foreign of systems exhibiting inverse response ok so let us look at one specific form of transfer function imagine that you have a process whose transfer function is given as G of s is equal to K 1 upon Tau 1 s plus 1 minus K two upon Tau two S Plus 1. do the individual transfer functions looks ah familiar well ah they do because K 1 upon Tau 1 s plus one is simply that the transfer function of a first order process with static gain K1 and time constant Tau 1 and now you have minus it is important you have a negative sign ah and again the transfer function for first order system K two divided by Tau two s plus one so ah K two as the static gain and Tau twice the time constant now if this is the case first of all let us see what is going on I have two individual transfer functions and therefore if I want to draw an input output type block diagram then what I can do is that I can have an input to the system which is given as U Bar of s b as we saw in the previous lecture I have added to with the negative sign so I have added negative of transfer function so we will take care of how to do that negative operations but I will have two streams here one will be given as K one upon Tau 1 s plus one the other one will be given as K two upon Tau two S Plus One I must have an Adder I will get an output Y Bar of s and then what I see is that when I get this adder the contribution from the top one k one Tau One S Plus K 1 upon Tau one Tau One S plus one is positive but the contribution from the negative from the bottom one is negative so I put a negative sign here ok so this is the input output block diagram for the system the overall transfer function Can Be Imagined to be Within this imaginary block which is shown here now if the transfer function looks something like this we would like to know the dynamical response of such a system so let us consider the case of a step input so step input of magnitude a so what will happen I can write Y Bar of s as a K 1 upon s Tau 1 s plus 1 minus a k 2 upon s Tau to S Plus One so this will be the Laplace transform of the output ah variable and when I invert I can I know all the terms already so its not very difficult for me to write y of T So y of T would be equal to a K 1 minus e to the power minus T upon Tau one minus a k 2 1 minus e to the power minus t upon Tau two ok and now I would like to know the dynamical behavior of the system let me ah see that in this I have four parameters I have K 1 I have K two I have Tau 1 I have Tau two I can set a as arbitrarily some value and I know the effect of K in general because since I have now K one and K two ah I would worry about K 1 and K two both but a I will set us one for my analysis so I would worry about K one k two Tau one and Tau two let us see the dynamical response of the system ah by plotting it on this plot I have F of X is equal to K one minus e to the power minus X upon c one let us call Tau one as c 1 minus K two 1 minus e to the power minus X upon C two ok so let us see the dynamical response clear this first to understand the dynamical response now we must realize that I have set a as Unity right so first of all when I make ah K one zero and K two also has zero I have simply the first term k 1 e to the power one minus T by Tau one first order response so is this the first order response to a step input function of magnitude Unity well it does perfectly see here start from here asymptotically reach one this is one fine now let me do one thing let me change K two from one to point five ok and have the same time constant Unity ok so time constant Tower and time constant Tau two ah both Unity but K1 is so my K 1 is 1 and K 2 is point five right when I have only the first order response I know that my ultimate value of y of T would be Unity what do I see here My ultimate value of y of T here is 0. 5 now this is strange because in every single case which we saw till now as T tends to Infinity you always got a response which would conform to the magnitude of your step input here what is going to happen is you are not reaching the applied input at all you are stopping somewhere in between but what that in between is let me see ah let me make K two s point two five ah point three for example I reach here point seven let me make this point nine I reach here at point nine so looks like looks like what is going on is that I have y of t at limit T tends to Infinity is K 1 minus K two ok this is going on I can adjust various values and you will see that this is always confirmed ah this is always confirmed in fact I can make ah K two greater than K one and you can see that this K 1 minus k 2 is always satisfied so multimeter response so the first point which I see specifically in this particular ah case is that as T tends to Infinity I do not reach my applied input it goes down and by how much K 1 minus K two now can I analytically get this solution turns out that it is possible because I have the solution as y of t is equal to a K 1 1 minus e to the power minus t by Tau one minus a K two one minus e to the power minus t by Tau two right so this is an exponential term decaying this is an exponential term decaying so therefore limit tends to Infinity y of t would be what the exponential terms will become zero so I have a times K 1 minus K two right and in our case a was equal to one right so therefore therefore what is going to happen my limit tends to Infinity y of T is equal to a k 1 minus a K two as simple as that if you do not have the other part of the transfer function you would have only this much a k one so this means what this means that if I have a transfer function which looks like this G of s is equal to e ah sorry K one K one upon Tau 1 s plus 1 minus K two upon Tau two S Plus One then if k 2 Tau to S Plus 1 is not there you would have reached the full response but it is opposing your response to reach the ultimate fate right so therefore therefore this is called the main mode of your system and this is called the opposition mode of your system ok so the first transfer function is called the main mode of your system and the second transfer function is called the opposition mode of your system ok lets see if we can see something more let us go back to our analysis let me set K 1 as 1 let me set k 2 as 0.
5 so that I know that ultimate response is going to be 0. 5 that is fine now let me tweak around the value of C two I will also keep one constant and let me change the value of Tau 2 right I know that ah Tau 1 and Tau two are positive so let me make it between 0 and 0 and say five ok now I have the value of k 2 as 0. 5 and now let me change the ah value of c one see what is going on I am keeping on increasing the value of C two sorry not c one C two that I am constant Tau two My ultimate value remains 0.
