Convolution Sum - Problems Part 1

[@SAISRIVIDYA66]

So thankful to u sir. U made me to crack my exam. Thanks sir.great explanation I ever have.

[@corneliatomson1506]

You are life saver! Love from Greece!

[@techjunkiejdb5476]

Greetings from india 🇮🇳

[@atifbashir7229]

Great video .... love from Pakistan

[@veereshkumar5551]

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[@techjunkiejdb5476]

Congratulations 😎

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[@techjunkiejdb5476]

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[@indhuindhu2354]

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[@techjunkiejdb5476]

Welcome 🙏🏻

[@keerthanrvarma]

I watched only your videos for all Chapters of Signals. I was shocked to see that same questions appeared in every unit. 🥳🥳🤩

[@techjunkiejdb5476]

Wow . That’s nice to hear . Which university?

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[@techjunkiejdb5476]

@@keerthanrvarma oh which college ?

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@@techjunkiejdb5476 Nmamit, Nitte in Udupi

[@techjunkiejdb5476]

@@keerthanrvarma that’s cool . Coincidence .

[@keerthanrvarma]

Thank you so so much.... All same questions were asked in our final exam today. I will pass only because of you. 😁

[@techjunkiejdb5476]

Welcome 🙏🏻

[@techjunkiejdb5476]

Share it with your juniors too 😆

[@keerthanrvarma]

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[@llll308]

Very nice video sir, cleared all my doubts...

[@vikramdedavath1351]

finally, I got it THANK YOU, SIR...

[@raichu987]

thank you so much sir !! Your guidance helped me pass my exam :)

[@techjunkiejdb5476]

Glad !

[@InfiniteInsightz]

Even me

[@reaamalmohamad2468]

thanks..it is very helpful

[@techjunkiejdb5476]

Welcome 🤗

[@farheenahmed969]

Very well explained, thanks a lot sir

[@Shaz-oo8lq]

I have one trouble every time. When deciding the cases for example case 1 when is less than 2 at 26:51 , how do you when to use [greater than and equal to] with [less than] or [less than and equal to] with [greater than]?? Thank you.

[@techjunkiejdb5476]

Less than is precisely taken since there will not be any sample to multiply with each other and hence the multiplication would be zero . Only when n=2 then only we will get some convolution values . Cos you will get sample to sample multiplication which will eventually have some value . Hope I made it clear

[@farrukhakhtar6613]

Thankyou Love from Pakistan

[@techjunkiejdb5476]

Welcome bhai

[@sharanpai4996]

I'm ain't gay..... But I love u dude 😍😍.... You saved me!!!!

[@techjunkiejdb5476]

Pai maaam namaskaru!! Javan jalve 😂

[@sharanpai4996]

@Killian Lachlan F.

[@aetrew5499]

Great explanation. My confusion is why did you multiply u[n-2] at 30:41, because we're considering values of n>= 2, hence u[n-2] = 1 for all n>=2 also there is a quicker way to get 1/2 - (1/2)^n using sum between two finite bounds formula.

[@techjunkiejdb5476]

It's convolution of two sequences. So the common area between two waveforms is considered. U(n) and u(n-2) if we multiply. The waveform will definitely start from u(n-2) . Because the value of u(n-2) is zero for n=0, n=1 .

[@avi-brown]

Thanks for this video. Can you explain the importance of changing the lower limit in the final example to zero, rather than leaving it as 2? Thanks!

[@techjunkiejdb5476]

We know the equation for summation of a series from 0 to N or 0 to infinity . That’s the reason why we have changed the limit !

[@AbdulHannan-hj8oq]

Explained very clearly !!

