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ОтветитьThe value of u(0) in Example 1 is unclear. If u(0) = infinity, then IV would be infinity, no?
Maybe u(t) is a step function at 0? It is not a generalized function?
Good Lecture . THANKYOU
ОтветитьThank you so much! You have helped me!
Ответить🔥🔥🔥🔥🔥
ОтветитьThere is a mistake in the initial value in the first example.
It's answer is 0 not 1
Seriously, its so weird having a college professor jump straight into how to derive these concepts without explaining the concept in terms of what its used for or how it relates to other things. These videos are what fills the gap for me, thanks for making this intuitive for a beginner!
ОтветитьIf final value theorem is not apllicable , then in all cases is that unbounded ?
ОтветитьIt's very helpful 🙂
ОтветитьIf we do Inverse Laplace Transform of F(s) = 1/(s+2) then it will give us f(t) = e^-2t and therefore f(t) violates the condition 1 i.e. f(t) = 0, t<0 as e^-2t is not 0 for t<0. Then how can we apply Initial value theorem for the question?? Pls clear this doubt.
Ответитьthanks a lot
ОтветитьInitial value? More like "Incredible; thank you!" These videos really are quite splendid.
Ответитьthanks Sir🙏❤
ОтветитьGood method and easy to understand
Again your's work is very Good
Uh r great sir
ОтветитьExcellent explanations sir
ОтветитьIn final value theorem 2nd example, how do you say F(s)=1/(s-2) pole value is 2? No clarity given about direct conclusion of RHS value?? Can you explain more..other than that quality of explanation is super.
ОтветитьI can't believe the fact that such amazing videos have such few likes!
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