 to all of you after a long break. In the previous lecture, we were discussing RC circuits. Before that, we had only considered circuits with resistors which have no memory. So that means that to evaluate what happens with a time varying signal, you just have to calculate point by point. Also, all the equations we got while analyzing resistive circuits were algebraic equations. Now, when you have a capacitor or an inductor, you end up with, when we have resistors only, we will have algebraic equations and when we have capacitors or inductors, we will end up with differential equations. So that is what we solved and we solved it for a couple of cases. The capacitor had some initial condition and the input was 0 and then we took a case where the capacitor initially had a 0 initial condition and we applied a step input. So now we will continue from there. We saw what happens in this case, C and the input V s goes from 0 to V p, I think I called it. Let me check. Yeah, 0 to V p at t equal to 0. Now, in this case, we saw that the capacitor voltage V c is given by V c of t equals V p times 1 minus exponential minus t by C. And the assumption, of course, was that the capacitor here is initially, that means for t less than 0 discharged. So one of you has raised your hand. So this was what we got, okay. So the general form of the expression turned out to be of this type, okay. It was of this type where Vc of t would be Vp, the applied voltage plus some exponential, okay. And this coefficient V0 had to be calculated from initial conditions, okay. Now in this particular case, we know that the initial condition on the capacitor is 0, okay. We will discuss more the business of initial conditions. And so at t equal to 0, Vp plus V0 will be the output voltage and that has to be equal to 0. So V0 turns out to be minus Vp, giving you this expression, okay. So if you start from a different initial condition, of course you will have a different value of V0, okay. So is this clear? Are there any questions on this? First of all RC circuit, how we analyzed it, the result that we have got, any of that stuff? When we had a 0 input under initial condition, this is what we analyzed first. We have C and R, okay. And the voltage across the capacitor turns out to be the initial condition of the capacitor times exponential minus t by RC, okay. So if we combine the two cases, that is, let me say that I have this circuit and the initial condition equals Vc of 0, okay. So what is the expression for Vc of t in this case? Please try to answer this. Here I am talking about a case where at t equal to 0, at t equal to 0, Vs jumps from 0 to Vp. But the capacitor is not initially discharged. At t equal to 0, it has an initial condition of Vc of 0, okay. So what is the expression for Vc of t? Please answer this. Please try to answer this one. So all you have to do is to take the previous expression that is Vc of t equals Vp plus V0 exponential minus t by RC, okay. And adjust the value of V0 for the appropriate initial condition, okay. So let me do it here. Then Vc of t is Vp a plus this whole expression. So Vc of 0 would be Vp plus V0. And this should be equal to Vc of 0, okay. So V0 would be Vp minus Vc of 0, okay. So the expression for Vc of t would be Vp plus Vp minus Vc of 0 exponential minus t by RC, okay. So this is the expression. I made a mistake in this. It should be the negative of this, Vc of 0 minus Vp. And here also it is Vc of 0 minus Vp, okay. So now this will spend some time discussing this equation. This is the general form of equation of the step response that is a response to a stepped input whether it be voltage or current for an RC circuit, for a first order circuit, okay. The RC circuit we have is an example of a first order circuit. By first order what I mean is the equation describing this is a first order differential equation, okay. So let us look at this somewhat carefully, okay. Now this RC by the way is the time constant and this minus 1 by RC equals P1. This is the characteristic frequency, okay. Now you have noticed that for all the cases we calculated, we end up with this exponential minus t by RC in the response, okay. Now this turns out to be a general property of linear systems that the response will consist of exponential of characteristic frequency times the time, okay. So exponential P1 t, this will form a part of the response of any linear first order system, okay. In fact, all linear systems have this behavior. If you do not have a first order system, if you have a higher order system, you will end up with multiple exponentials, exponential of P1 t, exponential of P2 t and so on, okay. So now this can be thought of as two parts. This part which is dependent solely on the input, this is known as the forced response of the steady state response and this part which has this exponential minus t by RC which is a characteristic of a circuit is known as the natural response or the transient response, okay. So the natural response of any linear system will be of the form of exponential minus t by RC. There are some small modifications which we will not go into but this is generally the case, okay. And the total response always consists of forced response which depends solely on the input that is the driving function of the forced state function and the natural response which depends on initial conditions and the forced state function and the natural response of the circuit of the form exponential minus t by RC, okay. So let me also write this. So this forced response of steady state response depends only on the input which is the forcing function, okay. And this natural response will always have this term exponential minus t by RC which by the way is also known as the natural mode, okay. So just like you would have a string with a characteristic frequency, right. If you block a string it will vibrate at some frequency