5 the only thing which is happening is that there occurs a change in the speed of response OK there occurs change in the speed of response you see here but then a funny thing happens here let me make it small it is important that you see what is going on here at this stage what happens is so let me plot it you have the response y of t you have the time t what is your ultimate response see this is tending to tending two point five which is expected because K one is one and K two is point five quite expected but what's weird here what is interesting in other words interesting here what you see here is that your enforced input was positive unit step input the enforced input was positive and although you and you put an ah unit step input your response was negative at least initially at least initially now what happens to your fate of the system as time T tends to Infinity well does not matter what was the initial dynamics of your system ultimately you reached 0. 5 which is which was anyway your analytical solution that is not a problem the problem is that initial Dynamics showed an inversion in your response let me emphasize why am I seeing an inversion in the response because let me plot on this particular plot all the type of responses which you have seen till today ok so this is the enforced input a is equal to one what was your first order response let us say this is your first order response if this is your first order response what would happen to your second order response this will become your second order response if this is your second order response what will happen to your third order response this will become your third order response then if you add zeros to your system what is going to happen probably this is going to be a response but if you have a lead lag system what will be possible response this can as well be a possible response so I have in the single plot made everything and then what I can observe is that as T tends to Infinity all of them tend to come here all of them tend to reach here first thing which has happened in this particular cases that you do not reach here you reach here and that is simply a k 1 minus K two but what is even more interesting to see here is that in all previous cases you always had right from the beginning the response of the system which went up ok but now for the first time you have the response which is going down the response which is going down now the system realizes that it is going in a wrong direction and therefore it inverts and then comes back again to catch up to this value that is well agreed thats not an issue but you still need to see one thing that you have and inverse response and why does that happen or rather when does that happen ok so let me look at the condition when it happens I have maintained constant K 1 I have maintained constant c 1 I have maintained constant K two and now I am tweaking around the value of C two which means I am changing the time constant C2 and when so as long as time constant Tau two is greater than k 2 you always have positive response C here the inverse response happens only when your Tau two becomes smaller than K2 ok but is that the the generic reason behind that or is that the generic condition let us let us try try to derive the condition ok so I have in my system this why of t is equal to what e K 1 1 minus e to the power minus T upon Tau one minus a K two one minus e to the power minus t by Tau two let me do one thing let me compare it against so compare it against y one of t which has only the first order Dynamics so it would be a K 1 minus e to the power minus t by Tau ok what I would like to know is the value of d y 1 by DT at T is equal to zero what is this e going to be equal to this is going to be equal to a K upon Tau that's it a k upon Tau at T is equal to zero and if a is greater than zero which means you have given a positive step then d y 1 by d t at T is equal to zero for a positive gain will always be greater than zero which means that if I plot Y versus T then here you will always have a positive slope always positive and this is your a so this is your direction of a positive and this is the direction of your response positive in every case now what happens here y of T is equal to a k 1 1 C to the power T by Tau one one minus a k two e to the power minus Tau two d by by d t is equal to a K one by Tau one minus K two by Tau two from here you can get the condition for inverse response right so now this d y by d t and this must be at T is equal to 0 right the response right at the time when you introduce the disturbance when d y by T is equal to divided by DT at time T is equal to zero is a k 1 by Tau 1 minus K two by Tau two so for this to be greater than zero the usual response K 1 upon Tau 1 would be greater than K two by Tau two and if this is less than 0 what would happen well the derivative will become negative the gradient would become negative that is the condition for inverse response k 2 by Tau two would become greater than K one by Tau one and this is the condition for inverse response now the question is that what does the inverse response Behavior depend upon it does not simply depend upon K one it simply does not depend upon K2 it simply 300. 1 Tau 1 and Tau two it depends upon this inequality where k 2 by Tau two as whenever k 2 by Tau two becomes greater than K 1 by Tau one you get an inverse response which means that before trying to catching up to the applied difference in even a k 1 minus a k two the system will go in the or respond in the opposite direction now you can do one more analysis under what condition do you get this inverse response and what kind of systems do you get this so let us look at the transfer function G of s is equal to K 1 upon Tau 1 s plus 1 minus K two upon Tau two S Plus One so I can simplify this as K one Tau two S Plus 1 minus K two Tau One S Plus One divided by Tau One S Plus One Tau two S Plus One which can be further simplified as K one Tau one minus K two sorry K one Tau two minus K two Tau one s Plus K 1 minus K two divided by Tau 1 s plus 1 Tau two S Plus One ok or this can further be simplified as K one minus K two K one Tau two minus K two Tau one divided by K 1 minus K two s plus 1 divided by Tau 1 s plus 1 Tau two S Plus 1.
right so we have already seen that the effective gain of the system is K 1 minus K two right so therefore I can write this as what K effective which is K 1 minus K two times what if I say that K one Tau two minus K two Tau 1 divided by key one minus K two as a quantity called Zeta then I can write this as Zeta S Plus 1 divided by Tau 1 S Plus One multiplied by Tau two s plus one ok so this is going to be a transfer function of your system now for an inverse response I know that K two by Tau two should be greater than K 1 by Tau 1.
