professor when you convert from h(-𝜏) to h (t-𝜏). The curve should shift right instead of left. This is because the shifting operation is performed on the 𝜏 component only but no the whole (t-𝜏) argument. My professor taught us to let another function g(𝜏) = h(-𝜏), then perform subtraction or addition to g(𝜏) so that h(-𝜏) becomes h(t-𝜏), in this case we perform subtraction: g(𝜏-t) = h (-(𝜏-t)) = h(t-𝜏) (note that the shifting is performed only on 𝜏 but not the whole argument). Finally, because of g(𝜏-t), -t means that the whole curve shifts to the right by t units. Therefore it should shift to right instaead of left.
Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}. Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0. I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
@@oguzhanbaser1389 Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}. Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0. I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
Which function,h(t-tau) is shifted towards right for different cases by varying the time t if it is shifted towards left then convulation will be 0 only
@@vaikh8450 Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}. Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0. I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}. Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0. I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
@@devanjanchakravarty5543 first shift then reverse also gives the same answer. while performing reverse operation we must take the mirror image about y axis (with respect to tau = 0 line), not with respect to tau = t. I think the shift should be to the right by t.
limit in 2nd case should be 0 to t....at 9:30
Great sir.....u cleared my confusion
Thanks a lot sir!!! Your videos are really helping me prepare for my exams Jai Siya Ram
I think in method-1,h(-T) should be shifted towards right to get h(t-T)
yes that is what am thinking
yes you are correct
You can think of T as a negative value to shift it towards left side.
nice teaching always
sir the limits of integration must be from minus infinity to 't'.
No limits are from 0 to t
No, you can (in case 1) immediately evaluate the integral from -infinity to infinity because there is NO overlap in signals in case 1 (t
professor when you convert from h(-𝜏) to h (t-𝜏). The curve should shift right instead of left. This is because the shifting operation is performed on the 𝜏 component only but no the whole (t-𝜏) argument. My professor taught us to let another function g(𝜏) = h(-𝜏), then perform subtraction or addition to g(𝜏) so that h(-𝜏) becomes h(t-𝜏), in this case we perform subtraction: g(𝜏-t) = h (-(𝜏-t)) = h(t-𝜏) (note that the shifting is performed only on 𝜏 but not the whole argument). Finally, because of g(𝜏-t), -t means that the whole curve shifts to the right by t units. Therefore it should shift to right instaead of left.
It is actually shifted to right but if you take t
we can write the given func as unit signals and then use u*u as ramp and get the answer directly using this method saves a lot of effort :)
Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}.
Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0.
I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
Cause when t goes to inf it is taken care of
There is a mistake at 9:36
In second case : for t>0 integration limit should be from 0 to t *not* from -infinity to infinity.
yes
same is for case 1...limit should be from -ve infinity to t
There is no mistake, both represent the same thing. The integral is investigated for different cases according to the t intervals which is important..
@@oguzhanbaser1389 Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}.
Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0.
I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
Thanks
Great videos keep up the amazing work!
SO helpful
Sir when will you start Fourier transforms?
I watched previous lecture but I have not understand yet why shift isn't to right , could you explain it more ,please ?
Which function,h(t-tau) is shifted towards right for different cases by varying the time t if it is shifted towards left then convulation will be 0 only
it should shift to right not left
why is the value of integration = t at 9:36?
the integral from 0 to t... When it's -inf to +inf, then the only shaded region available is 0 to t.
When t0, limits of integration reduce to 0 to t. This is why the value of the integration is t
Sir complete signal system as soon as possible
We can use trick # u(t+a) * u(t+b) = ramp(t+a+b) to solve question
Sir why you take limits as -ilinfinity to +infinity
because signal overlaping is from '0' to 't'
Same doubt
@@vaikh8450 Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}.
Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0.
I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
@@kritikabhateja110 thanks for rectifying my mistake but now I doesn't even remember what the Q was ? 😅
thanks a lot bro..
Why are we not taking 3 cases , {- infinity,0}, {0, t}, {t , infinity}.
Also IF we are only taking 2 cases {- infinity,0}, {0, infinity}, then in solution its showing y(t) = 0, t=0.
I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0
How invees Laplace of 1/s² is r(t) y not t
method 2 op , thanks sir
Sir ,in case no.2 how the result of output as "t" was came by the integrating w.r.t tau from -inf to +int
Limit should from 0 to t
See the overlap part
@@madhurgoyal2908 If this is the case m then why are we not taking one more condition t to infinty
In case II , t>0 range, why is the limit [-∞,∞]. Wouldn't it be [0,∞]? h(t) overlapped x(t) from 0.
confused about it. Hope you will help.
It will be further 0 to t which he forgot to write as -inf to 0 ,0 to t and t to +inf r 3 limits and for 1st and 3rd limits value of convulation is 0
@@vizviz72405 Then why in solution y(t) = 0, t=0 . As from t to infinity, y(t) = 0
good very helpfull
Plzzz cover more gate question sir
sir convulated u(t-1)+u(t-5)-u(t-3)with ramp function plzzzz......
sirr take h(t)as ramp and one is exponential signal plzzz convulated
Sir please method-1 confusion. I didn't get properly. Please explain me in a proper manner. From reversal I didn't get
Is answer y(t)=t.....for t>0.
8:20
Integration of zero is equal to a constant!
Yes for an indefinite integral, not for a definite integral
why you not put the limites .
Gate ques ans will be x(-t-t0)
Shifting is done wrong at 5:03 sec
You stupid, you should watch previous lecture
I watched previous lecture but I didn't understand yet why shift isn't to right , could you explain it more ,please ?
@@RahmaElsaeed-cf6ru first shift then reverse, you'll get the answer
Yes the shift should be to the right if t>0
@@devanjanchakravarty5543 first shift then reverse also gives the same answer. while performing reverse operation we must take the mirror image about y axis (with respect to tau = 0 line), not with respect to tau = t.
I think the shift should be to the right by t.
sir ne t negative li hai
i think it will resolve the confusion in the shifting part @ 5:03