chemistry 838 - hour exam 2

Tuesday December 7, 2004 - 1 -

Chemistry 838 - Hour Exam 2

Fall 2004

Department of Chemistry Michigan State University East Lansing, MI 48824

Name

Student Number

Question Points Score

Question 1 Number Systems 15

Question 2 Data Acquisition Systems 15

Question 3 Data Acquisition Systems II 15

Question 4 Computer Architecture 15

Question 5 Computer Architecture II 15

Question 6 National Instruments Hardware 15

Question 7 LABView 15

Question 8 LABView II 15

Question 9 Data Analysis 15

Total of 7 105

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 not allowed. A number of possibly useful tables are included at the end of the exam.

CEM 838 Fall 2004 Hour Exam 2 Question 1 Number Systems


Question 1 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

61453 F00D

1101101011000001 56001

44483 ADC3

1010101010101010 43690

b.) (7 points) Complete the following table with the appropriate 16 bit representations of the indicated signed integers.

Signed Integer Two’s Complement Representation

(Decimal) Binary Hexadecimal

31415 7AB7

-31415

5227 146B

-5227

CEM 838 Fall 2004 Hour Exam 2 Question 2 Data Acquisition Systems


Question 2 Data Acquisition Systems

Sample/Hold

(0 = sample, 1 = hold)

Out

Control

In In

ADC

CS

R

Mul

tiple

xer

Convert

Busy

d , ..., dn-1 0

a0a1a2

e1

e3

e5

e7

e0

e2

e4

e6

Dat

a

World Computer

Inte

rface

to I/

O B

us

Prog. Clock

Base Freq Select

Preload Reg

Counter

Enable Count

Latched Clock Out

Clear Latched Clock Out

Clock Out

Dat

a R

egD

ata

Reg

Dat

a R

eg

Enable Overflow CSR

Inte

rface

to I/

O B

us

Figure 1 – Acquisition System

Figure 1 illustrates a computer based acquisition system. The programmable clock has a n bit counter. This system is to be used to acquire data from an experiment in which two wave lengths of light are emitted from a sample that has been excited with a laser pulse. The intensities of the two wavelengths of emitted light are to be monitored as a function of time. The time course of such an experiment is illustrated in Figure 2. Isolation of the two wavelengths of light are achieved with a monochromator with two exit slits such as illustrated in Figure 3.



λ1

λ2

Time

Inte

nsity

Figure 2 – Time Course of a reaction with Two Emitters

Monochromator

PMT-V

PMT-V

ea

eb

λa λb

Figure 3 - Dual Slit Monochromator

a.) (2 Points) How would you connect the acquisition system to the experiment?

b.) (13 Points) List the steps the program would have to do in order to acquire the signals. Assume that a data point is to be taken every 379 milliseconds.

For this problem, ignore triggering, i.e. assume that the program will begin at the “proper” time. Notice also, that you are not to give particular computer instruction, rather general statements of what is to happen. For example, the following might be appropriate somewhere in your answer.



1. Read the ADC CSR

2. Check the Busy bit

3. Branch to Step 1 if Busy is true

CEM 838 Fall 2004 Hour Exam 2 Question 3 Data Acquisition Systems II


Question 3 Data Acquisition Systems II

In the previous question you were instructed to ignore triggering. For this problem, discuss in general terms how and why triggering might be applied to the experiment in the previous question. You should touch on manual approaches, software only approaches; pre-, mid-, and post- triggering; possible definitions of the trigger event, signal(s) to be used to observed the trigger event, etc.

CEM 838 Fall 2004 Hour Exam 2 Question 4 Computer Architecture


Question 4 Computer Architecture Figure 4, Table 1, and Table 2 illustrate aspects of the laser experiment used in class to illustrate computer architecture and computer interfacing. This question relates to these materials.

Interface Registers

LaserExpReg.cdr 21-Oct-2002

Status Register

Bus

y

b7 b6 b5 b4 b3 b2 b1 b0

D0

Control Register

SHU

TTER

FIR

E

CO

NVE

RT

b7 b6 b5 b4 b3 b2 b1 b0

D1 D0D2

Intensity Registerb7 b6 b5 b4 b3 b2 b1 b0

D1 D0D2D3D4D5D6D7

AD

C7

AD

C6

AD

C5

AD

C4

AD

C3

AD

C2

AD

C1

AD

C0

Figure 4 – Laser Experiment Interface



Table 1 – Code Example

Location Contents (word)

