When there is a shift by n-k then you should remember that the shift can either go to negative or positive values, when n is negative it will shift in left direction, and if it is positive then it will shift to the right. This is a very important point to remember.
@@SchruteDawg Then also if considering the time reversal then that is -k after that when we are time shifting then we get the h(n-k) standard delaying system because it shifts the time on positive axis. Even if you take any discrete value then by mathematically if you do then also it shift the positive axis and orgin changes to left by k. But whole graph shift on right direction that's why we comparing with standard delaying system Even sir has also told on lecture number 273.
Sir, I have the plot of u[-k] like you but when I plot u[n-k] I take 2 cases. First one n>0, so the plot of u[-k] is shifted left and second one n0 and y[n] = n+1 for n
Not an LTI system???? In the end the system we get is: u[n] --> LTI --> (n+1).u[n] The output is multiplied with the coefficient (n+1) which is a function of time n. The systems is linear but it's time variant. So it doesn't become an LTI system. What am I missing here???
I think you are doing time shifting wrong, the sequences will shift to right not left. Also, the same thing you have done on the continuous time convolution integral video. Thanks
Please could you help me with that (For the following X1(n) find its convolution when its lassing through linear filter with seqence [1,1,1,0,1,0]) As you can see in this problem he gave me only one sequence so how can i convolute it to get the output
No. Shifting the signal by n, results in h[k+n]. Reversing it results in h[-k-n], which is wrong. If you shifted it by -n instead, it would give you the right result.
@@vctropdiamond , on performing time reversal for h[k+n] with respect to independent variable k, it will give h[-k+n]. Then why we are doing reversal first and shifting next, it is because if you do reversal first by h[-k], once reversal done, is done for all values of k, so you can simply shift it with respective n value to get h[-k+n], so on changing n you already have h[-k], so you need to do only shifting. But if you do shifting first as h[k+n] and then reversal as h[-k+n], for every change in n value, you need to do both shifting and then reversal.
I understand the math works out that P = 1, but that makes zero intuitive sense. The sgn function spends half it's time at -1 and the other half of it's time at 1. The average of those is zero. How do we explain the contradiction that P = 1 ??? No one tell me its mod(sgn[n]), I know you do the mod part for the formula, but just looking at sgn[n], it looks like those values average out to be zero.
When there is a shift by n-k then you should remember that the shift can either go to negative or positive values, when n is negative it will shift in left direction, and if it is positive then it will shift to the right. This is a very important point to remember.
Been watching you for years now...helped me with digital logic and now.............STUPID SIGNALS.
Teacher I don't know how can I think u ...I really love u ..especially when u used the table Sol
All my friends of Taiz university are watching your videos ..u r our helper ..u r the helper of Engineers
Thank you sir very nice gide & very nice best explain discrete time convolution teaching video.👍
It was the neatest explanation on the subject I could find! Thanks for the video
around 9:43 where you said that "for other values of k > 2 u[n-k] is still 0" that will be one isnt it?
you're talking about n>= 2 right?
You have done FT of basic signals but
How to draw fourier transform mag and phase plot
please make videos on this topic
This was the best video I have seen about a subject since I got into college.
Last point was bit confusing but thanks a lot. I got an idea of this now!
That last bit, he multiplies by u[n] because it has to remain positive for everything. Pretty much is the n>= 0 it is multiplied by "1" or u[n].
we need 3x speed for quick revision 🤓🤗👌🙌
Exactly 😂😂😂
You saved my life.
the last bit was very confusing but thanks
lol no shit\
As far as i know, convolution expect the signal to move from left to right, that is u[-k-n], not from right to left, that is u[-k+n]
around 10:59 when you are saying tha u[k] = 1 ok i get that but how's u[n-k] also = 1?
u[n-k] is supposed to be 0
it's a constant input signal, the function varies between 0 and 1, but always equal to 1
thanks for your detailed explanation which opened up my mind on DT convolution sum😃
By precedence rule, shouldn't we do shifting first followed by reversal ?
