 आज्लाम उलेक्यं एक्होँ इंबस्याकया आपद्याई, आप आप रईजीन जीकाते है. अफ्ता आप आप ओ़गता क्यो फीलिटःवे वाड के भी अगता है, भी अब वो अप यहाँ दोब ताशकाउ पहलोग, जो के पहप्स थोड़ा सा दिफिकल्त हैं अबी तक जो भी जिन formats पे हमने बात की है वो आप देसिमल लंबर सिस्तम में यूँस करते रहें तो थोड़ा सा दिफिकल्त हैं, दिफिकल्त नहीं हैं बाद में दिसकस करेंगे तो आप दिफिकल्त नहीं कर सकते रहें वो आप दिफिकल्त नहीं है, वो आप दिफिकल्त नहीं कर सकते है अप वीरीबल्त दिफिकल्त नहीं करते है, दिफिकल्त नहीं कर सकते है, वीरीबल्त, वीरीबल्त, �´फ् SAP 16 కితిటితికూడి. కికికిదాల్టి ఇటికారికెపతెగాల్కనినితెగరరిదులెతెచియరసంగి. నాల్వం పటరవంమాయారిమాన్టానియారిని. हलत्ता. अछसाद तैसेर डबरी स्वनां, नहील रहठाउ, तब बवावा홍, फो लग़ा और मी बन बवूएखाउगी घवावान नहीर दत. गर्ड़्िए कुईशिम की प्रद्गार्था। रेक्डेफण नद्दर्खाँ। तो नच्द नमबर याजा कोईशिमनदा इसना साज्टी शिस्तम। ळर्गडेफण नमबर साज्टी यच्टी नमबर आपन चिर्ण। naments  DENNIS . Now to write this long ладно, 40-bit binary number in a ڡSICE and in short form you would use the hexadecimal number system. ॐ so basically hexadecimal number system allows you to convert long binary strings into short numbers of course written in the hexadecimal form . Convert karte kasein well you write saying, due to the string is divide the string of course the binary string into groups of 4 bits So for example, you have for example 4 bit 0, 1, 1, 1. Which is 7 so you would replace that with hexadecimal, 7, and so on. heensa decimal co, binary make I so convert Sn again, it is very easy. Let us suppose にicios你要 have the number 7 f hexadecimal 7 is going to be replaced by 4 binary bits zero, one, one, one and a f is going to replace by another set of 4 bits, one, one, one 1. Now you know how to converse from hexadecimal to binary and from binary to hexadecimal In the rial world we used use decimal number systems, so you should be able to convert word from hexadecimal to decimal आप ना के सी शाओक welcoming this की नम दूग लगां कस नहीं करनें नहीं कूढकर शिसब उ toget hundreds to a बोखिल उख़िन दूख इक मि Turner of the decimal number. Now if you use these hexadecimal numbers then of course you can perform multiplication using hexadecimal numbers, division, subtraction and addition of course. So we did some examples, I do not need to repeat that. Addition in hexadecimal would of course generate carry as rather when you add decimal numbers carry is generated there when you add binary numbers carry is generated there. Similarly if you subtract two hexadecimal numbers you might need to borrow. Let us look at the octal number system. The octal number system is a pace 8 number system. So each digit of an octal number system would represent values from 0 to 7. You cannot represent 8 in octal. So what do you do when you need to write 8? Basically you use two digits. The most significant digit would be a 1 and the least significant digit would be a 0. 9 कसेवोगा, 1, 1, 10, it is 1, 2 and so on. So the next set of 8 numbers decimal numbers would be represented using two digits of the octal number system. The most significant digit would be 1 and the least significant digit would be starting from 0 and going all the way up to 7. The next number would start with 2, 0, then 2, 1 and so on. Let us have a look at the octal number system. The first slide would show the octal representation of decimal numbers and the second slide would show the counting in octal. The octal number system has 8 values. It is a base 8 number system. So the values in the octal number system are from 0 to 7. So they in fact represent the first 8 values of the decimal number system as seen in the table. Binary equivalence of the octal numbers can be represented using 3 bits because we have a range of 8 numbers. So the binary equivalent 0, 0, 0 represents octal 0 and binary 111 represents octal 7. The table shows octal equivalence of decimal numbers 8 to 31. Octal number 7 has the largest value to represent decimal 8 a 2 digit octal number is used. The most significant digit is a 1 and the least significant digit is a 0. So the decimal numbers 8 to 15 are represented by the 2 digit octal combination. Similarly the decimal numbers 16 to 23 are represented by another 2 digit octal combination. The most significant digit is 2 and the least significant digits vary from 0 to 7. And similarly for higher octal decimal numbers. Why do we need to use the octal number system? The reason is that the hexadecimal number system is there. When we write large or long strings of binary numbers there are chances of error. So to represent long binary strings in a compact concise form you use the octal number system. But since we have the hexadecimal