2 = e 3 = -12 volts. b i = 1 then S i is closed. b i = 0 then S i is open. a.) (4 Points) What is the transfer function for this circuit? Thursday, December 28,

3 Question 1 Data Acquisition Systems b.) (4 Points) What is e out if all of the switches are open? c.) (4 Points) What is e out if all of the switches are closed? d.) (3 Points) What is the resolution, i.e. the minimum step size, of this DAC circuit? Thursday, December 28,

4 Question 2 Number Systems Question 2 Number Systems a.) (8 points) Complete the following table by filling in the missing entries. Each row of the table contains the representation of an unsigned 16-bit integer using three different modulo. Binary Decimal Hexadecimal DEAF DAC b.) (7 points) Complete the following tables with the appropriate 16 bit representations of the indicated signed integers. Signed Integer Two s Complement Representation Decimal Binary Hexadecimal A -314 Thursday, December 28,

5 Question 3 National Hardware Question 3 National Hardware Figure 2 is an abridged illustration of the front panel (pun intended) of the National Instruments BNC-2120 connection facility. Figure 3 is an abridged von Neumann representation of the National Instruments 6024E acquisition system that we used in CEM 838 this term. The facility of Figure 2 provides a convenient mechanism for connecting the signals that the NI software is capable of reading and writing. The abridged representation of the front panel of the BNC-2120 contains, among other things, the connection points for four types (two general types with two directions each) of signals that are accessible by the computer, i.e. the software. a.) On Figure 2 identify two examples of where each of the four types of signals would be attached. Do this by drawing an arrow to the actual point that the conductor (wire) carrying the signal of interest would be connected. i.e. differentiate among the various parts of each type of the connector on the BNC Identify each arrow with a unique label. Also, indicate where you would make a connection to common. b.) On Figure 3 indicate where the signals that you identified in Part a would be attached and indicate the direction of data flow for each. c.) Finally, complete Table 1 for your set of signals. Be sure your entries for Direction of flow of data clearly indicate whether the flow is into or out of the software, i.e. the CPU. [There are a few extra lines.] Table 1 - Catalog of Signals Your Label Signal type Direction of flow of data Com Common bi Thursday, December 28,

6 Question 3 National Hardware BNC2120.cdr 9-Nov-2000 T V Atkinson Department of Ch emistry Michigan State University NATIONAL INSTRUMENTS BNC-2120 FS GS FS GS FS GS FS GS FS GS FS GS FS GS FS GS National Instruments BNC-2120 Connection Accessory Thursday, December 28,

7 Question 3 National Hardware Figure 2 - National Instruments BNC-2120 NI6024E.cdr 7-Nov-2000 T V Atkinson Department of Chemistry Michigan State University Control Panel Display Registers Arithmetic Logical Unit CPU Registers Command Decoder Memory I/O Controller (AGP Video Adapter) Control Bus CPU Bus Address Bus Data Bus Bus Adapter Control Bus PCI Bus Address Bus Data Bus I/O Controller (PCI Video Adapter) Registers I/O Controller Registers Display Analog Input Analog Output Digital I/O Timing I/O NI 6024E Card Cable Analog Input Analog Output Digital I/O Timing I/O BNC-2120 Functions BNC-2120 World National Instruments 6024E Acquisition System Figure 3 NI 6024E Thursday, December 28,

8 Question 4 Computer Architecture Question 4 Computer Architecture Figure 4, Table 2, and Table 3 illustrate aspects of the laser experiment used in class to illustrate computer architecture and computer interfacing. This question relates to these materials. LaserExpReg.cdr 21-Oct-2002 Control Register Busy CONVERT SHUTTER FIRE b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 D 2 D 1 D 0 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 Status Register D 0 ADC 7 ADC 6 ADC 5 ADC 4 ADC 3 ADC 2 ADC 1 ADC 0 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 Intensity Register D 7 D 6 D 5 D 4 D 3 D 2 D 1 D 0 Interface Registers Figure 4 Laser Experiment Interface Thursday, December 28,

