TEJ2O_numbersystems

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=Number Systems=

A numeral system (or system of numeration) is a writing system for expressing numbers, that is a mathematical notation for representing numbers of a given set, using graphemes or symbols in a consistent manner. It can be seen as the context that allows the symbols "11" to be interpreted as the binary symbol for three, the decimal symbol for eleven, or a symbol for other numbers in different bases. [|(source)]

At first, it would seem like using any number system other than decimal is complicated and unnecessary. However, since the job of electrical and software engineers is to work with digital circuits, engineers require number systems that can best transfer information between the human world and the digital circuit world and those forms are usually binary, octal, or hexadecimal. It turns out that the way in which a number is represented can make it easier for the engineer to perceive the meaning of the number as it applies to a digital circuit. In other words, the appropriate number system can actually make things less complicated.



Assignment #11 - Number Systems
Go to the following website to answer these questions: [|BINARY TUTORIAL] Answer all questions and submit them to the handin folder

1. Why do you think that we came up with a base ten number system? Why not base eight, or base sixteen? What's so special about the number ten?

2. How many numeric symbols would we have to use? What would they be?

3. Fill in the blanks! What would the Simpsons' odometer look like here? 4. Using the above as an example, translate these base eight numbers into base 10 numbers! a) 62 in base eight equals _ in base ten. b) 146 in base eight equals _ in base ten. c) 2405 in base eight equals _ in base ten. d) 24134 in base eight equals _ in base ten.
 * Simpsons' Odometer (base eight) || Our odometer (base ten) ||
 * 314 || 204 ||
 * ? || 205 ||
 * ? || 206 ||
 * ? || 207 ||
 * ? || 208 ||
 * ? || 209 ||
 * ? || 210 ||
 * ? || 211 ||
 * ? || 212 ||
 * ? || 213 ||
 * ? || 214 ||
 * ? || 215 ||
 * ? || 216 ||
 * ? || 216 ||

5. Why do computers use binary numbers? (In other words, why do we care about learning binary?)

6. Take a look at the binary numbers 1, 10, 100, 1000, and 10000, along with their base 10 equivalents. What pattern do you notice? Why does this seem like a pretty natural pattern for BInary numbers?

7. Fill in the blanks of this binary odometer: 8. Convert these base 2 numbers to base 10: a) 1 is equal to _ in base 10. b) 1101 is equal to _ in base 10. c) 111110 is equal to _in base 10. d) 10101010101 is equal to _ in base 10.
 * Binary Odometer (base two) || Our odometer (base ten) ||
 * 11001100 || 204 ||
 * ? || 205 ||
 * ? || 206 ||
 * ? || 207 ||
 * ? || 208 ||
 * ? || 209 ||
 * ? || 210 ||
 * ? || 211 ||
 * ? || 212 ||
 * ? || 213 ||
 * ? || 214 ||
 * ? || 215 ||
 * ? || 216 ||
 * ? || 216 ||

9. Convert these to binary! a) 31 is equivalent to _ in base 2. b) 128 is equivalent to _ in base 2. c) 202 is equivalent to _ in base 2. d) 98 is equivalent to _ in base 2. e) 17 is equivalent to _ in base 2. f) 111 is equivalent to _ in base 2

10. Convert these to base 10 or binary! a) 1000111 is equivalent to _ in base 10. b) 255 is equivalent to _ in binary.



INTRO TEST REVIEW
History of Computers Computer Components Acronyms Counting (Binary, Hexadecimal, Decimal)

 =Number Conversions=

While useful in their own right, often numbers are more meaningful in another number system. For example, an electrical engineer who is working on a hexadecimal "programming layer", may want to translate something into a binary "machine layer", as such, they need to be able to convert quickly between the base 16 number system (HEX) to the base 2 number system (BIN, or 0b).

The three systems of interest for us are decimal (dec), binary (bin) and hexadecimal (hex).


