Lecture 58: Response of discrete-time systems continued...

So let us have some more observations on the  dynamical response of discrete time systems.   So let me have some comments before we close  stop it. So what we have in front of us is   some set of observations which are of special  importance when you consider the pulse transfer   function.

So the first thing is that a for an N  discrete systems in series. Till now we considered   a single system. So in the previous lecture,  we took the pulse transfer function of system,   which was first order and it was equipped with  a 0 order hold.

We determines the first transfer   function multiplied it with the z transform  of the input did the inverse z transform.   Now if you have N systems in series, what would  you do? Well, let us see what you would do   for a continuous domain system, you have, imagine   system of two sub systems in series.

So you have  an input u, it goes to the first system which has   the transfer function g 1 of s, it gives an  output y 1, output y 1 access the feed to the   second system which has the transfer function  g 2 of s, you get the ultimate output y 2.   So what are the different relationships which  we can write? I can write g 1 of s is equal to   y 1 bar of s divided by u bar of s and I have g 2  s is equal to y 2 bar of s divided by u bar of s   and therefore, from sorry y 1 bar of s.

So let me write this properly, this would be   y 2 bar of s divided by y 1 bar of s. From where  I can write g 1 s, g 2 s, is equal to y 1 bar of s   divided by u bar of s multiplied by  y 2 bar of s divided by y 1 bar of s,   which is equal to what? y 2 bar  of s divided by y u bar of s.  

So therefore, I can write g  overall simply g of the process   s, as g 1 of s, g 2 of s, and in fact, we did  make use of this fact in the previous lecture   by writing g as a product of g p and  g h, the process transfer function and   pole transfer function, is the same concept here. So therefore, this is a general observation that g   overall of s for an N system  would be g 1 of s g 2 of s,   up to g 1 of s. Well, now I can  do this I can make an analogous   case on the right hand side for a discrete time  system, you have u which is discrete now is fed to   the system, which has a transfer function g 1 of  z which acts as an output the discrete output y 1   fed to that is to second transfer  function first transfer function in fact,   g 2 z gives you the discrete output y 2.  

So I can try exact same steps here,  because if you remember the block diagram,   which we wrote in the previous case, our  expression was that y 1 of z in fact y of z   is equal to g of z multiplied by u  of z. So I can do this for all the   individual transfer functions from where  it is not very difficult to see that g   overall z would be equal to what? g 1 z g 2 z  up to g capital N Z.

For N systems in series,   your transfer function is simply the  multiplication of individual transfer functions,   this is applicable for continuous  as well as discrete time domain   fine. Then what else   for a N continuous processes in series with  Laplace domain transfer function as g 1 s,   g 2 s, and so on, Laplace domain transfer  function. The pulse transfer function has   given us this formula well did we  make use of this fact beforehand.  

Well, we did what this practice is that g  of z is the z transform of the individual   transfer functions in Laplace domain. So let us see if g   1 s, g 2 s, up to g n s are the Laplace domain   transfer functions then what  would be g overall or simply g?   Just now we develop this particular  expression this would be g 1 of s,   g 2 of s up to g n of s.

Now you would  like to determine the corresponding   pulse transfer function, how would you do  that? We said that you whenever you have g of s   you determine g of t.   If you remember the scheme which we wrote in  the previous lecture, you are in s domain,   you go to t domain, from where you go to z  domain.

So I get g of t, I discretize this,   I get g of n T and from g of n T I get g of  z, the flow which is been later on far right.   With some expertise and experience, I also  said that you can write this directly.   So what will I do?

I now have  g of s as this multiplication,   so I will get the overall transfer function  inward that to get the get g in terms of T,   discretize it and then again do this  transform, which basically means that g   z. Let me write this in this manner, g z   is what the Laplace transform, the z transform  of g s, you see this going from s domain   to z domain, which in other words means g of z as  the z transform of g 1 of s g 2 of s, g N of s.   So this is what which has been stated.  

What else? In general g z transform of G  1 s 1 small correction then it has to be   a multiplication not a comma. So G 1 of s,  G 2 of s uo to G N of s is not equal to G   1 of z, G 2 of z, G N of z and so on.

Let us try  to understand what this means. So I have this   overall transfer function   g of s, which is equal to g 1 of s and g 2  of s, up to g N of s. How do I obtain the   corresponding z transform?

Let  me write this again, S domain   to t domain to z domain alternatively  directly my flow is very clear. So how   do I go from g of s? Is a multiplication  of individual transfer function to g of z.  

What I do is this and how do I get g of  s, I have g 1 of s available with me,   I have g 2 of s available with me and so  on, I have g N of s available with me.   What do I do? I multiply them multiply,  multiply, multiply and then I do this,   for the multiplication I do a z transform, I  get g of z.

