Lecture 56: Response of discrete-time systems

Hello, and welcome to this last week  of instruction on this online course   on advanced process dynamics. We  are currently studying discrete-time   systems the dynamical response  of discrete time systems.   And in the previous week, we develop the  motivation behind that requirement of discrete   time systems, which becomes inherently applicable  when you use digital computers for your control,   how the discrete input is generated, what is the  requirement of conversion of models to discrete   time models and how is a discrete output from  the models converted to continuous output.  

We also introduced you to a new  mathematical concept called z transformation   similar to the Laplace transformation  in continuous domains z transformation   is expected to help us in understanding the  dynamical response of discrete time systems.   So, let us look into this particular aspect  of response of discrete time systems.   So, what we have in front of us is  something called a pulse transfer function.  

So, before we define what a pulse transfer  function is and how does it differ from   transfer function in Laplace transform domain.  Let us try to understand the larger scheme.   So, the scheme that we have in front of us  is that we have a situation where the input   output model equation is available with us.

So, typically, this would be a system in which   you have dynamical equation or set of dynamical  equations, if it is a higher order system and you   would have an output equation if it is a single  output system or a set of output equations if it   is a multiple output system. So, in general, you  may have a multiple input, multiple output system   following which you may have a system of dynamical  equations and a system of output equations.   Now, this dynamical equations as well as  output equations would be in terms of time.  

So, in state space domain, what you will do  is you would like to do the analysis of the   phase portraits if you are not in a position  to explicitly solve for the dynamical variable   output equation. But otherwise also you would  like to solve the equation and try to determine   how does the dynamical variable and the output  equation output variable change with time.   We took a special situation where we  were interested in the response of   the system rather than every possible  state that the system can attain.

We   were interested in the response of the  system and that was obtained in using   the Laplace transform of the equations.  So, rather than dealing with t directly,   we want transformed our system from t domain  to s domain via Laplace transformation.   Now, our intention was to determine the dynamical  response of the system subjected to some forcing   function, which you call the input function. 

So, if I have the input function, I should   in principle be in a position to solve for x  of t and y of t. And in this particular case,   now, I have y bar of s upon u bar of s and y  bar of s x bar has upon u bar of x which is   referred to as the transfer function of your  system, y bar of x divided by u bar of s.   So, the scheme is that from the model equations,  you determine the transfer function.

How would you   do that? You will take the Laplace transform  of the dynamical variable from where you will   get x bar of s divided by u bar of s. The Laplace  transform of the dynamical variable in deviation   variable form divided by the Laplace transform of  the input variable in deviation variable form.  

Similarly, you will do the Laplace transform  of the output equation from where you will get   after several rearrangements and simplification,  y bar of s divided by u bar of s which means the   Laplace transform of the output variable divided  by the Laplace transform of the input variable.   What do I require? I require the time  variation of my output from the system.  

And therefore, what I can do is given that I have  g of s which is y bar of s divided by s bar of s I   will identify the input function. I will take its  Laplace transform. So, I will get u bar of s.  

So, if I multiply g f s by u  bar of s, I will get y bar of s   which means I will have the Laplace transform of  the output variable and I can do an inversion. So,   inverse Laplace transform of y bar of s will  give me simply the time evolution of my system.   So, this is the larger scheme which we followed  for the analysis of continuous domain systems.  

Now, what we have done is that we have converted  our continuous system to a discrete time system.   So, again, the continuous input has been  converted to a corresponding discrete time input.   My continuous model has been converted  to the corresponding discrete time model.  

And as a result from discrete input and  discrete model, I will get a discrete output.   So, therefore, in principle, I  can define an analogous quantity,   which takes in the discrete input that  operates on the equivalent quantity,   which is a quantity which is equivalent to  the Laplace transform in continuous domain,   it would be operated by the discrete time  input and it would give you the output,   discrete output or continuous output  at this point of time, I am not saying   anything about that, all I am saying is that  this quantity will give you the output.   Now, the input is a discrete signal.

So,  therefore, instead of Laplace transform of   the signal you will have the z transform of the  signal. Similarly, when you described transfer   function in Laplace transform domain instead of g  of s. Now, since the entire system is in discrete   time domain, you will have a quantity g of z  because that whatever that transfer function is,   would be a functional z.

And that is precisely  what is the meaning of pulse transfer function.   So, let us see what is the meaning of pulse  transfer function. It is the function g of z   that relates the sampled input.

Now, this is important that g of z   takes the input as the sampled input. So, your  process may be continuous, but inevitably,   you will have to convert that continuous input  to a discrete input. So, the input to the   pulse transfer function would be the sampled  input which means it would be a discrete input   to the discretized output, the output also would  be discretized.

And that will follow the relation   output the z transform of the output y hat z  would be equal to the pulse transfer function   g of z multiplied by the sample input z transform  u bar u hat of z. So, equation one gives you the   definition of purse transfer function y hat  z is equal to g z multiplied by u hat z.   Now, one thing which is important is that you  have added the term pulse transfer function.  

