first solution: T=2*n*tpd where n=5 Hence T=10*tpd Since tpd=100 pico second T=10*100 pico sec.=1000pico second Now 1sec=10^12 ps(relation of second and picosecond) Hence 1ps=10^-12s 1000ps=10^-9s From this T becomes as T=10^-9 sec f=1/T =1/(10^-9) =1*10^9 hertz Again from relation of hertz and gegahertz 1 hertz = 10^-9 gegahertz. Hence ans is 1 Ghz

Sir, your flow of English is really impressive! :)

The 2nd problem was really impressing. I was kind of wandering what actually feedback does, and this gave me the answer. Anyway, the answer is 14 ns time period at first ouput and 12 ns after that. The key point is that after the 3rd NOT Gate the feedback sends the input signal. So when the first input goes from 3rd to 4th NOT Gate, simultaneously 2nd input goes to 1st. So it takes the second output to come out taking time of 1st to 4th NOT Gate, not the 0th to 4th. For who doesn't understand due to my short english, 1st 0(8 ns) 1st 1(6 ns) 2nd 0 (6 ns) 2nd 1 (6 ns) 3rd 0 (6ns) 3rd 1 (6ns) 1st output(14 ns) 2nd output (12 ns) 3rd output(12 ns) Thanks to the comments I also got wrong at first!

thank you so much to neso academy because i didn't even attended a single class in my college but you helped me a lot!!!!!!

Just suggestion, but you would be better off always starting with the previous H.W.'s answers.

12 ns imo... because from the feedback point to Y it will take 2ns but in that time the feedback will provide signal back to the first NOT gate... so what im saying is that by the time signal reaches Y...it has already passed the first NOT gate . So effectively there are 3 NOT gates namely 2,3 and 4 ...thus 12ns

well, my college is useless now.

question 2: 12ns. The question asks what is the TIME period of the square wave, it does not ask to say what you see at Y output. The first 3 gates create and determine the frequency and time period of the square wave. The last gate comes AFTER the feedback loop,it adds a small delay of 2ns.

Please add logic family videos as soon as possible before 15 dec. I understand now more clearly and free than my college facilities. Very very very very very very very nice 👍. Thank you for helping each of us

Q1-1GHZ,Q2-12ns .as only first three not gate is responsible for creating square wave so 12 ns is right answer.

Kind Attention Sir, Thank You so much for such a helpful videos. but sir i want the solution for the HW problems but cant access your website. kindly provide the solution for it.

remember 1 thing when signal passes from 3rd gate to 4th gate , at the same time it reaches as a input to 1st gate..... and when it passes from 4th gate to the output , at the same time it passes from 1st gate to 2nd gate........ in this way we have time period equal to 14ns for 1st cycle and 12ns for rest of the cycles..... hope u understood...:-)

Q1. 1GHz Q2. 12nsec. fourth gate is used to invert the square waveform generated by first three gates.

I suppose its 16ns because when Y is assumed to be zero then output of gate no 3 should be 1.Now after passing gate one it will be zero not affecting the output Y ,same for gate 2 but when signal passes through the gate 3 , 1 now becomes zero but this change will only affect the output Y only after passing through Gate 4.So in order to produce any change in Y.so it sjould be after passing a net of 8 gates. I am not too sure about it otherwise what difference will remain in this and the above question!

for 1. it is 1 GHz and for 2. in it's first input it takes 14 n sec for complete cycle and afterwards it takes 12 n sec. always and since time period means time taken for period motion to take place thus, 12 n sec. is the correct answer.

1:- 1ghz 2:- 1st o/p will take14ns and after that it will give time period of 12ns

2) c) 14ns , because if we start with 0 then after 6ns o/p to 3rd not gate will be 1 then by feedback after 6ns we will get 0 then this will go to 4th NOT gate producing 1within 2ns and hence a square wave and hence total T = 14ns

In order to access the answers to your homework problems, the website is showing 404 error. Kindly fix it ASAP ! Else your video work is awesome. Big thumbs UP!!!

Can I believe blindly this playlist as best ?

Amazing tutorial.Thanks a lot sir.

Sir the way you explain the subject is amazing sir thank you ,thank you very much sir .😊😊😊

I always feel at home whenever am here

Please provide answers for the home work problems also sir. Thank you

Thank you very much for providing us with all these well-made educational videos. They really helped me get through my last semester. I'll definitely recommend your channel to others!

12 nano second. At any instant we start applying input 0. We don't know what will be the output at Y, it can be anything 0 or 1. After 3nsec the output will be 0 and we can start drawing the square waveform.

