there are 3 main component in Basie computer architecture 1. cpu 2.memory 3. Phepherals CPU for processing memory for storing program and data peripherals are used in case of I/O interaction
Sir,at 25:24 the output of NOR gate will be 0 if all the three inputs are 1 and therefore the CS will enable for 0.Control switch (CS) will not enable when output is 1
You know 1 thing There are 2^11 memory locations in a single chip... There are 8 chips Hence total memory locations will be= 2^11*8=2^14 memory locations.... But we have 16 bit address bus that can provide 2^16 different memory locations.... Now question arises how to access 2^14 addresses from 2^16 combinations..... Now here 2 bits are extra and we can use them as enable....this is simple but why profeesor has messed all that IDK
there are 3 main component in Basie computer architecture 1. cpu 2.memory 3. Phepherals
CPU for processing
memory for storing program and data
peripherals are used in case of I/O interaction
Sir,at 25:24 the output of NOR gate will be 0 if all the three inputs are 1 and therefore the CS will enable for 0.Control switch (CS) will not enable when output is 1
If any of the three inputs is one along with all one..!!
Greetings from Europe
Sir which type of questions they will give in question paper not about this topic overall mpmc subject for some ieda
lagta h sirji khud hi padh rhe aur khud hi samajh rhe baaki hmse inko kuch lena dena na h
For which exam u teaching sir?
In this way teaching your students?
Ab samj me Aya why students from IIT has tendency to join in politics
Are kehna kya chahte ho
These are reason why people like Sundar Pichai are ruling Google 😂
okay 19:00 so this works when we have less memory
BASIC COMPUTER ORGANIZATION
you did a great job.
@@wood_pecker_ oh really ?
Aditya Singh Parihar Of course 😃
Aditya Singh Parihar to be honest I have no idea why I wrote like that.😂
@@wood_pecker_ hehe.. can relate to xd
Sir please add subtitles in English
You know 1 thing
There are 2^11 memory locations in a single chip...
There are 8 chips
Hence total memory locations will be= 2^11*8=2^14 memory locations....
But we have 16 bit address bus that can provide 2^16 different memory locations....
Now question arises how to access 2^14 addresses from 2^16 combinations.....
Now here 2 bits are extra and we can use them as enable....this is simple but why profeesor has messed all that IDK
Well you can use that mess to understand concept of memory folding !!
Not at all upto the mark! DNT know how IIT hiring such faculties
Same doubt bro
This lectures aren't good enough 😶
he literally destroyed my interest in this topic lmao
So boring 😭😭😭
Worst lecture ever 😢😢😢