 In the previous class, we have been discussing the response of linear time invariant systems. In fact, we took simple second order systems in order to illustrate the response. The basic interesting fact about the analysis of linear time invariant system is that there is the time response can be written down in terms of well known functions and because of that one can infer the stability and the dynamical properties fairly easily. So, the main important result which we got last time was that the response of a linear system is a sum of individual patterns or individual modes of response. So to make that a bit more precise, let us take a simple linear time invariant system without any input x is a vector, a is a matrix. Then the response of the system is given by it is in fact a summation of several modes. Of course, this particular response can be written down in case you have got the a matrix diagonalizable, which you will just kind of try to understand once more. If a is diagonalizable by a linear transformation of the variables x, in that case one should be able to write the response in this form. It is a summation of various patterns or modes. Lambda i's are the Eigen values. They are a property of the matrix a itself. In fact, determinant of lambda into an identity matrix minus a is equal to 0 yields a characteristic equations whose solutions in fact are the Eigen values of the matrix a. a remember is of course, an n cross n matrix. This particular characteristic equation will yield as a polynomial of order n and in general they can there will be of course, n solutions of a nth order polynomial. Now, importantly this p i are the columns of the Eigen vector matrix. So, if you recall what we did in the previous class, we were talking of a particular transformation x is equal to p y and a p such that p inverse a p yields a diagonal matrix. The columns of p are the right Eigen vectors p of course, has to be invertible and in we discussed this last time in case a is diagonalizable that is a has in fact n distinct Eigen values. In that case it is possible to find the columns of p such that p is invertible. So, in fact a is diagonalizable provided we get a p which is non-singular which also means that the columns of p should be linearly independent. In fact, if you have got n distinct Eigen values a is diagonalizable and as I mentioned sometime back if a is diagonalizable you can write the response in this form. The q i here the q i transpose is a row matrix is a row vector rather which is in fact the rows of p inverse the Eigen vector matrix inverse the rows of it are q i. So, the main this was the main result which we saw last time and the most important issue which we discussed was depending on the components of p i the components of p i you can observe or measure the relative observability of the pattern e raise to lambda i t in the various states. So, that was an important point which we discussed in the previous class also we saw that depending on the value of the initial conditions a mode could be excited weakly excited or not excited at all. So, this was the interpretation we could give to the to the the Eigen vectors. In fact, an important point which slipped my mind last time was that q i in fact also satisfies this relationship. So, q i is also known as the left Eigen vector. So, q i are not only the rows of p inverse, but they also satisfy this particular property. Now, what happens if the matrix is not diagonalizable. So, if the matrix a is not diagonalizable. In fact, if a has distinct Eigen values it is always diagonalizable, but if a has non distinct Eigen values then a may not be diagonalizable. So, that is an important point which you should know. So, for example, the matrix a which is 1 in fact is non diagonalizable if you take out its Eigen values which is very difficult to which is not very difficult to find out the Eigen values are determinant lambda i the identity matrix minus a is equal to 0 the Eigen values are 1 and 1. So, these are non distinct and this particular matrix is non diagonalizable. You cannot diagonalize this matrix because you cannot get you know p 1 and p 2 which satisfies a p 1 is equal to p 1 and a p 2 is equal to p 2 p 1 and p 2 it will turn out to be will be always in the same direction that is p 1 will be alpha p 2 alpha is any arbitrary constant. So, what you will find out is that because of this you will not get you will not be able to construct a p with p 1 and p 2 as its columns. So, that you can diagonalize this matrix. So, you cannot get the p to diagonalize the matrix. So, that is why this particular system is not diagonalizable. Now of course, there are examples in which you do have repeated non distinct Eigen values, but still you can diagonalize it. For example, this particular matrix is a kind of a academic example, this is a matrix which is diagonalizable in fact, it is already diagonal it also has got repeated Eigen values, but obviously this is a diagonalizable matrix it is a diagonal matrix in fact. So, there are exceptions. So, just remember that the correct statement is if a has non distinct Eigen values a may or may not be diagonalizable that is the important