Laplace transform of the unit step function | Laplace transform | Khan Academy

It's like you taught Laplace how to use his Transform !!!!!!!!!!!!!

Thank you!

this is an absolutely amazing explanation!!....but why do we shift the function to the right and multiply it by unit step function, what is the purpose of that?

I love your old math videos ! Best explanation ever ! It's pity you don't record it anymore.

I'm stuck Step Function 😬😬🤭

Hi Hello👋👋

Mr Sal I love your work, in this video I kinda disagree with how u simplified Laplace of the step function, u replaced x=t at the end but we know that X=t+c please help me out here

* Thanks a lot, for introducing this new -- (shifted) "unit step function"
* shifted is my own annotation, as i think μ_zero or just μ is a more fundamental function, with the general μ_c being shifted to right by +c i.e. the μ( t - c )

Anyways:
* this step fxn eases alot the task of defining functions which are stepped at only a few (say 1-3) points like the fxns for absolute, or signum, or other such
* eases a lot - or at least provides a new way to think about ...
* other than that, i don't think it helps much for other periodically partwise functions like floor, ceiling, fractional-part, modulus, remainder, etc.

Thanks mate keep going even if you upload it when i was just learning multiplication

Why do I have to watch these videos to complete my homework? Where is my school's lectures???

Isn't it discontinuous at t equals c?

Thanks a lot.

i didnt understand

Sir, You are a lifesaver. I salute you, Bro.

Can anyone explain the missing step of how f(t-c) = f(t) ??? I was following this course with no problems up until this point. It clearly confused a lot of other people too.

Jesus this was explained 34217923489231749872384293741982373982x better than how my DE professor explained it.

Its amazing how you and my professor teach the same exact thing but with you it just magically makes SO MUCH SENSE!

Thank you 😃

❤️

Are we assuming c>0?

Sir kindly explain in white board instead this.

"difFRENtial equations "

ur a crazy person,ur not saying unit step function instead saying second shifting property thereom in laplace........ dont watch this video

You are actually using Latin alphabet :D

x is the most fun variable, stay away from crazy latin alphabets

L{f(y) } if l putting e-sy ?

my hero

YOU ARE AMAZING

you are runing out of ink

Why is the laplace of f(x) = laplace f(t) when x = t-c ? Why does that seem wrong to me?

This helped more than a gruelling hour long lecture in college

what if i have both are unit step function instead of f of t... i thought i almost done but then my question slightly different

I want to to take the laplace transform of a heaviside function, how come i have to be lectured for 24 min? This is a 5 min tops video.

I DONT UNDERSTAND
How you said L(x) is same as L(t) when you substituted x=t - c

Awsummm Workkkk Sir, U know What Our Instructor Was Doing these things, Like things Are moving In air , But u Did, Insert those Concepts For What I was Thinking Thnx Again

Hi Sal

Thanks for the videos, could you please explain how did you assume F(X) is equal to F(t) in the end of the video because earlier you said x = t-c ?

Thanks again.

did u mean that unit step function is used to obtain equation for change in wave form?

1 no.

A super Duper help to me , thank you from the bottom of my heart :)

Why is the laplace transform L{u(t)f(t-c)} equal to e^(-cs)*L{f(t)} where f(t) is the unshifted function and not equal to e^(-cs)*L{f(t-c)} where f(t-c) is the shifted function? When you worked out the integral with the variable x, you got as an answer that L{u(t)f(t-c)}=e^(-cs)*L{f(x)} and since x=t-c, shouldn't L{u(t)f(t-c)}, when you substitute (t-c) back for x, be equal to e^(-cs)*L{f(t-c)} instead of e^(-cs)*L{f(t)}?

My DE's textbook, (Trench) uses these fucken insane latin alphabets with laplace and it pisses me off. This is much better, thank you :)