Lecture 47: Analysis of (p,q) order systems continued...

So, let us continue our analysis of  dynamics of p,q order systems. We defined,   in the previous lecture, a p,q order system  as a system in which the transfer function   has a polynomial of degree p in the denominator  and a polynomial of degree q in the numerator.   We have come across several examples in the past  few weeks, in which we had polynomial of degree   1 in the numerator, degree 2, polynomial of degree  1 in the denominator, degree 2 in the denominator,   2 in the denominator, 1 in the numerator, 1 in  the numerator, 1 in the denominator and so on.  

So, what we will do today is we would  like to understand that if I add one,   if I add a particular term in the  numerator, what happens to the dynamics,   which means that does the dynamic become faster  or slower if I add something in the numerator,   what happens to the dynamics and so on. So, let us take the examples one by one.   So, let us first take the case of a simple first  order dynamics.

So, my transfer function is g   1 of s. I am emphasizing this, the fact that it is  a first order system by putting 1 as a subscript,   is equal to K upon tau s plus 1.   And we saw that the system follows the  dynamics given as y of t is equal to for a,   as a response to a step function of a magnitude A,  A K times 1 minus e to the power minus t by tau.  

So, let us plot this. We have in fact studied, but  in order to compare this against the other ones,   let us plot this and see how  does the solution look like.   So, f of x is equal to, we will assume A  and K as unity, we know the importance of   those quantities.

So, 1 minus e to the power minus  x upon, we will denote the time constant tau as c.   We would be interested only in the first quadrant  because we know that the disturbance, the time at   which the disturbance happens can be set as 0. So, this is what happens to the dynamics,   and the effect of various quantities, in fact,  c in this particular case is known to us.  

I can animate this and appreciate that as time  constant increases, the response becomes slower,   not very difficult to see here. So, let me  set the time constant at a value say 0. 5.  

This is my dynamical response. Now, what I can do  is I can either add a term in the denominator or   I can add a term in the numerator. So, these  are the two possibilities which exist.  

So, let me do one thing, let  me start with this scheme   that I have a first order transfer function g 1  of s and this is given as K upon tau s plus 1.   Now, what I do is I add one more factor  in the denominator. So, I go here   and I get a second order system, a pure second  order system g 2 s and this would be K 1 K 2 upon   tau s plus tau s plus 1, tau 1 s plus  1 tau 2 s plus 1.

We have come across   this situation before, and we know that this  is a dynamical responsible non-interacting,   two non-interacting systems. So, when I have   such a transfer function it is given  as g 2 s is equal to K 1 K 2 upon   tau 1 s plus 1 tau 2 s plus 1. Then, the  dynamical response y of t is given as what?  

Let us see what we got previously. We  have A K 1 K 2 multiplied by 1 minus   tau 1 upon tau 1 minus tau 2 e  to the power minus t upon tau 1   minus tau 2 upon tau 2 minus tau  1 e to the power minus t by tau 2.   Let us see what happens to the dynamical  response by plotting this function.  

So, let me call this as c 1. So, c would  go over here and this would become c 1   from 0 to 5. I will set it at 0.

5. Now, we have,  we are back to the first order response. And now,   I will put the second order response.

The second  order response will be given as g of x is equal to   A K 1 K 2, all have been set  to unity, 1 minus tau 1 minus   tau 2. So, tau 1 upon tau 1 minus  tau 2. c.

So, c 1 upon c 1 minus c 2   upon c 1 minus c 2 e to the power minus x  by c 1, c 1 minus c 2 divided by 2 c 2 minus   c 1 e to the power minus x upon c 2. Let us see what has happened, let me   change this also from 0 to 1, 0 to 5, in fact,  and as we saw in the previous few lectures,   as I add more and more and more number  of interacting systems downstream,   the overall response of the system  becomes sluggish. Let us confirm   from here as well, let me animate these two.

The red curve on the top is the first   order response, the black curve at the bottom  is the second order response. And I will try all   possible combinations of c 1 and c 2 and see if  there is any way the black curve goes on top of   the red curve, which means the response of second  order system, pure second order system becomes   faster than the response of the pure first order  system. Let us see.

Let us try to animate this.   Now, the red one is on the top, we will wait for  various parameters. And the red one is still on   the top.

And does not matter how long you do  this, you will always find that the best that   can happen is that the two curves can come very  close depending upon one of, when, it will be the   situation where one of the time constants tends  to 0. So, the two solutions actually coincide. But   otherwise, what you see is that the response of  a first order, pure first order system is always   faster than the response of a  pure second order system.  

Now, so, what has happened here, and why has  it happened? Let us go back to our analysis and   see what we have done. So, in a general transfer  function, you have a polynomial in the numerator,   you have a polynomial in the denominator, you  can always factorize the numerator and the   denominator and therefore, we would give a name to  the solutions of this, these polynomials which are   in the numerators and the denominators.