Label Op Code I/D Ra Rb Operand

00001000 24 00 A0 18 START: LOAD D R2 NUMPNT

00001004 26 00 A0 20 LOAD D R3 POINT

00001008 22 00 A0 0C LOAD D R1 #1

0000100C 32 1F FF EC STORE D R1 CONTROL

00001010 10 00 00 00 NOOP

00001014 10 00 00 00 NOOP

00001018 10 00 00 00 NOOP

0000101C 10 00 00 00 NOOP

00001020 22 00 A0 10 LOAD D R1 #2

00001024 32 1F FF EC STORE D R1 CONTROL

00001028 10 00 00 00 NOOP

0000102C 10 00 00 00 NOOP

00001030 22 00 A0 14 LOOP1: LOAD D R1 #6

00001034 32 1F FF EC STORE D R1 CONTROL

00001038 22 00 A0 10 LOAD D R1 #2

0000103C 32 1F FF EC STORE D R1 CONTROL

00001040 32 1F FF F0 LOOP2: LOAD D R1 STATUS

00001044 92 00 10 40 BNE R1 LOOP2

00001048 28 1F FF F4 LOAD D R4 DATA

0000104C 38 E0 00 04 STORE I R4 R3

00001050 E6 00 A0 0C ADD D R3 #1

00001054 2A 00 A0 00 LOAD D R5 DELAY

00001058 EA 00 A0 14 LOOP3: ADD D R5 #-1

0000105C 9A 00 10 58 BNE R5 LOOP3

00001060 E4 00 A0 14 ADD D R2 #-1

00001064 94 00 10 30 BNE R2 LOOP1

00001068 00 00 00 00 HALT



Table 2 - View Of Memory Before Execution

Location Contents Logical Name

001FFFFC Last word

001FFFF8

001FFFF4 DATA

001FFFF0 STATUS

001FFFEC CONTROL

001FFFE8

001FFFE4

001FFFE0

0000A03C

0000A038

0000A034

0000A030

0000A02C

0000A028

0000A024

0000A020 data array

0000A01C

0000A018 00 00 03 E8 NUMPNT (10010)

0000A014 00 00 00 06 6

0000A010 00 00 00 02 2

0000A00C 00 00 00 01 1

0000A008 FF FF FF FF -1

0000A004 00 00 A0 20 POINT

0000A000 00 00 00 0F DELAY (1510)

00009FFC

00009FF8

Location Contents Logical Name

00001070

0000106C

00001068 00 00 00 00

00001064 94 00 10 30

00001060 E4 00 A0 14

0000105C 9A 00 10 58

00001058 EA 00 A0 14

00001054 2A 00 A0 00

00001050 E6 00 A0 0C Loop3

0000104C 38 E0 00 00

00001048 28 1F FF F4

00001044 92 00 10 40

00001040 32 1F FF F0 LOOP2

0000103C 32 1F FF EC

00001038 22 00 A0 10

00001034 32 1F FF EC

00001030 22 00 A0 14 LOOP1

0000102C 10 00 00 00

00001028 10 00 00 00

00001024 32 1F FF EC

00001020 22 00 A0 10

0000101C 10 00 00 00

00001018 10 00 00 00

00001014 10 00 00 00

00001010 10 00 00 00

0000100C 32 1F FF EC

00001008 22 00 A0 0C

00001004 26 00 A0 20

00001000 24 00 A0 18 START

00000FFC



a.) Discuss the memory location 0000A000. Include what the content (functional, not numerical) of the location is and how the content is utilized by the program. b.) Discuss the four program steps located in memory locations 00001030 through 0000103F. 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?



c.) The NOOP instruction is used in two different places in the program. Discuss the nature of this instruction. Why is this use of the NOOP less than ideal?

CEM 838 Fall 2004 Hour Exam 2 Question 5 Computer Architecture II


Question 5 Computer Architecture II

Latch GateLatch GateLatch Gate

A

LCa

La

GCa

Ga

B

Bus

LCb

Lb

GCb

Gb

C

LCc

Lc

GCc

Gc

FS04E2_1BitBus.cdr

Figure 5 – A Simple 1 Bit Data Bus System

Figure 5 illustrates a simple system consisting of a 1 bit bus connecting three 1 bit devices: A, B, and C. The initial contents of the devices are La(t0), Lb(t0), and Lc(t0). The following transfers are to be made in the order indicated. Lb copied to Lc. Lc copied to La La copied to Lb

a.) (11 points) In the manner used in class, list the sequence of steps that would be required to achieve the three transfers in the order listed.

CEM 838 Fall 2004 Hour Exam 2 Question 5 Computer Architecture II


b.) (3 Points) What are the final contents of the three devices in terms of the original contents, e.g. La(t0), Lb(t0), and Lc(t0)? La Lb Lc c.) (1 Point) What happened to La(t0)?

CEM 838 Fall 2004 Hour Exam 2 Question 6 National Instruments Hardware


Question 6 National Instruments Hardware A stopped flow apparatus provides the ability to thoroughly mix two solutions in a very short period of time. Fast reactions resulting from the mixing of the two solutions can thus be observed.