 * = **Counting** ||
 * Decimal (base 10) || Binary (base 2) || Hexadecimal (base 16) ||
 * 0 || 0 || 0 ||
 * 1 || 1 || 1 ||
 * 2 || 10 || 2 ||
 * 3 || 11 || 3 ||
 * 4 || 100 || 4 ||
 * 5 || 101 || 5 ||
 * 6 || 110 || 6 ||
 * 7 || 111 || 7 ||
 * 8 || 1000 || 8 ||
 * 9 || 1001 || 9 ||
 * 10 || 1010 || A ||
 * 11 || 1011 || B ||
 * 12 || 1100 || C ||
 * 13 || 1101 || D ||
 * 14 || 1110 || E ||
 * 15 || 1111 || F ||
 * 16 || 10000 || 10 ||



Converting between Decimal and Binary

 * To convert between Decimal and Binary you can either arrange the columns by their respective placeholders like this:

For the number 38(dec) code 256 128 64 32 16 8 4 2 1 code then plug in values like this: code 256 128 64 32 16 8 4 2 1 -           1  0  0 1 1 0

code
 * Or, you could work it through by the division/remainder method as such:

> Consider the number 1110. The steps below show how to convert this number to binary using repeated division. The 'R' stands for the remainder of the division.

__?__ __?__ __?__ __ 1 __ || __?__ __?__ __ 1 __ __1__ || __?__ __ 0 __ __1__ __1__ || __ 1 __ 0 __1__ __1__ ||
 * # First, we divide 11 by 2 to find the least significant digit (the rightmost digit). Since 1 is our remainder, the least significant digit in our answer is 1. || **11 / 2 = 5 R 1 ** ||
 * || Answer:
 * # Next, we take the result of the previous division (5) and divide by 2 again. Since 5 divided by 2 leaves a remainder of 1, the next digit of our answer is 1. || **5 / 2 = 2 R 1 ** ||
 * || Answer:
 * # Again we take the result of the previous division (2) and divide by 2. This time our division does not have a remainder, so we write a 0 as the next digit of our answer. || **2 / 2 = 1 R 0 ** ||
 * || Answer:
 * # One more division by 2 gives us the most significant digit (the leftmost digit) of our answer. Since 2 will not divide 1, our result is 0 with a remainder of 1. We know we are done when we get 0 as the result of our division. || **1 / 2 = 0 R 1 ** ||
 * || Answer:

A good way to organize this conversion is to list the divisions in table form as below.

code 11 / 2 = 5 R of 1 (rightmost digit) 5 / 2 = 2 R of 1 2 / 2 = 1 R of 0 1 / 2 = 0 R of 1 (leftmost digit) code


 * Reading from bottom to top, the final answer is 1011(0b). Remember that the first division gives us the least significant digit of our answer (furthest to the right), and the final division gives us the most significant digit of our answer (leftmost). Also, the result of the final division is always 0.

95 10 to base 2 (equals 1011111) 37 10 to base 2 (equals 100101)
 * Try this:**



Converting Binary to Decimal
To convert binary to decimal is relatively easy. All you have to do is list out the columns, then add up representative placeholders. Let's say for example we want to convert 101101(0b) to dec:

101010 2 to base 10 (equals 42) 1110 2 to base 10 (equals 14)
 * Column || 2 5 || 2 4 || 2 3 || 2 2 || 2 1 || 2 0 ||  ||
 * Binary column value || 32 || 16 || 8 || 4 || 2 || 1 ||  ||
 * Binary number || 1 || 0 || 1 || 1 || 0 || 1 ||  ||
 * || 32 || +0 || +8 || +4 || +0 || +1 || = 45(dec) ||
 * Try this:**



Converting Decimal to Hex
code 16^4 16^3 16^2 16^1 16^0 65536 4096 256 16 1 code You can also use the dividing approach as with binary, but with alterations. Here are the steps:


 * 1) Divide the decimal number by 16. Treat the division as an integer division (i.e. get rid of the decimal junk).
 * 2) Write down the remainder (in hexadecimal).
 * 3) Divide the result again by 16. Treat the division as an integer division.
 * 4) Repeat step 2 and 3 until result is 0.
 * 5) The hex value is the digit sequence of the remainders from the last to first.


 * Let's try the number 86
 * **NOTES** || **DIVISION** || **RESULT** || **REMAINDER (in HEXADECIMAL)** ||
 * Start by dividing the number by 16.

In this case, 86 divided by 16 is 5.375. So the integer division result is 5 (throw out anything after the decimal point).

