i think for this condition to be true, sigma has to be less than -1 and greater than 2, that way there are no intersections of R1 and R2, therefore the system is not stable nor causal so the answer should be: 1/3exp(2t)u(t) + 1/3exp(-t)u(-t)
Bhayya Myself Ritesh and I'm currently in 2nd semester ECE at NIT Kurukshetra. If.you don't mind can you share your notes of this chapter please?. Thankyou. My insta: hielboy_oo7
for neither causal nor stable -> ROC is LS (Left Sided) to LMP(Left Most Pole), so here sigma must be like sigma < -1 and sigma < 2 to make left most part shaded region. Thus corresponding h(t) will be -1/3 e^2t u(-t) + 1/3 e^-t u(-t)
This part 3 will have sigma2 as the ROC.There is no common region in these two. So I think no such Laplace transform will exist as there is no common region of convergence as we have an unstable and non causal LTI system.
this is not the correct explanation ,we have to make the system unstable so consider e^2t as e^-2t which means that sigma will be greater than 2 it will be right sided signal and in other case sigma should be left sided as it should be less than -1 so it will become as sigma will lie between-2 and -1 which will be the common region so which implies that jw axis is not included and it is in the left side so it's unstable and non causal.
Laplace transform exist only at the region of convergence. If we make both parts of the solution Left Sided Signal(LSS) then the Laplace transform ROC is sigma
sir i mainly dependent on you and i saw your all previous video carefully...i understand your logic that why u(t) is multiplied with e^at but i find google and any other website then i confused because there are no mentioned that this formula of unilateral or bi-lateral tutorial.math.lamar.edu/pdf/Laplace_Table.pdf ....now i see in the oppenheim book and i 100% clears that your channel is dealing with depth of concept more than other channels....thank you sir...
To the great neso academy We the viewers Are dependent on u We beleive in u We hope on u Please upload the videos very very fastly Per atleast two videos For every two days u are uploading only one Please as a viewer kindly requesting u to upload videos fastly Thank u Neso academy
can somebody pls explain why we are taking both the signals as left-sided. im asking this bcoz if both are left sided then the jw axis gets included which makes the system stable. but in our question we need to calculate h(t) for unstabe system
Both signals have to be left sided because the condition States not causal which implies anti-causal signal. For anti causal signals, The poles are Left sided. Now recall from the properties of Roc if we have a both sided signal i.e a signal that contains two Poles And Both Poles are Left sided. ROC IS ON THE LEFT SIDE OF THE LEFTMOST POLE. IN THIS CASE ROC
see for neither causal nor stable ROC will be to the left of the left most signal and should also not include JW axis.....so SIGMA will be less than -1 and since it is less than -1 it will also be less than 2...ok....so now you have 2 LEFT SIDED SIGNALS but h(t) for general case is right sided for both parts.....on changing to LEFT SIDED you will get the answer as posted by many guys above hope you understand bro
@@JJpunk434 If we are taking both as left sided signal ...the jw axis is getting included which makes the system stable...hence to make the system unstable sigma must be greater than 2......so please can you explain me this part??
H.W. Ans:
[(-1/3)*exp(2t)*u(-t)] + [(1/3)*exp(-t)*u(-t)]
How u solved can u explain please
Right...
Since... Sigma
this is also what i got !
i think for this condition to be true, sigma has to be less than -1 and greater than 2, that way there are no intersections of R1 and R2, therefore the system is not stable nor causal so the answer should be: 1/3exp(2t)u(t) + 1/3exp(-t)u(-t)
why dont we take alpha
ANS: [(-1/3)*exp(2t)*u(-t)] + [(1/3)*exp(-t)*u(-t)]….neither causal nor stable.....ROC will be left to the leftmost pole..!!
Bhayya Myself Ritesh and I'm currently in 2nd semester ECE at NIT Kurukshetra. If.you don't mind can you share your notes of this chapter please?. Thankyou.
My insta: hielboy_oo7
Your answer is wrong . For sigma
@@shashikantuike9495 No, the answer is right because for instability ,common Roc should not include jw axis and here sigma
for neither causal nor stable -> ROC is LS (Left Sided) to LMP(Left Most Pole), so here sigma must be like sigma < -1 and sigma < 2 to make left most part shaded region. Thus corresponding h(t) will be -1/3 e^2t u(-t) + 1/3 e^-t u(-t)
This part 3 will have sigma2 as the ROC.There is no common region in these two. So I think no such Laplace transform will exist as there is no common region of convergence as we have an unstable and non causal LTI system.
this is not the correct explanation ,we have to make the system unstable so consider e^2t as e^-2t which means that sigma will be greater than 2 it will be right sided signal and in other case sigma should be left sided as it should be less than -1 so it will become as sigma will lie between-2 and -1 which will be the common region so which implies that jw axis is not included and it is in the left side so it's unstable and non causal.
