I just wanan let you know how much I appreciate all your DE videos. I've been watching since Laplace transforms and you made stuff simple in 25 minutes where my Prof made it horribly complicated in 50. Thanks for clearing all this up for me, I think I actually have a chance to do well on my exam now. Now, time to get back to your videos. I'll probably reach vid #45 soon enough. THanks again
Why would anyone dislike this video? Was it hard for them to understand? Did they not like his approach/method? To me, it was very simple to follow and thus I learned something.
Somehow when Sal says it, it sticks. Hopefully we have youtube. The only place where half a million student can sit and world class watch a lecture. FOR FREE.
@@yash1152 Bro, that comment was from over 10 years ago. The comment system has changed multiple times since and the old reply chain stuff is all gone.
> _"The comment system has changed multiple times since and the old reply chain stuff is all gone."_ @@adituv8565 i know, i said to hard quote the comment to which a person is replying to in his/her own comment like i do.
@ 3:33, should't the sin function = a/s^2 PLUS a^2 ? I think the solution written is for the sinh function. I'm taking Differentials right now, so I noticed this. Thanks Khan!!!
You are really Awesome, thanks a lot for your help. Also, it would be so kind of you if you can pls tell me what software do you use for making videos? I really liked it.
u(t) is the unit step function, with its jump discontinuity at t=zero. u(t - a) is the unit step function, shifted a units to the right, in the time domain. L{u(t)} = 1/s L{u(t - a)*f(t - a)} = e^(-a*t)*F(s), where L{f(t)} = F(s). Both formulas are correct. One is just the original form, and the other is time-shifted, and generalized to apply to other functions as well, rather than just the individual unit step function.
Jose Cortes Breaking it into Partial Fractions and then Inverse Laplace will give you f(t) = 1 - (cos(√3/2 t )* e^(-t/2)) + (1/√3*sin(√3/2 t) * e^ (-t/2) )
I have to admit, this is the first time I got frustrated watching one of these videos because of how it was presented. Too much jumping around between the solution and the previous examples and substitutions, and it was hard to tell which was which in real time.
shoutout to this man for helping engineering majors for over 12 years
over 14 years now
I just wanan let you know how much I appreciate all your DE videos. I've been watching since Laplace transforms and you made stuff simple in 25 minutes where my Prof made it horribly complicated in 50.
Thanks for clearing all this up for me, I think I actually have a chance to do well on my exam now.
Now, time to get back to your videos. I'll probably reach vid #45 soon enough.
THanks again
You are right dear iBredgen
You saved me the night before the test, thanks a lot!!!
i'm still amazed at how easy this stuff is! If Sal taught me this stuff even in high school, I would def get it!
You are right PrinceFx!
thanks khan!! you just saved the day , 6 min of the clip explained so much that i could started solving problems
Why would anyone dislike this video? Was it hard for them to understand? Did they not like his approach/method? To me, it was very simple to follow and thus I learned something.
Those are two great reasons to dislike the video, thanks for sharing.
because it kinda becomes messy from 9 minutes = all over the place
PTSD from signals and systems? Lol.
When I win a nobel in the future I'll include khan academy in my thanks
When will you win?? Just curious.
Did u got it man?
Include me too! :D
Are ya winnin son?
Are you winnin son?
thank you so much man, this topic was starting to annoy me but now i figured it out thanks to your help
Watching this before my final exam❤
Somehow when Sal says it, it sticks. Hopefully we have youtube. The only place where half a million student can sit and world class watch a lecture. FOR FREE.
this guy.... I've no words!!!!!!!!
Hats off man
Man, you are the best on youtube !!!!!!
thanks a lot !!!
Wish you were my lecturer, makes so much more sense this way. thanks.
so useful to watch your videos. it is really helping me develop in my mathematics classes.
God bless you sir
Passed a quiz thanks to this video!!!!
You're the reason I'm not working at Mcdonalds right not
Well clearly you should if you can't spell "now".
ahaha, i am just kidding.
@@MsAsim123456 😂😂😂
Man , u made it so easy for me ..
You guys are absolutely killing it. So useful! Thank you.
You my friend are a gentleman and a scholar. Kind of a mathematical gangster. The offspring of a Jedi knight and a wizard.
Infinitely helpful. Completely forgot all the Laplace transformations!
If I understand it right, your U(t) is a step function which becomes 1 at t=0; his u_c(t) becomes 1 at t=c, so both those expressions are equivalent.
> _"if i understand it right, your U(t) ..."_
thats one of many reasons why i always quote with respect to what i am replying
@@yash1152 Bro, that comment was from over 10 years ago. The comment system has changed multiple times since and the old reply chain stuff is all gone.
> _"The comment system has changed multiple times since and the old reply chain stuff is all gone."_
@@adituv8565 i know, i said to hard quote the comment to which a person is replying to in his/her own comment like i do.
Jesus. I’m a third year EE. You have just made the light bulb turn on. Thanks so much!