The remainder is [86- (5*16)] = 6 || 86 / 16 || 5.375 (so 5) || **6** ||
 * Then, divide the result again by 16

(the number 5 on the RESULT column becomes the new division value. In this case, 5/16=0. So the integer division result is 0 The remainder is (5) || 5 / 16 || 0 || **5** || //Notice the little trick - for numbers less than 256, the very first division IS the answer: Result **5** Remainder **6 or 56**//
 * Stop because the result is already 0 (0 divided by 16 will always be 0) ||  ||   ||   ||
 * Well, here is the answer. These numbers come from the REMAINDER column values (read from bottom to top) ||  ||   || 56 16 ||

137 10 to base 16 (equals 89) 198 10 to base 16 (equals C0) 249 10 to base 16 (equals F9)
 * Try this:**



Converting Hex to Decimal
Is exactly like converting binary to decimal, simply list out the columns and add the results.

Let's take the number 1D 16 How about this one A63 16
 * Column || 16 3 || 16 2 || 16 1 || 16 0 ||  ||
 * Hex column value || 4096 || 256 || 16 || 1 ||  ||
 * Hex number || 0 ||  || 1 || D ||   ||
 * ||  ||   || 16 || +13 || = 29(DEC) ||


 * Column || 16 3 || 16 2 || 16 1 || 16 0 ||  ||
 * Hex column value || 4096 || 256 || 16 || 1 ||  ||
 * Hex number || 0 || A || 6 || 3 ||  ||
 * ||  || 2560 || 96 || +3 || = 2659(DEC) ||

D00D 16 to base 10 (equals 53261) C02 16 to base 10 (equals 3074 )
 * Try this:**



Converting Binary to Hex and Hex to Binary
This is the easiest type of conversion because we break the columns in hex down into 'octets' [|(also known as a byte)].

HEX to BIN
To convert we simply break the Hex into two separate column, then assign the binary values to the octet. To convert hexadecimal F8 to binary, write down the binary for F first, then the binary for 8.
 * F || 8 ||
 * 1111 || 1000 ||

So, the answer is 1111 1000.

This seems too easy, and it is. Use a calculator to convince yourself. Convert hex number 1A to binary.
 * 1 || A ||
 * 0001 || 1010 ||

BIN to HEX
Is the exact same as HEX to BIN, but in reverse. To convert, we break the binary value into octects, then convert each to hex separately. Let's convert 10110 to HEX. First break it into octects


 * 0001 || 0110 ||
 * 1 || 6 ||

So the HEX value is 16. What about the binary number 1101101


 * 0110 || 0110 ||
 * 6 || 6 ||

So the HEX value is 66. Pretty easy.



Assignment #12 - Conversions
1. Convert the following numbers from binary to decimal (show work in the email so I know you didn't use a calculator):

a) 100010001 b) 11110000 c) 110011 d) 01010110 e) 111111 f) 101011100011 g) 00001111

2. Convert the following numbers from decimal to binary (show work):

a) 98 b) 25 c) 255 d) 679 e) 37 f) 127 g) 1975

3. Convert the following numbers from hexadecimal to decimal (show work):

a) F b) 10 c) 2A d) 32 e) FAD

4. Convert the following numbers from decimal to hexadecimal (show work):

a) 14 b) 34 c) 99 d) 234 e) 758

5.Convert the following numbers from binary to hexadecimal (show work):

a) 1011 b) 1011001 c) 111000 d) 00111100

6.Convert the following numbers from hexadecimal to binary (show work):

a) F b) 13 c) 27 d) AAC



CONVERSION TEST REVIEW
Be able to convert from
 * dec to hex, dec to binary
 * bin to hex, bin to dec
 * hex to bin hex to dec



Assignment #13 - ASCII Entry
Write the answers to the following and submit them to me.

Why?

To become familiar with how a computer interprets your keyboard entries. To become familiar with the Standard ASCII character set. To practice converting from ASCII to Hexadecimal to Binary, and back. The attached ‘Chart’ and its associated detail describe the ASCII character set, and its associated controls. The chart, itself, is outlined with ‘Hexadecimal’ numbers across the top of the chart, and down the left side. The ‘intersection’ of the left side and the top row determine the value of the character at that intersection. For example: the character ‘A’ is at location ‘41’, as opposed to the character ‘a’ which is at location ‘61’. These locations represent a ‘Binary’ value, expressed in hexadecimal for convenience and ease of expression. Since a computer can process ‘only’ binary code, all input (from programs to simple keyboarding) must be converted to binary before the ‘processor’ can process it. This exercise will demonstrate the extreme detail the computer must deal with in order to convert your ‘computer input’ into something others can understand. For this exercise **you** are going to be the computer. You must convert the ‘Keyboard entry’ (Which is in ASCII) into binary, send it to the ‘Processor’ and have it processed then sent to the screen. You will convert a message into binary, then process and ‘decode’ your message.
 * The assignment**
 * 1) ** Convert a 5 word messages (of your choice) into Hexadecimal, then Binary. **
 * 2) ** Convert your "Received Message’ into Decimal, and Screen. (Using ALT+Keypad ) **

