Please change the bit positions from 1-7 -> 7-1 then it is meaningful otherwise miscommunication could happen to listeners. Also bits will start from 0 , 1, 2, .... and not directly from 1. Please correct this then this video will be awesome. Thanks.
I m surprised why some people dislike this video. U r really a great teacher. I like ur all videos.i hd subscribed this channel only because of u. I hv seen so many videos in this channel but i like specifically ur videos. Generally i don't comments but i m commenting u here to encourage.Do ur job very well. Truly ur explanations are very easy and comfortable to understand and ur expression is very innocent with a very beautiful face.just try give to little bit smile to make situation easy. It will help u also to b comfort to express the lesson proper. I m also a Lecturer Mtech (Digital communication). All the best.Go ahead. Thank you for ur video.
There is a small problem, since you started numbering the problem from left to right, you ended up with error bit at position 6. Thus by inverting the 6th bit from left your code can be corrected. The same problem, if numbered from right to left and solved accordingly, the error comes out to be 0. Thus this implies that a code detection and correction depends on whatever side we take the MSB and LSB as, but this shouldn't be the case. The correction process should work irrespective of the fact on whichever side we start. So what is the error??
Mam I've a query that in book I read that those codes having minimum distance 1 can't detect error and also can't correct errors bt at the same time there has been taken an example that 8421 code detects error and also correct the errors so how is it possible bcz 8421 having minimum distance of 1
i beg you please tell about The Network layer Network layer concepts, design issues, static and dynamic routing algorithms, shortest path routing, flooding, distance vector routing, link state routing, distance vector routing, multicast routing, congestion control algorithm, internetworking, Ipv4 Medium access layer Channel allocation problem, multiple access protocols, ALOHA, CSMA/CD, IEEE Standard 802 for LAN and MAN
Achin may b u r asking for digital data communication and wireless communication topics. Definitely we will forward your request to our teem. Soon will try to work on that.
Wrong from the start! Binary is a positional number system. You cannot number the bits right to left because the rightmost bit is the MSB. In your code word the two leftmost bits are check bits c[1],c[0].
Actually its easier to say Parity bits P3=1, P2=1, P1 =0 P1 =0 is determined by adding bits that are in positions 1, 3, 5, and 7. Those bits are 1+1+1+1 = 4 (an even number, in binary =100 ends with a 0) So that means our parity is even thus P1 =0 (even parity) . Parity in English means difference = so the number of one are even ( or balance out - or 'even out' if you like to say ) NB for P1 added bits for positions 1 (binary =001), 3, (binary 011), 5 (binary=101), 7(binary=111). The is a particular pattern in this list in that their last binary bits = 1 (i.e. ends with a 1 in last position) Similarly P2 we use position(s) that have a 1 in the second position --> 2(binary 010), 3(binary 011), 6 (binary, 110)
and position 7(binary 111) in that case P2 = (sum of bits in positions 2,3, 6 and 7) = 0+1+1+1 = 3 ( binary 011) This number is odd... so the number of 1s are not even (we have to 'even out' things here by making parity = 1) Thus P2 =1 (or you can use the value obtained --> 3 (011) ends with a 1, thus 1 becomes your parity bit)
1 in binary is = 0001, 3 is = 0011, 5 is 0101 and 7 is 0111 These numbers end with digit 1 (right most bit or least significant binary bit) P2 you do the same look for numbers with a 1 on the second position from the right you get 2 (which in binary is = 0010 ), 3 which is binary is = 0011, 6 which is binary is 0110 and 7 which in binary is 0111. Hope that helps @ Ananth S Karnavar I can only see you post on my gmail but here I had to copy paste. can't find it to reply
bro, convert 1,2,3,4,5,6,7 into binary form. then for e1, check which binary digit is having 1 at first position. for e2, check which binary digitt is hving 1 in middle position. for e3, check which binary digit is having 1 in last position. try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7
According to excess-3 code we need to add 3 to both digit in the decimal number then convert into 4-bit binary number for result of each digit. Therefore, = 15.46+33.33=48.79 =0100 1000.0111 1001 which is required excess-3 code for given decimal number 15.46.
