So, all the systems that we studied till now in the transform domain analysis, were the systems which were single input single output systems. We looked at various types of transfer functions and the effect of the poles and zeros in the transfer function on the dynamical response of the system. But what would happen if you have a multiple input multiple output system, what would happen to the transfer function, what would be the nature of the transfer function and how to analyze the system which is multiple input and multiple output?
So, let us analyze this particular problem today. So, we have the analysis of multiple input, multiple output systems. Have we come across such systems before?
To have a very quick look, we looked at the transient behavior during the stage operation and we took a particular case of a stage distribution column, and the mass balance over one particular plate is given by the equation, you can see here, equation number 1. And we realize that this is in fact a multiple input multiple output system. Please refer to one of the previous lectures in which we discussed why this is a multiple input, multiple output system.
And what were the dynamical equations for such a MIMO system that we considered previously? The equations are in front of you. So, I have an M input please note that I have an M input and P output system.
The number of dynamical equations that I have is capital N, and correspondingly for every output, I will help P number of output equations. So, these are the equations in front of us, and let us try to do transform domain analysis for this system. So, we also looked at that we can convert this system of equations to a system of matrix equations and various quantities with the dimensions are in front of you.
x under bar is the vector corresponding to the dynamical variable. y under bar. the vector corresponding to the output variable and A, B, C, D are the different matrices which you have in the system.
So, now, let us do an analysis of this system in transform domain. Now, before we can establish how to do this analysis of a relatively complex problem, let us first establish the method for a simple single input, single output system and then we will learn from that method and try to translate that to MIMO system. So, SISO system that we have is this.
I have d x by d t is equal to a x plus b u, and y is equal to c x plus d u. So, this is a first order single input single output system. How would I do a transfer function analysis or transform domain analysis for this system?
So, let me call this equation 1, and call this, equation 2. I can do suitable rearrangements in equation 1 right at this stage or it can be done later. For this particular case, let me do the rearrangement at this stage.
And I will write equation 1 as 1 over a d x by d t minus x is equal to b over a u. And now, what I need to do is I need to take the Laplace transform on both sides. So, I should get 1 over a s minus 1 times x bar of s is equal to b upon a u bar of s, from where I get the quantity x bar of s divided by u bar of s is equal to b upon a divided by 1 upon a s minus 1.
So, now this quantity x bar s divided by u bar of s is the Laplace transform of the dynamical variable divided by the Laplace transform of the input variable. What I would ideally like is, since I have a single input single output system, the transfer function which would give me the effect of input on the dynamical behavior of the output. So, what I need is the Laplace transform of the output variable divided by the Laplace transform of the input variable.
And output variable in this particular case is y. So, from equation 2, I can write y bar of s, I will take the Laplace transform of equation 2 on both the sides, is equal to c x bar of s plus d u bar of s. From here, I can write y bar of s divided by u bar of s is equal to c x bar of s by u bar of s plus d.
And I know the expression for x bar of s by u bar of s from equation number 3, I have that quantity as b over a divided by 1 over s minus 1. So, I can write this as y bar of s divided by u bar of s is equal to b c upon a divided by 1 over a s minus 1 plus d, which can be further simplified as d upon a s plus b c upon a minus d whole divided by 1 over a s minus 1. And this is y bar of s divided by u bar of s.
So, equation number 4 is my transfer function. the Laplace transform of the output variable divided by the Laplace transform of the input variable for my single input single output system. This is a SISO system.
What is the procedure that I followed? I started with d x by d t equation, took the Laplace transform on both the sides, got an expression for x bar of s divided by u bar of s, then I took the expression for the output variable, took the Laplace transform on both the sides, did rearrangements, used result from the previous expression of x bar s divided by u bar of s, got the final transfer function. And you can see that my final transfer function for this SISO system is a 1, 1 order system.
Because the transfer function is given as d by a s plus b c by a minus d divided by 1 over a s minus 1. Further simplification can be done if needed. So, I will now do the same analysis for a MIMO system.
