Laplace transform of f(t-a)u(t-a), the shifted unit step function

[@jackfernald8823]

Wow, that was the clearest explanation of this I've heard so far. I've been googling on how this works and nothing was making sense. Everything clicked. YOU'RE AMAZING!

[@pushprajbhardwaj4820]

Very helpful. Thanksgiving. Love from India 🌹🇮🇳

[@varkush399]

Thanks man! the very last part where you show how to deal with different inputs is clear and helpful. Way way way better than my cal prof's method.

[@ivanjosecaballero574]

i love you. this is my written homework right now

[@suekaur4638]

Your videos are so helpful! thanks for breaking everything down step-by-step!

[@blackpenredpen]

Sue Kaur you're welcome!

[@hypnovia]

NIce pun

[@faris338]

man your a life saver

[@Jatin-gm5zx]

Great video man I couldn't understand through my book but now I understand it completely ❤️❤️

[@thor0909Sanket]

Sir Can You solve some examples based on this method?

[@aneesak7365]

@jesslyn baolin f**

[@nonkanyisomntambo5763]

THANK YOU SO MUCH, YOU REALLY PUT EVERYTHING INTO GREAT PERSPECTIVE

[@ShAlAmAnAyA3]

Can't thank you enough for this video. So appreciated!

[@blackpenredpen]

You’re welcome!

[@Miguel-zb1ck]

Someone give this guy a medal!!

[@udulabandara4137]

Bruh, I do not regret bunking my lecs. Thank you loads

[@wonsko2135]

you are the best teacher!

[@afifakhanak]

the best teacher

[@karan_rathore]

Thanks, Very nicely explained

[@aashigupta2145]

Sir, I couldn't understand why did we not write e^-sa L{f(t-a)} as the answer in the first case. v = t-a was our substitution, shouldn't it be taken into account in the final answer as well?

[@yoelfeinmesser516]

V is a dummy variable, hence the result of the definite integral is the same in both cases and you don't need to change back

[@matthewfinis6723]

It looks like blackpenredpen is about to become just blackpen

[@TI_Ted]

absolutely exactly what i was looking for !

[@_DD_15]

You are one of my fave youtubers :)

[@darionmohammed9206]

You saved me

[@vtrandal]

Brilliantly done!

[@anahuaccontreras3232]

dude is rocking supreme

[@neeleshbhargava8145]

Will the graph for [e^(-st)]*f(t) show any shift along x-axis when drawn for [e^(-st)]*f(t-a)? And why? Thank you

[@guilhermemb9213]

thank you, that was really usefull and well explained!!

[@뀨뀨-e6l]

대학수업에서 증명을 안해줘서 궁금했는데 큰 도움이 되었습니다. Thank you.

[@flameon8185]

11:25 then it's only left with laplace of g(t) right ??

[@gameon8177]

well explained dude, thanks!

[@鲁迈萨·本塞吉尔]

Hello, what about integral of (exp(-s. ((v+a)^r)). f(t)) ?!

[@akshaias935]

sir you are amazing ,,,could you please suggest some ways to study maths well

[@jackpham7964]

My boy got that Preme' !!

[@blackpenredpen]

Yup!

[@eolvra]

too much clout

[@sitienlieng]

Could you please explain why everything before a is 0 on the graph?

[@qannayuri]

Best ive seen

[@daniellima5954]

Reaally helpful, man

[@gauravbangar3090]

thanks bro for explaining in detail

[@blackpenredpen]

You're welcome!

[@kbpost6947]

This method is easy 🤩

[@saya3237]

🙃thanks 😻

[@jiaStar0233]

謝謝 這是111年公職的考題 沒你的說明我真看不懂

[@Kay-dx8vm]

Can you pleased tell me how to solve u(t-2)*te^(-t) ?

[@potatoes8169]

what happens if a

[@williamadams137]

The graph of e^(-st)f(t-a) won’t be affected at x>a, where a could be any real number.

[@potatoes8169]

@@williamadams137 I mean whats the laplace transform of u(t+1)? is it 1/s ? and is L{(t+1)u(t+1)} = L{t+1} = 1/s^2 + 1/s. since u(t+1) = 1 for all t>0? (where L is the laplace transform)

[@ssdd9911]

can the unit step function be expreesed in terms of elementary functions?

[@carultch]

Depends on what you mean by elementary functions. If you include the sign function, then the answer is yes. u(t) = 1/2 + sgn(t)/2

[@zirma9296]

Outstanding!

