Laplace transform is so useful in control systems, I came back to this while I was deriving step response of a first order system. It's very interesting. God, I love math and everything. Thanks for this video! Great clarity in the content
Since laplace operator is linear, in example 4, can I just pull out the 1/s, take the laplace inv of that, and then multiply it by the laplace inv of 1/s^2+2? Or is that against the rules
Hello sir my name is kamleshdutt. Sir these days I am having bad days, I am not able to solve maths question. I feel like tearing the book and throwing it, please answer 2 days through the video.
Given: inverse Laplace 3/((s - 2)^2 + 25) Start by letting capital S = s - 2. Thus, we have: inverse Laplace 3/(S^2 + 25) Multiply by 1 in a fancy way, so we can make this look like the Laplace transform of sine: £{sin(k*t)} with S as the domain variable = k/(S^2 + k^2) inverse Laplace 3/(S^2 + 25) * 5/3 = 1/3 * inverse laplace 5/(S^2 + 25) thus: inverse Laplace3/(S^2 + 25) * 5/3 = 1/3 * sin(5*t) Now use the first shift theorem, to find the inverse laplace with lowercase s. £{e^(a*t) * g(t)} = G(s - a) Since we shifted s by -2, to get S, this means we'll have a multiplier function of e^(2*t). Thus our result is: 1/3*e^(2*t) * sin(5*t)
Yes. As a matter of fact, if you look at the Laplace transform of coshes and sinches, you'll see that the sign in front of the squared constant is negative instead of positive. This would imply that they are equivalents of cosines and sines with imaginary numbers as the frequency.
Channels like yours deserve all the tuition money I'm paying!
And you did it again! I was just studying this topic and you came up with a new video about it. Thank you!
Thank you for this high quality completely free content!
Laplace transform is so useful in control systems, I came back to this while I was deriving step response of a first order system. It's very interesting. God, I love math and everything. Thanks for this video! Great clarity in the content
I just finished my exam on differential equations wish I had this to watch a month ago. Thank you for the resource
u r literally best explaining with detail and revising it best wish we had more teacher like u
You are indeed a saviour
Cool. I was just thinking of Laplace Transforms earlier today and was wondering if you might have uploaded a video about it.
The math of the electrical engineers!! :)
Stunk
Since laplace operator is linear, in example 4, can I just pull out the 1/s, take the laplace inv of that, and then multiply it by the laplace inv of 1/s^2+2? Or is that against the rules
Taught me everything I needed to know about Laplace! tysm
Omg, I have finals the next week and I need to get this clear, thanks
My lecturer on TH-cam
Finally an actual math video
And that's how pros do it 😂 well done! These videos are very helpful!
i love your math videos!!!
what are the eigenvalues and eigenfunctions of the laplace transform operator?
thank you
Sir can u please do a full video about transformation of functions to graph, really need it..... 🙂
Thank u
Thx ^^
ecalpel ought be it's name
Hello sir my name is kamleshdutt. Sir these days I am having bad days, I am not able to solve maths question. I feel like tearing the book and throwing it, please answer 2 days through the video.
After watching his videos, I will never say cosine in my head the same again. lmao
pravoo dr.
what about (3/((s-2)2 +25) ??
Given:
inverse Laplace 3/((s - 2)^2 + 25)
Start by letting capital S = s - 2. Thus, we have:
inverse Laplace 3/(S^2 + 25)
Multiply by 1 in a fancy way, so we can make this look like the Laplace transform of sine:
£{sin(k*t)} with S as the domain variable = k/(S^2 + k^2)
inverse Laplace 3/(S^2 + 25) * 5/3 = 1/3 * inverse laplace 5/(S^2 + 25)
thus:
inverse Laplace3/(S^2 + 25) * 5/3 = 1/3 * sin(5*t)
Now use the first shift theorem, to find the inverse laplace with lowercase s.
£{e^(a*t) * g(t)} = G(s - a)
Since we shifted s by -2, to get S, this means we'll have a multiplier function of e^(2*t). Thus our result is:
1/3*e^(2*t) * sin(5*t)
how do you use math to solve problems?
You're gonna have to be more specific. That's like asking, "how do you use a kitchen to cook?"
If we let k be a complex number with a non zero imaginary unit, would be utilize sin or sinh
Yes. As a matter of fact, if you look at the Laplace transform of coshes and sinches, you'll see that the sign in front of the squared constant is negative instead of positive. This would imply that they are equivalents of cosines and sines with imaginary numbers as the frequency.
@@carultch awesome stuff! thanks
where's the residues?
Haha we are now doing the la place transformation for the differentials
You deserve my tution fees which i paid to clg
😂😂😂
aww man, he messed up in formula 6. the inverse is supposed to be sinh(kt). Where's my points Math sorcerer, lol
thanks , but ur s is weird
My lecturer on TH-cam