Laplace Transform Solution to a Feedback System
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- Опубліковано 7 лют 2025
- 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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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??
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!
This is continuous time. There are no "iterations". And yes, the function Y(s) appears twice in that particular equation, but it's just like having an equation like x^2 + x +4 = 6 ... it's the same "x" in both the places it appears. So, likewise, it's the same Y(s) everywhere it appears in the equation in the video. And for your question about the value of y(t) at t=0, and immediately afterwards, you might benefit from understanding a bit more about the Impulse Function - it's just a theoretical function that helps us to understand and model systems. Nothing in real life jumps instantaneously. See: "What is an Impulse Response?" ua-cam.com/video/WTmelRV_Yyo/v-deo.html and "How to Understand the Delta Impulse Function" ua-cam.com/video/xxGcI9WVoCY/v-deo.html
@@iain_explains Sorry for the late reply. Thank-you for the response it is very much appreciated. The aha moment for me, is that the feedback loop is just a visual representation of a differential equation. In this videos case the diagram is equivalent to the differential equation y'=x-ay in time domain.
You can confirm this by 1. taking the inverse laplace of the transfer function. 2 take differential of that. 3. Write a differential equation relating y' to y and you'll get what I just wrote. And it looks just like the diagram (assuming negative feedback loop).
Hi Prof iain, could you please direct me to where i should start watching regarding this topic. Thank you.
Check out my webpage, where all my videos are listed in categories, in the order I'd recommend. iaincollings.com
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??
Real (ie. causal) systems have a right handed ROC in the s-plane (from the "most positive / least negative" pole to infinity). This corresponds to the values of the "damping function" in the Laplace Transform that will stabilise the system. If that ROC includes the jw axis, then the system didn't need any damping in order to be stable. The jw axis is the "zero damping" case. It means the system is (already) stable.
@@iain_explains ok sir it refers to if a system is stable and its ROC include jw axis than even if we give positive FB to a system than it will be a stable system unlike the intigrator case shown in the video
anyhow thanks for the great video sir it cleared my most of my doubts.
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?
I think you're not quite understanding what the s-plane is. The function you mentioned does not depend on "s" (and actually, you've written exp^(-2), which is not even a function, it's just a number. But I presume you are meaning exp^(-2t) ). Hopefully this video will help: "Laplace Transform Region of Convergence Explained" ua-cam.com/video/SexBL1OlhhU/v-deo.html
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
I think you're confused about the stability of an integrator. It is _not_ a stable system. When the input is sin(t), the output will _not_ be cos(t) (which you claim). If you don't agree, then try doing a simple Google search on the term "bibo stability integrator".
@@iain_explains @Iain Explains Signals, Systems, and Digital Comms ok sir if integrator is not stable then how it is practically implimentable means we can make a intigrator circuit by using opamp and its practically possible
Unstable systems are easy to implement in practice. Think of what happens when a microphone gets too close to an amplifying speaker at a concert, or in a lecture theatre where the lecturer has a wireless microphone and walks in front of a speaker that is amplifying the signal. The sound gets louder and louder and turns into a screech. That is an unstable system.
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 !
What do you mean "according to the control theory"?
@@iain_explainsam taking about control system subject sir, in control systems if we increase the gain of a system by multiplying a gain k than the stability of the system decreases..sorry for going on to control systems that might be little offtopic but they all are related
Sorry, but I'm not sure you're understanding. I have a PhD in Signal Processing and Control, so I know about Control Theory. The point of my question was to highlight to you that you can't simply say "according to control theory such-and-such happens", without actually specifying which exact system you are referring to. Just simply by increasing the gain in a system doesn't necessarily make the system unstable. For example, I can increase the volume on my TV, but it doesn't mean the speaker output becomes unstable. Anyway, what I say in my video is correct, so perhaps you might need to think more about exactly where your confusion lies.
@@iain_explains ok sir I got your point and the example too which you just explained, but as you have said I should particularly mention the exact system which I have actually
" mentioned" in my "first comment" . moreover thanks for answering I will look in to it more and will try to fix it 😊
Can you do a video to explain what is a gaussian codebook please :)
Great suggestion, thanks. I've put it on my "to do" list, with a high priority. It's a question I wondered about too, when I first heard of it.
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.
That's the formula for _negative_ feedback (ie. when the summation has a negative for the feedback path). But in any case, it matches with my formula, if you change the "+" to a "-" in your formula (since I am considered just a standard adder). It really doesn't matter. You can consider negative feedback with a "positive" feedback impulse response (your approach), or positive feedback with a "negative" feedback impulse response, ie. when my "a" value is negative (my approach). If we use your formula, but switch it to a negative in the denominator (to match with my "positive" feedback summation), then you get (1/s)/(1-(a/s)) = 1 / (s-a) , which is that same formula that I got.
hello professor lain, just a small doubt if we give input to integrator a cosine wave (cost-----Laplace tx-------s/s^2+1) 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 !
No, just because one particular input waveform leads to an output that stays bounded, doesn't mean the system is stable. Other input waveforms may result in unbounded outputs (which is the case for the integrator - ie. any waveform with a non-zero constant offset (zero frequency) component).