[@aneesak7365]

Poli saanam

[@Pooja14able]

what if unit function is shifted by a time varying function like a(t)?

[@usairummirza9617]

please put this in the ordenary differential playlist

[@alfabargh]

liked your video! that's helpful :)

[@AXg_46]

thank you 🙂

[@ju-uq9zi]

can you please prove the laplace transform of the zeroth orderband the nth order bessel function of the first kind

Genius!

[@blackpenredpen]

: )

[@muhammadafzaalkhan9277]

Great.upload video on frobenious method when roots are differ by unity and when cant solve by reduction method for second solution.How to solve the second solution.please!please!please! just do this question.

[@SISKCERTWaJaVlogs]

isnt it also called the heaviside function

[@leewinlan8084]

thank you! :)

[@0019jacky]

Brilliant!

[@rma1698]

Thank you !

[@blackpenredpen]

you're welcome!

[@SamuelAndradeGTutos]

What is the u(t)? Is it any function?

[@lucasm4299]

Samuel Andrade Unit Step Function

[@0019jacky]

Heaviside functions, just look it up in your text book

[@chintanviradiya7046]

Laplace of Intigration 0 to t integration o to t sinau du du kese solve hoga

[@jyaswanth9688]

Thanks bro

[@siddharthasankarmaitra5529]

It is difficult to convince an examiner by h(t-a) stand for f(t). Because for f(t) between t=0 to t =a, h(t-a) remains 0. Another process that clicks me, if at all correct, is that👇

[@mvelasebafana7076]

y is everything before a equal to zero?

[@mvelasebafana7076]

thank you man

[@JossinJax]

Why is anything before 'a' zero?

[@ianmoseley9910]

Jossy Beth definition of the step function?

[@dashzevegp7685]

Thanks

[@benjamingoldstein14]

Why doesn’t it matter if we’re using v or t?

[@Mystery17S]

Great

[@squiddy5020]

I'm sorry but can you activate the subtitle because some of your viewers aren't english speakers. Thank you

[@jonydaoud2209]

legenddd

[@blackpenredpen]

thanks!

[@ehsanali5119]

thank you

[@ahmedal-riyahee3094]

Hi Could you please solve the following 1)L{u(t^2-5t+6).u(t)} 2)L{u(t^3-t^2-5t+6).u(t-0.5).u(t)} I know it's creazy quizzes.. It will be nice if you can help me.. Thanks allot ☺️

[@carultch]

Part 1: Given: L{u(t^2 - 5*t + 6) * u(t)} Find the roots of t^2 - 5*t + 6. Where it is positive, u(this) will equal 1. Where it is negative, u(this) will equal zero. t^2 - 5*x + 6 = (t - 3)*(t - 2) Thus its roots are t=2 and t=3. Because this expression of t is positive for the t^2 coefficient, this means it is concave-up, and is thus positive when x3, and negative in between these values. Thus, our expression initially jumps to 1 at t=0 for the remaining u(t) function. It then drops down to 0 at t=2, and then jumps back up to 1 at t=3. Construct a linear combination of the unit step function to reflect this: u(t) - u(t - 2) + u(t - 3) Take the Laplace transform of each term, to find our result: 1/s - e^(-2*s)/s + e^(-3*s)/s Part 2: Given: L{u(t^3 - t^2 - 5*t + 5)*u(t - 0.5)*u(t)} Find roots of the cubic. It has a rational root at t=1. Use polynomial division to find quadratic factor: (t - 1)*(t^2 - 5) Thus, its roots are at t=-sqrt(5), t=1, and t=sqrt(5). We don't care about the negative root, since the remaining u(t) is off at that point. Since the cubic is increasing after its final root, this means it starts positive at t=0, becomes negative at t=1, and switches back to positive at t=sqrt(5). Multiplying with u(0.5) this means the function doesn't start until t=0.5. So it starts equaling 1 at t=0.5, switches to equaling zero at t=1, and returns to equaling 1 at x=sqrt(5). Construct a linear combination of unit steps to match this behavior. u(t - 0.5) - u(t - 1) + u(t - sqrt(5)) Take Laplace transform of each term for our result: e^(-0.5*s)/s - e^(-s)/s + e^(-sqrt(5)*s)/s

[@leewaston]

genius

[@Qbabxtra]

Isn't it!

[@mertafacan1171]

zort

[@mertafacan1171]

@Osman Navdar midterm2=31