So, now, I will do an analysis for a MIMO system, and let me take an example of two input, two output system. So, I will take an example of two input two output, and then we can generalize this for multiple input multiple output systems. So, the equations for two input two output systems will look like this that I will have 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 x 2, first dynamical equation.
Similarly, the second dynamical equation would be 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. And this must be u 2. So, x 1 x 2 u 1 u 2 x 1 x 2 u 1 u 2.
So, the corresponding output equations would be 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 21 u 1 plus d 2 2 u 2. I will need to determine the transfer function.
So, let me follow the same method, equation 1, equation 2, equation 3 and equation 4. So, I will do the Laplace transform of equation 1, from where I would get s minus a 1 1 x 1 bar of s. See what I have done, I have taken the Laplace transform of the derivative and on the other side, I had x 1 I took it to the left-hand side.
Similarly, minus a 1 2 the Laplace transform of 2 is equal to b 1 1 u 1 bar of s plus b 1 2 u 2 bar of s. I hope this is not very difficult to understand. Similarly, now, I will do the Laplace transform of equation 2.
So, I have minus a 2 1 x 1 bar of s plus s minus a 2 2 x 2 bar of s. This would be equal to the Laplace of the right-hand side b 2 1 u 1 bar of s plus b 2 2 u 2 bar of s. We actually have come across the situation before in Laplace Transform Domain analysis.
And during that time, what we did was, I said that you would need to solve for x 1 bar, x 2 bar, get the quantities in terms of u 1 bar, u 2 bar. Instead of doing that, let us make an observation. The observation that we make is that these are two equations, which can be converted to a matrix equation.
Let us see if that is the case. I can write this as s minus a 1 1 minus a 1 2 minus a 2 1 s minus a 2 2 multiplied by x 1 bar of s x 2 bar of s, would be equal to what, on the right hand side. I can write the equation b 1 1 b 1 2, the matrix b 1 1, b 1 2, b 2 1 b 2 2, multiplied by u 1 bar of s u 2 bar of s.
So, now, what I have is instead of the Laplace transform of individual dynamical variables, I have a vector in which the Laplace transform of each component has been taken on the left-hand side as well as on the right-hand side. I can do one more thing. I can further do this splitting and let us see if this is correct.
What I can do is s times 1 0 0 1 minus a 1 1 a 1 2 a 2 1 a 2 2, whole quantity multiplied by x bar 1 s x bar 2 s is equal to b 1 1 b 1 2 b 2 1 b 2 2, whole multiplied by u 1 bar of s u 2 bar of s. And now, I make certain observations. I have a matrix a 1 1 a 1 2 a 2 1 a 2 2.
Remember, that in previous analysis, we used to declare this matrix as the matrix A double underbar. Similarly, we used to declare this matrix b 1 1 b 1 2 b 2 1 b 2 2 as B double underbar. So, therefore, I can write my equation simply as this s multiplied by what is this?
This is an identity matrix, s multiplied by identity matrix minus the matrix A times. Now, I can declare this as the vector x underbar overbar s. I hope the notation is clear what is the significance of underbar, what is the significance of overbar.
We have been doing this over and again write from lecture 1. And this would be u underbar overbar s. So, I will have our s I minus A multiplied by x underbar overbar s is equal to B double underbar u bar of s.
In other words, I can write x, single underbar overbar s is equal to, I have matrices on the left hand side product operations that have a matrix on the right hand side. So, I can do a transfer. So, I can write this as s I minus A inverse B, u overbar underbar s and let me call this equation 5.
So, what does equation 5 give me? Well s multiplied by I minus A, its inverse, again multiplied by B is going to be a matrix and it is going to be multiplied by the input vector, the Laplace transform of the input vector. And if I do a inverse Laplace transformation, I will get the dynamical variable.
So, in certain sense I get an idea about the transfer function involving the dynamical variable and the input variable. That is not my ultimate aim. So, let us proceed forward.