that is called one of the modes of a string. The exponential minus t by RC is the mode of the first order RC circuit, okay. So all linear system response can be classified into the forced response on the natural response or equivalently the steady state response and transient response. And another terminology here particularly in relation to differential equations in mathematics is this is the solution to homogeneous equation and this is the particular integral, okay. So these are the same thing and these all mean the same thing, okay. So this is a general property of all linear systems and of course we are now looking at a first order RC circuit, okay. Now we can also rewrite this as VP times 1 minus exponential minus t by RC plus VC of 0 exponential minus t by RC. Okay. So you can see that this part is proportional to the input, okay and it does not have any initial condition term in it. In fact this is known as the zero state response, okay. So this is the response you would have brought if the state of the capacitor, the initial condition on the capacitor would have to be 0, okay. And this part which depends only on the initial condition is known as the zero input response, okay. Now the reason for this terminology is obvious. The zero state response is the response you get when the initial state is 0 and the zero input response is the response you get when the input VP is 0. By the way I have to constant once more that we are only considering a case of a piecewise constant input. What I mean by that is the input is not arbitrarily varying with time. It varies from 0 to VP, okay. It jumps from one particular constant value to another particular constant value. What I am particularly warning against here is that if let us say VP were to do this, VP were doing something of a sort, okay. You should not substitute VP of t in this expression, okay. That is completely wrong. You will get some integral with VP inside it. You can't simply substitute VP of t. This is for a step of VP, okay. Now this is supposed to be a linear system and the response we classified into different parts, okay. So we have a number of things. First of all we have the total response which is VP of t and we have the transient response. We have the steady state response. We also have the zero state response and the zero input response. And all of this is for a system which we know is linear. At least we know that it's made of linear components, resistor and capacitor, okay. So now my question is what does linearity mean in this case, okay. That is one way to express linearity is to say that some response is proportional to some input, okay, or something is proportional to something else. Now please try to answer this. Like is the total response linear with the input? Is this linear? Yes or no? You can type your answers into a chat window or I can post the same question in a different way. We have all these different types of responses or the total response was classified into these different things. Let me put down these things. Which of these is linear with the input? You can also indicate your choices on the poll that has appeared. Now I have listed five responses. All either the total response or part of the total response of a post-over RC circuit. My question is which of these is linear with the input to the RC circuit? And of course we are only considering step inputs at this time. Please try to answer this. In fact all you have to do is to look at the expressions for these, okay. So the expression for the total response is VP plus VC of 0 minus VP exponential minus T by RC. The transient response is VC of 0 minus VP exponential minus T by RC. The steady state response is VP. The zero state response is VP 1 minus exponential minus T by RC. And finally the zero input response is VC of 0 exponential minus T by RC. So now from this you should be able to very easily tell which of these is directly proportional to the input. Now I have got some responses. Clearly the first one is not proportional to VP because of this VC of 0 term. So this is also not proportional to VP because of this VC of 0 term. So the steady state response and the zero state response these are proportional to VP. So these will be linear with the input. And similarly this one I did not ask that question but this part the zero input response will be linear with the initial condition. So this is what linearity means in this particular context. The total response will not be linear with either the input or the initial condition in general. If VC of 0 happens to be 0 that is a different matter that is a coincidence. Otherwise the total response of the transient response will not be linear with either the input or the initial condition. The steady state response will be linear with the input because it depends only on the input and the zero state response also depends only on the input and it will be linear with the input. And the zero input response will be proportional to or linear with the initial condition. So that is what linearity means in this context. I mentioned this because we know that in linear system the response is proportional to the input. If you have a resistive circuit you change an input then the response will change in proportion to that input. I am assuming a circuit with a single input. Now in case of these circuits which have memory elements like capacitors or inductors the situation is a little more complicated and it is linear in this particular sense. Now we can also go ahead and interpret this equation further. First of all if I plot this let us say it starts from some vc of zero and reaches a steady state of vp because at t equal to zero you can see that this vc of t will be vc of zero and at t equal to infinity and the exponential has died out it is equal to vp. So the response goes as an exponential and finally reaches vp. So there are different