Let me repeat I had these individual   transfer functions available with me, I multiplied  them first and then whatever product I got,   which is g of s I took the transform of  it z transform. I can do 1 more thing,   I have g 1 of s, g 2 of s, g N of s. Now I can do  this thing first, I from g 1 of s I get g 1 of z,   from g 2 of s I get g 2 of z and from  so on from g N of s I get g N of z.  

And then I do this thing I take this I multiply.   So this is taking z transform, taking z transform,  taking z transform and then I multiply them,   what do I get? g 1 of z, g 2 of z, up to g N of z,  which is the correct method.

Multiplication first   followed by z transformation or z transformation  first followed by multiplication are these   in other words, are these 2 operations equivalent  can you do 1 operation followed by other or second   operation followed by the first, would you get  the same answer in general you will not get the   same answer, the only method to get this answer  the pulse transfer from overall pulse transfer   function correctly is to first multiply your  transfer function in the Laplace domain itself   and then do z transformation in the end. And why does this happen? There is an elaborate   proof which is not in the scope of this current  discussion, we will not go into those details   in this particular course.

What needs to be  remember this, get the overall transfer function   in Laplace domain first by multiplication and do z  transformation later, not the other way around.   Now we all always looked into a single input  single output process, did this a transformer   analysis, but for the transform domain analysis  we did the transform domain analysis for multiple   input multiple output systems as well, so  is it possible that we do this analysis   for multiple input multiple output discrete  systems? Turns out that yes, it is possible.  

Now when you do a multiple input multiple  output MIMO analysis of discrete time systems,   you get a system of difference equations. Let us quickly see how do we get it for single   input single output system and then can we  extend this. So for a single input SISO system   we can write the corresponding equations  as a d x by d t is equal to a x plus b u,   this is the dynamical equation, let us call  this as equation 1 and y is equal to c x plus d,   this is the equation for the output variable. 

So to determine the corresponding discrete time   equations for this SISO system, let us first  discretize equation number 1 we can write   x n plus 1 minus x n upon T  is equal to a x n plus b u n.   Let us do some simplifications this is x n  plus 1 upon T is equal to 1 upon T plus a   x n plus b u n, which can be further  simplified to x n plus 1 is equal to t times   1 plus a t upon T plus b T times u n, in other  words x n plus 1 is equal to, okay so we forgot   x n here so it should have been x n. So  1 plus a T times x n plus b t times u n.  

So I can make a general observation that  a T b are all constants. So if I set   1 plus a T as a 1 and b T as b 1, then  I can write my difference equation for   the dynamical variable as x n plus  1 is equal to a 1 x n plus b 1   u n, let me call this equation 3. And what is the meaning of equation 3?  

If you know x n the value of the  dynamical variable at any instant of time   with an index n and the corresponding value of  the input function at the same instant of time,   then the value of the dynamical  variable at the next instant of time   would be given us x n plus 1 is equal to a 1, a  constant times x n plus v 1 another constant times   u n. And then the way we converted equation number  1 to the discrete form from equation number 2, I   can simply write y n is equal to c x n plus d u n,  let me refer this equation as equation number 4.   So the value of the output variable at any given  instant of time can be obtained by multiplying   the value of the dynamical variable at that  instant of time by c and followed by addition   of the multiplication of d with the value of the  forcing function at that given instant of time.  

So let us see how this scheme works. You have u  and u would be known at, u at 0, u at T, u at 2 T,   and so on, and all of these would be known,  let us call these as u 0, u 1, u 2 and so on.   The initial condition, because equation  number 1 is an initial value problem.

So   all of these are known. Since equation  number 1 is an initial value problem, x   at 0 would have be known, so this is x 0, and this  is also known. So if I know u 0 and I know x 0,   then from equation number 3, x 1 would be what?

a  1 times x 0 plus b 1 times u 0. And since x 0 and   y 0 are known, you now know x 1 at time t, this  is known. Now what is the meaning of equation 4?

y   at 0 would be equal to c times x 0 plus  d times u 0, which means y at 0 would be   known because x 0 and u 0 are known. Now since x 0 and u 0 are known,   you can determine you could determine x 1.  So therefore, from x 1, which is known now   and u 1 which is always known, you can determine  y 1 as c x 1 plus d u 1 and this way you can   continue this whole procedure.

So the basic method  is that you start with the dynamical variable,   the dynamical equation, discretize it and we  what we learned is that from equation number 3,   your value of the dynamical variable at next  instant of time can be known from the value of the   dynamical variable and the forcing function at the  current instant of time. So x n plus 1 would be   a 1 x n plus b 1 u n, a 1 and b 1 are constants.  Further y n, the value of the output variable at   a given instant of time can be calculated from the  value of the dynamical variable and the value of   the input function at that given instant of time,  so now this helps me propagate my system.  

