integration of a function gives the area under the function. If you plot e(-st) vs t, area under it is a function of t, which can be calculated by integrating e(-st) which yields 1/s
It has much more value for systems than for signals. You can investigate frequency response and transient behaviour once the Laplace transform is found for a system and you stay within roc in your studies of the system.
but ... I don't understand the result in 11:47 after of integrate the result is : -1/(s+1) + 1/(s+2), then x(s) = -1 / (s+1)(s+2) ?? Someone can review this please?
Loving this whole series. Really filling in some gaps from years ago.
Very much appreciate this lecture, MIT.
totally understandable I love it I am just so curious about learning all this stuff
Great lecture! Thanks Prof. Freeman.
Awesome work MIT. Keep on uploading please!
Is there anyone noticed that the lecturers voice and that sound he does with his lips reminds of heath ledgers joker ?
thanks MIT
Great ,Thanks MIT
Thanks
thank you MIT....(y)
44:38. someone explaine for me why area under e^(-st)--> 1/s?
integration of a function gives the area under the function. If you plot e(-st) vs t, area under it is a function of t, which can be calculated by integrating e(-st) which yields 1/s
Wow!! Amazing!!!!
awesome
Laplace Transform converges (gives finite value) in ROC. How is this information (the finite value of LT) help us anyhow?
It has much more value for systems than for signals. You can investigate frequency response and transient behaviour once the Laplace transform is found for a system and you stay within roc in your studies of the system.
awsome
but ... I don't understand the result in 11:47
after of integrate the result is : -1/(s+1) + 1/(s+2), then
x(s) = -1 / (s+1)(s+2) ??
Someone can review this please?
check the signs. It is not exactly as you write
(s+2)-(s+1)/(s+1).(s+2)=1/(s+1).(s+2)
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It should be 5. Laplace Transform instead of Z Transform, the order is wrong I guess.