ways to interpret this. If you look at this part of it this is the difference between final or initial and final values. And this whole term if you look at it it says that the difference between the initial and final values decays with time. So you have the final value plus any difference that was there between the final and initial values will decay with time. So that is one way to interpret this. We do all of this so that it turns out that for the step like inputs for first order systems after enough practice you will be able to simply look at the circuit and write the response. Of course we need a lot of practice to do this. Initially you should do it systematically but you should be able to do after a while and this part of course is the final value. So this vc of t in general this turns out any v of t in a first order system and by the way it could be i of t also any current as well. I have demonstrated this with the capacitor voltage in a particular circuit but it turns out that any branch quantity either voltage or current in any first order circuit can be represented in this form for a step input. So this is v final plus v initial minus v final times exponential. It is minus t by rc but to make it more general I will write minus t by tau where tau is the time constant. So this is the general form and like I said exactly the same holds for a current as well my final plus i initial minus i final exponential minus t by tau. So I hope this is clear. In fact you would have seen this type of formula but it has some basis. Essentially what it says is that it expresses it as the final value plus the difference decaying over time. Now we also had expressed the expression in a slightly different way vc of t is vp 1 minus exponential minus t by rc plus vc of 0 exponential minus t by rc. It is exactly the same expressions but divided differently as 0 input and 0 state responses. So now again writing it in a more general form v of t would be v final 1 minus exponential minus t by tau plus v initial exponential minus t by tau. So compared to the previous expression this gives you a slightly different interpretation. Exactly the same final solution of course but only the interpretation is different. Here you think of this as this term building up to the final value that is if you just look at this part of it it starts from 0 because 1 minus exponential is 0 and then it builds up to v final. And this you can think of as any initial condition that is there on the capacitor decaying with time. So t equal to 0 you will only have an initial part and it decays with time. So this is a total interpretation. You can think of the total response as one part which is building up to the final value and another part which is the initial part decaying with time. And like before this also holds for currents. We will quickly take some examples and see. So this also holds for currents. So with this you should be able to write down the step response of any first order circuit. We will also take examples of circuits with inductors and evaluate the step response without necessarily having to go through the differential equation. As usual in the beginning when you are not very fluent with this I would recommend writing down the differential equation identifying the term properly and then writing down the solution. As you get more and more practice you should be able to do this more conveniently. To be able to do this we have to identify. There are three quantities in this. Three things we have to identify that is v final of the final value v initial of the initial value and also the time constant. So we will say how to do this. Now the time constant is the same for a given circuit regardless of which quantity you pick. For instance let me take the same example RC and let us say that this has an initial condition of vc of 0 and the input steps from 0 to vp at t equal to 0. So now we have been looking at vc of t. What I would like you to do is to give me the expression for IR of t, the resistor current. Please try to do this. Please calculate the expression for IR of t, the current in the resistor. I have got a response that says it is vc of t by r that is not correct because the voltage across the resistor is not vc of t. So please try to do this. Give me the expression for IR of t. Obviously to one way to find the current is to find the voltage across the resistor and divide it by r. Alternatively you recognize that the current through the resistor is the same as the capacitor current and if you know the capacitor voltage you should be able to calculate the capacitor current. You can do it either way. I think there are a number of responses. Clearly voltage across the resistor is given by vp minus vc of t. So IR of t is vp minus vc of t divided by r which in turn can be written as vp minus vc of 0 divided by r exponential minus ty rc. So again you see this exponential minus ty rc appearing in the expression. And also let us try to relate it to the initial and the final condition. First of all if the initial voltage across vc is vc of 0 then the initial voltage across this r so let me write it like this vr is in this direction. So vr initially would be or let me write the IR directly. So I initial is when I say initial it is just after the step is applied. So it is vp minus vc of 0 divided by r. Now a useful special case to consider is when vc of 0 equals 0. In that case you know that the capacitor voltage is 0 initially. The input voltage here jumps up the capacitor voltage cannot change. So the voltage across r will be equal to vp that is what we see. If vc of 0 is 0 the initial current is vp by r and that is what will charge the capacitor. We also saw that that will be the slope of vc that will be related to the slope of vc. And what is I final? I final is what does it mean to have steady state in a circuit with capacitors and a