Based on 2^3 combinations we take 421 000-0 001-1 010-2 011-3 100-4 101-5 110-6 111-7 For e1 we take the reference 1,3,5,7 because the binary weight 2^0 consist 1's in this positions and similarly for e2 and e3
At 2:54, That switch from E3, to E1 was super confusing. You should have numbered them in ascending order and just redid the video rather than scratching out (1,3,5,7) E1, and replacing it with (4,5,6,7). Makes sense now but you can't just correct from a fatal flaw like that and move on. You've lost your viewers. You either need to explain why you're changing your course or recast the video. You should also explain Even Parity rather than assuming it, perhaps before you start the lessen.
She explained it in her previous video Among all numbers from 0 to 7 Consider the numbers 2,3,67 2 - 0010 3 - 0011 6 - 0110 7 - 0111 We can observe that all of them have '1' in 2nd place from right So they come under E2 Similarly which have '1' at 1st place from right comes under E1 n which have '1' at 3rd place from right comes under E3
Basically counting up to 7 in binary, you will see there are 4 numbers which include 1 value 00(1), 01(1), 10(1) and 11(1), these represent E1, thus E1 = 1,3,5,7. Same applies for E2 but this time they must include the 2 value so again: 0(1)0, 0(1)1, 1(1)0 and 1(1)1 which then E2 represents 2,3,6,7. Following the same principle but must include the 4 value, we have 4,5,6,7 representing E3. Hope this wasn't too confusing
Arnab Bag plz refer video with title Hamming code generation example with even parity Under hamming code there i explained how we can take p1 p2 p3 same is applicable to take e1 e2 e3
bro, convert 1,2,3,4,5,6,7 into binary form. then for e1, check which binary digit is having 1 at first position. for e2, check which binary digitt is hving 1 in middle position. for e3, check which binary digit is having 1 in last position. try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7
For those who are wondering about formation of E1, E2 and E3. Follow this trick : bro, convert 1,2,3,4,5,6,7 into binary form. then for e1, check which binary digit is having 1 at first position. for e2, check which binary digitt is hving 1 in middle position. for e3, check which binary digit is having 1 in last position. try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7 Hope it helped
Check out Digital Electronics courses at: bit.ly/3aIM1ur
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Bit designation should be MSB to LSB i.e7 6 5 4 3 2 1so Error at 010 position.
Please change the bit positions from 1-7 -> 7-1 then it is meaningful otherwise miscommunication could happen to listeners. Also bits will start from 0 , 1, 2, .... and not directly from 1. Please correct this then this video will be awesome. Thanks.
i watched so many videos on hamming correcting codes , but i wasn't understood any of them except ur video
How to write e3 e2 e1 values
watch this ruclips.net/video/UY0VpqyJ3U4/видео.html
With pen or pencil
Thank you so mach I passed in my control system sub because of u thankyou
Mam why u numbered bits from left to right...
We usually write from right to left...
Can u explain
THIS IS HAMMER CODE NOT USUAL BINARY CODE AND FOR HAMMING CODE REFERENCE IS TAKEN FROM LEFT TO RIGHT NOT USUAL RIGHT TO LEFT IN BINARY CODE
@@ekanshdhawan8854 nice reply bro
But he asked 1yr ago
I m surprised why some people dislike this video. U r really a great teacher. I like ur all videos.i hd subscribed this channel only because of u. I hv seen so many videos in this channel but i like specifically ur videos. Generally i don't comments but i m commenting u here to encourage.Do ur job very well. Truly ur explanations are very easy and comfortable to understand and ur expression is very innocent with a very beautiful face.just try give to little bit smile to make situation easy. It will help u also to b comfort to express the lesson proper. I m also a Lecturer Mtech (Digital communication). All the best.Go ahead. Thank you for ur video.