To complete the exercise you must convert the binary characters back to ASCII, but instead of just converting with the chart, you will input the ‘Coded message’ into a text editor (or word processor) in ‘Decimal’ form by using the ‘ALT’ key and the ‘Number Pad’ on your keyboard. Refer to the following example: Message: **Hello out there.** Translating : H=48 e=65 l=6C l=6C o=6F SPC=20 o=6F u=75 t=74 SPC=20 t=74 h=68 e=65 r=72 e=65 .=2E 01001000 01100101 01101100 01101100 01101111 00100000 01101111 01110101 01110100 00100000 01110100 01101000 01100101 01110010 01100101 00101110 72 101 108 108 111 32 111 117 116 32 116 104 101 114 101 46 Hello out there.
 * HINT**:
 * Into Hexadecimal**
 * Into Binary**
 * Into Decimal**
 * Then using the ‘ALT’ key and keypad**
 * ** Decimal ** || ** Binary ** || ** Hex ** || ** ASCII ** ||  || ** Decimal ** || ** Binary ** || ** Hex ** || ** ASCII ** ||
 * 0 || 0000 0000 || 0 || NUL ||   || 49 || 0011 0001 || 31 || 1 ||
 * 1 || 0000 0001 || 1 || SOH ||  || 50 || 0011 0010 || 32 || 2 ||
 * 2 || 0000 0010 || 2 || STX ||  || 51 || 0011 0011  || 33 || 3 ||
 * 3 || 0000 0011 || 3 || ETX ||  || 52 || 0011 0100 || 34 || 4 ||
 * 4 || 0000 0100 || 4 || EOT ||  || 53 || 0011 0101 || 35 || 5 ||
 * 5 || 0000 0101 || 5 || ENQ ||  || 54 || 0011 0110 || 36 || 6 ||
 * 6 || 0000 0110 || 6 || ACK ||  || 55 || 0011 0111 || 37 || 7 ||
 * 7 || 0000 0111 || 7 || BEL ||  || 56 || 0011 1000 || 38 || 8 ||
 * 8 || 0000 1000 || 8 || BS ||  || 57 || 0011 1001 || 39 || 9 ||
 * 9 || 0000 1001 || 9 || HT ||  || 58 || 0011 1010 || 3A || : ||
 * 10 || 0000 1010 || A || LF ||  || 59 || 0011 1011 || 3B || ; ||
 * 11 || 0000 1011 || B || VT ||  || 60 || 0011 1100 || 3C || < ||
 * 12 || 0000 1100 || C || FF ||  || 61 || 0011 1101 || 3D || = ||
 * 13 || 0000 1101 || D || CR ||  || 62 || 0011 1110 || 3E || > ||
 * 14 || 0000 1110 || E || SOH ||  || 63 || 0011 1111 || 3F || ? ||
 * 15 || 0000 1111 || F || SI ||  || 64 || 0100 0000 || 40 || @ ||
 * 16 || 0001 0000 || 10 || DLE ||  || 65 || 0100 0001 || 41 || A ||
 * 17 || 0001 0001 || 11 || DC1 ||   || 66 || 0100 0010 || 42 || B ||
 * 18 || 0001 0010 || 12 || DC2 ||  || 67 || 0100 0011 || 43 || C ||
 * 19 || 0001 0011 || 13 || DC3 ||  || 68 || 0100 0100  || 44 || D ||
 * 20 || 0001 0100 || 14 || DC4 ||  || 69 || 0100 0101 || 45 || E ||
 * 21 || 0001 0101 || 15 || NAK ||  || 70 || 0100 0110 || 46 || F ||
 * 22 || 0001 0110 || 16 || SYN ||  || 71 || 0100 0111 || 47 || G ||
 * 23 || 0001 0111 || 17 || ETB ||  || 72 || 0100 1000 || 48 || H ||
 * 24 || 0001 1000 || 18 || CAN ||  || 73 || 0100 1001 || 49 || I ||
 * 25 || 0001 1001 || 19 || EM ||  || 74 || 0100 1010 || 4A || J ||
 * 26 || 0001 1010 || 1A || SUB ||  || 75 || 0100 1011 || 4B || K ||