In q2: Shouldn't the first cycle be 20ns at first and then 12ns? Explanation: We start with 0. Then after the first 4 time periods, Y is still 0. Then during the next 3 time periods, Y remains 0 and after the 7th time period, Y=1 and then Y falls back down to 0 after the next 3 time periods. So it should actually be (7+3)tpd * 2 = 20ns for the first cycle and then 12ns for the next cycles.

answer to the 2nd que is 14nsec .there are 3 not gates in the feedback loop so applying 2nTpd= 2*3*2ns = 12ns and there is a delay given by the 4th gate of 2ns . adding them we get : 14 nsec

HW Question 2 Answer:- 12 n sec T= 2n tpd n= no of NOT gates involved in feedback Circuit , ( which are 3) tpd = 2 n sec Already given Hence T = 2*3*2 n sec T = 12 nsec

Great work have a nice day for you.

Sir why it is not buffer if it have feedback then what reason behind that could u explain feedback plz sir

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in q 2 .. we get square wave a both after three and four buffer ..the T will be defined by first three buffers only =12ns the fourth one will only give us the delayed square wave having same frequency... so ans is 12ns..

1. 1Hz 2. 16 ns Because ,when may be the feedback is provided the input have to paas through all four gates to get output. And each gate will take 2 ns. It is simply like above problem. And it requires to pass through 8 NOT gates to complete on cycle. T=8 *2 ns T=16 ns I think it is mostly correct.

a general approach: square wave generator will work as the same period calculation, but the last not gate provides a time shift to positive direction. anyone agrees?

Magnificent knowledge

I am happy for this service

Why it is not buffer sir ? I didn't understand

Sir, are these all videos in order? I find some topics are not in sequence!...

Answer of the 1st question is b and the answer of 2nd is b as well try to go through each gate and mention the time passed each time u pass the gate. and u will get 12ns as the answer

why the first one is not a sq wave generator while 2nd is sq wave generator

I feel like this course is not for a complete beginner who haven't even heard the term Logic gate in their life hahaha.

Here feedback is only for generating square wave(in this situation only), we only consider total output delay, so for each state we get 8nec delay and so total time period is equal to 16nsec. So q2 ans:: 16nsec, think about it.

for the 2nd one the answer will be 14tpd or 28 ns say you have 0 as input so after 4tpd Y is still at 0 but after 3 tpd the feedback comes into play and feeds 1 back to 1st NOT gate so after 4 tpd that 1 has come out of the 1st NOT gate as 0..this passes another 3tpd to become 1 at Y so you have 4+3 = 7 tpd for the signal to change from 0 to 1 so a cycle will be of 7+7 = 14 tpd = 28 ns

According to me the answer of question 2 is 12ns cuz the question is saying to find the time period of generated square wave .. So first three NOT gates are making square wave ... So we have to calculate with respect to these 3 gates... So answer will be 12 ns... Plzz tell me if i m wrng @nesoacademy

i think second one will be 12ns because consider y=0 then the input for that gate is 1 if we analyse this then from 0 to 6ns waveform is high 6ns to 12ns it will be low so time period is 12ns .now this square pulse is passed through not gate which will just convert 1 to 0 and 0 to 1 this will not ofcourse change time period or frequency so 12ns is correct

Hi Sir.Your explanation is very good.Thank you.I got option 'B'' for first prob and option 'C' for second prblm.Could u plz solve the second prblm and upload it.

For second one answer is 12ns. Why because after the third not gate we can generate a square wave and also there is no feed back for the 4th not gate. But the delay is 2ns already We know that if any signal may be transmitted to the delayed device that will produce that output after that delay only. Hence in the given problem they are asking calculate the time period of the generated signal. Hence after passing the generated signal through the NOT gate we will get an output in the form inverse after 2ns time delay but time period is same. Hence don't consider the 4th NOT gate of while calculating the time period of generated square wave. Hence ans:12ns I hope it is useful to everyone

ques 2 ans is 12ns square wave generator is not gate in series and last is gate is just to delay the o/p. hence 2*3*2=12 ns

In q1. Even without the feedback isn't the circuit going 2 give the same results? Why is not a buffer, if it has feedback.?

dear ne so ....plz provide the sol of h/w problem as well

Sir plzz combine all videos in one shot of 2-2.5 hr according to the topic

I didnt understand in first example why the circuit was not buffer ... someone pls explain ... and why feedback was the reason to not being a buffer.

why is example 1 not squar wave generator?

