1. Lecture 1: Course Introduction Overview of Computers and their Applications in MAE B Burlingame 2 September, 2015 Heavily based on work by Dr. Buff Furman, SJSU

2. The Plan for Today Welcome to ME 30 Learning objectives Highlights from the syllabus (aka, ‘greensheet’)   Overview of computers and their applications in MAE   Binary representations of phenomena Bit, Bytes, and numbering schemes Base conversion

3. My Background Industry experience  Microsoft Corporation - Hardware Engineering Consumer Electronics Development UltimateTV Xbox Xbox 360 Kinect Xbox One, Kinect II HoloLens California Native  Grew up in Manteca, CA (one of those places where they still grow food) Personal  Hobbies: build things, fix things, literature, cycling, scuba diving Why Engineering?  Engineering is the highest form of creation open to man. We get to harness the possibilities crafted by science to creatively enhance life’s experience Why am I in this class in the middle of a work day?  To help foster as many awesome engineers as possible  The more brains we have to create a more interesting world, the more interesting the world shall become

4. Syllabus Highlights and Where to Find Things Syllabus available at  https://sjsu.instructure.com (once I obtain access) https://sjsu.instructure.com  Eng 213, Wed 5:45 (starting next week) Course information  Course goals and learning objectives  Textbook Programming in C (4th Edition) by Stephen G Kochan  Lab Kit Availability Pending. Needed in Week 6  Policies and protocol  Grading  Resources  Course schedule Back

5. Announcements Homework 1 posted tonight, due in two weeks in class  Generally, two weeks per homework covering the lesson of the day and the following week’s lesson All homework and labs should be printed and a hardcopy turned in, a digital copy should be turned in to instructure If you have a laptop, bring it to lab next week

6. Focus of ME 30 Solving problems with computers  Learn the process for formulating a computational solution to a problem  Practice the process Gain familiarity with other software used by MEs and AEs for technical computing  Matlab  Excel Learn structured programming using the C language  Get prepared for learning about mechatronics

7. Where and How are Computers Used in MAE? 1 Technical and Personal Communication  Email  Word processing (reports, procedures, etc.)  Presentations  Conferencing Computation and Analysis  Arithmetic  Equation solving  Data analysis and visualization  Multiphysics and numerical modeling sin(r)/r

8. Where and How are Computers Used in MAE? 2 Design  Solid modeling (ME 165)  Finite Element Analysis (FEA) (ME 160)  Dynamic modeling and simulation (ME 147, ME 187, ME 190)  Design tradeoffs and analytical modeling of mechanical components and systems (ME 154, ME 157) Gathering Information  Web searches  Patent searches  Databases  Vendor and other websites  Datasheets

9. Where and How are Computers Used in MAE? 3 Testing and Experimental Work  Data acquisition from sensors (ME 120) Laboratory and Factory Automation and Control  Dynamic systems (ME 187, ME 190) Instrumentation and Product Design  Embedded controllers (ME 106, ME 190) MRP, Inventory, and Document Control  Drawings  Procedures  Bill of Materials (BOM)  Specifications  Engineering change orders (ECOs) http://www.plm.automation.siemens.com/en_us/products/teamcent er/solutions_by_product/mechatronics_process_management.shtm l http://las.perkinelmer.com/Content/Images /smallImages/janusLabAuto.jpg

10. The Bottom Line Mechanical and aerospace engineers use computers widely You need to know how to use computers to be successful ME 30 will help you especially in the area of computation and analysis  Exposure to Matlab and Excel  Focus will be on the C language I hope it will also whet your appetite to learn about mechatronics Next -->

11. Networking Introduce yourself to someone you DO NOT know Find out one thing about them that they like to do in their spare time Two minutes! Back

12. Binary Representations Computers only contain two values  On/Off (High/Low, 0/1, etc.) Binary number only have two value 0/1  Base 2 numbering system  A bit: A number in base 2 (binary digit) How computers do what they do?  Humans are the translators  The machine just presents information Think, what is a letter?  Just a squiggle, humans give meaning

13. Binary Representations Everything in a PC is encoded in binary  Computers are finite, so are their representations  RGB for colors 24 bit color (8 bits for red, 8 for green, 8 for blue) TIFF  Music Midi – Instrument + note MP3 – Sound waves compressed  Cruise Control Voltage level going to an actuator