piecewise constant voltage? Steady state means that the voltages in the circuit will be constant with time. So if the capacitor voltage remains constant with time then there is no current through the capacitor. So it is like an open circuit. So clearly the current through the resistor is the same as the current through the capacitor. So I final will be equal to 0. And you again see that whatever we have here this response confirms to the general formula which is I final plus I initial minus I final exponential minus t by tau. Because I final is 0 we have only the, this part is not there and we have only this part. This is okay. So every quantity I mean this is a very simple circuit but every quantity will correspond to or will conform to a formula this or this. By the way these two formula are the same the only difference is the terms are grouped differently and interpreted differently that is all. Is this fine? Now we should also try and see how to evaluate the initial and the final conditions by looking at the circuit. Let me first take the final condition. Final condition is after the steady state is reached and a steady state because we are looking at constant inputs means that voltages are constants. So now if a capacitor voltage is constant any capacitor voltage it means that derivative is 0 and that means that the current is 0. The current through the capacitor is 0. Now if the current through the capacitor is 0 that means that it can be replaced by an open circuit. So replace the capacitor by an open circuit and find the quantity whichever quantity is of interest to you. So this is how you would find the final value. Let us do it for the case we took. This is our circuit RC and let us say I am interested in the capacitor voltage VC. Now to get the final value what I would do is to open circuit the capacitor that is the capacitor is not there. So clearly the voltage that appears here equals VP and this would be the final value. Now this circuit is very simple but the same principle applies to all the circuits. Similarly let us say I am interested in the resistor current then again I will take the case of input stepping from 0 to VP to calculate the final value I open circuit the capacitor. Clearly this current would be 0. So this current would be 0 and that indeed is the final value of IR. We will take a slightly more complicated example later. So is this part clear? Any questions about what we discussed so far? What we did was to extend our discussion from last class. There we had mathematically evaluated the solution to the differential equation. Now here what we have done is to identify certain characteristics of the solution to first order system with a step input and we can use those characteristics directly to write down the solution with a little bit of experience. So if you have any questions I will take them now. It appears that things are clear. So we have to identify the time constant. So there is a question but it looks like it is incomplete. I am not able to understand the question here from LG Industries. We also have to identify the time constants that I will come to later. In case of our circuit which has only R and C it is very easy. So let me call those Vs. So first of all you deactivate the source. Sources determine the resistance across the capacitor and finally let us say the resistance is R in this case and finally the time constant tau will be R times C. Now this is very simple and silly for this case because we have only R and it is very obvious what the resistance across the capacitor is. When I say across the capacitor what I mean is between these two terminals. Now this principle applies even to very complicated circuits as long as you have only one capacitor. Let us say I have a circuit with a number of resistors and I will show only the capacitor like this connected to terminals 11 prime. I also have a number of sources. I will show one example here. So now what I have to do to find the time constant is first of all I null this circuit that means that any independent source is set to 0. If it is a current source it is an open circuit. If it is a voltage source it is a short circuit and I have these two terminals 11 prime across which the capacitor is connected. So what I have to do is to determine the resistance of the circuit looking into 11 prime. Now this we have done many times. It is basically the terminal equivalent resistance looking into the terminals 11 prime. So this RTH is what we have to determine and once we do that the capacitor value is C the time constant term will be given by RTH times C. So you can very easily verify that it is true for R's very simple circuit. Looking between these two terminals to the left side we have only R but even if you have a very complicated circuit it could be true. You would have to find the equivalent resistance looking this way and the time constant would be RTH times C. Is this clear? So we know how to find the final value and we know how to find the time constant. We also have to find the initial value. When I say initial value it will be just after the step is applied. That also we will proceed to do. So if we have any questions please interrupt. I am not getting many questions today. So that means that I will be understanding everything or probably understanding nothing. So if you are understanding everything it is good but if you have difficulty understanding something please ask the questions on the chat window and I will answer them. So RTH. RTH is the notation I use for the Thevenin resistance looking into 11 prime. So if you have a circuit with two terminals 11 prime then when you deactivate the circuit and find the resistance that is the Thevenin resistance across those two terminals. So that is RTH. So now we have to find the initial condition. Again I am talking about the initial value of any variable not just the