There is only one thing which is left  that you have y is equal to c x plus d u,   from where you wrote y n is equal to c x n plus  d u n. Which means to determine the value of the   output variable at a given instant of time,  you need the value of the forcing function as   well as the dynamical variable at that instant  of time. Again we since we are having a SISO   system single input single output system, can we  devise a method such that I know only I need to   feed only the input variable and I get the output  variable directly.

Well, that is possible, I have   let me refer this to as equation 1, refer this  to as equation 2, and x n plus 1 is equal to a 1,   x n plus b 1 u n, this is what we obtain by  discretization of the dynamical equation.   So let us see how I can get the relationship  directly between y n u in the discrete form.   So from equation number 1, I can write d  y by d t is equal to c times d x by d t   plus d times d u by d t.

So let me discretize this  equation now therefore, the denominator d t would   get cancelled out. So I can write y n plus 1  minus y n is equal to c x n plus 1 minus x n   plus d u n plus 1 minus u n, I hope that you  are able to see that throughout the left hand   side and the right hand side, you will have a  division by capital T and we have cancelled it.   So I can write y n plus 1 as y n plus c x n plus  1 minus c x n plus d u n plus 1 minus d u n.  

And what is x n plus 1 from equation number  1? Equation number 3 it is even x n plus b 1   u n. So let me make use of this.

So y  n plus 1 is equal to y n plus c times   a 1 x n plus b 1 u n minus c x  n plus d u n plus 1 minus d u n,   from where I can write y n plus 1 is equal to y  n plus I will collect the terms having x n, so c   a 1 minus 1 times x n plus c b 1 minus  d times u n plus d times u n plus 1.   Well, after doing this rearrangement,  I have gotten rid of x n plus 1,   but I still have x n which means, I need to  know the value of the dynamical variable at   the current instant of time and this is what I do  not want. So from equation number 2, I can write   x n is equal to 1 upon c y n  minus d upon c u n.

So I can   substitute equation number 5 in equation  number 4 and what I get is this,   y n plus 1 is equal to y n plus c times a minus 1  times now x n is 1 upon c y n minus d upon c u n,   this is for x n plus c b 1 minus d times  u n minus plus d times u n plus 1.   And we can do further rearrangements  to basically see that y n plus 1   can be written as some quantity alpha, where  alpha will be in terms of a, b, c, d times y n   plus some quantity beta times u n plus  some quantity gamma times u n plus 1.   Now see what has happened, when I have obtained  equation number 6.

If I know the value of the   output function at the current instant of time  and the value of the forcing function at current   instant of time and the next instant of time,  which will always be known, you can know the value   of the output variable directly without having to  calculate the value of the dynamical variable.   So x 0 sorry, u 0. So let us see the  scheme now, u 0 is known, u 1 is known,   u 2 is known and so on, all of these are known,  because you have the function forcing function.  

Because you have an initial value problem x 0  also would be known. And since x 0 is known,   y 0 is equal to c x 0 plus d u 0 this  would be known. And since use 0 is known,   u 1 is known and y 0 is known, y 1 is equal  to alpha times y 0 plus beta times u 0 plus   gamma times u 1 is known and therefore, in  this manner, we can propagate our system.  

So this was the way to convert single input single  output system to a difference equation form.   So now let us extrapolate this to  obtain the relationship for MIMO system.   So let us consider a second order 2 input,   2 output system, and then we can generalize it  to n-eth order m input p output system.

So how   will you deal with second order input-output, 2  input, 2 output system? My equations would become   d x 1 by d t is equal to a 1 1 x 1 plus a 1 2  x 2 plus b 1 1 u 1 plus b 1 2 u 2 d x 2 by d t   is equal to a 2 1 x 1 plus a 2 2 x 2 plus b 2  1 u 1 plus b 2 2 u 2, these are my dynamical   equations the output equation are y 1 is  equal to c 1 1 x 1 plus c 1 2 x 2 plus d 1   1 u 1 plus d 1 2 u 2 and y 2 is equal to c 2 1 x  1 plus c 2 2 x 2 plus d 2 1 u 1 plus d 2 2 u 2.   Now we need to discretize this system.

From  a single input single output system, we   saw that your dynamical equations can be  described discretized in the form x n plus   1 is equal to a 1 times x n plus b 1 times  u n. Now you have 2 dynamical variables x 1   and x 2, but if we write this in this form  subscript x 1 x 2 in our current notation,   this will become the time. So you  want to avoid this confusion.  

