Erdm 95 summation has to happen only once . See the lower limit is -infinity and upper limit is infinity . Since the function has two different values for n= 0 ., We need to split the summation like how we split the intergral . Only the change here is if u summate from -infinity to -1 then the next sample to be summated should be from 0 to infinity . Or as u have asked if u summate from - infinity to 0 then u need to summate from 1 to infinity. There is no mandate that u need to summate with the same limit which I have taken here . It can be summation from -infinity to -2 and -1 to infinity . U just have to make sure that u cover all the points on the integer number line. Hope you got it .
1 like and 1 subscriber from my side..........saved my day man
Thank you for the example, really saved my homework assignment.
thank you for saving my semester
very good explanation which I have ever seen till now. sir plz solve more convolution problem
ranjitha p I have solved some problems in convolution sum and integral in the same video series . Check out .
Thank You for the whole series, I Promise if i scored ranks in GATE or IES I will give You credit too
Welcome buddy
Even me also
Don't feel what I have commented! Because with a effort you have done this but some steps are very fast and difficult to understand
sir , can the answer for Q 10 i,e..del(4-2n) be 0.5e^(-iw2) , i used one more property of del ie..,1/|a| del(n). here only the weight has changed!!!
Thank you so much sir. it is helpful to me. Pls solve more problems sir.
Pavan Kumar under which header ?
Techjunkie Jdb under dtft on point of signal and system exam.
Techjunkie Jdb . Sir pls send link of more problems under dtft video
Thankyouuuu!!❤🙏
Welcome 🤗
You videos are great ,but the only problm is with the focusing of your camera which keeps changing.
Please do anything for that
I will take care in future
1:43 why from -infinity to -1 . İt has to from -infinity to 0 ? I didn't understand this part , can you help me sir ?
Erdm 95 summation has to happen only once . See the lower limit is -infinity and upper limit is infinity . Since the function has two different values for n= 0 ., We need to split the summation like how we split the intergral . Only the change here is if u summate from -infinity to -1 then the next sample to be summated should be from 0 to infinity . Or as u have asked if u summate from - infinity to 0 then u need to summate from 1 to infinity. There is no mandate that u need to summate with the same limit which I have taken here . It can be summation from -infinity to -2 and -1 to infinity . U just have to make sure that u cover all the points on the integer number line. Hope you got it .
Can't we solve question 10 from dirac[n] >>>1?
Shift right by one ? U mean
@@techjunkiejdb5476For question 10
Dirac[n] >>>>1
Dirac[-2n]>>>>1.X(e^-jw/2)
Dirac[4-2n] >>>>1.(e^(4jw)).X(e^-jw/2) Is it a correct solution?
why do you combined dtft of two domain given that there are different function per domain. how do you justify that.
Where have I combined ?
@@techjunkiejdb5476 I was talking about the dtft result but I see you have just used the definition of DTFT.
thank you
Welcome
Good job
nice explanation
thank you so much
very informative
Thank you sir
Good one!!
น้ำตาจะไหล ฟังไม่รู้เรื่อง
good one !!!!
Hard to understand
Ok
أوجد DTFT للإشارة التالية X (n) = u (n + 3) _ u (n_ 4) = (1،1،1،1،1،1،1)
صار 6 اشهر
thank you
most welcome
Thank u sir
Welcome