point which you should note. Now one aspect which we should discuss is if it is non diagonalizable what is the response. For example, suppose I have a matrix which it turns out that the Eigen values are lambda lambda lambda 1 lambda 3 and lambda 4. There are 4 Eigen values of a 4 by 4 matrix a suppose this is non distinct they are you know Eigen values are repeated. So, a may or may not be diagonal let us assume that it is not diagonalizable. In that case I will just without really proving it we can show that you can find a p matrix which does not get you to a diagonal matrix, but gets you to a near diagonal matrix which is also called a Jordan matrix. So, you have got lambda 1 0 lambda 0 0 0 0. So, if a matrix is not diagonalizable and has as Eigen values as shown we can actually get it in this form this is called a Jordan form. Now this is a simplified matrix this is a simplified kind of transformed matrix which of course, one can get the time response quite easily. So, please recall that in case you have a system in which a is diagonalizable your response is x of t is equal to p e raise to diagonal matrix into p inverse into x of 0, but in case your system is not diagonal is not diagonalizable then the response can be different. Remember here e raise to this diagonal matrix is in fact this is a diagonal matrix like this. So, if your matrix is diagonalizable your response is as shown here of course, the expanded form of this response is in fact, what I discuss sometime back that is this is the expanded form. If your matrix is not diagonalizable then your response x of t is equal to p e raise to j t p inverse where j is a raise to lambda t t e raise to lambda t this is taking the example which we discussed some just a while ago that is you have got e raise to j t is this. So, remember that in case you do get non distinct Eigen values you can still write down the response, but remember now you are getting the response in terms of not only exponential functions, but you have got time coming here t into e raise these kind of terms will come into your response. So, that is the important difference you can of course, have more than 2 repeated Eigen values, but we will not go further and deeper normally in our power system analysis one will not come across situations will be there will be more than you know you know you have 3 Eigen values which are non distinct and is not diagonalizable that is very fairly rare situation most in most cases which we will consider you will have all Eigen values distinct. So, it is important to remember this now one important point which you should note is that in this particular course we do require you to understand the response of linear systems of course, it is not practicable in this course to give you a full exposition of everything about linear system theory. So, it is good if you refer to some control systems linear control systems books for more details about the response of linear time invariant systems. So, for example, you can look at the book by Ogata state space analysis of control systems 1967 of course, there are many other good books. So, this is just an example. So, this is an example of a book you could refer to now of course, and associate problem which we will come across is what happens when you have got a system which is forced that is it has got an input. So, if you have got a system x dot is equal to A x plus B u and you are not observing all states, but you are observing one output which is a which is basically a combination of the states. So, such a kind of a system has got both an input and an output. So, I will just write it down x dot is equal to A x let us assume there is only one input u and you are also observing a certain variable you know which could be a combination of the states c is a vector x is our state. So, you could have a combination of states is a multiplication of a row with a column we will assume u and y are scalars. So, this is a single input single output system in such a case what is the response of x we had written down the response when A was a scalar some in some lectures before. The response of the system when A is a coupled matrix is x of t is equal to e raise to A t x of 0 plus 0 to t e raise to A t minus tau B u could be a function of time in the convolution integral d t is should actually be d tau. So, there is a minor error here y is simply c of x t y of t plus d of u u of t. So, this is our response of the system remember e raise to A t is a shorthand for p e raise to p inverse we have already discussed what this is in case it is diagonalizable or it is equal to p e raise to j t p inverse if it is non diagonalizable this is the Jordan form matrix. So, just remember this formula which you have for the forced response. So, it is quite easy to take out the forced response as well it is not too difficult especially if you are a simple inputs like sinusoidal inputs or step inputs this x trying to evaluate this is not really a major problem. Now, before we will just write this particular equation and this particular equation in full form. So, if I expand this let us assume that A is diagonal and I expand this what you will get is y of