Ramp Response of a System

[@harshvardhansingh8428]

Impulse resp = 1/RC * e^ -t/RC * u(t) Step resp = 1 - e^ -t/RC ramp resp = t- RC + RC*e^ -t/RC

[@eda-un8zr]

Hi, can you explain impulse response? I'm a little stuck and it can lead me.

[@harshvardhansingh8428]

@@eda-un8zr sure , I am here on Instagram @hv6537

[@eda-un8zr]

@@harshvardhansingh8428 i wrote you but i guess you didnt see :) anyways, i solved it!! Thanks tho

[@curiouscoder9566]

yes correct

[@BiswajitDas-bp7qr]

@@eda-un8zr please provide the solution 🙏 mail: [e-mail]

[@mathsop5583]

Hi sir All subject's videos are best Thankyou 🙏

[@Partha_swarnakar]

Egarly waiting for upcoming lectures💞

[@karlm9584]

Very clear thank you

[@PunmasterSTP]

Ramp response? More like "This information is tops!" Thanks again so much for making and sharing so many high-quality lectures.

[@bretthaddin8071]

Thank you sir

[@sanskarkumar6484]

thanks Sir🙏❤

[@Семён-т9с7т]

Спасибо. Очень полезные видео.

[@楊柳樹-u8i]

excellent!

[@bhukyaakhil86391]

Impulse resp = 1/RC * e^ -t/RC * u(t) Step resp = -1/RC*2(1 - e^ -t/RC) ramp resp = -1/RC(t-e^ -t/RC+1 )

[@ss-tp6mz]

please post the home work problem solution

[@francescoejlli8985]

What's the name of the Digital blackboard you use?

[@tanpreetsingh271]

Impulse resp = 1/RC * e^ -t/RC * u(t) Step resp = [1 - e^ -t/RC]*u(t) ramp resp = [t - RC + RC*e^ -t/RC] *u(t)

[@maruthimaruthi3720]

What is full stack web development & full stack gaming development.? what we need in it.?

[@namratasaini2065]

which is more stable.... impulse response or ramp response?

[@sushrutmdhakate9874]

Bhai , ek baat batana , inverse laplace nikalte waqt kabhi kabhi sir u(t) multiply karte hain aur kabhi nahi . To kisiko pata hain ki kis condition me u(t) multiply karna hota hai ?

[@swastikmishra7209]

u(t) ka value t>=0 time me 1 hota hai

[@bhanderi_anisha]

Can anyone explain answer of homework problem ??

[@PunmasterSTP]

What did you get for the answers? If you let me know then I can double-check and explain.

[@carultch]

@@PunmasterSTP Start by finding the transfer function of the filter. It is a simple voltage divider where the output voltage is the input voltage times the ratio of impedances. Impedance of capacitor: Zc = 1/(C*s) Impedance of resistor: Zr = R Transfer function H(s) = Zc/(Zc + R) = 1/(C*s)/(1/(C*s) + R) = H(s) = 1/(C*R*s + 1) Output Y(s) = X(s)*H(s), where X(s) is the input Laplace transform. For an impulse, X_imp(s) = 1 For a unit step input, X_step(s) = 1/s For a unit ramp input, X_ramp(s) = 1/s^2 Thus, the output transfer functions are: Y_imp(s) = 1/(C*R*s + 1) Y_step(s) = 1/(C*R*s^2 + s) Y_ramp(s) = 1/(C*R*s^3 + s^2) Take inverse Laplace to find time-domain responses: y_imp(t) = e^(-t/(R*C))/(R*C) y_step(t) = u(t)*(1 - e^(-t/(R*C)) y_ramp(t) = R*C*e^(-t/(R*C)) - R*C*u(t) + t*u(t)

[@PunmasterSTP]

@@carultch Thanks for sharing! I mostly reply to comments like the original one because I want to encourage students to take the lead and try to figure out things first. I've tutored off and on for many years, so I think it's kind of from force of habit.

[@carultch]

@@PunmasterSTP No problem. I'm trying to come up with a transfer function that demonstrates the concept of steady state errors on step/ramp/parabolic inputs, and that eliminates its error when you have the proper number of stand-alone s-terms. The function in question: G(s) = 9/(s^2*(s + 2)) In a simple unity negative feedback loop, with the ramp input 1/s^3. This one simplifies nicely, but the exponential terms grow like crazy instead of decay to zero. The ones that don't simplify nicely, will stump Wolfram alpha. I found your video while trying to search for ideas on what I could do.

[@PunmasterSTP]

@@carultch That sounds good. Just to be clear, I’m not associated with Neso Academy. I just like to watch some of their videos.

[@sushrutmdhakate9874]

1. 1/CR x e ^ -t/CR 2. e^-t/CR -1 3. e^ -t/CR -1 -t

[@lokendrasinghlodhi718]

Only first is correct ! plz do the caln again.

[@meralkilicarslan8659]

1.dirac(t)-(1/ح)u(t)e^(-t/ح) 2.u(t)e^((-1/ح)t) 3.-(1/ح)u(t)e^(-t/ح)+حu(t)