So instead of subscript we will use superscript  for the variable, so superscript x 1 x 2,   so superscript x 1 at the current  instant of time would be n,   x 2 at current instant of time would become x 2  n, in the next instant of time after T, after T   this will become x n plus 1 the same variable  1 and x n plus 1 the same variable 2, so this   is how the system is hopping to the next state. So  therefore, what would be my discretized equations   I hope you are able to see this that  you will have x n plus 1 is equal to   a 1 1 x n plus a 1 2 x n plus b 1 1 u n plus b  1 2 u n except that now you need to account for   the indices corresponding to the variable. So for  first equation for first variable first variable,   this will become second variable, this will become  first variable, this will become second variable.  

Similarly, x n plus 1 for the second variable  would be a 2 1 x n from the first variable plus   a 2 2 x n from the second variable plus b 2 1  u n from the first variable plus b 2 2 u n from   the second variable. So this is the discretized  form of the dynamical variable. The corresponding   form for the output equations will be  simple y n for the first variable would   be c 1 1 x n for the first variable plus  c 1 to x n for the second variable plus d   1 1 u n for the first variable plus d 1 2  u n for the second variable.

And finally,   y n for the second variable would be c 2 1 x  n for the first variable plus c 2 2 x n for   the second variable plus d 2 1 u n for the first  variable plus d 2 2 u n for the second variable.   So, these are the discrete equations or difference  equations for your discrete system. And you can   follow the exact same protocol for evolution of  the system that if you know x 0, you know u 0,   you can determine y 0 from x 0 and y 0 for the  first and you will determine this for all the   variables and substitute in equation 1 and 2 and  after you get the values from equation 1 and 2,   you can subsequently substitute the values  in equation 3 and 4 to get the evolution of   the output variable.

Alternatively, you can take  a longer path and come up with explicit expression   for y n in terms of x in terms of u n, the way we  did for the single input single output system.   You can model multiple input multiple output  system using the difference equations which   we wrote here. And then for multiple  input multiple output discrete system,   you can determine a discrete transfer function  matrix.

Let us see the genesis of this discrete   transfer function matrix. For that let us make a for a 2   input 2 output system, let us make  a block diagram. So you have input   1 u 1 you have an input 2 u 2, go to the  system and you have the ultimate output y 1   and you have an ultimate output  y 2 from the system.

Now y 1   sorry u 1 is acting on the system to give you  output as y 1. So I will have a block here   and this gives you y 1, the effect of input  u 1 on the output y 1. How would you do that?  

It would be via a transfer function  g of z. Similarly, u 2 would act on   a transfer function to give you an output y  2 g of z. But now u 1 may act on y 2 on may   affect y 2 as well, so therefore, what I need  to do is this I will have the way y 1 u 1 at   via some transfer function g such that  the combination of the effect from   the block down and the block up will  give you the overall effect.

So,   I have an addup here, so plus plus. So I have u 1 to understand this, I have u 1   which is acting on a block g of z which gives  you the effect on output y 1. Similarly,   it gives the effect of effect on y 2 via another  block.

Well I can draw an analogous situation here   g z and let me   make another adder here, plus,   plus. So this is the   overall flow of different effects. I can  do one thing; I can make a large block   like this and what I can see is that   there are a lot of operations which are going  inside this dashed block but all I can see   from outside is that I have 2 inputs and have  2 outputs.

So, what I can do is, I can assign a   block which has 2 inputs and which  has 2 outputs, u 1 u 2, y 1 y 2   and inside that is a set of operations. Now what does the top post block give me? g z,   it gives you the effect of input 1 on output  1.

What does the next block gives me? It gives   me the effect of input 1 on output let us  follow the arrow it goes to 2 output 2.   Similarly, the third block from top gives me the  effect of u 2 on y 1 and finally, the 4th block   gives me the effect of u 2 on y 2.

So therefore,  I will have a transfer function matrix g   2 cross 2, whose elements would be the individual  transfer functions g 1 1 z, g 1 2 z, g 2 1 z and g   2 2 z. So in this manner, it is possible  to determine the transfer function matrix   of a multiple input, multiple  output discrete time systems.   So, what we understood today  is that it is possible that   you have a multiple input multiple output system   and how to deal with such a system?

Well, you deal  with such a system in an analogous manner as you   did with the continuous systems. In continuous  systems, you had the transfer function matrix,   in this case also you have a pulse transfer  function matrix where the elements individual   elements of the transfer function matrix would  give you the individual effects of the inputs on   the individual outputs on the system. So,  we will stop here today and we will take   an interesting case of analysis of stability of  systems in the next lecture.

Till then goodbye.