t is equal to c times i is equal to 1 to n this is the response of x of t plus of course, other terms if input u is if the input u is 0 these other terms are 0. One interesting point here is that if c p i is equal to 0 then that but the mode or the pattern will not be visible in y t. So, if c is such that c p i is equal to 0 this is not visible although it may be visible in the states, but it still may not be visible in y. So, y is actually a combination of states these other terms of course, are dependent on u. So, this is an important point. So, not visible there is a proper technical term called observable. So, if c p i is equal to 0 then the ith mode is not observable in this input output y. Similarly, since x of t is equal to sigma of i is equal to 1 and just continue here. So, what we have we can just write this a bit differently. In fact, we can take out p i e raise to lambda i t common and then in the bracket we have got 0 to t. This whole thing is summed q i you will have e raise to lambda minus lambda i tau q i transpose b u tau d tau. It is a bit messy here, but I have just since the variable which is getting integrated is tau I have kind of taken out this e raise to lambda i t outside the integral. So, what we see one critical point if q i transpose b this is 0 for some i in that case term corresponding to lambda i t in this response is dependent only on initial conditions. If such a situation occurs we also say that this particular mode i is not controllable by the input u. So, please remember these important concepts in case q i transpose b is equal to 0 the e raise to lambda i t term in the response of the states is dependent entirely on the initial conditions and cannot be changed by the input u. So, that is one important point which you should note. So, we have through this rather straight forward analysis given you a very intuitive kind of notion of observability and controllability. We shall actually come back to these concepts in a multi machine systems where we shall see that there are many modes and certain modes are not observable at certain places certain modes are not controllable by controllers which are located at certain locations. So, this is a very these concepts are pretty important the concept of observability and controllability in linear control systems. Of course, our analysis of the same has been very intuitive and sketchy hopefully we will revisit these things and it will become clear at that point of time that will be sometime occurs much later in the course. Now, although we have been doing the analysis of linear time invariant system there are few computational issues which we should worry about because eventually the power systems which are going to the realistic power systems or systems we are going to study are going to be fairly large. So, one or two simple points and important points which you should bear in mind are that if I want to compute Eigen values I had mentioned that one of the ways of doing it is solve this equation determinant lambda i minus a is equal to 0. It turns out that this is not very easy to do if your size of a becomes larger. In fact, if you have got for example, a matrix of size 20 by 20 you will find that it is becomes almost next to impossible to actually solve it by actually computing the characteristic impedance characteristic equation. In fact, computing the characteristic equation the coefficients of the characteristic equation themselves become pretty tedious. So, there are in fact, iterative methods like power method or the qr method we will of course, not be discussing all the methods or in fact, we will not really be going into Eigen value computation when e is a large system I can recommend that you look at any book on matrix computations where this particular aspect is covered. But remember one small point that whenever you need to compute the Eigen values of a matrix and if that matrix is large you have to take recourse to some iterative methods it is not feasible to do it by actually computing the coefficients of the characteristic equation by actually evaluating this that to be very very computationally intensive. Of course, we shall take a bit of the easy way out in this particular course otherwise we will get our hands will get totally occupied in trying to understand all the computational methods and the theory behind linear systems and so on. We lose track of what we really wish to do that is understanding a power system. So, from now onwards I will also take help of mathematical software to do the computations. So, some of the software which can help you to do this which are quite widespread and used often a psi lab this is of course, freely downloadable software and also MATLAB which is quite popular. So, if you use these one of these two tools or probably any equivalent tools it should be fairly straight forward to do most of the computations which we will do in this particular course. So, rather than spend more time on trying to give you more and more results and mathematical results and so on. Let us instead take a more pragmatic view let us take a simple system and numerically compute the Eigen values. We will take a slightly