Lecture 46: Analysis of (p,q) order systems

Welcome back. We are studying the dynamics of  systems in transform domain. In the previous week,   we came across a situation where we try to  understand the step response of a system having   two tanks, and we considered two cases.

In one  case, where these two tanks were non-interacting,   we got a second order transfer function  where the denominator of the system,   where the denominator of the transfer function  rather was a second degree polynomial.   Whereas in the second case when the transfer  function, when the systems were interacting, then   the transfer function had a first order polynomial  in, a first-degree polynomial in the numerator   and second degree polynomial in the denominator.  Now, following the conventional definition,   we identify the order of a system from the  degree of the polynomial of the denominator   of our transfer function.

Here, when you have a first-degree   polynomial in the numerator and second-degree  polynomial in the denominator, the question of the   order of the system becomes a little tricky and  we need to have a little different definition. So,   that particular concept gives rise to the p, q  order systems, which we are going to study today.   So, today's lecture will involve  the analysis of p, q order systems.  

But before we understand the meaning of a  p,q order system and associated dynamics,   let us consider one specific  example from a physical system.   Now, this is a system where you have a  mixing tank and you have two streams coming   in the tank, mixing takes place and then  you have one stream going out of the tank.   So, let us look at the system and various  assumptions which we can make about the system.  

So, we have a tank   which is a mixing tank, so it would be equipped  with a stirrer, and you have two inlet streams.   The first stream on the left-hand side  gives you the feed as A with a volumetric,   corresponding volumetric flow rate of F A. And  similarly, on the right-hand side, you feed B   and the corresponding volumetric flow rate  of B is F B.

There is a certain level of the   liquid which would be maintained in the mixing  tank. And at the outlet you have an arrangement   such that a part of your inlet stream B  can be bypassed. So, this is a bypass.  

So, now, we are mixing A and B. So, for this  particular problem, we would be tracing the   concentration of B. So, let concentration  of B in the tank be C B tank.

This is the   concentration of B inside the mixing tank. We  would assume a constant volume of the system.   So, we would have the notation v which denotes the  constant volume of the liquid inside the tank.  

Now, I have a situation where a fraction of my  inlet stream B goes to the tank and the rest   is passed as a bypass directly to the outlet  of the system. So, let us say r which is a   fraction of what goes, of what is fed as F B  is sent as a bypass and therefore, 1 minus r   would go to the tank . Further, what we  can say is that the concentration of the   species B in the feed is C B F, F  stands for feed.

Is there anything   else which we would need to specify? Yes, what we also need to specify is that   out of the tank, there is a stream which is  coming out and the outlet from the tank is,   has a volumetric flow rate, F out. So, F  out is what is coming out of the tank and   you have a junction, let  us say the junction J here,   the outlet from the tank.

So, the material  which is coming out of the tank after mixing   and the bypass, they meet at the junction  J and therefore, the concentration of the   species B downstream to J would be  different and let me call that as C B out,   let me call that as C B out . So, now, I need to do the analysis of   the dynamics of this system, Before we do that,  do this analysis let us make some assumptions.   So, first assumption that I am making is that F A,  the volumetric flow rate of A is very very large   compared to the volumetric flow rate of B.

F A  is very large compared to F B. First assumption.   And then you see that I have not written any  densities anywhere in my system either on the   figure which is given on the left-hand side or the  figure which I have drawn the right-hand side.  

So, this obviously should give you an idea  that you have a constant density system.   But let, before I do that, let me  say that the density is a rho A   of the stream which is coming from the left  hand side, density is rho B of the stream is   coming from the right hand side. Then you have  rho tank which will be equal to rho tank here.  

And then after mixing takes place in junction you  have a rho out . This is, these are the details.   But then what we can assume is here, that all  of the densities are similar, we cannot say   that the densities are same but the densities  are similar.

So, therefore, rho A is similar to   rho B, which is similar to a rho tank,  which is similar to a rho out. This is   another assumption that I have made. Now, based  on this diagram and these assumptions, I can now   develop my model equations which I would like  to use for the analysis of my system.  

So, let us quickly try to write the model  equations. Very quickly, the diagram is like this.   C B tank, v rho, the densities are constant  everywhere and let us call that as rho, simply.  

So, I have here F A, here F B, C B feed,  a part of that goes as their bypass.   So, this is 1 minus r, this is C B out,  So, if this is my system, I can do one,   overall mass balance. So, I can write overall mass  balance over the tank.

So, overall mass balance   over the tank. What it would give   me? Densities are constant everywhere, so I can  write the time rate of change of v d v by d t   multiplied by rho, density is constant is  equal to F A rho plus 1 minus r rho F B   minus, this was written as F out,   F out times rho.

So, this is the  overall mass balance for the streams   which are coming in and going out. Rho is constant  throughout so it will get cancelled and for   F A very very large compared to F B, I can write  this. And constant V, v is equal to constant.  