9 Question 4 Computer Architecture Table 2 Code Example Location Contents (word) Label Op Code I/D Ra Rb Operand A0 18 START: LOAD D R2 NUMPNT A0 20 LOAD D R3 POINT A0 0C LOAD D R1 # C 32 1F FF EC STORE D R1 CONTROL NOOP NOOP NOOP C NOOP A0 10 LOAD D R1 # F FF EC STORE D R1 CONTROL NOOP C NOOP A0 14 LOOP1: LOAD D R1 # F FF EC STORE D R1 CONTROL A0 10 LOAD D R1 # C 32 1F FF EC STORE D R1 CONTROL F FF F0 LOOP2: LOAD D R1 STATUS BNE R1 LOOP F FF F4 LOAD D R4 DATA C 38 E STORE I R4 R E6 00 A0 0C ADD D R3 # A 00 A0 00 LOAD D R5 DELAY EA 00 A0 14 LOOP3: ADD D R5 # C 9A BNE R5 LOOP E4 00 A0 14 ADD D R2 # BNE R2 LOOP HALT Thursday, December 28,

10 Question 4 Computer Architecture Table 3 - View Of Memory Before Execution Location Contents Logical Name 001FFFFC Last word 001FFFF8 001FFFF4 DATA 001FFFF0 STATUS 001FFFEC CONTROL 001FFFE8 001FFFE4 001FFFE0 0000A03C 0000A A A A02C 0000A A A020 data array 0000A01C 0000A E8 NUMPNT ( ) 0000A A A00C A008 FF FF FF FF A A0 20 POINT 0000A F DELAY (1510) 00009FFC 00009FF8 Location Contents Logical Name C E4 00 A C 9A EA 00 A A 00 A E6 00 A0 0C Loop C 38 E F FF F F FF F0 LOOP C 32 1F FF EC A F FF EC A0 14 LOOP C F FF EC A C C 32 1F FF EC A0 0C A A0 18 START 00000FFC Thursday, December 28,

11 Question 4 Computer Architecture a.) Discuss the memory location 0000A018. Include a discussion of what the content of the location is and how the content is utilized by the program. b.) Discuss the two program steps located in memory locations and C. Include a description of the operations performed when the CPU executes these instructions. What is the purpose of these instructions in the context of the experiment? Thursday, December 28,

12 Question 4 Computer Architecture c.) The NOOP instruction is used in two different places in the program. Discuss the nature of this instruction. What use is made of this instruction in this experiment? Thursday, December 28,

13 Question 5 Data Acquisition Systems Question 5 Data Acquisition Systems ProgClk.cdr 7-Oct-2000 T V Atkinson Department of Chemistry Michigan State University 10 MHz Osc. /10 /10 /10 /10 /10 /10 /10 10 MHz 1 MHz 100 KHz 1 KHz 1 KHz 100 Hz 10 Hz 1 Hz Multiplexer (Switch) Selected Frequency A 2 A 1 A 0 Gate Enable Count Programmable Clock Counter (n-bits) Gate Enable Overflow Clock Overflow Clock Out Freq Reg CSR Preload Reg (n-bits) CSR I/O Bus Interface Not Shown: Control signals for strobing information into registers. Figure 5 Programmable Clock Figure 5 illustrates a programmable clock. Discuss how the programmable clock can be used in at lease two different styles of data acquisition systems. Include a discussion of the steps that would be required of the software to control the programmable clock and achieve the desired operation. The discussion can be very general, you don t have to provide software program steps, merely discuss what the steps would be in plain English. You don t have to include the details of the other components of the data acquisition systems, but you must describe how the various output(s) of the clock would be used in the various styles of acquisition systems. Thursday, December 28,

14 Question 5 Data Acquisition Systems Thursday, December 28,

15 Question 6 Data Analysis Question 6 Data Analysis Intensity Wavelength Figure 6 - Sample Spectra Figure 6 is an example spectrum of a chemical system that has two features. The chemical system is stable and any number of spectra maybe obtained. Discuss strategies for data acquisition and/or data analysis that would result in a higher quality spectrum. Thursday, December 28,