larger system we have been always considering second order systems which are Eigen values Eigen vectors are easy to compute. We will take a slightly more complicated example it is a third order system not really very complicated, but we would require some computational help others will become very tedious to do. So, rather than just talk in terms of you know abstract some random a matrix we will just take a simple engineering example which will bring out not only how the method for getting the response, but will also tell us some interesting ways of interpreting the results you know all the Eigen value and Eigen vector information has to be interpreted correctly. So, let us consider an example there is a voltage source connected in fact, you can look at this as some kind of equivalent of a pulse transformer. Of course, I am not given very realistic values of a transformer it is a kind of this is a kind of leakage reactance the resistance of the wire you can consider this as a magnetizing react reactance and also this capacitor really denotes the capacitance of the widening. So, this is a kind of a near realistic example it is a model a toy model of a pulse transformer I have chosen the values in such a way that it will aid us. So, in that sense I have not given exactly realistic values if you just look at this particular example without even having to solve it you will notice that this is a very large inductance compared to the leakage which is 10 milli Henry this is 100 micro farad and this is 0.1 ohm. So, this is a kind of a situation in which there are in fact, components which are fairly differing in size there they are not of the same size this leakage is much smaller than this and so on. In fact, the main reason why I have given you this example we shall actually pursue this example even further after this class is over is that this particular system has got some certain characteristics it is in fact, what is known as a stiff system. So, before I tell you what is a system etcetera let us just look at the steady state. In fact, if this particular system has is not excited at all in case it is not excited at all it is easy to see that is if the voltage here is 0 or I keep this open the equilibrium is when what is the equilibrium of this the equilibrium point of this is obtained by setting the rate of change of the states of the system to be 0. So, let us for example, choose the state suppose I l 1 is a state I l 2 is a state and we see is a state. So, this is the initial conditions of these have to be given if I want to tell you what the response is going to be, but if you for example, if I when is for this particular circuit if the input u is 0 that is I connect this to a voltage source which has got voltage 0. In that case it is not very difficult to see that in steady state if d i l 2 by d t is equal to 0 it also means that the voltage here is equal to 0. So, v c will have to be 0. So, the equilibrium if input voltage is 0 input is 0 is the v c is equal to 0 I l 1 is equal to 0 and I l 2 is equal to 0 this is the equilibrium for no input for for input u is equal to v i is equal to 0. So, if the input is 0 in that case this is the equilibrium it is not very difficult to find that out we can do it formally that is I l 1 1 by d t is equal to this is 10. So, minus 2 10 milli Henry's this is 10 milli Henry's this is 0.1 ohm. So, you will have if you apply this k v l in this loop you will get is equal to or in other words l d i by d t plus 0.1 into I l 1 plus v c is equal to v 1. So, I have just got things some things on to this side when I applied k v l. So, this is the basic equation this is the second equation and the third equation is of course, 100 micro farad is equal to of course, if v i is equal to 0 that is the input I make if the input were to be 0 then the steady state value is obtained by putting all these things equal to 0. So, if v i is equal to 0 this is equal to 0. So, v c also will be equal to 0 if v c is equal to 0 and v i is equal to 0 it means I l 1 is equal to 0 if this is set to 0 then I l 1 is equal to I l 2 which means I l 2 also is 0. So, if v i is equal to 0 then this is the equilibrium condition this is obtained by setting all the derivatives equal to 0 that is it. So, this is if you do not excited at all that is all you excited with a 0 voltage source this is the equilibrium condition for the system. However, if as I have shown here I excited with a 1 volt source in that case I can write this compactly in this fashion to save some time I will omit all the algebra and write this equation. So, for t is equal to 0 this is a and this is b into u this is b into u. So, this is basically our system and this is the system after t greater than 0 and let us assume that at t is equal to 0 the initial conditions are all 0 that is all the currents and the voltage are actually 0. So, let us assume at t is equal to 0 the initial conditions remember in all linear dynamical systems or all dynamical systems we need to be given some conditions like the initial conditions in order to tell what the future response is going to be. So, let us assume the initial conditions are the initial conditions only a time