number system so octal number system is not very frequently used. Anyway since we are discussing the octal number system let us look at converting from decimal to octal to decimal and octal to binary and binary to octal. So let us first start by looking at the conversion from binary to octal. How will it be? Well that is what we used for the hexadecimal conversion. You write out the binary string and divide it into groups of 3 bits. Not 4 bits because this is a base 8 number system. And remember binary bits represent a single digit of the octal number system. So you divide the binary string into groups of 3. Then what do you do? You just replace that binary group of 3 bits by its equivalent octal number. So 011 here it would be replaced by octal 6. 101 here that would be replaced by octal 5 and so on. Octal numbers can be converted into binary. How? Again we will use the same method which we used to convert from hexadecimal. So you have the number let us say 7 6 octal. So 7 is going to be replaced by its binary equivalent 111. 6 is going to be replaced by its binary equivalent 110. So you have a 6 bit binary equivalent of a 2 digit octal number. We will see its slides and examples later. Let us consider converting octal number system into decimal number system. Again there are two methods. One is the indirect method. The indirect method would be you convert from octal to binary and then binary to decimal. The other standard method is the sum of weights method. So you have an octal number, you write out an expression. When you solve it you would get the decimal equivalent. The base system is 8. So base number 8 use over all. How do you convert from decimal to octal? Again you have the indirect method where you convert to binary and from binary to the other number. The standard method is the division by 8 or the repeated division by 8 method. So you have a decimal number which you are trying to convert into octal. What do you do? You divide it by 8. You get a quotient value, you get a remainder value. You note down the remainder, you divide the quotient again by 8. You keep on doing this until the quotient value becomes 0. That means end of conversion, you just write out the remainders and that would give you the octal representation of the decimal number. Now you can in fact perform all arithmetic operations using octal number system as you did with the hexadecimal number system. You can add two octal numbers, you can subtract octal number, multiply, divide. The same rules apply in the decimal, binary and hexadecimal. When you add a carry might be generated. When you subtract you might need to borrow something from the next higher digit. So let us look at the examples of these conversions, the addition and the subtraction. Consider the binary to octal conversion first. Consider the 21 bit binary string which is to be converted to octal. The 21 bit binary string is divided into groups of 3 bits starting from the least significant bit. 7 groups of 3 bits are formed. In the next step each group of 3 bits is replaced by the equivalent octal digit. The result is a 7 digit octal number. Now consider another example. Consider the 13 bit binary string which is to be converted into octal. The 13 bit binary string is divided into groups of 3 bits starting from the least significant bit. Now 4 groups of 3 bits are formed and an incomplete 5th group with a single bit is formed. What we do, we add 2 most significant 0 bits in the 5th group to make it a complete group. Now each 3 bit group is replaced by the equivalent octal digit to complete the conversion. So the original 13 bit binary number is now represented as 1, 3, 3, 5, 1 and octal. Let us now consider the octal to binary conversion. The octal number is 1726. Each octal digit is replaced by its equivalent 3 bit binary number. Thus octal 1 is replaced by 001. Octal 7 is replaced by 111. Octal 2 is replaced by 010. And octal 6 is replaced by 110. Now let us consider the sum of weights method to convert an octal number 4037 to decimal. An expression in terms of the base number 8 and weights is written as can be seen. The expression is solved to give 4 sum terms 2048, 0, 24 and 7. The 4 terms are added to result in the number 2079, the decimal equivalent of octal 4037. Now let us consider the conversion of a decimal number into an equivalent octal number. The decimal number is 2079. It is divided by 8 which results in equation value of 259 and the remainder 7. The remainder 7 is noted and the equation 259 is divided by 8. 259 divided by 8 results in equation value 32 and a remainder value of 3. The remainder 3 is noted and the equation 32 is divided by 8. 