capacitor voltage. There is another question. Can we use RL network for this? It is not clear to me. I am talking about circuit with resistors and inductors. Yeah the same principle applies. I will come to those examples a little later. Exactly the same principle applies because with a single inductor it is also a first order circuit. Now let me look at how to calculate the initial condition or the initial value. When I say initial value it is value just after t equals 0 plus. So again let me take my simple circuit example. The input jumping from 0 to vp. I have a resistance r and I have a capacitance c initially charged to vc of 0. So the principle we use to find the initial value is the following that a capacitor voltage cannot change instantaneously unless there is infinite current. In fact we will come to cases where there is indeed infinite current. But initially we will assume that there is no infinite current. So the capacitor voltage does not change at that instant. So for that particular instant capacitor can be replaced by a battery whose value equals the initial condition. So that means that instead of this vc of 0 voltage does I have a battery or maybe let me use the usual symbol for the voltage source vc of 0 and this is vp just after t equal to 0. So now clearly we see that any quantity we take. So this is the network from which we calculate the initial condition. So you replace the capacitor by a battery and then you will do the circuit analysis. So let me write the dom first and you determine initial value from this circuit. So first let me say that I am interested in this voltage. That was the capacitor voltage. Clearly the initial value equals vc of 0. Again it will become more clear if we take a slightly more complicated example. And if I am interested in the resistor voltage the initial value of that equals vp minus vc of 0 because this is now when you remove the capacitor this is just DC circuit analysis. Circuit analysis with DC sources. So it is easy that is why we choose to do this. And let us say I had I was interested in the inductor current and the initial value for this sorry the resistor current not the inductor current. The initial value for this is vp minus vc of 0 divided by R. So now we know how to calculate all three parameters that appear in the general equation for the solution to the first order system with a step input. What are the three things? The initial value, the final value and the time constant. The initial value is obtained by replacing the capacitor with a battery equal to the initial condition value. If the initial condition is 0 then you replace the capacitor by a short circuit. So then you find whatever quantity you want. It could be a current, it could be a voltage. Whatever quantity for which you are trying to write the expression that is that you find from this new circuit and that is the initial value. How do you find the final value? We are looking at constant inputs. So in steady state all the voltages and currents will be constant. If capacitor voltages are constant its current is 0. So we can simply open circuit or remove all the capacitor from the circuit and from this new circuit we can analyze the final value and finally the time constant. How do we find this? We look at the terminals across which the capacitor is connected. We find the equivalent resistance looking into it after nulling all the independent sources. So that gives you the time constant. So we have all three. So we should be able to simply write down the expression whether it is voltage or current V final plus V initial minus V final exponential minus T by tau. So is this part clear? Any questions about this? Okay, so now let us take an example which is only very slightly more complicated than what we had. And of course I would like you to participate and answer questions that way you will get practice in doing things like this. So let us say I will again take the same kind of input, input stepping from 0 to VP. Now I could calculate any quantity I want. So first I will try to okay let us try to calculate these two things. This is VC and another one is IR2. So first let us calculate VC. Let us say this is what we are interested in. So we need the three things, the initial value, the final value and the time constant. So please give me the initial value for VC of T. Let me, this is rather because we have already calculated the capacitor voltage. Let us do it for IR2, the current through resistor R2 of T. And let us say the capacitor has an initial voltage, initial condition VC of 0. So now please give me the initial condition for IR2 of T. And the algorithm how to do this, the initial condition means just after the step is applied and I have told you how to do this. You replace the capacitor with an appropriate battery and do this. So please give me the expression for IR2 of T, the initial value of that. What is the initial condition? What is the initial condition IR2 of 0? So that is correct. It is IR response from IR3. It is clearly IR2 of 0 is VC of 0 divided by R2. In fact, for this case, you do not even have to replace it by a battery and analyze because if we have VC of 0 here, that appears directly across R2. So initially the current here would be VC of 0 by R2. So now let us get to the final condition, which is also typically denoted as IR2 of infinity. What would this be? So again, I outlined the algorithm all you have to do is to follow that. Of course, it is not a good idea to blindly follow the algorithm without understanding it. So if you have difficulty with the logic, then get back to me and I will explain it again. So again, there is a response and for the initial condition, we replace this with a battery and that voltage appeared across R2. For the final condition, we have to open circuit the capacitor. So we will be left with this VP, which is the value just after T equal