Convolution (Solved Problem 1)

[@gladiatorsblog2080]

limit in 2nd case should be 0 to t....at 9:30

[@KrishnaYadav-sn9zy]

Great sir.....u cleared my confusion

[@MdRaiyanRaziBEC]

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.

[@Talha80777]

yes

[@alihassnain2513]

same is for case 1...limit should be from -ve infinity to t

[@oguzhanbaser1389]

There is no mistake, both represent the same thing. The integral is investigated for different cases according to the t intervals which is important..

[@kritikabhateja110]

@@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

[@BhavikRaghav14905]

Thanks

[@saisunayan1701]

I think in method-1,h(-T) should be shifted towards right to get h(t-T)

[@Huzaifakhan-nc9dx]

yes that is what am thinking

[@trivendrasinghbisht6009]

yes you are correct

[@meetshah3400]

You can think of T as a negative value to shift it towards left side.

[@BhavikRaghav14905]

Thanks a lot sir!!! Your videos are really helping me prepare for my exams Jai Siya Ram

[@pratiushanand1598]

sir the limits of integration must be from minus infinity to 't'.

[@BhanuPrathap]

No limits are from 0 to t

[@danielvanderstruis1516]

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

[@hardikjain-brb]

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 :)

[@kritikabhateja110]

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

[@hardikjain-brb]

Cause when t goes to inf it is taken care of

[@rohithpokala]

Sir when will you start Fourier transforms?

[@c.cody.3886]

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.

[@sirjantudu7553]

It is actually shifted to right but if you take t

[@Parthj426]

nice teaching always

[@SaikatPodder]

Sir complete signal system as soon as possible

[@dwayneanthony5412]

Great videos keep up the amazing work!

[@RahmaElsaeed-cf6ru]

I watched previous lecture but I have not understand yet why shift isn't to right , could you explain it more ,please ?

[@vizviz72405]

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

[@c.cody.3886]

it should shift to right not left

[@inspireeshwar2225]

why is the value of integration = t at 9:36?

[@markmaroki4841]

the integral from 0 to t... When it's -inf to +inf, then the only shaded region available is 0 to t.

[@christopherkintz7821]

When t0, limits of integration reduce to 0 to t. This is why the value of the integration is t

[@greenemerald3194]

We can use trick # u(t+a) * u(t+b) = ramp(t+a+b) to solve question

[@pedhey]

SO helpful

[@akashjagtap2784]

Sir why you take limits as -ilinfinity to +infinity because signal overlaping is from '0' to 't'

[@vaikh8450]

Same doubt

[@kritikabhateja110]

@@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

[@vaikh8450]

@@kritikabhateja110 thanks for rectifying my mistake but now I doesn't even remember what the Q was ? 😅

[@sadatlucas4045]

thanks a lot bro..

[@GimmeMonie]

8:20 Integration of zero is equal to a constant!

[@scottsimmons9296]

Yes for an indefinite integral, not for a definite integral

[@SandeepChaudhary-vx9zy]

Plzzz cover more gate question sir

[@ayushagarwal69]

method 2 op , thanks sir

[@zvirus0074]

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

[@madhurgoyal2908]

Limit should from 0 to t See the overlap part

[@kritikabhateja110]

@@madhurgoyal2908 If this is the case m then why are we not taking one more condition t to infinty

[@ShaikAfsana-gg4ue]

good very helpfull

[@abhaykondru3570]

How invees Laplace of 1/s² is r(t) y not t

[@saisumanth8064]

sirr take h(t)as ramp and one is exponential signal plzzz convulated

[@kunalchouhan453]

sir convulated u(t-1)+u(t-5)-u(t-3)with ramp function plzzzz......

[@asmaulhosna7201]

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.

[@vizviz72405]

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

[@kritikabhateja110]

@@vizviz72405 Then why in solution y(t) = 0, t=0 . As from t to infinity, y(t) = 0

[@kritikabhateja110]

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

[@beeramaishwaryalakshmi8934]

Sir please method-1 confusion. I didn't get properly. Please explain me in a proper manner. From reversal I didn't get

[@AbdulJabbar-nm6yf]

why you not put the limites .

[@anonymousperson2032]

Gate ques ans will be x(-t-t0)

[@yogipro183]

Is answer y(t)=t.....for t>0.

[@sedlyfsedlyf7011]

sir ne t negative li hai

[@sedlyfsedlyf7011]

i think it will resolve the confusion in the shifting part @ 5:03

[@santanumondal6926]

Shifting is done wrong at 5:03 sec

[@Stevotube12]

You stupid, you should watch previous lecture

[@RahmaElsaeed-cf6ru]

I watched previous lecture but I didn't understand yet why shift isn't to right , could you explain it more ,please ?

[@devanjanchakravarty5543]

@@RahmaElsaeed-cf6ru first shift then reverse, you'll get the answer

[@GimmeMonie]

Yes the shift should be to the right if t>0

[@engineer3961]

@@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.