32 divided by 8 results in equation value of 4 and a remainder 0. The remainder 0 is noted and the equation value 4 is again divided by 4. This results in equation value of 0 which indicates the completion of the repeated division by 8 method. The remainder 4 and the 3 remainders noted earlier represent the 4 digit octal number 4037. Let us now consider octal addition. Consider the addition of octal numbers 7602 and 4771. The 2 octal numbers are added by first adding the 2 least significant octal digits 2 and 1. The sum of the 2 octal digits is 3. The next 2 octal digits 0 and 7 are added and the result is 7. The octal digits 6 and 7 are added next. The answer in decimal is 13 which is represented as 15 in octal. Thus the sum is octal 5 and the carry is 1 which is carried over to the most significant digit. Adding the most significant octal digits 7 and 4 along with the carry results in decimal 12 which is equivalent to octal 14. The sum is 4 and carry of 1 is carried over to the fourth octal digit position. The sum is 14573 octal. Consider the subtraction of octal 4771 from octal 7602. The 2 numbers are subtracted by first subtracting the least significant digits octal 1 from octal 2. The difference is octal 1. Next octal 7 is subtracted from octal 0. A borrow is required from the next most significant digit. Borrowing a 1 means subtracting octal 7 from octal 10 which is decimal 8. The difference is octal 1. The octal number 7 is subtracted from octal 5 instead of octal 6 in the second digit position as a 1 had been borrowed from octal 6. To subtract octal 7 from octal 5 requires another borrow from the most significant digit 7. Thus subtracting octal 7 from octal 15 results in a difference of 6. The most significant digits are subtracted to result in 2. The difference of octal 7602 and 4771 is octal 2611. Up till now we have looked at 4 different ways of representing binary numbers. One of them was the floating point, the remaining were the normal binary. There are again several other alternate ways of representing binary numbers. Let us have a look at those alternate methods. The first method is the excess code. The second method is the BCD code and the third alternative way of representing binary numbers is the gray code. What is the excess code? Excess code or biased code is used by the floating point numbers. Why did we use the biased number or the code with the floating point numbers? We used it to represent the negative as well as the positive exponents. We had 2 options either we increased the exponent bit by 1 to have a positive or negative sign or just to keep it as 8 bits and use the excess code method. So let us consider an example before we look at the actual excess code. Consider the number minus 1, 0 and 1 and decimal. How would you represent this in binary? Sign binary use karna hoga kyunke negative number bhi hai, positive number bhi hai. So positive 1 is represented as 0, 1. 0 signifies positive and 1 is the magnitude. The 0 number itself is represented as 0, 0 and negative 1 would be 1, 1. The most significant one signifies the negative number and the least significant one represents the magnitude of negative 1. Now if you look at the 3 binary representations, you cannot compare them. Now 1, 1 actually means 3 in unsigned binary. So it would have been better if minus 1 had been represented by bits 0, 0. 0 had been represented by bits 0, 1 and 1. Positive 1 had been represented by 1, 0. So you have a smooth scale through which you can compare. So 0, 0 is of course less than 0, 1 which is in fact a 0. So the excess code is used to give a uniform scale to positive number and negative numbers. Let us consider an example. We are going to be using an excess 8 code to represent numbers between minus 8 to plus 7. Consider the decimal number range positive 7 to minus 8. These positive and negative decimal numbers can be represented by the 2's complement representation as can be seen in the table. The magnitude of positive and negative numbers can not be easily compared if the numbers are represented in 2's complement form. The decimal number range plus 7 to minus 8 is represented using an excess 8 code that assigns 0, 0, 0 to minus 8, the lowest number in the range and 1, 1, 1, 1, 2 plus 7, the highest number in the range. Now excess code 8 or excess 8 code is obtained by adding number to the lowest number minus 8 in the range such that the result is 0. So now the number which you add to the lowest number minus 8 to give you a 0 is the number 8. So that is how you get excess 8 code. Minus 7 is represented