to 0. So open circuit the capacitor and do this. Once we open circuit the capacitor, the value here, the current here would be VP divided by R1 plus R2. So the final condition is VP divided by R1 plus R2 because once the capacitor is open circuited, I mean no current flowing through that, it is VP by R1 plus R2. And finally, we need the time constant. What is this going to be? So again, I outlined the algorithm for this. You have to look at the resistance that appears across the capacitor with all the independent sources being null. So please do that and let me know what will be the time constant tau. So what we need to do is we have this circuit here. So first, I identify the terminals across which the capacitor is connected, that is this much. And also I said V has to be 0. When V has become 0 and voltage source is 0, it is a short circuit. So I will end up with this R2. I have to find the resistance looking into 1, 1 prime. And clearly that resistance is R1 parallel R2 because you see that R2 R1 appears in parallel with R2. So the resistance here is R1 parallel R2 and the time constant is nothing but the product of the resistance and the capacitor C. Time constant tau is R1 plus R2 times C. So with this we will be able to construct the expression for IR2 of T. So IR2 of T is the final value plus the initial value minus the final value which decays over time, which is basically V s by R1 plus R2 plus, this should be IR2 of infinity. IR2 of 0 is Vc of 0 by R2 minus Vs by R1 plus R2 exponential minus T by R1 parallel R2. By the way, this notation for R1 parallel R2, I think all of you are familiar with, R1 parallel R2 times C. So this is the expression for IR2 of T. In fact, you can write down the differential equation and make sure that this satisfies this. I encourage you to do that especially in the initial parts when you do not have enough practice. So this is by identifying a general part of in the solution to the first order differential equation. For a step input, we can write down the any expression. We wrote it for IR2, I could write it for the current through R1 or the current through the capacitor or any quantity in this circuit. Are there any questions at this point? So one question is, where are we taking tau as R1 parallel R2? Tau is not taken as R1 parallel R2. The time constant tau is the Thevenin resistance looking into 11 prime times C or Tx times C which is as I said, the Thevenin resistance looking into this is R1 parallel R2 times this C. So I do not know if this is a confusion with notation. This is R1 parallel R2 times C which is the same as R1, R2 by R1 plus R2 times C. Now another question is how to find IR1 of T and if it is the same? Now clearly it will not be the same as IR2 because IR1 will be IR2 plus the current through the capacitor. So it will have a different expression and you can evaluate it. You can find the initial condition on IR1, the final condition on IR1 and the time constant will be the same because you see that the procedure to evaluate time constant does not depend on where what input is there because we null the input and what output we are taking. That is what variable we are considering. The time constant is the property of the circuit. So the time constant will be the same for that. So you can evaluate IR1 by yourself by following this algorithm. Any other questions? So you can try it for any other circuit as well. So if you want some examples you can, for instance you can try this and you can calculate any value. For instance you could calculate Vc of T or you could calculate IR2 of T and again assume that the input jumps from 0 to Vp. Now also the input does not have to be a voltage. It can be a current. Let us say I have some IS and I have a capacitor and resistor in parallel. Let me take this very simple case and let me say that the initial condition here on the capacitor is Vc of 0 and the current steps from 0 to Vp. So let us try to do this very quickly. For this case that is this problem I encourage you to try the upper problem by yourself. This problem, so what is the and let me say what I am interested in is this particular voltage which is the voltage across the capacitor Vc of T. So what is the initial value of Vc of T? Please give me some responses and I know that to construct the total response I need the final value of Vc of T and also the time constant. So first please give me the initial value of Vc of T. This is very very easy in this case. So clearly we are looking at the voltage across the capacitor. So this Vc of T at T equal to 0 is Vc of 0 itself. So this is Vc of 0 and what is the final value of Vc of T? So if we open circuit this capacitor for the final value all of this IS will flow into the resistor. So the voltage across this is the current times the resistor. So the final value of Vc is simply Ip times R and what is the time constant? Again it is very simple in this case. You have to null the independent source that is you have to null the current source. So if you null the current source that is you open circuit it. So this part goes away completely and across the capacitor you just have a resistor R. So the time constant is the resistor you have across the capacitor times the resistance value. So it is Rc. So the expression for Vc of T is the final value plus the initial value minus the final value times exponential minus T by Rc. So please think about these things. If you have any doubts or something needs to be further clarified we can discuss them in the next lecture. So what we have done is to detect a general pattern in the response of a first order system. In the next lecture we will extend it to RL circuits. It is a very simple extension because we have already done Rc where we have spent so much time on it. It is quite easy to do that and then move on to second order circuits. Okay. Thank you. I will see you on Thursday.