Now, let us now try to determine the Laplace transform for the output variables, so output equation. So, then our input equations were 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 u 1, d 1 1 u 1, d 2 1, in fact, u 1 plus d 2 2 u 2.
So, we will take the Laplace transformation. So, I get y 1 bar of s is equal to c 1 1 x 1 s bar plus c 1 2 x 2 bar of s plus d 1 1 u 1 bar of s plus d 1 2 u 2 bar of s. And similarly, y 2 bar of s is equal to c 2 1 x 1 bar of s plus c 2 2 x 2 bar of s plus d 2 1 u 1 bar of s plus d 2 2 u 2 bar of s.
Again, by looking at these equations, I can see that these equations can be converted to matrix equation. So, I can declare this as y 1 bar of s y 2 bar of s is equal to the matrix C on the other side c 1 1 c 1 2 c 2 1 c 2 2 x 1 bar of s, x 2 bar of s plus the matrix d, d 1 1 d 1 2 d 2 1 d 2 2 u 1 bar of s u 2 bar of s. Let us use our matrix notations now, to convert this equation to compact form.
I can write y 1 bar of s y 2 bar of s vector as y underbar overbar s, the vector in which the Laplace transform of the components have been taken, is equal to c it is the same vector, same matrix which we have been using multiplied by x underbar overbar s, plus again D u underbar overbar s. So, what does this equation give me? This equation will give me the Laplace transform of each individual input, output variables in the form of a vector, which is in terms of the dynamical variable, and input variable.
I already have the expression for x underbar overbar s. So, this can be written as y underbar overbar s is equal to c double underbar, x was s I double underbar minus A underbar, double underbar inverse B times u under bar overbar s plus D double underbar u single bar overbar s. From where, I can write y underbar overbar s is equal to, now, a huge matrix operation what would that be c, double underbar s I double underbar minus A double underbar inverse B or take D also plus D all multiplied by u underbar overbar s.
And what is interesting about this final expression, you have this final expression again which is similar to what you get for a transfer function, except that on the left-hand side you have the vector of Laplace transform of output, on the right-hand side you have the vector of Laplace transform of the input multiplied by a large complex operation which ultimately is a matrix. So, therefore, therefore, you call this as the transfer function matrix. Important.
That, for a single input single output system you get a transfer function. For multiple input multiple output system, you get a transfer function matrix. Now, let us see what kind of matrix is this.
I can write this G double underbar s to emphasize that it is a transfer function matrix. This is c multiplied by s I minus A inverse B plus D. Let us see what all are the different quantities that I have, what are the dimensions?
A and I both are N cross N. From our previous analysis, we know this. So, therefore, this and then inverse also would be N cross N.
So, I have this as N cross N system. See, in this particular case you c was the coefficient matrix for the output equation with coefficients of the dynamical variable. So, I have P number of output variables and N number of dynamical variables.
So, this would be P cross N, the dimension of C is P cross N. So, therefore, this is P cross N. B comes from your input.
I have an M input system. So, this is N cross M. So, overall, this is going to be a P cross M system.
And again it is not very difficult to see that D is a P cross M matrix. So, overall, you have a P cross M matrix. So, your transfer function is a P cross M matrix.
So, very quickly, for a single input single output system you have a transfer function whereas, for an M input P output system you have P cross M matrix. So, what we learned today is that you can handle multiple input multiple output system the way you handle a single input single output system. You take the Laplace transform of the input of the dynamical variable, you take the Laplace transform of the input, output equation from where you get the overall transfer function.
And this is basically the recipe for a single input single output system. Instead of one equation you will have multiple equations for a multiple input multiple output system and therefore, the final quantity that you get would be a matrix which you call a transfer function matrix. So, we will stop here today.
And from the next, in the next lecture will take up case where we would be in a position to understand how we can do our transformation of one type of analysis, which is state space domain analysis to the other type of analysis which is the transformed domain analysis. Till then, goodbye.