Relax, Signals and Systems Class my dude
@ 3:33, should't the sin function = a/s^2 PLUS a^2 ? I think the solution written is for the sinh function. I'm taking Differentials right now, so I noticed this. Thanks Khan!!!
John Buckner it's a + but it's not clear enough,he also said "plus" while speaking.
write equations tidy and in order
You my friend are a gentleman and a scholar. Kind of a mathematical gangster
Thanks a lot for uploading these Laplace and Inverse Laplace videos. These really helped me understand the topic :)
2:23 Summary of Laplace Transform properties we have looked so far
Thank you for the lesson one again.
the capital F makes it super confusing, try G,H to replace would be extremely helpful
Tq Dr. Nasir
Nice work Sir👍
Thank you Sal. I love you.
You are really Awesome, thanks a lot for your help.
Also, it would be so kind of you if you can pls tell me what software do you use for making videos? I really liked it.
thank you very much of this wonderful example and explanation
uh saved my time
thank you very much
My textbook says this: L{U(t-a)f(t-a)} = e^(-at)F(s). In other words, it says U(t-a) and not U(t). Which is right?
u(t) is the unit step function, with its jump discontinuity at t=zero.
u(t - a) is the unit step function, shifted a units to the right, in the time domain.
L{u(t)} = 1/s
L{u(t - a)*f(t - a)} = e^(-a*t)*F(s), where L{f(t)} = F(s).
Both formulas are correct. One is just the original form, and the other is time-shifted, and generalized to apply to other functions as well, rather than just the individual unit step function.
Why is the laplace transform of t to the n the reciprocal of the integral power rule?
great video sir .....
Great video.
very helpful ,, thanx o lot 4 ur work :D
your amazing!
you are really great!!!!!!
Should have defined the inverse laplace transform at the beginning
15 yrs ago is crazy
OMG this video created in 2k9 and am watchin it in 2021 lol..
excellent
try to solve this inverse laplace transform: (s+1)/s(s^2+s+1)
Jose Cortes Breaking it into Partial Fractions and then Inverse Laplace will give you f(t) = 1 - (cos(√3/2 t )* e^(-t/2)) + (1/√3*sin(√3/2 t) * e^ (-t/2) )
Given:
(s + 1)/(s*(s^2 + s + 1))
Complete the square on the irreversible quadratic:
s^2 + s + 1 = (s + 1/2)^2 + 3/4
Set up partial fractions:
A/s + (B*(s + 1/2) + C)/((s + 1/2)^2 + 3/4)
Heaviside coverup at s=0 finds A:
A = (0 + 1)/(covered*(0^2 + 0 + 1)) = 1
Result so far:
(s + 1)/(s*((s + 1/2)^2 + 3/4)) = 1/s + (B*(s + 1/2) + C)/((s + 1/2)^2 + 3/4)
Let s = -1/2, so the B disappears and we can find C:
(-1/2 + 1)/(-1/2*((-1/2 + 1/2)^2 + 3/4)) = 1/(-1/2) + C/((-1/2 + 1/2)^2 + 3/4)
-4/3 = -2 + 4/3*C
C = 1/2
Let s = +1/2, so the coefficient on B becomes 1:
(1/2 + 1)/(1/2*(1 + 3/4)) = 1/(1/2) + (B + 1/2)/(1 + 3/4)
12 = 14 + 4*B + 2
B = -1
Partial fraction result:
1/s + (-1*(s + 1/2) + 1/2)/((s + 1/2)^2 + 3/4)
Modify the expression, so it resembles standard transforms:
1/s - (s + 1/2)/((s+1/2)^2 + 3/4) + 1/sqrt(3) * sqrt(3)/2/((s+1/2)^2 + 3/4)
1*L{1} - L{cos(sqrt(3)/2*t)*e^(-t/2)} + 1/sqrt(3) * L{sin(sqrt(3)/2*t)*e^(-t/2)}
Inverse Laplace result:
1 - cos(t*sqrt(3)/2)*e^(-t/2) + sqrt(3)/3*sin(t*sqrt(3)/2)*e^(-t/2)
is u subscript 2 (t) pretty much the same as u(t-2) please someone answer!!!!
I'm confused, what is this?
Kaash ap musalman hotay to ap jannati hotay.
you better if you write like that: F(s)=2.g(S).e^-2s
What program does he use?
🐐🐐🐐
@aligsaysai junk comment
@Pickulus wtf LMFAOOOOO
Soo complicated to learn last day :/
بلاس وبلاس مافهمت
I have to admit, this is the first time I got frustrated watching one of these videos because of how it was presented. Too much jumping around between the solution and the previous examples and substitutions, and it was hard to tell which was which in real time.
Specially from 9 minutes... My ADHD got flipped
#69 comment
You seem to be prone to jump around a bit. Stay closer to your notes.
you confuse me even more
unlearn and learn
it is very unorganized, previous videos where way better
Thank you!
thank you very much. very clear explanation
Thank you so much