14. Binary to Decimal Binary number only have two value 0/1  Base 2 numbering system  A bit: A number in base 2 (binary digit) Decimal numbers have ten values 0 – 9  Base ten numbering system Think scientific notation  6.02E23 = 6 * 10 23 + 0 * 10 22 + 2 * 10 21  01010101 2 = 85 10 = 1 * 2 7 + 1 * 2 6 + 1 * 2 4 + 1 * 2 2 + 1 * 2 0 Range?  Largest 8 bit value: 11111111 2 = 255 10  Largest 16 bit value: 1111111111111111 2 = 65535 10

15. Negative Values Recall:  The computer only deals in 0/1  Humans provide context Sign bit  The first bit represents sign  11010101 = -213, 01010101 = 213 1’s Complement  Flip the bits (change 1’s into 0’s)  10101010 = - 213, 01010101 = 213

16. Negative Values (Cont) 2’s Complement  Take 1’s complement, add one  01010100 Start (84 10 in this example)  10101011 Flip bits  10101100 Add one (notice the carries)

17. Binary to Hexadecimal Hexadecimal (base 16) is a convenient numbering system  0 – 9, A – F (A = 10 10, B = 11 10, etc Observe 2 4 = 16, allows four bits (a nyble) to be grouped 1100 2 = 12 10 = 0xC 16

18. Binary to Hexadecimal Convert Binary to Hex 1100001110011111b

19. Binary to Hexadecimal Convert Binary to Hex 1100 0011 1001 1111 b (grouping 4 bits) 12 3 9 15 d C 3 9 F h 12 * 16 3 + 3 * 16 2 + 9 * 16 1 + 15 * 16 0 = 50,079

20. Representations Decimal (default)BinaryHexadecimal 1080’011011000x6C 108d0110 1100b6Ch 108 10 0110 1100 2 6C 16 108 dec0110 1100 bin6C hex Group by 3s with a comma Group by 4s with a space or an underbar Generally don’t group Spoken short form

21. Decimal to binary Many ways, this is how I do it 120  120 < 128, 0 in 7 th position  120 > 64, 1 in 6 th position 120 – 64 = 56  56 > 32, 1 in 5 th position 56 – 32 = 24  24 > 16, 1 in 4 th position 24 – 16 = 8  8 = 8, 1 in 3 rd position 8 – 8 = 0  Once we hit zero, all other values are zero  0111 1000 Why do I do it this way? It is quick when I don’t have a calculator. Your mileage may vary BinaryDecimal 11 102 1004 10008 1000016 10000032 100000064 10000000128

22. Binary (Bitwise) And - & &01 000 101 Y X 194 & 225 = 227 1100 0010 (194) & 1110 0001 (225) -------------------------------- 1100 0000 (192) Binary And is commonly used with “bitmasks” A binary & used with a bitmask can ensure that a given bit is turned “off”

23. Binary (Bitwise) Or - | |01 001 111 Y X 194 | 225 = 227 1100 0010 (194) | 1110 0001 (225) -------------------------------- 1110 0011 (227) Binary OR, also, is commonly used with “bitmasks” A binary | used with a bitmask can ensure that a given bit is turned “on”

24. Binary (Bitwise) Xor - ⊕ ⊕ 01 001 110 Y X 194 ^ 225 = 35 1100 0010 (194) ⊕ 1110 0001 (225) -------------------------------- 0010 0011 (35) Xors are commonly used to switch the values of selected bits or to test for inequivalence C uses ^ for Xor

25. Bitshifts ( > Right Shift) A bit shift shifts bits in a direction Left shift shifts bits to the left and simulates a multiply of 2 Right shift shifts bits to the right and simulates an integer division of 2 Bits outside the range are lost 23 << 3 = 184 (left shift by 3) 0001 0111 shifted = 1011 1000 23 >> 2 = 5 (right shift by 2) 0000 0101 = 5 (note the last two bits are lost)

26. True and False Generally, false is zero and true is not- zero Usually, all comparisons which are true generate a 1 (23 > 4) = 1 This comes up a lot!

27. Modulo Division (%) Integer division which provides the “remainder” from an integer division Assuming integer operations  123 % 6 = 3  123 / 6 = 20 (note, the fractional portion will be dropped) Recall: False is 0, and True is not 0  What does X % 2 give for all odd numbers?

28. Order of operations – C & Matlab Operators * / %Multiplication, division, modulo (integer) division + -Addition, subtraction >Left bitshift, right bitshift &Bitwise and ^Bitwise xor |Bitwise or

29. References Nicholas, P. (1994). Digital Design. Minneapolis/St. Paul: West Publishing Company