as a 1 because 8 is added to minus 7, the result is 1. So you write 0, 0, 0, 1 which is the excess 8 notation. The resulting code which you get is known as the excess 8 code or biased 8 code. We have just seen an example of excess 8 code. Excess 8 code is in fact a binary code which has been shifted by 8. The excess or biased 127 code which we use to represent the exponent part and the 32 bit floating point representation is in fact again a binary code which has been shifted by 127. Now let us look at another representation, the BCD code. Now you must have traveled via train. So when you go to a railway station you see those huge clocks, digital clocks where time is shown as decimal digits. You must have also seen these clocks and buses. So now the clock circuit is a digital circuit which sends out or displays time on a 7 segment display. Now let us suppose you need to display the time 12 o'clock. So how do you display the 2 digits? Well you have to send out some binary code to display the digit 2 and send out some binary code to display the digit 1. Now since you are displaying decimal numbers 0 to 9 you cannot go beyond that. If you need to display 12 then you are displaying 2 digits. Each digit can change from 0 to 9. So now how many bits you require to display a single decimal digit from 0 to 9? Well you require 10 different codes, binary codes. So how do you represent decimal digit 0? Bits 0 0 0 0. How would you represent the highest decimal number 9? Binary number 1 0 0 1. Now you can get more combinations if you use 4 bit binary. So you can also have a number 1 1 0 0 which is in fact 12. But you do not use binary codes higher than 1 0 0 1. So the BCD representation only uses the first 10 binary codes starting from 0 0 0 0 up to 1 0 0 1. Now as I have said where are they used? These binary codes or the BCD code is used to display decimal numbers. Now if you learn to use these BCD codes then you can of course represent decimal digits using these BCD codes. You can in fact perform additions and subtractions using BCD numbers. Let us have a look at an example which shows us how we can add two BCD numbers. Let us first consider the BCD code. As I had said earlier BCD code represents the decimal digits 0 to 9. So a 4 bit BCD code 0 0 0 represents decimal 0 and 1 0 0 1 represents decimal 9. A 4 bit binary code can represent up to 16 different values. The first 10 binary codes represent the decimal numbers 0 to 9. The remaining 6 codes 1 0 1 0, 1 0 1 1, 1 1 0 0, 1 1 0 1, 1 1 1 0 and 1 1 1 1 are considered to be invalid BCD codes. Consider the addition of 2 2 digit BCD numbers 23 and 45. BCD numbers are added using binary addition method resulting in 68. Now consider the addition of 2 2 digit BCD numbers 23 and 49. Adding least significant digits 3 and 9 results in 12 represented as 1 1 0 0. Now 1 1 0 0 is an invalid BCD digit. Secondly adding 3 and 9 in decimal should result in 2 as the sum and 1 as the carry. The BCD result 0 1 1 0, 1 1 0 0 is equivalent to 62 which is incorrect as the carry generated by the addition of the least significant digits has not been considered. When addition of BCD digits results in an invalid BCD number or a carry the number 0 1 1 0 which represents decimal 6 is added to the result. Reconsidering the addition of BCD 23 and 49 the addition of digits 3 and 9 results in 1 1 0 0 an invalid BCD number. A 0 1 1 0 which is decimal 6 is added which results in 0 0 1 0 and a carry which is added to 0 1 1 0 to result in 0 1 1 1. Now the BCD number 0 1 1 1 0 0 1 0 represents 72 which is the correct answer. We have just seen the BCD code. BCD code represents decimal digits from 0 to 9. So the binary code 1 0 1 0 which represents decimal 10 is invalid. So only the first 10 binary codes are used. BCD in fact stands for binary coded decimal. The third representation the grey code representation. Let us have a look what or rather where do we use this grey code. Now if you look at the binary code when you let us suppose count from 2 to 3 how many bits change. Only the first or the least significant bit changes. Let us suppose you are counting from 7 to 8 how many bits change. Well 7 is 0 1 1 1 and 8 is 1 0 0 0 4 bits have changed. Now in certain electromeganical applications of digital systems you would like to keep the bit change to 1 when you count up or down to the next successive value. The grey code is a binary code which allows you to represent successive numbers where the bit change is limited to 1. For example when you count from 3 to 4 or from 3 to 2 the changes in bits would be limited to 1. As I have said before you need to