It in itself gives you an idea that you are  dealing in discrete time domain. You are no more   in continuous domain. So, therefore, you do not  use to emphasize that you are sampling a discrete   signal and giving output also as a discrete  signal by processing using discretized models,   you use the term pulse transfer function and what  is the reason behind use of the term pulse because   we said that how would you sample the input  you will sample the input using a sampler there   will be a switch and therefore, in a sense, your  continuous signal would be in the form of pulse,   ideally it should be an impulse in  that case, where delta t tends to 0,   but realistically, it would be a pulse.

So, we are in general using the term   pulse transfer function. So, it is important to  realize that we have transfer function g of s for   continuous domain analysis and we have a pulse  transfer function g of z in the discrete time   domain analysis and the transfer function will  involve in this particular case, the z transform   rather than the Laplace transform. Let us see  if we can make any more comments on this.  

So, now, when you have a transfer  function in continuous domain g of s,   the basic utility of this entire exercise was that  you could develop a simple input output model.   So, you could diagrammatically represent  the system using a block diagram, where   you have a single input as you can see,  in this diagram, you have a single input   single output, input being u, the output being  y. Now, there used to be a box.

Box basically   means a set of mathematical operations and in the  previous case, u bar of s as an input to the box   would give y bar of s and the box would  have the transfer function g of s.   Now in an exact analogous manner, you can make a  block diagram for pulse transfer function as well.   So, what is the meaning of pulse transfer  function?

Well, if you apply the pulse transfer   function to the input, you will get output. So,  you can draw a block diagram which looks like   this. You have u hat of z which acts as the input  to the box, the box has mathematical operations,   that particular operation in this case is the  z transformation of the input output model   discretize model that you have and the output that  you will get would be y hat of z.

But, in case of   Laplace transformation, well you had a continuous  time input, you had a continuous time output.   So, therefore, there was not much of an  issue regarding the location at which you   are determining your functions, because it is  a continuous system, this function was defined   at all locations of time or all instances of time.  But in this case, we actually have in the previous   lecture, seen the importance of the sampling,  all of your analysis depends upon the sampling   time.

And we also said that the z transform  is not unique, it depends upon the location.   So therefore, if you change the sampling  interval, there is a transform will change.   Therefore, it is important that when  you define a pulse transfer function,   you have a statement here on the screen, the input  and output signals must be sampled synchronously,   which means that if you are having, so let me do  it diagrammatically that, if this is u for input   and you are doing it sampling it here at 0, T,  2T, 3T and so on.

Your output will also be at 0,   T, 2T, 3T and so on the same time and also at  the same rate, of course, this goes without   saying that, if the input is at a time interval  T, capital T, the output also should be at capital   time interval T. So, this is the general  idea about the pulse transfer function.   So, let us see if we can have something about  the transfer function pulse transfer function   in a little greater detail.

So, the first  thing which you would like to know is the   no hold transfer function. Let us try to remind  ourselves of the scheme that we are following.   So in a continuous domain  analysis, you have u of t   from where you get u bar of s, you do the  Laplace transform.

And then what you do is   you feed it to g of s, the transfer function when  you apply g of s to u of s, what you get is y bar   of s from where you get y of t, very simple. So, you have a continuous input and you get a   continuous output simple. Now, you  implement a digital control system, so,   it is imperative to convert this  entire business to discrete time domain   and how does this scheme change now?

Your  continuous input will pass through a sampler;   we make use of the analogy with a switching  element. So, you will pass it to the sampler.   And what would be the output from the sampler?

The  output from the sampler would be a discrete input.   Now, what you would do is you  would feed this discrete input   to another box, which comes as the transformation  we do not know the notation for it, but it would   do the transformation of the discretized model. So, this is the process model   and in fact continuous process model.

Now, I am  feeding my input to a discrete process model. So,   I will have to do this   discrete process model. Well, I will feed this  discrete input to my discrete process model   and if I am printing it to my discrete process  model, obviously, I will get a discrete output   and then what happens, the discrete output is  sent to a hold element if you remember what   we studied in the previous lecture.

This is  sent to a hold element and the hold element   would convert these discrete outputs to a  continuous output and this is what you get.   So, this is the larger scheme. And now, what we have done here is that we have a   no hold situation, no hold pulse transfer function  which means what I send in the continuous signal.  

It is processed by the model but my  output is still discrete. So, therefore,   I have this situation   with me. So, the input is discrete.

Well input  has always got to be discrete because you are   doing z transformation and the only input which  z transformation will work on is discrete input.   So, the input is discrete, but the output is no  more continuous. In fact, you have not allowed   the output to be continuous at all.

And therefore,  your output also is discrete and it goes through   some mathematical jugglery with in this box and  that mathematical jugglery would be called the   no hold pulse transfer function. So, now the question is how would   you determine the no hold transfer  function? Well, it is very simple.  

The way to determine the no hold transfer  function is this that you will have a   rather than continuously let us  write Laplace domain transfer   function. Let us imagine that you have  a Laplace domain transfer function,   which is available to you. And that is g of s,  you would like to determine the no hold transfer   function for this system.