These are the assumptions which I have made. I can write this that F A is equal to F out.   So, very small amount of B is entering as the  inlet flow rate of B.

So, therefore, essentially   that outlet from the tank is equal to the  inlet of A. This is what this equation means,   So, equation 1. Now, I can  write the overall mass balance   at the junction J.

So, this is the junction  J. I can write this overall mass balance at   the junction J. And this would be equal to what?

Rho times d v by d t. This is the volume of the   junction and junction has no volume. So, this term  will become equal to 0.

This is equal to F out   times rho plus r times F B rho minus  rho F outlet, F outlet is here, F   outlet. Ultimate flow rate out of this system.  And from here I know that F out is equal to F   A and F A is much larger than F B therefore,  I can write and considering the volume of   the junction as practically zero, I can  write F A is equal to F out is equal to F   outlet So, this is my equation 2 and these two  equations have come from overall mass balance.  

Now, I can do a similar component balance.  So, let me write the component balance.   Component balance over the tank My system is  this, I have a constant volume I have C B tank,   F A,   junction J.

So, what would be the concentration of  B coming out of the tank? This would be simply C   B tank. What would be the flow rate  from the overall mass balance?

Overall   mass balance, I have established that this would  be equal to F A, which is equal to F out.   Now I have F B C B feed. This is r, this is  1 minus r and this is C B out, and this is   F A which is equal to F A outlet.  

This is also established. So, let me make  the component balance over the tank for B.   So, the time rate of change of B in the tank  will be given us for constant volume v d C B tank   by d t, d C B tank by d t   is equal to, from stream A no B is coming, from  stream B, B is coming at a concentration of   C B F but only fraction 1 minus r is going there.

So, therefore, I can write this as 1 minus r F   B. This would be the volumetric flow rate  multiplied by the concentration, so C   B F. This is the amount of B which is coming in  the tank.

And what is leaving out of the tank   would be minus of the concentration  of B in the, at the outlet of the,   coming out of the tank is simply C B tank, and  the flow rate is F A. So, this would be F A   multiplied by C B tank. Fine.

So, what I see from here is that I have a   dynamical equation in which my dynamical variable  is C B tank, and if I have understood the method   of doing a transfer function analysis, that, then  I can rearrange this equation to a suitable form.   So, let me do that rearrangement. I can write  this as v d C B tank by d t plus F A C B tank   is equal to 1 minus r F B C B feed. 

And to have the transfer function,   we would convert it to a standard form and  this would be equal to v upon F A d C B tank   by d t plus C B tank is equal to 1 minus  r multiplied by F B C B feed upon F A.   Now, equation 1 seems to be in  a standard form because I have   d C B tank by d t with some coefficient  plus C B tank with coefficient unity   which is of the form tau d y by d t plus  y, is equal to, now on the right hand   side I have 1 minus r F B C B feed by F A,  which means I have on the right hand side,   k times u, where u is the input function for the  system and F C B upon F A can be identified as a   term u. Can I write anything  more?

Well, in the previous   overall mass balance I wrote the overall mass  balance around the tank and also at the junction.   Let me do the component balance. So,   component balance at the junction,   So, at the junction, you have the inlet which is  coming, which is the outlet from the tank.

So,   therefore, I can write this as F out and F out  is equal to F A. So, F A multiplied by C B tank   plus from the other   side, the fraction of, the stream from F B which  is the bypass stream, is coming. So, therefore,   this would be r times F B.

And concentration  of B there is C B feed. And this must be equal   to the outlet, what is going on at the outlet.  At the outlet F A outlet is equal to F A.

So,   I can write this as F A multiplied by C B out. From where, I can write C B out is equal to   C B tank plus r multiplied by F B C  BF divided by F A. Equation number 2.  

So, let me try to understand what   the two equations 1 and 2 give me. Equation 1  gives me, equation 1 is a dynamical equation,   it gives me how does the concentration  of B inside the tank change as I change   the volume, as I change the flow rates as I  change the feed concentration of B in the stream B   and also the ratio of the fraction which goes  to the tank and one which goes to the bypass.   But now, my ultimate output from the  system is C B out because I am mixing   the two streams at this junction J.

So, therefore,  my ultimate outlet is C B out. And how do I get C   B out? I get C B out from equation number 2 which  is nothing but the concentration which is coming   from the tank plus the effect of the stream  which is coming directly to the junction.  

Now, let me do one thing. Since I have  put the two equations in a standard form,   let me try to do a transfer function analysis and  if I do that, I would prefer my system to be in   deviation variable form. So,  therefore, I say that I have,   I now declare that let us say that C B tank  minus C B tank at steady state is a variable x.  