t is equal to 0. So, the response is if this is a and this is b u you will get sorry e raise to a t 0 0 0 0 0 0 0 0 0 0 0 0 0 this is the initial condition plus 0 to t e raise to a t minus tau into b u tau. So, this is after time equal to 0 after time is equal to 0 this is how it looks like. So, this turns out to be we will cut a long show result this will come out to be 0 and if you evaluate this I will skip the steps I will just write down the final answer i 3 minus e raise to minus a t into 10 10. So, I just evaluate this particular matrix this particular this is of course 0. So, I just evaluate this particular matrix this particular this is of course 0. So, if I actually evaluate this from 0 to t I will get i 3 minus e raise to a t into 10 10 and 0. So, which of course can be written down we know what e raise to a t of course is a shorthand for p e raise to lambda 1 lambda 2 of course this is a 3 by 2 3 matrix. So, this is what we get. So, this is our response now of course the point is what is in fact we are assuming here of course it is diagonalizable this needs to be actually checked. So, we need what is lambda 1 lambda 2 lambda 3 and also we require this matrix p. So, this is what we really require in this particular system. So, this of course has to be obtained from a now for very simple second order systems I showed you how you can take out the Eigen values and Eigen vectors we did a very simple example in the previous class, but if you really want to do an example in which is a third or fourth order matrix is used to useful to use some software. And here I will show you this computation using the software psi lab note down psi lab is freely downloadable it is available at this point of time at this particular web address w w w dot sky lab dot org. So, I will show you this computation on psi lab. So, just keep this in mind what we are doing if we start up psi lab this is what you will get. Now, I will not really tell you all about the syntax of using psi lab and so on what I will try to you know just show you the commands and the syntax is fairly intuitive. So, you should be able to pick it up on your own. So, what I will do here in order to take out the Eigen values and Eigen vectors I first enter the matrix a this is nothing, but then this is the first row then I enter the second row this is the syntax of entering the matrix then you have got the third row. So, this is your a matrix the Eigen values and Eigen vectors of a can be obtained from a command called spec of a that will give you actually the Eigen values associated with the system. So, the Eigen values associated with the system are roughly minus 5 plus a complex number this is a complex number roughly this is approximate I have just rounded off. Now, the point is these are the Eigen values one important point you are seeing is that you are getting Eigen values which are not real numbers they are complex numbers even if they are complex numbers I told you the stability criterion is that real of lambda should be less than 0. So, this is in fact a stable system because the real parts are all less than 0 one important point which you notice here again is that although these are complex number they seem to appear in complex conjugate pair. In fact, a general result which can be stated is that if a is a real matrix then its Eigen values are real and or whenever they are complex they will appear in complex conjugate pairs. So, you can have real Eigen values and or complex conjugate pairs your final response as I shall show you in some time is in fact quite is going to be real it is not going to containing imaginary terms at all. So, please remember that your Eigen values whenever for real matrix you will get Eigen values as real and or complex conjugate pairs as you are seeing here. One more thing which you should notice here remember we are trying to do a study if interpreting the results as well this Eigen value and the magnitude of these Eigen values are very much different. In fact, you will see that this particular Eigen value is small in fact a small Eigen magnitude Eigen value is associated with slow rates of change. So, you will find that your response will contain one pattern which is relatively slow and one pattern which is corresponding to a high magnitude Eigen value the rate of change of that pattern will be fairly fast. So, in fact a larger Eigen value is associated with faster rates of change. So, that is one important point which you should note these of course, Eigen values are three distinct Eigen values. So, our response will be a superposition of modes in fact. So, we do not have to worry about Jordan form or what happens how do you actually solve with repeated Eigen value this is a luckily we have got a system with three distinct Eigen values. Now, I mentioned some time ago that your response is this. Now, we also require in order to get the response to find this value p, p is the Eigen vector matrix. In fact, there is a command in psi lab which allows you take out the Eigen vector matrix that is p. So, if you do this p in fact gives you the right Eigen vector matrix. In fact, you get the columns 1 2 and 3 remember column 1 itself is