ECE 3561 - Lecture 1 1

Binary number system

Department of Electrical and Computer EngineeringThe Ohio State University

ECE 2560

ECE 3561 - Lecture 1 2

Today

Number systemsTo and from base 10AdditionSubtraction (made easy)Multiplication

THIS LECTURE IS REVIEW MATERIAL

We live in a base 10 world

Why base 10? Could have been base 5 or base 20. We can thank Ug! the caveman.

In base 10 we have 10 symbols 0 1 2 3 4 5 6 7 8 9

In any number base system you have n symbols Base 2 – 0 1 Base 8 - 0 1 2 3 4 5 6 7 Base 16 – 0 1 2 3 4 5 6 7 8 9 A B C D E F

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Other number bases

Number system baseBase 10 § § § § § § § § § § § § §0 1 2 3 4 5 6 7 8 9 10 11 12 13Base 2 § § § § § § § § § §

0 1 10 11 100 101 110 111 1000 1001 1010

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Base 5 and Base 8

Base 5 (would have digits 0 to 4) § § § § § § § § § § § § § 0 1 2 3 4 10 11 12 13 14 20 21 22 23

Base 8 (octal) § § § § § § § § § § § § §0 1 2 3 4 5 6 7 10 11 12 13 14 15

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To and from base

Base 10 to binary (base 2) Number in Base 10 1910 = ? Procedure (Integer division)

Divide by 2 19/2 = 9 r 1 9/2 = 4 r 1 4/2 = 2 r 0 2/2 = 1 r 0 1/ 2 = 0 r 1 So the binary of 1910 is 1 0 0 1 1

(In general for any number base you divide by the number system base and use the remainders)

More examples?

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Another

How about 13910 = ?Again divine by 2 each time

So have 1 0 0 0 1 0 1 1More Examples?

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number div 2 remainder139 69 169 34 134 17 017 8 18 4 04 2 02 1 01 0 1

Base 2 to base 10

In first example have 1 0 0 1 1In binary the digits have the following

weightValue10=a4*24+a3*23+a2*22+a1*21+a0*20

Value10=a4*16+a3*8+a2*4+a1*2+a0*1

So here1 0 0 1 1 = 1*16 +0*8 +0*4 +1*2 +1*1 = 19

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2nd example

Had 1 0 0 0 1 0 1 1 (written msb to lsb)

Value of positions is (lsb to msb) 1,2,4,8,16,32,64,128,256,512,…The powers of 2

Value of the number above = 128 + 8 + 2 + 1 = 139

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Binary addition

Follow the same rules as base 10 carries -> 1 1 1 1 1 1 0 0 1 0 0 1 1 1 (7) 0 0 0 1 0 0 1 0 1 1

(11) 1 1 0 1 1 1 0 0 1 0 (18)

More examples?

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Binary subtraction

Set the problem

And we need to borrow – step 1

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1 0 0 0 0 160 0 0 1 1 3

21 0 0 0 0 160 0 0 1 1 3

Binary subraction

The next steps work through to

To allow answer to be done

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Binary multiplication

Just like multiplication in base 10

More examples?

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1 0 0 0 0 16 x 0 0 0 1 1 3 1 0 0 0 0 1 0 0 0 0 . 1 1 0 0 0 0 48

2’s complement

For 4 bits can represent values -8 to 7

In general can represent -2n-1 n 2n-1 - 1

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0 0 0 0 0 -0 0 0 0 01 0 0 0 1 -1 1 1 1 12 0 0 1 0 -2 1 1 1 03 0 0 1 1 -3 1 1 0 14 0 1 0 0 -4 1 1 0 05 0 1 0 1 -5 1 0 1 16 0 1 1 0 -6 1 0 1 07 0 1 1 1 -7 1 0 0 1 -8 1 0 0 0

Arithmetic with 2’s complement

Add 5 and -3

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1 1 0 1 0 1 5 1 1 0 1 -3 0 0 1 0Carry out = 1

Finding the 2’s complement

To generate the 2’s complement of nSay n is 3

3 in binary (4 bits) is 0011 Procedure 1) Take the 1’s complement, then add 1

1’s complement – complement all bits 0011 1100 +1 = 1101

2) Starting at the lsb (rightmost) bit Keep the 1st 1 and then complement the rest of the

bits. Can easily see on previous example.

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Subtraction via 2’s complement

14 – 6 (need 5 bits to represent in 2’s complement

form) 01110 – 00110 or

01110 + 11010 (i.e. 14 + (-6))

01110 +11010 01000 and a carry out of 1 value is 8

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Operating in 2’s complement

In general can represent -2n-1 n 2n-1 - 1

So with 4 bits can represent values of -8 n +7

So with 8 bits can represent values of -128 n +127

So with 16 bits can represent values of -32768 n +32767

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A look at the MSP430

The chip – The MSP430F2003

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The pins

Vcc and Vss – Power and Ground being supplied to the chip. The data sheet specifies the tolerance on the voltage supply.

P1.0-P1.7,P2.6,P2.7 are for digital input, grouped into 2 digital ports, P1 and P2

TACLK, TA0, TA1 are associated with the Timer_A. More details to come.

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The Pins (2)

A0-, A0+ up to A4- and A4+ are inputs to the analog-to-digital converter. There are 4 channels and each has it own + and – input. Another pin, VREF is the reference voltage for the converter.

ACLK and SMCLK are outputs of clock signals and can be used to supply external devices with a clock.

XIN and XOUT are the connections for a crystal. RST’ – is the active low reset signal (could also

be designated _RST or /RST)

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The Pins (3)

NMI – the nonmaskable interrupt. Interrupts allow an external device to assert a value on this pin (a low) that causes the processor to halt operation after completion of the current instruction and ‘service’ the interrupt. When service is complete a software instruction allows resumption of normal operation.

There are no maskable interrupts

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The Pins (4)

TCK, TMS, TCLK, TD1, TD0 and TEST form the full JTAG interface used to program and debug the device.

SBWTDIO and SBWTCK provide the Spy-By-Wire interface which is an alternative to JTAG and uses only the 2 pins.

NOTE: each of the pins serves multiple purposes. The actual mode for each pin will vary across application and within a given application the use can be multiplexed.

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Assignment

Read Chapter 1 and 2 READ FOR UNDERSTANDING

Bring your questions to class Assignments will be due 2 classes after assigned to the

drop box on Carmen. No paper submissions – all are electronic.

Next time – the internal structure of the MSP 430 and start assembler coding

Go to ti.com – order launchpad – get Code Composer Studio

Quiz next Monday at start of class. Go to wikipedia.com and read on von Neuman architecture.

Write a 1 page summary and submit to drop box, HW1