Then I say that C B out minus C B  out at steady state is a variable   called y. y is the output variable. If you remember, long back, we actually did   this analysis in state space domain and we  said that we can have a dynamical variable   and we can also have the specification of output  variables.

So, in this particular case, it is not   difficult to find out that y is your output  variable. And then on the right-hand side of   equations 1 and 2, I see FB, C B F F by F A.  All of these would act as the input functions.  

So, therefore, let me call F B C BF upon F A  minus the same quantity at steady state F B C B F   upon F A steady state is equal to u. If this is  the case, then I can write equation 1 as v over   F A d x by d t plus x is equal to  1 minus r u, equation number 3,   and y is equal to x plus r times u, equation  number 4. So, now I have two equations in the   deviation variable form, and then let me try to  determine the transfer function of the system.  

So, this is v by F A. So, let me call this as  v by F A d x by d t plus x is equal to 1 minus   r times u, so, equation number 1, and y is  equal to x plus r times u, equation number   2. So, when I have equations of this form what  do I do?

Well, I can take the Laplace transform   on the on both the sides. So, from equation  number 1, I can get a v upon F A s plus 1   of x bar of s is equal to 1 minus r u  bar of s, from where I can simply write   x bar of s divided by u bar of  s is equal to 1 minus r upon   v by F A s plus 1. Is this a transfer function?  

Well, yes, it is a transfer function, but  what kind of transfer function is this,   what information does it give to us? This is x bar   upon u bar. So, this means that this transfer  function gives us the dynamics of the system when   you are tracing the change in the concentration  of B in the tank.

So, you have here C B tank.   So, as you change C B, as you change the variables  of u which involve C B feed F B upon F A, how does   C B tank change? What is the response of C B tank? 

This is what this transfer function gives us.   But then, you also had this here, C B out.  You would like to know, if I change various   quantities in the input function, how does the  ultimate concentration out of the system change.  

That can be obtained by doing a transfer function  analysis from equation number 2. So, I can write y   bar of s is equal to x bar of s plus r times  u bar of s, from where I can write y bar of s   upon u bar of s is equal to x bar of s upon u  bar of s plus r. And from equation number 3,   I can write x bar of s upon u bar of s 1  minus r upon v by F A s plus 1 plus r.  

And this is equal to what? So, y bar of s upon  u bar of s is equal to 1 minus r plus v r upon F   A times s plus r divided by v upon F A s plus 1,  from where I can write y bar of s upon u bar of s   as v r upon F A times s plus 1 divided by v upon  F A s plus 1. And let me call this equation number   4.

And if I look at the transfer function  given in equation number 4, what I see is   that I have y bar, which means the ultimate  output, the outlet concentration from the   system divided by the input to the system. So, therefore, now you can, instead of having   an indirect idea of how does the concentration  change inside the tank and then how does it   affect the outlet, you can have the idea directly  as how does the outlet concentration change as   a function of different quantities in the input.  That is given by transfer function of equation 4.  

But lastly, what we can see is that if I declare  v by F A as a quantity, and I told that this seems   to be the time constants of the system, so, if I  declare it as a constant quantity tau, then in the   numerator, I have some different quantity. And let me call v r by F A   as zeta. If that is the case, then I can write y  bar of s upon u bar of s is equal to zeta s plus 1   upon tau s plus 1, equation number 5.

And let us  compare equation number 5 with equation number   3. So, I have x bar of s upon u bar of s,  and this is equal to 1 minus r divided by   tau s plus 1. This was equation number 3. 

So, now I need to ask this question that,   can I comment on the dynamics of the system? Well, I have two variables which I can identify   in the system. If I am tracing the  dynamical response of my system   for the change in the concentration of B inside  the tank, then from equation 3, it is quite simply   a first order system because 1 minus r can be  identified as the static gain of the system,   tau, the time constant of the system  is v by F A, first order dynamics.  

But what I see from the dynamical response of  the outlet concentration as a function of various   input quantities, then I have the transfer  function as zeta s plus 1 divided by tau s plus 1,   one polynomial of degree one in the  numerator and one polynomial of degree one   in the denominator, it is not a simple first  order transfer function. And therefore,   we declare such transfer function  as a system as 1,1 order system.   What is the meaning of 1,1 order?

1, the first  1 stands for the degree of the denominator   and the second 1 stands for the degree of the  numerator. So, therefore, a p,q order system   is a system which has a transfer function  in which the degree of the denominator   is p, and the degree of the numerator is q, the  order of the numerator, the denominator and the   numerator as p and q must be taken care of. So, now, we understand that we can have situations   where different polynomials of different degrees  can in fact appear in the numerator as well as   the denominator of the transfer function.

What we  would like now to know is the dynamical response   of such systems. In other words, what would be  the effect of adding a term in the denominator,   what would be the effect of adding  a term in the numerator and so on.   We will take this particular concept in  the next lecture.

Till then, goodbye.