Designing a Lag Compensator with Root Locus 

And Nise!

Jacob Hiller oh yes definitely!

Yesss, I have noticed the same problem.. The step response is the convolution of the unit step with the impulse response.. If impulse response remains the same, how come the step response be so different?

​@@sahhaf1234 notice the magnitude of the time axis. the long time period magnifys the small difference in the impulse response.

I think while finding the dominant poles, the criterion should not be just that whichever pole is most near to the origin, dominates. One should also check whether there are zeros very near to it, as they can cancel out the effect of that pole.

Ah...new zero and pole have canceled out each other mostly with some effects left which slow down the dynamic response. Is that what you meant? It kind of makes sense. Thank you!

Peng Bai Ya right.

Brian Douglas mentioned some things about how the location of the poles affect the system's response to an input. I forget where it was, but try among these videos: ru-vid.com?search_query=brian+douglas+drawing+root+locus As for how frequency, time and s-planes are related, look up the Laplace transform. If memory serves, that is used to go from the time domain to the s-domain (since it has a factor of exp(st) in it or something). The relationship between the s-plane and frequency is by setting s = jw where j is the imaginary unit and w is omega, the frequency in radians per second. Not a perfect explanation but hopefully it can be a start.

Good question and I think the response doesn't change because of the fact that theta(p) - theta(z) = 0 Secondly, the pole zero combine do not become a dominant pole, instead they help you to maintain your root locus that is why the impulse response didn't change because the R-locus remains the same

take a count that lag compensator practically doesn't has impact over the open loop gain, because theta(p) - theta(z) = 0 and its gain on the compensator its approximately 1, thats why doesn't change the root locus, thus they are not dominant.

Late response, but I didn't find the 2 replies to your comment very satisfying. Because after all, we *are* adding a new closed loop pole that is closest to the imaginary axis (hence, a dominant pole). In fact, this is why the settling time for the step response is so much longer now, we've got a new closed loop pole with a low magnitude real part which is causing the system to take its time! But then why is the impulse response settling time so close to the lead only controller? The contribution of the lag compensator actually does have a long settling time in the impulse response, but the magnitude contribution is so small that it doesn't even look like it's there! Notice how the lag/lead impulse response is slightly larger than the lead compensator. This discrepancy gets smaller over time, ie it's that long time constant we've added slowly dying off. For a more concrete view of this, try plotting the impulse response of a few versions of (s+z)/(s+p) while maintaining the z/p ratio. For small values of z&p, you'll find very low magnitude impulse responses with large settling times. For larger values of z&p, you'll find higher magnitude responses with shorter settling times! So, by adding a small z&p pair as in our lag/lead compensator, we'd expect the contribution to the impulse response to be small, but long lasting (as long as we haven't moved around the other closed loop poles).

@@thevirtualsphere Also a late thanks to you!

Yeah I have the exact same doubt. does'nt the dominant poles change because of adding the lag compensator.

Hi Jason, since you're placing your lag pole close to the lag zero they will 'cancel out' and you will still maintain the dominant poles as they were before introducing the lag compensator. Hope that helps

I have the same quesiton.. I think they dont cancel out because theyre not exactly on top of each other.. i cant fully understand this

+Uday Santosh Raju V I also have this question ! If you find the answer through other resources in future than please don't forget to post it here, too. thanks in advance !

+Milan Shah mmm... Yes, technically by putting the lag pole closer to the imaginary axis than the original dominant poles, you have made of it the dominant pole. When he says that the dominant pole doesn't move too much, he is talking about the original dominant poles. Using Matlab you can find that the position of the poles for the closed loop transfer function for the uncompensated system(for the Open Loop TF in the minute 6 of the video) are in p1 = -0.6111+ 0.42673i and p2 = -0.6111 - 0.42673i. The position of the the poles for the compensated are in p1 = -0.6120 + 0.4278i , p2 = -0.6120 - 0.4278i and p3=-0.00549. As you can see, the position of p1 and p2 have changed very little. Plotting also the root locus in matlab you will see that there is practically no difference. The only one that you have is that now in the compensated system you have one additional locus branch in the real axis that goes from the lag pole to the lag zero. Hope it helps!

+Uday Santhosh Raju V I think there is also this little trick in control theory where if you have a pole zero pair which is very close (called a dipole pair I think), they basically "cancel out" like they would if there were in the same position (so they negligible effect on transient response) so we can approximate our system to be a second order system. The exact math though, would be kind of complex.

+Prabhpreet Dua Well, in this case, if you take the step response of the compensated and the uncompensated closed loop in Matlab(for the Open Loop TF in the minute 6 of the video) , you will see that the transient response of both systems is different. The settling time is way different(around 10sec for the uncompensated, 800s for the compensated(!!)), the rise time is also very different. You can see that also in the graph that he shows in the minute 10:25, where both the rise and the settling time are way higher in the compensated system. As he says, the problem didn't have any time settling time requirement so that increases didn't matter that much. The objective was to decrease the steady state error and the Lag Compensator did its job. The Lag compensator moves the root locus of the uncompensated system to the right(in the opposite direction to the Lead Compensator). The purpose for wanting the original root locus from not moving very much from its place is in order to not lose stability. I really don't think that you approximate the system to a second order equation!

Based on your next video, it looks like that's exactly what's going on. The lag compensator reduces the gain at higher frequencies which offsets the effects of including your additional DC gain that addresses steady-state error.

He has mentioned to place it near the imaginary axis but you cannot place it very near as the practical values for components that make system become impossible to achieve. So it can be taken as thumb rule to place the zero at 1/50 of the dominant pole real part i.e. real of (-3+2i)/50

-1 is uncompensated close loop pole -3 is designed dominant pole location we want to make locus like P(D)=-3 not -1 I hope that this is some help for u

mhah lol

+Don-Roberts Emenonye as you are thinking -1 and -3 are no more our poles. In the last part of this video we already designed lead compensator to set poles to -3+2i and -3-2i

That can be explained from the definition of root locus and basics. root locus is the location of closed loop poles as k value varying from 0 to infinity. -1, -3 are the closed loop pole locations for k = 0 , we need our dominant close loop poles at -3+2j and -3-3j, we should adjust this k value such that we get our desired closed loop poles. i.e., for some k, closed loop poles will be -3+2j and -3-3j not at -1,-3.

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-S4DhKn1fRAI.html or ru-vid.com/group/PLUMWjy5jgHK3-ca6GP6PL0AgcNGHqn33f