Assuming that you can  convert this process model to a discrete model   that is a step which we have already learned and  that should not be an issue at this point of time   for this analysis. So, how  would I determine no hold   pulse transfer function and let me denote  that by g of z at this point of time.   So, I need to do this from g  of s, I will obtain g of t.  

Now, this is strange, because  at no point of time till today,   we have done the inversion of  the transfer function itself,   because we were interested in determining  the dynamical response of the system. So,   in continuous domain, we used to multiply it with  the Laplace transform of the input function. Here,   we are not doing that we are converting g of t. 

So, therefore, g of t is a continuous function   of t. So, therefore, what I would do is now,  I would sample, so, I would do sampling.   So, what I would get?

I would get g  at T, g at 2T, g at 3T and so on.   Now, this would act as the input to my  z transform and therefore, from here   using the usual definition I can determine  g of z and this would be the no hold   pulse transfer function. Let me repeat  the procedure, I started with g of s,   I inverted it determine various   points, the value of the continuous  function g of t at various sampling points   and then use those for determining g of z.

Let us take an example and see if we get this.   So, let us say I have a first order  process and I want to determine the   no hold pulse transfer function. So, I have g   of s is equal to k upon tau s plus 1.

So, this is  equal to k upon tau 1 upon s plus 1 upon tau, this   is g of s from where I can get g of t is equal to  what k upon tau e to the power minus t by tau.   Now, you can as well use this continuous  function directly to write from here,   g of z and to emphasize that this is a  no hold pulse transfer function. I am   writing N H is equal to what, k upon tau and  what is your transfer function z transform,   sorry no transform function, z for transform  e to the power minus t by tau is simply 1   divided by 1 minus e to the power  minus t by tau times z inverse.  

We have written directly. You can as well do one  thing you can write g of 0 is equal to k upon tau   g of t is equal to k upon tau multiplied  by e to the power minus capital T by tau   and so on. So, these are the very sampled inputs  for your z transform.

So, therefore, I can write   g no hold z is equal to what, the formula was  summation from n is equal to 0 to infinity,   f of n t z to the power minus n. So, therefore, k upon tau z to the power 0, 0   plus k upon tau e to the power minus  capital T by tau z inverse plus k by   tau e to the power minus 2T by tau z to the  power minus 2 and so on from where I get the   same answer using the geometric progression  summation k upon tau 1 upon 1 minus e to the   power minus t by tau z inverse because that  is the ratio of two subsequent terms in the   series the progression. So, this is how you  determine the no hold transfer function.  

Now, you have in general a hold because  you would like to ultimately get the   output as a continuous function. So, instead of  analyzing the output only as a discrete I am now   implementing a hold and what I will do is I  will using the hold I will convert that discrete   signal to a continuous signal. So, what does the  hold element do?

So, you have a discrete signal   coming from   the process. It is fed to your hold  element and you get a continuous signal.   Now, what you need to do is this that how do  you get the discrete signal?

Well, you have the   discrete input   which would always be the case because you are  dealing with z transformation. It goes through   the process, the   process generates discrete output and this is  fed to a hold element and hold element gives you   continuous output. So, now, if I need to define  a transfer function between the continuous output   and the discrete input, I have two boxes here, I  have two blocks and using the usual principles,   I know that the overall transfer function would  be simple multiplication of the individual   transfer function blocks.

So, therefore,   my pulse transfer function for  zero order hold. So, g of z   the pulse transfer function for a zero  order hold, will have what? Will have   the transfer function for the hold multiplied  by the transfer function for the process,   the transfer function for the hold multiplied  by the transfer function for the process.  

And what is that which I now need to  know transfer function of the process   will come from your discrete model. For example, in the previous case we   looked into the first order process and then  we try to convert it to a discrete so, that   g of p will be known to you, what you need  to know is you need to know g of h that the   transfer function corresponding to hold and  what does the hold do? The hold does is this   so, this is h of t versus t the hold  keeps your output to that constant value   and then puts it down to zero that  is the hold.

So, for this particular   process, what would be h of s? The h of s would be  for this pulse for the rectangular pulse which is   1 upon s the first one minus  1 upon s e to the power minus   T s. In other words, h of s is equal to 1  upon s 1 minus e to the power minus T s.  

So, this is the Laplace transform   of the hold element and the Laplace transform  of the first order system for example,   like the previous one is this, so, this  is h of s and then g of s is what K upon   tau s plus 1. So, this is g of s and let us be  consistent with the notation let me write this as   g of H, Laplace transform the hold and the  Laplace transform of the process. So, g of p   for the process.

So, what would you do  to remind the zero order pulse transfer   function? Will do g H multiplied g p and repeat  the same procedure as you did previously.   In the interest of time, we will stop  here today and continue this discussion of   pulse function for zero order hold  in the next lecture and also see   the dynamical response of a discrete  time system given that you have the   pulse zero order function hold function  available to you.

Till then, goodbye.