Laplace Transform Solution to a Feedback System

Explains how to find the Impulse Response of a feedback system using Laplace Transforms. Gives an example with an integrator in the forward path.

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[stanleyche]

hey, professor first let me thank you for this wonderful intutive video. Now my question is that an integrator as a system is a marginally stable system as its impulse response is bounded but not absolutely integrable so what is the difference between a marginally stable system and a stable system in terms of the damping? A stable system does not require external damping in order to be stable what about a marginally stable system and if it requires external damping than how is it different from the unstable system??

[iainfraser]

Hi Professor Iain,

Thank-you for your great informative videos. Very much appreciated.

I've searched the web far and wide but cannot find an answer to this question. A loop must start somewhere and presumably iterate infinitely from there. Using the example in this video, say the input is a impulse function. At the t=0 the x(t) will be infinite and y(t) must be 0. Is that correct? And then after the first iteration, the integrator will have output a so y(t)=a and x(t)=0 so the error signal will be a. Thus the second integration will have y(t)=2a and x(t)=0.... so on and so forth.

In the derivation of the feedback loop formula, Y(s) appears twice in the equation and is treated as if they are the same value. Am I correct in thinking that in the time domain this would not be possible since the y(t) used for input would be from the previous iteration and the output y(t) would be for current iteration and therefore are not the same. So my understanding is that in the s-domain the Y(s) terms refer to the same function hence why you can derive a formula for the transfer function by solving for Y(s)/X(s).

However, the inverse laplace of impulse multiplied by 1/(s-a) is e^at. If my understanding of initial conditions are correct shouldn't the output signal y(t=0)=0 and immediately after that should be one? But e^at at t=0 is 1. So it's as if the output starts at the exact same time as the input. That doesn't make sense, surely it has to iterate around the loop at least once before the output changes?

Thank-you very much in advance!

[hesokaheso]

Hi Prof iain, could you please direct me to where i should start watching regarding this topic. Thank you.

[rishabhkumar]

sir according to bibo stability integrator should be stable system
suppose we give input Sint (which is bounded) then we will get cost(which is also bounded) this it is a BIBO stable system so why it's not stable in terms of ROC please correct me where I am wrong
thankyou

[ranchordaschancad]

my doubt is that in video on "ROC for Laplace transform" you mentioned that right side of jw axis corresponds to the damping of signal and thus we can find Fourier transform by converting a signal in to finite energy.
but here in this video you are considering the right side of jw axis as positive FB aren't the above two concepts contradictory??

[aarzoosingh]

hello sir hope you doing fine
as you have mentioned that the pole should be -a suppose a=1 thus pole for a system will be at -1
sir my question is that if we apply input exp^(-2) to the above system then why will not get converged output though we are applying an input with decaying exponential means sir i know its mathematically defined but is there any intution behind it?

[stringstoparadise]

hello, sir suppose if we have an open-loop system (1/s an integrator which you have mentioned in the video) if we multiply it with gain k and give it negative feedback than the poles will move towards the negative side and the system will become stable, and the transfer function will look like (k/s+k) and accordingly higher will be the value of the K higher will be the stability as the poles will be more -ve (farther away from Jw axis).
everything is ok but according to the control theory as we increase the gain the system becomes unstable, but in the above case with the increase in gain in the presence of -ve FB the system is getting more stable so sir why is this contradiction in the concept coming please correct me if I am wrong somewhere.
thank you !

[tuongnguyen]

Can you do a video to explain what is a gaussian codebook please :)

[jamesbra]

I thought feedback meant that the transfer function was H1(s)/(1+H1(s)H2(s) if you add their frequency responses then they'd be in parallel.

[stephenflatham]

hello professor lain, just a small doubt if we give input to integrator a cosine wave to intigrator(laplace = 1/s) than o/p would be (1/s)*(s/s^2+1) which will finally give (1/s^2+1) and taking the inverse Laplace will get sint, which conclude that a bounded input cost pass through intigrator gives sint means its stable system. how can this be justified (i have referred a question in research gate where people are claiming it to be a stable system ), please reply i am badly confused !