In other words, we would like to   know the zeros of the numerator and the  denominator. Except that the zeros of the   numerator are called zeros and the zeros of  the denominator are called poles. So, I have   the zeros of the numerator as zeros, we will  call them as zeros.

And zeros of the denominator   will be called the poles. So, what has happened  as I have g 1 from g 1 to g 2, g 2, I have   added a pole. I have added a pole. 

What is the effect of adding a pole?   Now, I can observe this and write  an inference that addition of a pole   makes the response   slower, always slower. You keep on adding a  pole and the response will keep on becoming   slower and slower and slower.

What about  the effect of adding a 0? So, now, I have   added 0. So, I will have now g 1 1 s.

So,  I am emphasizing that I have one polynomial   of degree 1 in the numerator, the other one  in the denominator, and this would be what,   K times zeta s plus 1 upon tau s plus 1.   So, you have minus 1 over theta as 0 and  minus 1 over tau as the pole of your system.   I had only pole minus 1 over tau in my pure first  order system, I made a 1,1 order system, and now I   have the transfer function given like this.

So,  when I have g 1,1 as K zeta s plus 1 upon tau   s plus 1 as the transfer function, I have my  response given as, how would you determine   the response for a step order system, for  a step input g 1,1 which is K zeta s plus 1   upon tau s plus 1 multiplied by A upon s, the  transfer function and the Laplace transform   for the step input of magnitude A. You will do a  partial fraction, invert the Laplace transform.   When you do that the answer that you  would get is what I am writing here, A K   multiplied by 1 minus 1 minus zeta upon  tau e to the power minus t upon tau.  

This is the response that you would  get. Let us try to plot the response and   try to understand the dynamics. So, let  me close this and let me do one thing.  

Let me try the function here, g of x  is equal to what, A K unity 1 minus   zeta upon tau. So, let me call zeta as d  upon c 1 e to the power minus x by c 1.   So, let us look at this, what has happened,  I have added one 0 to otherwise pure   first order system.

To make the system look a  little less cluttered let me clear the text here.   Now, d also can be set as positive so, from  between 0 and 5. Now, see you had given the system   a step input in this particular case of magnitude  1.

So, what is your ultimate response of the   system is time t tends to infinity, you are going  to unity, not a problem there. And, for a first   order system you asymptotically reach the above. What is interesting to see here is that now you   have some, you have something very interesting  that you, your response is going farther than 1.  

So, let me do one thing, let me animate the  effect of d and let us see what is going on.   You are going up, and then you also see  that you are going down. So, one case   where you are going beyond the applied  input, the current case, you can see here.  

And then I will do this and what  you see here is that you, in fact,   again, asymptotically reach there, you do not go  beyond the applied input and the response looks   somewhat like a first order response. So, if I just close this and if you see the nature   of the plot, you will see that what has happened  is this looks very similar to a first order   response, except that now you have a, you have an  intercept here. So, you have a y-intercept here.  

You will never get a y-intercept in case of  first order response. So, let me animate this.   So let me get rid of this.

Let us  just see the first order response.   You never get a y-intercept. You always  start with 0, and reach asymptotically   the applied input.

There is no  intercept at any point of time.   But what actually happens in case of a 1,1 system  is that you have an intercept. See here, you have   an intercept.

And the magnitude of the intercept  depends upon the ratio d by c 1, the ratio zeta   it depends upon the ratio zeta upon tau. The  larger the zeta by tau, the larger is the   intercept. What is the physical significance  of the intercept?

Right in the beginning when   you give the system a disturbance, the system is  so enthusiastic that it starts responding. So,   what is the problem with the first order response?  Let us look at just the first order response.  

If I look at just the first order response,  let me get rid of g of x here, this is,   the red which you see is our first order response.  So, let me make a small time constant here.   And what you see is a first order response which  looks like this that this is your applied input   and this is your response.

The response  tries to catch the applied input.   But there is always a difference unless you reach  t is equal to infinity. The response is below   the applied input and therefore, I can mark this  area, I can mark this area and this area is called   reluctance area.  

Why is it called reluctance area? Because  this area will indicate whether the system is   instantaneously adjusting to the input or  is reluctant to go and settle to the new,   to the input or to the new steady  state. The larger the reluctant area,   the slower is the response.

So, which  means that the system is reluctant to   adjust to the new steady state.  And when would that happen? That   would happen when your time is large.

So, let me increase the time constant.   Now, the reluctance area has increased now.   The reluctance area is this now.

Larger time  constant, this is what is going to happen to   the reluctance area. But then, when you provide  a system with a 0, then the reluctance reduces,   the system becomes enthusiastic. Let us see that.