restrict the bit change to 1 when you count from 1 number to the other successive number. Let us have a look at an example which explains the utility of the grey code. But before that I would like to mention that the grey code is not a positional code. Binary code we said was a positional code because each bit represents a certain value. Grey code is not a positional code. Now let us have a look at the example. Let us consider the 3 bit binary code and the 3 bit grey code representing the decimal numbers 0 to 7. The binary code is a positional code and the 3 bits starting from the least significant bit have the weights 1, 2 and 4. The grey code is not a positional code. However the change from 1 value to the next using the grey code guarantees a single bit change. So as you can see counting from 0 to 6 the change between successive representations or successive numbers is limited to a change of only 1 bit. Consider the example of shaft encoders that best explain the utility of the grey code. The diagram shows a disk connected to the shaft of a rotating machine. The shaded areas on the disk indicate conducting strips connected to plus 5 volts. The non shaded area indicate non conducting strips. The 3 stationary brushes A, B and C touch the surface of the rotating disk. The 3 brushes are connected to 3 LED lamps through wires. As the disk rotates the brushes come in contact with the conducting area and the insulated area. The 3 LEDs display the position of the rotating shaft in terms of 3 bit numbers. Consider the disk on the left. If the disk on the left rotates in the anti clockwise direction by 45 degrees the brush A comes in contact with the non conducting strip. Thus LED connected to brush B lights up indicating binary 010. Consider the disk on the left. If the disk on the left rotates in an anti clockwise direction by 45 degrees the brush A comes in contact with the conducting strip at 5 volts and turns on the LED indicating binary 001. If the disk continues its rotation and after another rotation of 45 degrees brush B comes in contact with the conducting strip and brush A comes in contact with the non conducting strip. Thus LED connected to brush B lights up indicating binary 010. Thus at any instant of time the LEDs indicate the angular position of the rotating shaft in steps of 45 degrees. Assume that the 3 brushes A, B and C are not aligned properly and brush B is slightly ahead of brushes A and C. Now if the disk rotates 90 degrees from its start position brush A is in contact with the conducting strip brush B due to its misalignment is in contact with the conducting strip and brush C is in contact with the insulated strip. Thus when the disk rotates the LEDs will show a 001 followed by a 011 for a short duration when the disk rotates from 90 degrees to 91 degrees and then it would display 010. Thus due to misalignment the count value jumped from 1 to 3 and then back to 2. Consider the disk shown on the right the conducting and non conducting strips follow a grey code pattern 0000, 001, 011, 010, 110, 111, 101 and 100 representing decimal numbers 0 to 7. Now even if the brushes are misaligned the LEDs would always display the correct count value. We have just looked at an example of the grey code, we might be looking at other examples of a grey code during the course of the study. Up to now we have been looking at only numbers, positive numbers, negative numbers, numbers having fraction part, numbers having an integer part. Now we said in the beginning in the first lecture that computers can be used to write documents. You have been writing programs, so in your program you have text information as well as numbers. So how do you represent capital letter A, small letter B, number 0, number 9, full stop, comma, question mark. So all these characters have to be again representing in binary because as we know digital systems only recognize binary. Now the representation which is used to represent text punctuation mark is known as an alpha numeric format or the alpha numeric representation. The most commonly used representation is the ASCII representation, the American code for information interchange. It is a 7 bit format, so how many codes do you think it can represent? 