Hi sir, Why xor and xnor are called arithmetic gates even though they don't perform any arithmetic operations

There are 2 not gates connected in series. For first not gate logic 0 is the input. Second not gate output is Ored to input of second gate only. What is the optput sir

2nd ans is 6ns that is a option.because the square wave is generated by the feedback circuit which has only 3 diode and 4th diode will take its delay time and just reverse the input signal nothing more than that

Problem 1: 1 GHz for problem 2 is it 12 nsec?

first problem- 1GHZ.. second problem for simplification take the point after 3 NOT gates as X. now make a graph of X. After that to get the output Y take the graph of X, each time X changes add a delay and invert the value and there you have Y. you'll see X and Y have the same time period only Y is 2nsec lagging. Ans is 12nsec

nice explanation

What is the 2nd answer? Last gate propagation delay has to be counted?

Thx sir but kindly discuss h.w problem in next lecture

in the last hw problem the output would be same as input as there are 4 not gates so when we give 0 at the input after 8ns 0 will be obtained at the output while at 6ns the feedback will provide the signal 1 to input and during the time of 2ns in which the first 0 input will cross the last not gate this 1 will cross the first not gate and after crossing the rest of 3 not gates i.e after 6ns more the signal 1 will be recieved at the output and will last for 6ns till the next signal is recieved at output so conclusion is the time period is 12ns but initially there was a time lag during the start of generation of the square wave. correct me if i am wrong.

Need answer for the second question

🔴H/W problem 1: Solution: frequency f = 1/T Given~ n= 5, tₚₔ= 100ps = 100 × 10^(-12) s So, f = 1/(2 × n × tₚₔ) = 1/(2 × 5 × 100 × 10^(-12)) = 1/(10^(-9)) = 1 GHz ✔️ans: Option (b) 1 GHz ................................................................. 🔴H/W problem 2: Solution: Time period T = 2 × n × tₚₔ Given~ n = no. of NOT gates in feedback loop = 3, tₚₔ= 2 ns So, T = 2 × n × tₚₔ = 2 × 3 × 2 = 12 ✔️ans: 12 ns

great explation for problem , I have question how i can draw tpd Idont really catch the idea could you help me?

Thank You Sir

sir what if we took 1st input as 0,in that case after 5th Tpd aslo we get output 0 then signal would be low.

The first tpd question. How come the first 1 wasn't represented in the time diagram?

In example 2 being explained, please tell Why it is astable multivibrator?

thank you, really helpful.

sir plzz upload the solution of 2nd problem

Odd priority me kuchh change hogq???

can you explain about cmos using all logic gates in one video it is very useful

Thanks Sir😃😃

Tpd = 2nd 2*ntpd= 2*3tpds= 6tpd Each tpd = 2ns Finally 6*2= 12 Ghz I understood like this

Do you have the bubble pushing example video?

how many transistors are in NOT , AND ,OR gate?

Do you have any playlist for computer architecture and logic design?

Can someone expalin how to get the answer to question number 2 ?

i don't think the second circuit generates a square wave at all....or if it does can someone please explain?

Great video ! One question though. Why do we multiply by two when we have 5 NOT gates if the function repeats itself after 5tpd and not 2*5tpd ? Thank you

what is the answer for second home work problem. please send answer sir.

Thanks a lot...I owe to u sir.

So nice

In this video : logic gates are the physical device that perform logical operation on logical input and produce single logical output. three types basic: AND , NAND , universal: NAND,NOR , arithmetic : XOR,XNOR NOT -->O

yes q1 ans is 1Ghz and q2 ans is 12 ns

Is there any possibility to adding of noise to the parity bits??

What is the case if noise occurred in parity bit?

2nd question Answer is 12nsec. T=2*n*tpd //n=3,tpd=2nsec T=2*3*2nsec T=12nsec

I get choice b for both

at 12:44 how they made diagram , they said time after first tpd y is zero , how

Sir,can you tell me the how to convert the last answer psec to Ghz ??????

not gate me more input kiss type se de sakte h

Sir can you please upload the videos on 8085 microprocessor

@nesoacademy The website is not functional in many ways. I can help

Please explain 2nd question

Where do we get the answers to HW problems? Ur website is showing a 404 error at that page.

Please post your solution of 2 question

In a logical system consisting of a feed backloop ,,,,,,, the processing time will be....... 2 x Numbers of NOT operators inside the feedbackloop System x timeDelay.

1. None of the above. Y changes every 100 ps because the fifth inverter changes every 100 ps. It does not wait 500 ps to change. That gives it a period of 200 ps, and a frequency of 5 GHz. 2. None of the above. Y, again, will change every 2 ns because that's the t_pd of the final inverter. In other words, the final inverter receives a new value every 2 ns. That gives Y a period of 4 ns. Or am I missing something? Also, your website is giving 404 errors when I try to access the homework solutions.