16 Question 7 Computer Architecture Question 7 Computer Architecture Figure 7 and Figure 8 illustrate two registers of a system that utilizes an 8-bit data bus to communicate among the various registers of the system. The system uses a master register as part of all communications, i.e. to move information from register A to register B, the information would be moved from A to the master and then from the master to register B. Figure 7 A Simple 8 Bit Data Bus System - Slave Thursday, December 28,

17 Question 7 Computer Architecture Figure 8 A Simple 8 Bit Data Bus - Master a.) Discuss the nature of the address bus. What are the consequences of the choice of n? Thursday, December 28,

18 Question 7 Computer Architecture b.)what is the Decoder and what does it do? How would it vary from one slave register to another? c.) Discuss in general the steps that must occur to transfer the contents of a Slave Register to the Master Register. These steps would, of course, be orchestrated by the Bus Controller. Thursday, December 28,

19 Question 8 Strings in LabView Question 8 Strings in LabView Figure 9 is the block diagram of a VI that takes a string of characters and displays them as a series of 8 bit unsigned integers. The display is achieved with a series of four arrays that are identical in data type and representation. Figure 10 and Figure 11 show the help information for the functions used in this VI. Figure 12 shows some of the properties of the elements of the arrays. Figure 12 results from doing a [Right-Click] on any of the elements of the arrays and selecting Properties. Figure 13 shows part of the resultant Front Panel for a given input string. Notice that display shows you 9 elements of each array. The index displays of the four arrays are set to show four different parts of the complete string. Furthermore, you will see that the display for each array has one element that is in common with the display for the next array. This overlap is provided to instill confidence that you are seeing all of the information. Figure 9 - String VI - Block Diagram Thursday, December 28,

20 Question 8 Strings in LabView Figure 10 - String Length - Help Figure 11 - String to Byte Array - Help Thursday, December 28,

21 Question 8 Strings in LabView Figure 12 - Properties of Array Elements a.) (4 points) For all elements of each of the four arrays and in the space immediately below each array element, place the index (in decimal) for that element. b.) (4 points) For all elements of each of the four arrays and in the space immediately above each array element place the hexadecimal equivalent of the contents of that element. c.) (4 points) For all elements of each of the four arrays and in the space immediately above the hexadecimal equivalent place the ASCII or ANSI character, i.e. the printable symbol. If the character is not printable, give the name of that character. Thursday, December 28,

22 Question 9 LabView Figure 13 - String VI Front Panel (Partial) d.) (3 points) Reproduce the input string in the box below. Include any formatting and blank lines included in the string. The left side of the box is to be treated as the left margin. And the text cursor begins in the upper left corner of the box. [Hint: the end of line indicator is a single character.] Question 9 LabView Identify the LabView objects in Table 4 by giving the name of the object, e.g. Random noise generator ; recording the data type of the object, e.g. U8, DBL, ; checking the Object Type, Thursday, December 28,

23 Question 9 LabView i.e. function(f), control (C), indicator (I), Constant (Co), variable (V); and checking whether the object is a source and/or sink of data. Table 4 LabView Objects Object Outline Color Object Name Data Type Object Type Data F C I Co V Source Sink Lavender Orange Orange Orange Blue Black Blue Orange Orange Thursday, December 28,

24 Appendix Appendix Table 5 - Powers of 2 n DEC OCT HEX Thursday, December 28,

25 Appendix Table 6 - ASCII Character Codes Character Octal Dec Hex Char Octal Dec Hex Char Octal Dec Hex Char Octal Dec Hex <NULL> ` <SOH> 1 1 1! A a <STX> " B b <ETX> # C c <EOT> $ D d <ENQ> % E e <ACK> & F f <BEL> ' G g <BS> ( H h <HT> ) I i <LF> A * A J A j A <VT> B B K B k B <FF> C, C L C l C <CR> D D M D m D <SO> E E N E n E <SI> F / F O F o F <DLE> P p <DC1> Q q <DC2> R r <DC3> S s <DC4> T t <NAK> U u <SYN> V v <ETB> W w <CAN> X x <EM> Y y <SUB> A : A Z A z A <ESC> B ; B [ B { B <FS> C < C \ C C <GS> D = D ] D } D <RS> E > E ^ E ~ E <US> F? F _ F DEL F Thursday, December 28,