2 raised to power 7 gives you 128 unique codes. Now 127 or rather 128 character ASCII code is divided into capital letters, small letters, numbers, punctuation marks and some control characters. Let us have a look at the ASCII table. Consider some of the ASCII codes that represent the numbers 0 to 9. The ASCII code 011000 which is also equivalent to 30 hexadecimal represents the number 0. The ASCII code 0110001 which is also equivalent to 31 hexadecimal represents the number 1. Similarly the ASCII code 0110001 which is equivalent to 39 hex represents the number 9. How do we represent the capital alphabet letters and the small alphabet letters? Well the lower case alphabet letters a to z are represented by ASCII code 1100001 which is 61 hex which represents small letter a up to 1111010 which is 7a hexadecimal and it represents the lower case character z. The ASCII code 1000001 which is equivalent to 41 hex represents the upper case alphabet a and the code 1011010 which is equivalent to 5a hex represents the capital character z. The 32 control characters are represented by ASCII code 00000 which is 0 hex up to 0011111 which is equivalent to 1f hex. These control characters are not seen on the screen. They are simply executable codes. The 7 bit ASCII code is known as the alphanumeric code because it is used to represent alphabet characters as well as numbers. Now 128 different characters is not enough. You would like to represent or write some other characters some graphical characters. So we have another version of the ASCII code which is known as the extended code. The extended ASCII code is an 8 bit code. So how many codes can we uniquely represent using the extended ASCII code? Well it is 2 raised to power 8 or 256 unique codes. The 128 codes which we have just looked at is a standard whereas the extended version the remaining 128 codes are not a standard. Different when this use different versions of the extended set. Now even an 8 bit ASCII code which allows you to represent up to 256 different characters is not enough. On your computers you must have seen some Urdu characters or Arabic characters. So how do you represent such characters? There are other languages, Japanese, Chinese. Well there are other standards. One of them is the Unicode. The Unicode is a 16 bit code. So how many characters can you represent a Unicode? Well 2 raised to power 16. So nearly 64000 different Unicod characters can be represented using the Unicode. There are other codes. Now let us look at error detection. In the first lecture we mentioned that digital systems are very reliable systems. When you store information or when you transmit information there are chances that the information is going to be corrupted. Now what do you do? Well in order to have secure information you need to have a mechanism to detect errors if they have occurred and perhaps try to remove those errors. Now one simple method of detecting errors is using the parity bet. Let us consider an example. Let us suppose you and your friend set up a piece of wire, a communication link. And you send some numbers let us say 0 to 7 to your friend across that wire link. You agree that each number is going to be represented in an 8 bit binary format. So you would be sending 8 bit numbers to the other end. Your friend at the other end would read the number and send back its 2's complemented form. Now let us suppose you send the number 7 and your friend should be receiving the number 7 of course in binary. Now how would you ensure that your friend has received the correct number? It is 7 and it is not 8. To ensure correct exchange of information between you and your friend you decide that you would only be sending numbers which have an odd number of 1's. So that means you would send the number 1 how many 1's does it have an odd number of 1's. You could send 2 how many 1's does it have again an odd number of 1's. You cannot send 3 because there are 2 1's so it is an even number of 1's. Well this particular requirement restricts the number which you can exchange. Well why not use a better method. The sender you count the number of 1's in your message. If it is an even number of 1's you append an extra bit and set it to either 0 or 1. So that the total number of bits add up to an odd number. So for example you are sending the number 3 how many bits do you have you have an even number of bits. So you append a parity bit which is known as the odd parity bit because you would be setting that parity bit to 1. So that the total number of bits add up to an odd number. So you would now be transmitting 9 bits the 8 bit number 2 rather 3 and the parity bit. Your friend would receive the 9 bits he would count the number of bits in that message. So if it is an odd number of bits then that means the message is correct no bit has been corrupted. If the number of 1's in the message turns out to be an even number that means some error has occurred. So now you as a sender would be generating 8 bit numbers. Counting