Discrete Time Convolution (Tabular Method) 

y(n) = (2,-3,4,-2,1)= 4 is the value where the 0 is place

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y[n]={2,-3,4,-2,1], 4 is at n=0.

Never heard of such a straightforward method before, and WOOOW! 😍

A lot simpler method than Graphical Represent method. Thanks sir.

Thank you so much sir, Im taking signals and systems 1 atm, wouldnt be able without ur videos, thank u again and love ur channel

what do u do when the number of values for X1 and X2 are different, u can just put 0 for the the less numbered signal but how do we know if it should be on the left or right?

Thank you Dr. so much 💕 Your explanation was great👌🏻

y[n]= {2,-3,4,-2,1} ↑ Is it correct, Doctor?

Thank you so much for this video and the other video about discrete time convolution. They were great!!

great video indeed !! but a little doubt that is as convolution is performed here, don't we need to flip one sequence ???? then do the rest of the thing

thank you so much. my professor makes this as convoluted as possible in class

You have motivated me to practice and learn Digital Signal Processing, your job is very great and excellent thank you so much. Appreciate it from you. Keep up and please keep up the good work. 🙏🙏🙏😃😃😃 You really motivate me to always learn my lessons with you.

This method is a lifesaver. Why don't they teach this in my uni

This video Save my life

Please complete analog electronics also

Thank you sir that was very helpful ❤

really they done a very hard work......... thank u..

Very good content really so helpfull

that's very helpful

Excelente aula!!!!!

Is it necessary that no of rows should be equal to no of columns

y(n) = { 2, -3, 4*, -2, 1} ,,,,,,tnx sir

Thank you very much sir..... Please I want to ask if y(n)={-2,-1,0,3,2,1,8} and x(n)={4,-1,1,2} find the system response h(n) How would I go about it...... thank you

Thank you, and what is the procedure if one signal has 5 values and the other 4 values And dont have the information of n=0 (no arrow pointing)

Thank you so much 💕

Thank you brother my doubts is clear how to find origin for different situation .❤️❤️

great work broo. i love u

Thanks so much

should I do padding if the size is unequal?

Y[n]={2,-3,4,-2,1} (2

this help me a lot. thanks

is this method only applicable only when we are given values of discrete system or we can also use this method for given equations of x[n] and h[n] ?

Thanks

This was on the FE exam for electrical engineers...

thank you sir

Y[n]=(1,2,4,6,8,8) then intersection of x1[n],x2[n] is 8 then which one is X(0) is please give me reply

y[n]={2,-3,4,-2,1} n=0 at 4

If n=0 is not given or represented in given problem then how to find it

y【n】 = { 2,-3,4,-2,1}

Then what is Matrix method ?

Hello Neso academy. Is there a reference (a book/article/paper etc) which i can cite in my report (i need citation) for the above tabular method. Thank you.

y(n)= (2,-3,4,-2,1). 4 is the valu of zero.

thank youu !

Good

Y[n]={2, -3, 4, -2, 1} where 4 is element when n=0;

thanks,the method work in the exam but the professor only give me 25% ,he didnt accept this

is it tabular or matrix

N value is taken as per our wish

this isnt te tabular method. this is the matrix method

Sorry but this doesn't seem to work in all case. Try the following example, it fails: x[n] = d[n] + 2d[n-1] - d[n-3] , h[n] = 2d[n+1] + 2d[n-1]. If you do the tabular method you end up with y[n] = -d[n+4] - 2d[n+3] + 2d[n+1] + 3d[n+1] - d[n-1]. That answer is incorrect, the true answer would be y[n] = 2d[n+1] + 4d[n] + 2d[n-2] + 2d[n-1] - 2d[n-4]. If I found one example of where it doesn't work, there's bound to be infinitely more.

I thought this was matrix method

Its a matrix method not tabular..

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LoL

y[n] = { 2, -3, 4, -2, 1 } ⇧

Thanks a lot