 * 27 || 0001 1011 || 1B || ESC ||  || 76 || 0100 1100 || 4C || L ||
 * 28 || 0001 1100 || 1C || FS ||  || 77 || 0100 1101 || 4D || M ||
 * 29 || 0001 1101 || 1D || GS ||  || 78 || 0100 1110 || 4E || N ||
 * 30 || 0001 1110 || 1E || RS ||  || 79 || 0100 1111 || 4F || O ||
 * 31 || 0001 1111 || 1F || US ||  || 80 || 0101 0000 || 50 || P ||
 * 32 || 0010 0000 || 20 || space ||  || 81 || 0101 0001 || 51 || Q ||
 * 33 || 0010 0001 || 21 || ! ||  || 82 || 0101 0010 || 52 || R ||
 * 34 || 0010 0010 || 22 || " ||   || 83 || 0101 0011 || 53 || S ||
 * 35 || 0010 0011 || 23 || # ||  || 84 || 0101 0100 || 54 || T ||
 * 36 || 0010 0100 || 24 || $ ||  || 85 || 0101 0101  || 55 || U ||
 * 37 || 0010 0101 || 25 || % ||  || 86 || 0101 0110 || 56 || V ||
 * 38 || 0010 0110 || 26 || & ||  || 87 || 0101 0111 || 57 || W ||
 * 39 || 0010 1111 || 27 || ' ||  || 88 || 0101 1000 || 58 || X ||
 * 40 || 0010 1000 || 28 || ( ||  || 89 || 0101 1001 || 59 || Y ||
 * 41 || 0010 1001 || 29 || ) ||  || 90 || 0101 1010 || 5A || Z ||
 * 42 || 0010 1010 || 2A || * ||  || 91 || 0101 1011 || 5B || [ ||
 * 43 || 0010 1011 || 2B || + ||  || 92 || 0101 1100 || 5C || left slash ||
 * 44 || 0010 1100 || 2C ||, ||  || 93 || 0101 1101 || 5D || ] ||
 * 45 || 0010 1101 || 2D || - ||  || 94 || 0101 1110 || 5E || ^ ||
 * 46 || 0010 1110 || 2E || . ||  || 95 || 0101 1111 || 5F || _ ||
 * 47 || 0010 1111 || 2F || Right slash ||  || 96 || 0110 0000 || 60 || ` ||
 * 48 || 0011 0000 || 30 || 0 ||  || 97 || 0110 0001 || 61 || a ||
 * ** Decimal ** || ** Binary ** || ** Hex ** || ** ASCII ** ||||||||||  ||
 * 98 || 0110 0010 || 62 || b ||||||||||  ||
 * 99 || 0110 0011 || 63 || c ||||||||||  ||
 * 100 || 0110 0100 || 64 || d ||||||||||  ||
 * 101 || 0110 0101 || 65 || e ||||||||||  ||
 * 102 || 0110 0110 || 66 || f ||||||||||   ||
 * 103 || 0110 0111 || 67 || g ||||||||||  ||
 * 104 || 0110 1000 || 68 || h ||||||||||  ||
 * 105 || 0110 1001 || 69 ||  i   ||||||||||   ||
 * 106 || 0110 1010 || 6A || j ||||||||||  ||
 * 107 || 0110 1011 || 6B || k ||||||||||  ||
 * 108 || 0110 1100 || 6C || l ||||||||||  ||
 * 109 || 0110 1101 || 6D || m ||||||||||  ||
 * 110 || 0110 1110 || 6E || n ||||||||||  ||
 * 111 || 0110 1111 || 6F || o ||||||||||  ||
 * 112 || 0111 0000 || 70 || p ||||||||||  ||
 * 113 || 0111 0001 || 71 || q ||||||||||  ||
 * 114 || 0111 0010 || 72 || r ||||||||||  ||
 * 115 || 0111 0011 || 73 || s ||||||||||  ||
 * 116 || 0111 0100 || 74 || t ||||||||||  ||
 * 117 || 0111 0101 || 75 || u ||||||||||  ||
 * 118 || 0111 0110 || 76 || v ||||||||||  ||
 * 119 || 0111 0111 || 77 || w ||||||||||   ||
 * 120 || 0111 1000 || 78 || x ||||||||||  ||
 * 121 || 0111 1001 || 79 || y ||||||||||  ||
 * 122 || 0111 1010 || 7A || z ||||||||||  ||
 * 122 || 0111 1010 || 7A || z ||||||||||  ||