Thanks a lot for this video, i just got aproved in a exame !
but u[n-k] is 0 at 0 not 1 at 0. we shifted u[-k] by positive n and therefore it is no longer 1 at 0 but 0 at 0.
u[n-k] = 1 if n-k>0 by definition wish means k
Thank you❤
how you shift n = +2 , n= +3 to the right?
Are not the supposed to be going left
because we time reversed in the previous step, now the axis is (-k), not (k). therefore its a shift to the right.
@@SchruteDawg Then also if considering the time reversal then that is -k after that when we are time shifting then we get the h(n-k) standard delaying system because it shifts the time on positive axis.
Even if you take any discrete value then by mathematically if you do then also it shift the positive axis and orgin changes to left by k.
But whole graph shift on right direction that's why we comparing with standard delaying system
Even sir has also told on lecture number 273.
Please do another video explaining that ending multiplication. It is confusing
how u(n-k) is plotted without considering n is +ve or -ve
Very well explained. Thank you so much
at 11:20 you say u[k] = 1 when k is 3 but that isn't true is it?
Waiting for this topic thanks
Sir...can u please provide some more examples pdf or videos on this concept?
Sir, I have the plot of u[-k] like you but when I plot u[n-k] I take 2 cases. First one n>0, so the plot of u[-k] is shifted left and second one n0 and y[n] = n+1 for n
Ig u[n-k] ,for n>0 u[-k] will shift towards right and left ,for n
@@sudpn2861 what???
This is the only video that I haven't understand much
exactly
u[-k+n] is time shift towards the right not the left. u[-(k-n)]=u[-k+n]. Thanks
Ya so whole answer becomes opposite
Can we do this convolution example with tabular method ?.
Why this topic is so tough i have a exam tommorow and i know nothing
You try to give the video more brightness it will be great if you do
Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
fantastic
Not an LTI system????
In the end the system we get is:
u[n] --> LTI --> (n+1).u[n]
The output is multiplied with the coefficient (n+1) which is a function of time n. The systems is linear but it's time variant. So it doesn't become an LTI system. What am I missing here???
I think you are doing time shifting wrong, the sequences will shift to right not left. Also, the same thing you have done on the continuous time convolution integral video. Thanks
Please follow the lectures related to the basics of time shifting th-cam.com/play/PLBlnK6fEyqRhG6s3jYIU48CqsT5cyiDTO.html
Please could you help me with that
(For the following X1(n) find its convolution when its lassing through linear filter with seqence [1,1,1,0,1,0])
As you can see in this problem he gave me only one sequence so how can i convolute it to get the output
shouldn't the signal be first shifted by 'n' and then time reversed
No. Shifting the signal by n, results in h[k+n]. Reversing it results in h[-k-n], which is wrong. If you shifted it by -n instead, it would give you the right result.
@@vctropdiamond , on performing time reversal for h[k+n] with respect to independent variable k, it will give h[-k+n].
Then why we are doing reversal first and shifting next, it is because if you do reversal first by h[-k], once reversal done, is done for all values of k, so you can simply shift it with respective n value to get h[-k+n], so on changing n you already have h[-k], so you need to do only shifting. But if you do shifting first as h[k+n] and then reversal as h[-k+n], for every change in n value, you need to do both shifting and then reversal.
The last part of the video is very poorly explained.
hmmmmmmmmmmmmmmmmm
2x speed is better
I understand the math works out that P = 1, but that makes zero intuitive sense. The sgn function spends half it's time at -1 and the other half of it's time at 1. The average of those is zero. How do we explain the contradiction that P = 1 ??? No one tell me its mod(sgn[n]), I know you do the mod part for the formula, but just looking at sgn[n], it looks like those values average out to be zero.
Tomorrow is your paper right?
I guess its wrong. The output must be y(n) = n*u(n)