26 Appendix Table 7 - ASCII Control Characters <NUL> Null <SOH> Start of heading <STX> Start of text <ETX> End of text <EOT> End of transmission <ENQ> Enquiry <ACK> Acknowledge <BEL> Bell (audible signal) <BS> Backspace <HT> Horizontal Tabulation <LF> Line Feed - go to new line <VT> Vertical tabulation <FF> Form Feed - go to new page <CR> Carriage return - return to left margin <SO> Shift out <SI> Shift in <DLE> Data link escape <DC1> Device Control 1 - XON <DC2> Device Control 2 <DC3> Device Control 3 - XOFF <DC4> Device Control 4 <NAK> Negative Acknowledge <SYN> Synchronous idle <ETB> End of transmission block <CAN> Cancel <EM> End of medium <SUB> Substitute <ESC> Escape <FS> File Separator <GS> Group Separator <RS> Record Separator <US> Unit Separator <DEL> Delete Thursday, December 28,

27 Appendix Table 8 - ANSI Character Set The ANSI character set consists of the ASCII character set plus the set of characters in this table. Char Octal Dec Hex Char Octal Dec Hex Char Octal Dec Hex Char Octal Dec Hex A0 À C0 à E A1 Á C1 á E A2  C2 â E2 ƒ A3 à C3 ã E A4 Ä C4 ä E A5 Å C5 å E A6 Æ C6 æ E A7 Ç C7 ç E7 ˆ A8 È C8 è E A9 É C9 é E9 Š A ª AA Ê CA ê EA B « AB Ë CB ë EB Œ C AC Ì CC ì EC D AD Í CD í ED E AE Î CE î EE F AF Ï CF ï EF B0 Ð D0 ð F ± B1 Ñ D1 ñ F ² B2 Ò D2 ò F ³ B3 Ó D3 ó F B4 Ô D4 ô F µ B5 Õ D5 õ F B6 Ö D6 ö F B D F B8 Ø D8 ø F ¹ B9 Ù D9 ù F9 š A º BA Ú DA ú FA B» BB Û DB û FB œ C ¼ BC Ü DC ü FC D ½ BD Ý DD ý FD E ¾ BE Þ DE þ FE Ÿ F BF ß DF ÿ FF Thursday, December 28,

28 Appendix Chemistry Hour Exam Question 1 Data Acquisition Systems...2 Question 2 Number Systems...4 Question 3 National Hardware...5 Question 4 Computer Architecture...8 Question 5 Data Acquisition Systems...13 Question 6 Data Analysis...15 Question 7 Computer Architecture...16 Question 8 Strings in LabView...19 Question 9 LabView...22 Appendix Question 1 The following are true for the 4 bit DAC shown in Figure 1 1*R 0 = 2*R 1 = 4*R 2 = 8*R 3 = R = 16*R f R offset = R f V offset = 6 e in = e 0 = e 1 = e 2 = e 3 = -12 volts. e out = ( e R in n 1 f i = 0 bi ) V R i offset R R f offset Again, assuming the following. then R i = 2 i R e out R n f = ein 1 ( R i= 0 i b 2 ) V i offset R R f offset e out R n f = ein 1 ( R i= 0 b 2 i i ) V offset e n 1 = ( 12* 16 i= out b i 0 1 i 2 ) ( 6) Thursday, December 28,

29 Appendix 1 16 a.) e = ( 12 0) ( 6) = 6 out b.) e out = ( 12 15) ( 6) 16 = ( 11.25) ( 6) = = e out 1 c.) e resolution = ( 12 ) = Parameter Value R f /R V offset e in R R f offset 1 volt 0.5 volt Parameter Value 2 value R R R R Question 2 Thursday, December 28,

30 Appendix Thursday, December 28,

31 Appendix Thursday, December 28,