Construction of Mealy Machine
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- Опубликовано: 12 янв 2017
- TOC: Construction of Mealy Machine
Topics discussed:
1. Construction of a Mealy Machine that produces 1's Complement as the output of any given binary input string.
2. Construction of a Mealy Machine that prints 'a' whenever the sequence '01' is encountered in any input binary string.
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Axol x Alex Skrindo - You [NCS Release]
• Axol x Alex Skrindo - ...
heard that voice , checked if i'm on right channel or not !
You, Sir, are a legend for doing this. Your videos on these topics are soo good, the best out there on youtube, probably the best on the internet.
Watch TOC videos of RavindraBabu Ravula. 😁😁
Ben Eater is better, he doesn't read every sentence twice.
@@gateprep8076 Neso Academy >>>>>>>>>>>>
This sub was new to me...after i hav been watching ur lecture series..the confidence is also multiplied itself.u r d best online teacher i hav ever felt.
Bunch of thanks for the crystal clear explanations❤❤❤
Hail the *king* 👑 he makes things simpler ❤
Thanks a lot, your really helped me with my exam tomorrow. Wish you the best xoxo
Thank you so much, amazing explanation, you made it very lucid to grasp
Tu padhi Mata kara rey chupke chupke 😂
@@krushnashinde1623friend ko pakad liya kya pdhte hue?
Thankeo so much sir..i appreciate you for this great work..
Why did you make a DFA for a machine accepting 01 at the end only ? Why not anywhere in the string ?
Yes, even I didn't get this
It prints 'a' whenever 01 encountered ,even if it is on the middle of the string or starting of the string or ending of the string . 000001 prints 'a' one time because 01 encountered at last of string . 0101000011110000001 prints 'a' three times as 01 encountered in starting position ,third position and last position. Hope you understand.
The DFA accepts strings with 01 at the end but doesn't deny 01 in the middle.
Thank you sir 👍
Good explanation sir
Thank You💙
I have a doubt why we construct transition C(1/b) --> A,
rather than not construct self loop at C for this transition C(1/b --> C)..? is it fine(Self loop)..?
Same doubt bro
for ex 2 I was able to draw mealy machine in 2 states with:
A: 1/b(A) ,0/b (B)
B:0/b(B),1/a(A)
here i am representing input/output(state reached). Correct me if i am wrong.
You are absolutely right.
I did the same as you. I pause the video before the solution, so I can practice, then, I got the same solution that you. I thought that I was wrong ...
actually the DFA given by sir is easy for those who just know about what a DFA is...as mealy or moore machine have no final states but only outputs.....if the DFA given by sir is minimized then your DFA can be obtained....both are correct but yours is minimized...nothing else...you can check it......
Me too
Tom is final exam n i thought of skipping moore n mealy thinking its difficult bt who knew jst spending 20 mins watching ur vdo wud make me attend tis question in exam fr sure.😭
your way of construction is very satisfying. thank you sir :)
Instead of having a C state, couldn't you just have the 1/a transition go back to A?
Indeed, I think he didn't ignore the 'C' state because in the Moore example he did the same machine and the difference between the Moore and Mealy machines became clear, but he should've said that you could ignore the 'C' state anyway.
Yeah. We can do it with just two states. Output can be displayed when 1 is encountered and goes to starting state
Exactly that what i did make
in 5:20 could we have 2 states A and B? Untill B is the same but when we get an 1 we go to A with output a
Yes when you do the state minimization you get actually 2 states
i like your teaching technique sir #fann
sir at state c we can use a selfloop because u say that the string should be any where .... and at state c our string o1 is comming in starting..so why we cant use selfloop of 1 and 0 at state c...plzz reply sir
Bcoz when u reach final state u must accept the string 01 ....bcoz in this sir first explain dfa and then mealy machine and in dfa u only accept the string which is given when you are on final state...
bring back the other guy
thanku sir💓
Can't this be done just with 2 states?
A +1-> A, A+0-> B, B+0->B, B+1->A (outputs 'a') .
can it?
I also have same doubt , I think it can be done
actually your solution is minimized version of the DFA given by sir...
@@michabinda7390 Definitely
What a background music 😂😂 still waiting part 2 on Kamran Bagraa😂😂❤❤❤
in 1st example, why is A not marked as final state?
ok got it! Mealy machine doesn't accept or reject, that's why.......You should have explained that point also!!! Tell each and every point. Don't leave gaps!
