Convolution (Solved Problem 1)

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 and y(t) = t , t>=0.
I didn't get where did t to infinity portion went ? If its 0 then why in solution we took , t>=0

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 :)

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.

These subject is very boring, embarrassing 🤯🤯 headache WTF

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.

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

thanks a lot bro..

method 2 op , thanks sir

Great sir.....u cleared my confusion

sir ne t negative li hai

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

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

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

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

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

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

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

good very helpfull

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

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

Shifting is done wrong because h(- τ+t) indicates folding and delay operation. So it is obtained by shifting h(- τ) towards right!!

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

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

why you not put the limites .

Plzzz cover more gate question sir

Great videos keep up the amazing work!

Sir complete signal system as soon as possible

Sir when will you start Fourier transforms?