The two solutions flow together into a mixing chamber. The turbulence of the flow mixes the two solutions together. The reaction begins as soon as the two solution mix and continue as the mixed solution flows out of the chamber. If the flow is stopped, the solution that is in the mixing chamber at the time the flow is stopped will have just been mixed. Thus, the reaction can be studied by observing the solution in the mixing chamber as time progresses.

Figure 6 illustrates a simplified design of an apparatus that could be used for such an experiment. Two syringes are ganged together and driven by a piston that can be pushed in or pulled out in small steps. The direction of piston travel is controlled by the signal Direction, which has the value 0 for pushing and the value 1 for pulling. A rising edge on Step causes the piston to be advanced one increment in the direction given by Direction. A valve (A or B) allows a syringe to be connected to the corresponding reservoir or to the mixing chamber. Valve B is shown in the mixing chamber position. Valve A is shown in the reservoir position. Each valve is controlled by a controller (not shown) that responds to a signal, Valvei. Valvei = 0 causes valve i to be placed in the mixing chamber position. Valvei = 1 causes valve i to be placed in the reservoir position. A series of experiments can thus be performed by refilling the syringes from the reservoirs between pushes into the mixing chamber. Also not shown is an exhaust port coming out of the mixing chamber out of which the solution flows during a push.

The selection of the wavelength to be passed through the monochromator is made by setting the analog voltage ewavelength as shown below, where m and k are known constants.

Wavelength = m* ewavelength +k

In addition, the temperature of the mixing chamber is to be monitored with a thermistor, Rtemp. The analog voltage, Vtemp, is to be computer controlled in order to be able to turn the thermistor off when the temperature is not being measured (to avoid self heating) and to adjust the sensitivity of the temperature measurement.

a.) (8 Points) Computer control of the stopped flow experiment is to be achieved with the NI-6024E and the BNC-2120. Design the interface by indicating on Figure 7 where each of the signals listed in Table 3 are to be connected. Do this by drawing an arrow to the actual point that the conductor (wire) carrying the signal of interest would be connected. Identify each arrow with a unique label.

Also, indicate where you would make a connection to common.

b.) (7 Points) For each of the interface signals, indicate in Table 3 whether that signal is an input (to the computer) or and output.



Table 3 - Signal Direction

Signal Input/ Output

Direction

Step

ValveA

ValveB

Vtemp

ewavelength

etemperature

eLight

Mixing Chamber

Solution A Valve A

Valve B

ToReservoir

A

ToReservoir

B

Solution B

DirectionStep

Syringe Body

Syringe Plunger

Piston Positioner

Piston

FS04E2_StoppedFlow.cdr

Monochromator

Emitted Light

WavelengthSelected Light

ewavelength

PistonControl

eLight

etemperature

PMT-V

Vtemp

Rtemp

R1

R2

Figure 6 – Simplified Stopped Flow Apparatus



FS04E2_BNC2120A.cdr

1 Res

ANALOG INPUTS

TIMING I/O

ANALOG OUTPUTS

FUNCTION GENERATOR

DIGITAL I/O

SCREW TERMINAL TOBNC CONVERTERS

Frequency Selection

BNC

PWRBNC-2120

NATIONALINSTRUMENTS

BNC

ACH0 ACH1

ACH3

ACH3

ACH5

ACH7

ACH2

ACH4

ACH6

DAC0

Sine Triangle

Amplitude Adjust Frequency AdjustLo LoHi Hi

100-10kHz 1K-100kHz 13k-1MHZ

Sine/Triangle TTL Square Wave

DAC7

USER1

CLK

OutputsInputs

PFI1

PFI10/TRIG1

QUADRATUREENCODER

96 Pulses/rev

PFI2

PFI3

PFI4

PFI5

PFI6

PFI7

PFI8

PFI9

UP/DN

AI-TRG2

AI-CONVERT

CTR - SOURCE1

CTR - GATE1

AO - UPDATE

AO - WFTRIG

AI - STARTSCAN

CTR - SOURCE0

CTR - GATE0

CTR - OUT1

CTR - OUT0

FREQ OUT

+5V

DGND

USER2

DAC6

DAC5

DAC4

DAC3

DAC2

DAC1

DAC0

DGND

DAC1

BNC

FS GS

FS GS

FS GS

FS GS FS GS

FS GS

FS GS

FS GS

TempRef

ThermoCouple

2

3

4

Figure 7 - National Instruments BNC-2120

CEM 838 Fall 2004 Hour Exam 2 Question 7 LABView


Question 7 LABView Figure 8 is the block diagram of an example VI. Figure 9 is a view of the Front Panel for the VI after execution.