10100 sign magnitude is 00100
10100 1's complement is 01011
10100 2's complement is 01100
sir, please upload videos on microprocessor and microcontrollers
Super sir
Thank you sir
Cool!
Sir, if on input 1 c goes to c is that ok or not??? at timestamp 5.18
I think its right
Does Mealy and Moore Machine having Dead State
Why no dead state only self transition
Another example on melay machine with table
6:43 could you also say "let's print c because i don't care what it prints here?"
No, because the Language is {a, b}.
@05:12 what if we write 𝛿(C,1)---->C/b!?
For 0110 we are not getting final state C ,bases on your construct DFA
Nah it prints a like see it would give a once it reaches 01 but won't print a for 10
i dont know why but i was a little disappointed in hearin a different voice from the usual lectures i have been seeing
so true
me too am dissapointed
Why isn’t this in digital electronics playlist?
micoprocessor 8085 ka lecture plz upload karvaiye
dfa in ex 2 can not accept string '011'.
right
There is no factor of acceptance here...we are just concerned about the output the machine would produce...
shouldn't we use the double circle around 'C' to signify it being the final state?
Not required, the question is about getting any 01 in between the strings and whenever that happens you should print 'a'
sir
if we writing 1/a it is like a moore machine,input/out put?
Give example on melay machine
What subject is this, it's not the general digital electronics
Sir you sound very similar to mahesh babu(telugu actor),
why is a 'b' if it go from c to b?
7:00 is also a 'b' when c goes to a
example 2 was wrong
you have explained for the string ending with 01
and the question was that the string should contain 01
Mohan Sai Aree stupid , see properly he actually gave a loop af 1 on A. so we can start string any of the ways.
But still why are we supposed to end it with 01, the question doesn't ask for it, this just reduces the no. of possible valid input strings.
Abhijeet Singh
there is no final state so, the final state can be A/B/C ..isnt it ?
so it doesn't have to end in C with ending string 01
Yes, the sequence can be encountered anywhere within the states. So this is like a question of DFA design where your string contains substring '01'.
Or even better, remove state C completely. And use only two states A and B. On getting 1 on B, come back to A with output 1.
sir why don't you upload all branches videos
On getting input 1 on C state why u just go to state A cant we use the self loop in C on getting 1.???
if you loop the C, how exactly would you re-check for more inputs?
nice sir
In first example, shouldn't A be the final state also?
This is not DFA design.
In Mealey models, you don't specify whats your starting or final states. You just care about your input and output.
@@w1ndro1d But start symbol is necessary i guess, unless how'd you know where to start from???... Also can u explain why there's no final state...why isn't it necessary here????
@@somyajitnath6303 start symbol is required but no final state indication is necessary when you construct Mealy machines. You just write input and transfer flow accordingly, unlike in DFA.
does mealy machines contains output null value
This is nice...
Dont we need to double circle the final state in these machines??
No not...there is no final state in Mealy Machine, since it focuses only on the outputs it produce.There is no factor of acceptance as such !!!
Sir, can you please tell me how to design a bank locker system using sequential circuits? The logic is...when we enter our bank locker code, then within 5seconds the bank manager must enter the password which matches our bank locker. If both passwords matches then only the bank locker opens and led must glow green. Otherwise if the bank manager didn't entered his password or our key mismatched then red led must glow and locker does not opens
kipah huai mahmah ei
last example is not suitable in a case when we repeat 01 again in our input like 0101101
so this machine gives us wrong output so chek it again
How does it gives a wrong output ???
according to your example, the output should be bababba and that's what the machine is giving...
In the last example a third state C is not required. We can solve it in two states, did anyone else also come up with this?
then you must have formed an NFA...cuz using DFA its not possible.....i could be wrong ...but i tried and couldnt
Moore machine string are 012
Sir what is DFA?
amogh mp Deterministic Finite Automata
28
couldn't you do Ex-2 with just 2 states?
No
Thanks for your video excellent work.But i have this question for you sir?
module patternMealy(input logic clk,
input logic reset,
input logic a,
output logic y);
typedef enum logic {S0, S1} statetype;
statetype state, nextstate;
// state register
always_ff @(posedge clk, posedge reset)
if (reset) state
bhai mila kya answer
do you get the answer??
@@Tanwar59 abe chatgpt hai ab toh kuch dimag lagale😂.
@@Tanwar59 usko nahi mila toh tujhe toh mil sakta haina chatgpt will help
here