....leacturer? but your English writing skills?
She is everywhere 🌹🌹🌹🌹🌹 may Allah bless you
How can we take the refrence locations
We can use 8421 code then take referral location
There is a small problem, since you started numbering the problem from left to right, you ended up with error bit at position 6. Thus by inverting the 6th bit from left your code can be corrected.
The same problem, if numbered from right to left and solved accordingly, the error comes out to be 0.
Thus this implies that a code detection and correction depends on whatever side we take the MSB and LSB as, but this shouldn't be the case. The correction process should work irrespective of the fact on whichever side we start.
So what is the error??
If you will take it from LSP than there will be error at 2nd place not on 6th place so the answer will be same @SourinRoy
Mam why we are taking E3 position 4,5,6,7
Thank u so much Api :)
love from Bangladesh
Why we should consider E3 values as 4567?
Write truth table for e3 e2 e1 in binary form from 0 t0 8 4=100 5=101 6=110 7=111 e3 is msb bit all 1s are coming in 4 5 6 7 position
@@ayyappareddy4461 thnx bro
How did u take E3 values as 4567?
Mam bit postion should be from LHS or RHS we r getting different answers from each side which one is current plz tell us🙏
Mam
Thank you so much
Really help this class for me.this class too late to me
Same question was in our sem exam, thanks mam for teaching us........
was the answer correct? actually i got the error position as 2
Explain is not good but i got something thank you for your video ma'am
Thanks ma'am .. very well EXPLAINED 😊
Good explanations...
Only we should consider the multiple s of 2 in parity bits ????
Why we not take e1e2e3 instead of e3e2e1
Nice explanation
Mam how you take references as eror2as 2367
nice explanation mam good...
thank madam easly explained
Akhila Vedakrishna r u understand all....
Nice ..,it's clear tq
Mam I've a query that in book I read that those codes having minimum distance 1 can't detect error and also can't correct errors bt at the same time there has been taken an example that 8421 code detects error and also correct the errors so how is it possible bcz 8421 having minimum distance of 1
Thank you.. inverse of ratio is also tru e
i beg you please tell about The Network layer
Network layer concepts, design issues, static and dynamic routing algorithms, shortest path routing, flooding, distance vector
routing, link state routing, distance vector routing, multicast routing, congestion control algorithm, internetworking, Ipv4
Medium access layer
Channel allocation problem, multiple access protocols, ALOHA, CSMA/CD, IEEE Standard 802 for LAN and MAN
Achin may b u r asking for digital data communication and wireless communication topics. Definitely we will forward your request to our teem. Soon will try to work on that.
Wrong procedure
Don't Beg .. Just request :) :) ...
3:25 how would be come to know how to use that order???
Thank you
How did you know the reference data for e3,e2 and e1??
Super mam, we need source and error control... For collage level reference....
But only thing , how you consider that E1 E3 E3....
Wrong from the start!
Binary is a positional number system. You cannot number the bits right to left because the rightmost bit is the MSB. In your code word the two leftmost bits are check bits c[1],c[0].
Thank You So much!! :)
Thank You So Much ...This video is properly learning me the concept of hamming code...👍
Great Explanation Mam made lot of things clear now, Don't Mind the Comments, Continue.
Thank you maam , you are like my god mother,🙏
beauty with brain...great tutorial
hawas
Hawas
hawas
😂
@@ajaysangwan1519 chutiye, chain kyu toda?
Too much thanxx I m doing mistake yet ............now its ok
I'm having a doubt that why e1,e2,e3 are changed to even when they are in odd...why don't we take them in odd itself mam
hello... we should take numbers 1234 for hamming code from left to right or right to left..
kiran kumar you can take it from any side .you will get the same answer
Niranjan Naidu thanx brooo😙
For the people who are saying "we can't understand and worst explanation" see the last 2 previous videos... Then u can understand :)
As u said that error will be only in the 2^n format
But the error is in 6 position the the formula 2^n is wrong or what?????
