Impulse Response and Step Response

[@itzm1kea]

this man is one of the best teachers/professors I have ever seen

[@Labroidas]

God bless you, Professor Strang, thank you so much for everything you are doing for us students and for the world!

[@eefunhuang6958]

Well said about the relationship between impulse response and step response!

[@Joe_Yacketori]

8:16 His voice was underdamped.

[@ejminava407]

lmao

[@realhumphreyappleby]

Lol

[@wallacechan2339]

Absolutely fantastic. Thank you Professor Strang.

[@sossupummi]

delta function's derivative delta'(t) at t=0 is not 1 but infinite? delta is supposed to be a function with infinitesimally small width and infinite height while keeping the area equal to 1

[@iwonakozlowska6134]

The step response r(t) is not exactly the integral of impulse response g(t) but it is the convolution of impulse response with the step function g(t)*u(t). r(t) is the same as presented in this video but under condition that C=1 , in other case we need to divide r(t) by C. By the way C=s1xs2.

[@nicholasborrego1640]

Go away

[@DRACOBUCIO]

r(t) is the convolution of impulse response with the step function g(t)*u(t), and that is the integral of g(tau) d(tau) from -inf to t. And in case of f(t) is causal signal, then the integral limits are from 0 to t. So... at the end of the road, r(t) is the integral of g(t).

[@oussamarap2759]

@@DRACOBUCIO can you explain a little bit more

[@DRACOBUCIO]

@@oussamarap2759 Well, basically if you use a LTI system, x(t) -> y(t). Then x'(t)-> y'(t) or integral{x(t)} -> integral{y(t)}. That is because the integral or derivative is a linear operation. I would like to explain more intensive but you can send me message if you need a demo in order to satisfy your curiosity.

[@PSM1974]

@15:05 the step function r(t) is not the integral of g(t) as given above…….there is a 1/(s1s2) term missing in the denominator of r(t) and the +1 in r(t) equates to 1/(s1s2)…so what gives? ….are there some special initial conditions that are needed to obtain r(t) as given above?…we can let s1 = 1/s2 and that will give r(t) above, but this seems a bit contrived…I’m looking for a formal definition/proof….anyone? Cheers

[@DRACOBUCIO]

14:20 the most importan question in this video (my opinion). How the impulse response and the step response are related?

[@carultch]

The step response is the integral of the impulse response.

[@navidmohammadzadeh2141]

Amazing and clear. Thank you!

[@MrAngryCucaracha]

but the formula for r(t) is not the integration of the formula for g(t). Am i making a mistake?

[@kevinnejad1072]

It would be if the initial conditions are set correctly. To see why that's the case, watch kzbin.info/www/bejne/e3TWnaCrfNGoY7c where prof. Strang explains about step function and delta function. In fact the whole video is about properties of these two functions.

[@Y747Y]

The intrinsics of the materials must have negative real part of eigen value, otherwise the function will not goes to 1 when time goes to infinity.

[@debajyotichoudhuri7896]

I am not sure how has r(t) converged to 1 asymptotically?. The added part in r(t) seems to not be converging to if s_1, s_2>0.

[@freegg123]

If s1 and s2 are greater than 0, it indicates an unstable system meaning the output would increase exponentially rather than decaying to equilibrium, and of course it does not coverge to 1.

[@vajk7]

Thank you for the inspiring nice videos, it's a pleasure to listen your lectures. Is it possible that the initial condition for the step response is r(0)=1, instead of r(0)=0?

[@myvideo911]

it's ok. the aim is to decide the coefficient.

[@getusel]

That is impossible because a ramp function starts from zero and grows linearly.

[@ohmakademi]

thank you very much.

[@SimmySimmy]

Can I get output response(particular solution) under any arbitrary excitation to the system if I know its impulse response?

[@SimmySimmy]

So the complete response would be the natural response(source-free) + excitation response?

[@wickbron8964]

Is unit step response equal with step response

[@PenningYu]

They are different. That's why he multiplied H by C.

[@roshan8853]

Can someone guide me to see how to find the s1-s2 on the denominator?

[@jonathansum9084]

This professor explains thing little bit ahead. You should watch these videos first. www.khanacademy.org/math/differential-equations/second-order-differential-equations After watching the videos on Khan, please solve the c1 and c2 in these two equations: y(t)=c1e^(s1*t) + c2e^(s2*t) y(0)= c1e^0 + c2 e^0 y(0)=c1+c2=0 thus, c1=-c2 Same with y prime y prime = come on, you know how to take the derivative y prime(0) = c1*s1+c2*s2=1 y prime(0) = -c2*s1+c2*s2=1 (why do i changing the c1 to -c2? read the "thus, c1=-c2" :D) Now you can solve these as a high school math: -c2*s1+c2*s2=1 c2(s2-s1)=1 c2= 1/(s2-s1) c2 = -1/(s1-s2) what is c1? c2=-c1 let's plug in the c1 and c2 into the y(t)=c1e^(s1*t) + c2e^(s2*t). you will get y(t)= [e^(s1t)-e^(s2t)]/(s1-s2)

[@roshan8853]

Thank you! I appreciate the effort you went to

[@jonathansum9084]

Did you get the 1/s1-s2 before I post my reply?

[@T4l0nITA]

Thank you so much.

[@14598175]

What I wouldn't give to move your mic away by about 4 inches!

[@lucass8610]

epic

[@Amine-gz7gq]

Laplace Transform FTW