the number of 1's setting the parity bit to 1 or 0 so that the total number of 1's in the message comes out to be odd. Transmitting the number to the other end. Your friend at the receiver would be counting the number of 1's in the message including the parity bit. If it turns out to be odd then the message is correct. If it turns out to be even then the message is incorrect. Then perhaps he can send you a signal requesting you to send or we send the message. You could use the even parity method. In the even parity method you and your friend decide that the number of 1's which you would be sending would be an even number. So the procedure is the same. You would generate a number and 8 bit number. You would count the number of 1's if it is an odd number. Then you would attach a parity bit set to 1's so that the total number of 1's come out to be even. So you would be sending these 9 bit messages at the other end. The receiver and your friend would be counting the number of 1's. If they are even then the message is correct. If they are not even the message is incorrect. Then we either use the odd parity method or the even parity method to detect 1 bit errors. Let us have a look at an example which explains this parity bit method of detecting errors. Consider an example of odd parity error detection. The original data which is to be sent or which is to be stored is 100110. With odd parity the total number of 1's should be an odd number. So how many 1's do we have in the original message and even number. There are 4 1's. So you append a parity bit and set it to 1. So the total number of 1's now becomes an odd number. Now let us suppose this 9 bit information is stored or transmitted to a remote location. And at the remote location the message has changed to 110111010. The most significant bit is the parity bit. How many 1's do we have in the message? 6. Since we are using the odd parity scheme the total number of 1's in the message should have been odd. So this indicates a single bit error. Now let us consider that a 2 bit error occurs. So we send the original message with the odd parity bit 110011010. Let us suppose the 2 bits which have been underlined change. What do you get at the receiver end? The total number of 1's is still odd. So have we detected an error? No. So if a 2 bit error occurs we are not able to detect an error. Now let us consider the third case. We send the original message with the parity bit and a 3 bit error occurs. The 3 bits which change due to an error are underlined. Now how many 1's do we receive at the receiver end? The total number of 1's which have been received in the 9 bit message are 4. It is an even number. So that means an error has occurred. But we are not sure if a single bit error has occurred or a 3 bit error has occurred. But we know that an error has occurred. So in summary by using a parity bit we are able to detect 1 bit error, a 3 bit error, a 5 bit error or a 7 bit error. But we are unable to detect even number, even bit errors. For example 2 bit error occurs we are not able to detect that. If 4 bit error occurs we are not able to detect that. We have just looked at an example of odd parity which is used of course to detect a single bit error. We said using the parity bits we cannot detect even bit errors. We end today's lecture here. Before we end let us quickly summarize the topics which we have discussed. We started with the octal number system. We said it is a base 8 number system and it is basically used to represent large binary strings in a short concise form. We talked about alternate ways of representing binary numbers. We talked about the excess code which we use in the floating point implementation. We talked about BCD code which is used to send decimal digits in binary. Again applications are to display decimal digits on 7 segment displays. We talked about a grey code which allows only one bit change when you count to the next successive value, the upper value or the lower value. We also talked about alpha numeric codes, the ASCII code. The alpha numeric or the ASCII code allows you to represent numbers, alphabet characters, punctuation marks and some control characters. Towards the end we looked at an error detection scheme which uses the odd parity method or the even parity method to detect one bit, three bit or five bit errors. Let us stop for today. With today's lecture we have finished our discussion on number systems. As I have been saying before, keep practicing with the binary number system. You should be able to relate to binary numbers as you do with decimal numbers. See you next time.