Boolean Logic
Logic gates are primarily implemented electronically using diodes or transistors, but can also be constructed using electromagnetic relays (relay logic), fluidic logic, pneumatic logic, optics, molecules, or even mechanical elements

These basic elements include:

AND

 * [[image:http://upload.wikimedia.org/wikipedia/commons/thumb/b/b9/AND_ANSI_Labelled.svg/120px-AND_ANSI_Labelled.svg.png width="156" height="65" caption="And gate"]] ||
 * And gate ||

And who's truth table looks like this:
 * **INPUT** || **OUTPUT** ||
 * A || B || A AND B ||
 * 0 || 0 || 0 ||
 * 0 || 1 || 0 ||
 * 1 || 0 || 0 ||
 * 1 || 1 || 1 ||

OR

 * [[image:http://upload.wikimedia.org/wikipedia/commons/thumb/1/16/OR_ANSI_Labelled.svg/200px-OR_ANSI_Labelled.svg.png caption="OR gate"]] ||
 * OR gate ||

Who's truth table looks like this: A B || **OUTPUT** A + B ||
 * **INPUT**
 * 0 || 0 || 0 ||
 * 0 || 1 || 1 ||
 * 1 || 0 || 1 ||
 * 1 || 1 || 1 ||

NOT (inverter)


Who's truth table looks like this: A || **OUTPUT** NOT A ||
 * **INPUT**
 * 0 || 1 ||
 * 1 || 0 ||

NAND

 * [[image:http://upload.wikimedia.org/wikipedia/commons/thumb/f/f2/NAND_ANSI.svg/200px-NAND_ANSI.svg.png caption="NAND"]] ||
 * NAND ||

Simply not AND
 * **INPUT** || **OUTPUT** ||
 * A || B || A NAND B ||
 * 0 || 0 || 1 ||
 * 0 || 1 || 1 ||
 * 1 || 0 || 1 ||
 * 1 || 1 || 0 ||

NOR

 * [[image:http://upload.wikimedia.org/wikipedia/commons/thumb/6/6c/NOR_ANSI.svg/200px-NOR_ANSI.svg.png caption="NOR"]] ||
 * NOR ||

Not OR
 * **INPUT** || **OUTPUT** ||
 * A || B || A NOR B ||
 * 0 || 0 || 1 ||
 * 0 || 1 || 0 ||
 * 1 || 0 || 0 ||
 * 1 || 1 || 0 ||

Network Applications
In networks, these often apply as octets pass through logic gates. For example, as the octet 10110010 and the octet 01011011 pass through an OR gate, the resulting table looks like this:


 * **OR Application** ||
 * Octet One || 1 || 0 || 1 || 1 || 0 || 0 || 1 || 0 ||
 * Octet Two || 0 || 1 || 0 || 1 || 1 || 0 || 1 || 1 ||
 * Result || 1 || 1 || 1 || 1 || 1 || 0 || 1 || 1 ||


 * **AND Application** ||
 * Octet One || 0 || 0 || 1 || 1 || 0 || 1 || 1 || 1 ||
 * Octet Two || 1 || 1 || 1 || 1 || 0 || 0 || 0 || 1 ||
 * Result || 0 || 0 || 1 || 1 || 0 || 0 || 0 || 1 ||


 * **NAND Application** ||
 * Octet One || 0 || 0 || 1 || 1 || 0 || 1 || 1 || 0 ||
 * Octet Two || 1 || 0 || 0 || 0 || 1 || 1 || 0 || 1 ||
 * Result || 1 || 1 || 1 || 1 || 1 || 0 || 1 || 1 ||



Assignment #14 - Boolean Logic Questions
Answer the following and submit it into the handin folder

1. Apply **__AND & OR__** to the following pairs of binary numbers: 11101010 : 11101110 01011111 : 10111011 11111000 : 11110101 2.Develop a truth table for the circuit listed below.

Circuit 1 inputs are 1,0 Circuit 2 inputs are 1,1 Circuit 3 inputs are 1, 0, 1, 1