complex. Column 1 is complex column 2 is also complex because they correspond to complex Eigen values and the column 3 is real. In fact, the column 3 corresponding to the Eigen vector associated with the real Eigen value. In fact, we should not just let this go as it is one important point you will notice is that the Eigen vectors associated with the complex conjugate pairs turn out to be complex conjugate themselves. So, you see this particular Eigen vector is minus practically 0 and this is almost point 1 into j j or i i is actually square root of 1 that is j I mean I have been using j the software here uses i. You will find that the Eigen vector component here is the complex conjugate of this. So, this is an important property of real matrices. So, even your Eigen vectors turn out to have certain properties. One more interesting point is I will just write this Eigen vector matrix. This is the Eigen vectors and roughly this is roughly very roughly this is j into point 1. This is practically 0 this term here other the second term is practically 0 and the third term is practically 1 roughly this is minus j 0.1 0 1 and this is minus 0.7 minus 0.7 and 0.06. So, please have a look at this screen again the computer screen I just approximated these columns just rounded it off I mean rather I should say approximated it rather than round off this is practically 1 and this is a very small number here. So, this is how I get p you can concentrate on the sheet which I am writing on you will get a p matrix of this kind. So, if I numerically evaluate now I have got lambda 1 lambda 2 and lambda 3 lambda 1 lambda 2 lambda 3 and I also have got p 1 p 2 p 3 and the p matrix. So, I can actually take out p inverse. So, p inverse is take out the inverse of p matrix. So, I just print this out. So, this is the inverse of the p matrix the columns of this this is column 1 column 2 column 3, but if you look at the row that will give you q i the you know q i transpose. So, I have also computed p inverse in this particular case, but let us now evaluate the time response of x. This is what I eventually managed wanted to take out p this and p inverse. So, if you finally compute the response I will omit the algebra what you will get is i l 1 of t i l 2 of t and v c of t is equal to this is the final response 10 minus 10 e raise to minus 0.1 t plus 0.1 e raise to minus 5 of t sign of 1 0 0 5 t and this is 10 minus 10 e raise to minus 0.1 of t. Please write down just try to work it out on your own with psi lab. I am just writing down the response directly this is of course, making certain approximations in since I have rounded off certain terms. So, this is your final response. So, your i l 2 t is 10 minus 10 and so on. Now, one important point is that lambda 1 and lambda 2 which were in fact, a complex conjugate pair of here seem to be after lot of algebra seem to be giving you a sinusoidal term here. Remember that things get so arranged because of the fact that Eigen values appear in complex conjugate pairs the complex Eigen values are in complex conjugate pairs. You will find that you should be able to make these simplifications and your responses in fact will turn out to be in terms of sines and cosines. There is no complex number in the response eventually because you will be able to combine all the complex terms appropriately in order to get these terms. Now, one of the important points after a bit of you know approximating the Eigen values and Eigen vectors these are the this is the Eigen vector and these are the Eigen values. One thing you will notice here is that the there is a exponentially decaying component of the response there is a force component because of the input which you have given and there is also a component here which is oscillatory in nature it is a decaying oscillation. Similarly, here so because of the fact that you have got a complex conjugate pair of Eigen values the mode which you get will be oscillatory there is a oscillatory mode and there is also an exponentially decaying mode. Another important point which you notice is that the oscillatory mode is practically not observable in I L 2 that is a very very important point. So, what you are finding is that in this particular system for all practical purposes of course we did make a few rounding of kind of approximations, but for all practical purposes there is no the there is no observability of this oscillatory mode in this I L 2. In steady state I L 2 turn out to be 0 I L 2 is 0 V c also turns out to be 0 if I put t tending to infinity V L and I L V L V c and I L 2 becomes 0 and I L 1 becomes 10. Does it if you look at this if V i is 1 volt what are the steady state conditions here? In steady state remember an inductor becomes a short a capacitor becomes open circuited. So, what you have really is if this is a short this is open circuit this is a short your current through here will be simply 1 divided by 0.1 which is 10 amperes this current here this particular capacitor here in steady state becomes open circuited this inductor here becomes a short circuit. So, what you will find is eventually you will have 10 amperes flowing here and going into this, but the voltage