Figure 8 - Block Diagram

CEM 838 Fall 2004 Hour Exam 2 Question 7 LABView


Figure 9 – Front Panel

Identify the LABView objects in Table 4 by giving the name of the object, e.g. “Random noise generator”; the data type; checking the Object Type, 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 Type Data Object Outline

Color Object Name Data TypeF C I Co V Source Sink

C Orange

E Orange

F Orange

H Orange

J Orange

K Black

M Black

N Orange

O Orange

P Black

Q Orange

CEM 838 Fall 2004 Hour Exam 2 Question 8 LABView II


Question 8 LABView II Objects A and B in Figure 8 are two parts of one instance of one type of LABView object.

a.) (2 Points) What type of object is this?

b.) (3 Points) What is the order of execution of the two parts?

c.) (3 Points) What defines when the transition between the two parts occurs?

CEM 838 Fall 2004 Hour Exam 2 Question 8 LABView II


Object D is another type of LABView object.

d.) (2 Points) What type of object is D?

e.) (2 Points) Which (if any) other objects are always associated with D?

f.) (3 Points) Describe the operation of Object D.

CEM 838 Fall 2004 Hour Exam 2 Question 9 Data Analysis


Question 9 Data Analysis

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

400 410 420 430 440 450 460 470 480 490 500

Wavelength

Inte

nsity

Figure 10 - Sample Spectra

An computerized experiment produces spectra of a chemical system that has two features. Figure 10 is an example spectrum produced by the experiment. The chemical system is stable and any number of spectra maybe obtained. Discuss how each of the following topics could be applied to the experiment. The goal of the experiment is to find the position, line shape, and amplitude of all features.

a.) (5 Points) Averaging Successive Scans

CEM 838 Fall 2004 Hour Exam 2 Question 9 Data Analysis


b.) (5 Points) Smoothing

c.) (5 Points) Curve Fitting

CEM 838 Fall 2004 Hour Exam 2 Appendix


Appendix

Table 5 - Powers of 2

n DEC OCT HEX 0 1 1 1

1 2 2 2

2 4 4 4

3 8 10 8

4 16 20 10

5 32 40 20

6 64 100 40

7 128 200 80

8 256 400 100

9 512 1000 200

10 1024 2000 400

11 2048 4000 800

12 4096 10000 1000

13 8192 20000 2000

14 16384 40000 4000

15 32768 100000 8000

16 65536 200000 10000

17 131072 400000 20000

18 262144 1000000 40000

19 524288 2000000 80000

20 1048576 4000000 100000

21 2097152 1000000 200000

22 4194304 20000000 400000

23 8388608 40000000 800000

24 16777216 100000000 1000000

25 33554432 200000000 2000000

26 67108864 400000000 4000000

27 134217728 1000000000 8000000

28 268435456 2000000000 10000000

29 536870912 4000000000 20000000

30 1073741824 10000000000 40000000

31 2147483648 20000000000 80000000

32 4294967296 40000000000 100000000



Table 6 - Instruction Set

Op Code Mnemonic I/D? Description

0 HALT N Halt CPU operation

1 NOOP N Do nothing for one instruction cycle

2 LOAD Y Load contents of Operand (i.e. Memory Location) into Register Ra

3 STORE Y Store contents of Register Ra into Operand (i.e. Memory Location)

4 MOVE N Move the contents of register Ra to register Rb

5 AND N The logical AND of the contents of registers Ra and Rb is calculated and stored in Register Ra

6 OR N The logical OR of the contents of registers Ra and Rb is calculated and stored in Register Ra

7 INV N Invert the contents of Register Ra

8 NEG N Negate (take two's complement of) the contents of Register Ra

9 BNE N Branch to the address of the operand if Register Ra is not equal to zero

A BEQ Y Branch to the address of the operand if Register Ra is equal to zero

B JMP Y Branch to the address contained in the Operand

C RET Y Branch to location after last CALL instruction

D CALL N Branch to the address contained in the Operand and store next address for return

E ADD N Contents of Register Ra is added to the contents of Rb and the result is stored in Register Ra

F MUL N Contents of Register Ra is multiplied by the contents of Rb and the result is stored in Register Ra



Chemistry 838 - Hour Exam 2 .........................................................................................1 Question 1 Number Systems............................................................................................2 Question 2 Data Acquisition Systems .............................................................................3 Question 3 Data Acquisition Systems II..........................................................................6 Question 4 Computer Architecture ..................................................................................7 Question 5 Computer Architecture II ..............................................................................12 Question 6 National Instruments Hardware.....................................................................14 Question 7 LABView ......................................................................................................17 Question 8 LABView II...................................................................................................19 Question 9 Data Analysis ................................................................................................21 Appendix..........................................................................................................................23

chemistry 838 - hour exam 2

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