[@kaursingh637]

sir excelent -please solve convolution of ramp with delta -thk u

[@techjunkiejdb5476]

It will be same ramp . At t=o

[@sushmakammar9694]

Thank u very much sir .. it's really gonna help many of us :)

[@techjunkiejdb5476]

Pleasure :)

[@malavikadborkar6209]

great job sir 👌✊

[@gajendrakrsharma485]

Thanks nice .... lecture

[@ritwikadas7258]

I am confused. @19:18 y(n) for n>0 shouldn't be 0? Because these were no values for n>0 in the first pic???? Similarly like the first example for n

[@techjunkiejdb5476]

There are two sequences. First one u(-k) has no values after k>0 . But u(n-k) has values for as per k=n , so when n >0 the product wouldn't be zero . Since the summation is from -infinity till zero.

[@aqila3006]

thank you so much, you have explained it very well

[@techjunkiejdb5476]

Thank you :)

[@liliisweetaa4962]

YOU ARE PERFECT

[@afaankhan3336]

sir why did you multiplied it by u(n-2 ) at the end? at 30:41

[@techjunkiejdb5476]

It's convolution of two sequences. So the common area between two waveforms is considered. U(n) and u(n-2) if we multiply. The waveform will definitely start from u(n-2) . Because the value of u(n-2) is zero for n=0, n=1 .

[@nishantdesai3021]

Thank you for the video

[@rounaksunvaiya6219]

Thanku for your guide

[@abilashshankarpalli9301]

Ur too much man!! Awesome explanation

[@ramdevrc5605]

excellent sir👌👌

[@bhargavib7718]

Very helpful

[@soman6133]

yah, absolutely prefect..

[@techjunkiejdb5476]

Thanks

[@aman6487]

Sir, how to find convolution of cos(nπ)u(n) and (1/2)^n u(n) It gives two sequences, how to converge??

[@techjunkiejdb5476]

Aman Sharma follow the same procedure what I have done . U will have only two conditions one for k0

[@aman6487]

@@techjunkiejdb5476 for k0 kth term depends upon on if k is odd or even so how to find kth term here

[@techjunkiejdb5476]

Aman Sharma cos (kpi) expansion has to be done . In terms of exponential terms

[@aman6487]

@@techjunkiejdb5476 ok sir, I'll try

[@techjunkiejdb5476]

Aman Sharma if u can’t then I’ll help u

[@adishshetty6568]

well explained

[@abdalsalamedrees4099]

Please how can I contact you

[@techjunkiejdb5476]

[e-mail]

[@amitjaiswal7179]

Well done sir

[@amitjaiswal7179]

Sir Plz also upload problems by using their properties

[@techjunkiejdb5476]

sure :)

[@bahmansokhansanj4639]

Perfect man

[@samikshakamath8877]

explanation so easy 💜

[@shivamnayak3893]

Thanks sir

[@satheeshakumar2954]

Best videos to studu

[@srinivassai712]

U can explain properties of laplace tranform

[@techjunkiejdb5476]

I will, in my upcoming lectures :)

[@mvamsis.someswarao7031]

thank u

[@shylesh2428]

Thanks so much i cracked gate exam

[@veereshkumar5551]

Nem.mane hatira park Alli sir night Alli bandidhe

[@ramahanuman8271]

At 18.33 when n is >0 it is zero right? But r doing again summation .

[@techjunkiejdb5476]

There is a multiplication of two sequences . So it's not zero . Overlapping space is between -infinity to zero.

[@herishbadal3719]

To the point

[@techjunkiejdb5476]

Thanks

[@bhagyashree8424]

sir,how to recognize the signal is finite or infinite

[@techjunkiejdb5476]

Bhagya Shree if signal exists for short duration of time then it's finite otherwise infinite . U(t) is infinite . Sine cos tan is infinite . Exponential is also infinite .

[@darkrose3344]

when n=o?

[@techjunkiejdb5476]

Which problem ?

[@janpan4761]

when you can not explain sth do not teach

[@techjunkiejdb5476]

Jan Pan thank you . When u can't type a sentence , better not to comment .

[@bhargavib7718]

Very helpful

[@ahmerbaig6972]

well explained