here is equal to 0. So, the steady state in case you excite the circuit with a non 0 input is going to be that I L 1 and I L 2 are going to be 10 amperes and V c is going to be 0. So, this is 10 and 10 and this is going to be 0 as t tends to infinity. So, this is an important thing which you should be able to do by setting t tending to infinity you should be able to take out the steady state final that is the final values of these terms. Now, what you really are seeing here in fact this particular example we do not we would like to interpret the results appropriately because the Eigen vectors are in fact giving you some idea about the observability of modes. If you look at p which is the right Eigen vector you see that the second state that is I L 2 the component is practically 0 that is why it is not observable. You will not see the oscillatory mode observable in the second state which is I L 2. Similarly, actually if you see here in V c you will not see the exponentially dying mode to the same extent as you will see it here. In fact it is almost 10 times more seen in these two states than it is seen here. So, that is what we really got in the final response if you focus on the sheet which I have written down here. You will find that whereas the coefficient of this e raise to minus 0.1 t is 10 here it is only 1 here. So, you will see that relatively you can observe this particular mode more in these currents than in the voltage. Of course, voltage and current are incompatible because the units are slightly different. So, you one should take care in interpreting the results. You cannot compare 10 amperes with 1 volt. So, that is one issue which you should remember. So, what I said was only you know kind of giving us some relative feel. Now, although we have written down the response it is a good idea to simply plot the response. We will just plot the response using psi lab and you can of course, see how the response looks like because it is sometimes difficult to picture if you are new to studying dynamical systems how the response is going to look like. So, if you for example, see what is the response of I L 1 or what is the response of V c. So, I will just try to you know write down the response. So, let us assume that I will evaluate the response at some time steps. So, I will I will try to evaluate the response at the time steps t. So, at discrete time steps I will evaluate the time response and plot it. So, the time response of course, of I L 1 is given by 10 minus 10 star is a product of I L 1. So, time plus 0.1 into exponent of minus 5 t and of 1005 minus t. So, I L 1 I have evaluated I L 1. I L 1 is simply this formula which is evaluated out here. I simply evaluated the formula which I had mentioned before. So, what I will do is plot time versus I L 1. So, what you see is this is the waveform of I L 1 it practically if you look at what it contains. It is basically containing mainly this term e raise to minus 0.1 t it is very easy to see this particular term. Remember that e raise to minus 0.1 t has got a time constant of 10 seconds. So, around 4 times that time constant you will find that the system here is settling to its final value. An important point is that you are not able to see much of the oscillatory response. Let us just look at in contrast V c. V c is nothing but I will plot V c. What you see is if this is I L 1 this is what V c is. So, V c of course has got us this oscillation here this initial oscillation in fact is that 1000 t cos of 1000 t which you are seeing here and there is a also a component which is decaying. So, you find that the response of course is a superposition of the two modes and both modes in this case are fairly visible. So, you see this oscillatory mode in the beginning of course the rate of decay of the oscillatory mode is more e raise to minus 5 t decays faster than e raise to minus 0.1 t. So, this is basically the response which you get. Now, if of course I plot I L 2 I L 2 is simply 10 minus 10 star x 1 star t. So, if I plot I L 2 is what I get it is almost actually kind of is plotted over I L 1. So, I L 1 and I L 2 practically have the same response. So, you see that different modes are seen in different extents in the different states. In this particular example we could really interpret the presence of the complex Eigen value pair, the faster and slower responses and also in some ways interpret the Eigen vector. In fact, in this particular example I have chosen specifically to illustrate a very very important modeling approximation which we will see in the next class. This particular system is in fact what you call a stiff system which has a combination of a slowly moving transient e raise to minus 0.1 t and a faster relatively faster oscillation which also decays faster. So, these kind of systems are called stiff systems and whenever you have stiff systems we may have problem solving them numerically we shall see later numerical integration may be a problem using simple methods, but we also would be able to make some good modeling simplifications. So, with that we stop today's lecture in the next lecture I will introduce you to some very important modeling principle.