@@kshitijwadhwa4277 how u can say that if the system is unstable than it bound to anticausal?
Laplace transform exist only at the region of convergence. If we make both parts of the solution Left Sided Signal(LSS) then the Laplace transform ROC is sigma
sigma < -1 and sigma > 2 and according to this plot of s-plain, in h(t) have correction.
I understand well.!
H.W. Ans:
Roc: (Ó
If sigma2?
@Ashiqa Jacobs, ROC:lies only on (ó
We should not consider exp(2) here
there should be a common part (as explained in the stable system condition)
@@Kavithomas_Engineer stable
Great explainassion....
for non stable and non causal sigma
This video cleared all my doubts
thank you, much appreciated
great work
signa < -1 , sigma
Sir ur lecturers r really fantastic
sir at 7:23 in inverse laplace transform no one multiply u(t) with e^at then why we multiply...
sir i mainly dependent on you and i saw your all previous video carefully...i understand your logic that why u(t) is multiplied with e^at but i find google and any other website then i confused because there are no mentioned that this formula of unilateral or bi-lateral tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
....now i see in the oppenheim book and i 100% clears that your channel is dealing with depth of concept more than other channels....thank you sir...
To the great neso academy
We the viewers
Are dependent on u
We beleive in u
We hope on u
Please upload the videos very very fastly
Per atleast two videos
For every two days u are uploading only one
Please as a viewer kindly requesting u to upload videos fastly
Thank u
Neso academy
ok
Sir we know roc should not include any pole but in case ii ) roc including of pole 2 how could be explain that way
We must see only the common ROC
Sir give us...wt is correct answers of Homewrk problems in the next cls... We wll check it rght or wrng
Hey man thanks! I am actually confused in something what will be laplace of y(t) and dy(t)/dt
Ans = -(1/3)*exp(-t)*u(-t)
h(t)=-(1/3)e^-2t u(-t) +(1/3)e^-t u(-t) ,,,sig
Sigma
Hw problem answer is sigma < -1 and sigma
ANS: [(-1/3)*exp(2t)*u(-t)] + [(1/3)*exp(-t)*u(-t)]
can somebody pls explain why we are taking both the signals as left-sided. im asking this bcoz if both are left sided then the jw axis gets included which makes the system stable. but in our question we need to calculate h(t) for unstabe system
Both signals have to be left sided because the condition States not causal which implies anti-causal signal. For anti causal signals, The poles are Left sided. Now recall from the properties of Roc if we have a both sided signal i.e a signal that contains two Poles And Both Poles are Left sided. ROC IS ON THE LEFT SIDE OF THE LEFTMOST POLE. IN THIS CASE ROC
Sir,please upload the remaining videos quickly because I dont understand what my professor is teaching
[(1/3)exp(-t)-(1/3)exp(2t)]u(-t)
If sigma
there is no common ROC between -1 to 2 where JW axis lies ,common ROC part comes below sigma
h(t)=(1/3)*(e^2t)*u(t)+(1/3)*(e^-t)*u(-t). Is this answer correct?
No
Sir upload videos very fastly
sigma
Why it is not sigma>0;sigma
Sir your pace of uploading the lectures has gone down even more... Please resolve the issue... Actually it's a big issue
In q1, why it can't be [(1/3)exp(2t)u(-t)-(1/3)exp(-t)u(t)]
Can someone explain 3. option?
see for neither causal nor stable ROC will be to the left of the left most signal and should also not include JW axis.....so SIGMA will be less than -1 and since it is less than -1 it will also be less than 2...ok....so now you have 2 LEFT SIDED SIGNALS but h(t) for general case is right sided for both parts.....on changing to LEFT SIDED you will get the answer as posted by many guys above hope you understand bro
@@JJpunk434 thank you very much
@@JJpunk434 then what about the jw axis? doesnt it become a stable signal if we do so?
@@JJpunk434 If we are taking both as left sided signal ...the jw axis is getting included which makes the system stable...hence to make the system unstable sigma must be greater than 2......so please can you explain me this part??
@@ritwikasanyal9939 if you take this as a case then there will be no common region ,which means no ROC which further implies no laplace transform.
h(t)= 1/3{e^-2t u(-t) - e^-t u(-t)}
Sigma
Sahi answer toh bata do sir
Why it is not sigma>0:sigma