I don't understand how is coming E3=1,E2=1,E1=0. please tell me
Actually its easier to say Parity bits P3=1, P2=1, P1 =0
P1 =0 is determined by adding bits that are in positions 1, 3, 5, and 7. Those bits are 1+1+1+1 = 4 (an even number, in binary =100 ends with a 0)
So that means our parity is even thus P1 =0 (even parity) . Parity in English means difference = so the number of one are even ( or balance out - or 'even out' if you like to say )
NB for P1 added bits for positions 1 (binary =001), 3, (binary 011), 5 (binary=101), 7(binary=111). The is a particular pattern in this list in that their last binary bits = 1 (i.e. ends with a 1 in last position)
Similarly P2 we use position(s) that have a 1 in the second position --> 2(binary 010), 3(binary 011), 6 (binary, 110)
and position 7(binary 111) in that case P2 = (sum of bits in positions 2,3, 6 and 7) = 0+1+1+1 = 3 ( binary 011) This number is odd... so the number of 1s are not even (we have to 'even out' things here by making parity = 1) Thus P2 =1 (or you can use the value obtained --> 3 (011) ends with a 1, thus 1 becomes your parity bit)
@ Ananth S Karnavar
glmstar how do you determine that the values 1,3,5,7 are for p1..? I mean, why take 1,3,5,7 ?
1 in binary is = 0001, 3 is = 0011, 5 is 0101 and 7 is 0111 These numbers end with digit 1 (right most bit or least significant binary bit)
P2 you do the same look for numbers with a 1 on the second position from the right you get 2 (which in binary is = 0010 ), 3 which is binary is = 0011, 6 which is binary is 0110 and 7 which in binary is 0111.
Hope that helps @ Ananth S Karnavar
I can only see you post on my gmail but here I had to copy paste. can't find it to reply
bro, convert 1,2,3,4,5,6,7 into binary form.
then for e1, check which binary digit is having 1 at first position.
for e2, check which binary digitt is hving 1 in middle position.
for e3, check which binary digit is having 1 in last position.
try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7
samajh me ache se aata
Thanks mam for this wonderful teaching 🙏🙏
Thankyou so much!
Mam please say how to take e3,e2,e1
Very nice explanation madam , u have good teachings skills thank you
mam ap kaha rahte ho
Mam..if u take E1,E2,E3 then error bit is different know mam...can u explain it mam.....
The combination of digits0-7 follows e3 e2 e1 so we take error in this order only
How to take reference numbers ??
the lecture was really good thanks😇
You have e3,e2,e1 (110) mean 6 why can't we consider e1,e2,e3 (011) mean 3
Why are we traversing the array from MSB to LSB?
Traversing from LSB to MSB is giving us a different error bit...
Thank you. Can you please explain how to convert a fraction to excess 3 code
According to excess-3 code we need to add 3 to both digit in the decimal number then convert into 4-bit binary number for result of each digit. Therefore,
= 15.46+33.33=48.79 =0100 1000.0111 1001 which is required excess-3 code for given decimal number 15.46.
Madam,very good explanation, keep it up,. this lectures are helpful very much to us. thank you
Wow excellent, you tought very welli understood in a few minutes
how to consider reference variable .
how reference place came mam
dhorun amai 9 biter kno code dilo tkhn returndency bit 5 hoba to.MNA 9+5 =14 hoba tai to.am I right????
Ur english is not good .. but thanks to u I get it ... thanks again
How the reference is taken medam
Based on 2^3 combinations we take
421
000-0
001-1
010-2
011-3
100-4
101-5
110-6
111-7
For e1 we take the reference 1,3,5,7 because the binary weight 2^0 consist 1's in this positions and similarly for e2 and e3
mam how you take the reference (1,3,5,7). Is there is any rule
watch the previous video. It's explained there.
Amazing thanks heaps
Thanks a lot
What to do if it was when odd parity ?