So, now I   have shown the response of a 1,1 order system  which means you have added a 0. And when you   add a 0 and you increase the magnitude of  the constant which exists in the numerator,   then the system becomes more and more and  more enthusiastic to an extent that you   can in fact have the response which is  larger than the enforced input. But the   system then realizes that I have done  something more than what was expected   to do, and therefore, then the system comes  back asymptotically to the new steady state.  

So, therefore, such systems so, systems where  you have g of s is equal to K zeta s plus 1   tau s plus 1 are called lead-lag systems. Why?  Lag can be understood pretty simply because   the first order system, will result in a lag  of the response.

But since you have added a 0,   the system also has a lead. So, therefore,  tau would be called lag constant,   and zeta would be called the lead constant.   And therefore, the ratio zeta by tau will  determine whether the initial intercept would be   greater than the magnitude of the  applied step input or lesser than that.  

So, when this is greater than 1, when your,  this is greater than 1, what you will observe   is that if this is the applied input, this is y,  this is t then, you will go up and come down.   And for zeta less than tau, zeta by tau  less than 1, what you will observe is   you will have y, you will have t, which is your  applied input, then you will asymptotically go   like this. So, this ratio lead constant to lag  constant will give you this kind of behaviour.  

And now, the question is, why has this happened?  What have you changed? You have added a 0.

So,   addition of a pole slows down the response,  addition of a 0 fastens the response,   it makes the system enthusiastic. So therefore, now I can look into   this system. I can have g 2,1 s.

So, I am  looking at p,q order system, p is is 2 here,   q is 1 here. So, if this is K times zeta s  plus 1 upon tau 1 s plus 1, tau 2 s plus 1.   Well, this is the transfer function of  two interacting systems, if you saw.  

We did not plot the response, we will do it now.  What is the response y of t? The response y of t   is given as this, let me write the response,  we will follow the exact same procedure,   multiply it by A by s as the Laplace transform  of the step input, the partial fraction,   and then ultimately do the inverse Laplace.

So, A K multiplied by 1 minus tau 1 s, tau 1   minus zeta upon tau 1 minus tau 2 e to the power  minus t by tau 1 minus tau 2 minus zeta upon tau 2   minus tau 1 e to the power minus t by tau 2. This  is the response function. And let us plot this.  

So, let me do this. So, let me change this  particular quantity here by a simple, so it is, if   you put zeta is equal to 0, you will get basically  a second order system. You have just added one   0 to the system.

So, g of x now is c 1  minus d. And this will become c 2 minus d.   And what I will do is, let me get rid of  this function and let me plot this function.  

This is the response.   Now, what are the some characteristic features  which we can see? Again, I have c 1, c 2 and d,   I can adjust them, I will adjust c 1, you get  a behaviour, I adjust c 2, you get a behaviour,   and then I can adjust d, let us see  what happens to d.

When I make d as   0, when I have made d as 0, you get a pure second  order response, same as what you got previously.   Now, as I start increasing the value of d, which  means I start increasing the value of the lead   constant, what happens is, see, as I am increasing  this, the black curve is starting to come up.   In the previous case, second order was a first  order and the black curve would never go up,   the red curve would be on the top, the  black curve would be on the bottom.

Now,   I have a situation where you can have the  black curve on the top of the red curve.   First, that by adding a 0, you have overcome  the issue of sluggishness of the second order   system. Not only that, since you have added a  0 for your 1,1 system, you in fact, could go   above the enforced step.

This is what you  can see here. Let me make it even larger.   The system become, has become very enthusiastic,  it has gone even larger than, even higher than   the enforced input.

But then it realizes,  that it has probably done more than what was   expected, and it then comes down as it  settles down to the new steady state.   But since you have a combination  of zeta tau 1 and tau 2,   in none of these cases, let me animate all of  this, in none of these cases you will see that   your response has an intercept. So, you can have a  case where you have the response which conforms to   the second order response, where your lead time  constant is very poor, very small in magnitude.  

So, essentially you have the first order  response on top of the second order response.   When you start increasing the lead time constant,  the system may become enthusiastic it may   overshoot but it would eventually come down. So,  therefore, the response becomes faster than the   first order response, but in other cases, you will  start with 0 and you will settle asymptotically to   the enforced input.

So, this is what we saw  that there are very interesting features of   the response of the system which is, in which we  have identified the system as p,q order system.   By simply looking at the dynamical equation,  you may not be in a position to know   such exotic responses which the system  offers. So, it is important that you   look at the transfer function.

And  by looking at the transfer function,   you may expect certain response or certain  qualitative behaviour of the system.   The systems which we considered till now  were single input single output systems, but,   as I made a mentioned previously, in reality, you  may come across, in majority of the occasions,   the systems which are multiple input and  multiple output. So, what we will do is   in the next lecture, we will take the case where  you have multiple input multiple output system,   and how can we do a transfer function analysis  of such systems.

Till then, goodbye.