Thank you 😌
Mam ap pehle E1 kyu nhi li E3 kyu li? E1, E2, E3 lene pe 011 aa raha h mean 3rd place pe error h
VERIFY AND CORRECT:
bit numbers in E3E2E1 start from 000.
If E3E2E1 = 110 = 6 (in decimal) ; It means the 7th bit is in error
How to take reference places in errors bits
Love u mam 💚💜💙
Having an exam tomorrow and this helped me a lot
How take E1,E2,E3
At 2:54, That switch from E3, to E1 was super confusing. You should have numbered them in ascending order and just redid the video rather than scratching out (1,3,5,7) E1, and replacing it with (4,5,6,7). Makes sense now but you can't just correct from a fatal flaw like that and move on. You've lost your viewers. You either need to explain why you're changing your course or recast the video.
You should also explain Even Parity rather than assuming it, perhaps before you start the lessen.
Thanks for explaining mam
But how we consider reference bits like
2367 for E2 ...?
She explained it in her previous video
Among all numbers from 0 to 7
Consider the numbers 2,3,67
2 - 0010
3 - 0011
6 - 0110
7 - 0111
We can observe that all of them have '1' in 2nd place from right
So they come under E2
Similarly which have '1' at 1st place from right comes under E1
n which have '1' at 3rd place from right comes under E3
@@raamatulasi728 lol u saved my time tanks
how it will become 110=6 ? can u anyone explain?
chandu anugula its 1*2²+1*2¹+0*2^0=4+2=6
(110) is the binary of 6, which shows the error place......
Thanx mam
Thank you mam
Mam how E1 e2 E3 are formed please explain
Basically counting up to 7 in binary, you will see there are 4 numbers which include 1 value 00(1), 01(1), 10(1) and 11(1), these represent E1, thus E1 = 1,3,5,7. Same applies for E2 but this time they must include the 2 value so again: 0(1)0, 0(1)1, 1(1)0 and 1(1)1 which then E2 represents 2,3,6,7. Following the same principle but must include the 4 value, we have 4,5,6,7 representing E3. Hope this wasn't too confusing
can we find error by seeing it without using this process, where the error is
Nice mam superb
madam can you explain exactly how E1,E2,E3 are formed
Arnab Bag full video series is in www.tutorialspoint.com under video tutorials for digital electronics. there you can find full data for hamming code
There is no video about it!
Arnab Bag plz refer video with title Hamming code generation example with even parity
Under hamming code there i explained how we can take p1 p2 p3 same is applicable to take e1 e2 e3
bro, convert 1,2,3,4,5,6,7 into binary form.
then for e1, check which binary digit is having 1 at first position.
for e2, check which binary digitt is hving 1 in middle position.
for e3, check which binary digit is having 1 in last position.
try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7
ia to may samasgia ager e3=1 e2=1,e1=0 kasa agia. tom janthiti to bataw
Thank you for your video. But you have not explained how you get reference bits. So your explanation is cloudy
everything is fine but how can you confirm places of e1,e2,e3,please explain
We have to subtract 7 and 4
How can I find even parity or odd parity
Sakshi Shirpurkar I can help you to understand
how to correct code when parity not mentioned whether it is even or odd
For those who are wondering about formation of E1, E2 and E3.
Follow this trick :
bro, convert 1,2,3,4,5,6,7 into binary form.
then for e1, check which binary digit is having 1 at first position.
for e2, check which binary digitt is hving 1 in middle position.
for e3, check which binary digit is having 1 in last position.
try it u will get e1= 1,3,5,7 e2=2,3,6,7 e3=4,5,6,7
Hope it helped
I think u interchanged e3 n e1
Thank you...😚
v clear explanation
thanx great
how to correct if there are 2 errors?
thank u mam
Are u play the role of tara in indori ishq.. Mam.. 🙄
I didn't understand that how to take reference places
Madam do a video for 1110101 plzzz
Right Order => 7 6 5 4 3 2 1
Wrong Order => 1 2 3 4 5 6 7