 In the previous lecture, we discussed in detail opamp circuits, we looked at a couple of example opamp circuits and also I emphasize that opamp has to be used in negative feedback and I also mentioned how to find the signs of the opamp in any circuit such that it is in negative feedback. Okay, an opamp is basically a voltage controlled voltage source with a very large gain because the gain is very large and it is a negative feedback, it forces the input signal of the opamp to be a very small value which can be considered to be 0 and if the gain of the opamp goes to infinity that is the idealized version of this which we call the ideal opamp in that case the input difference voltage of the opamp can be is assumed to be 0 okay and there is no input current into the opamp. So based on these things you can analyze any opamp circuits that you have okay. So if there are any questions regarding what we did in the last lecture please go ahead now otherwise we will start with today's topics. So far we have looked at circuits that have resistors and control sources and so on and all these are known as memoryless elements. What is meant by this is that the current at instant T depends only on voltage at instant T that is it does not depend on the voltage at any other instant okay or similarly if you have a voltage to current relationship the relationship is exactly the same and for all instants and more importantly the voltage at some instant is related to current at that instant only okay similarly for control sources the control quantity is related to the controlling quantity instantaneously only at that instant. Now if we have capacitors and inductors we know this these properties are not true if you have a capacitor the voltage across the capacitor depends on not only the current at that instant but on the history of the current. Similarly if you have a current through the inductor it depends not only on the voltage across the inductor but on the history of the voltage across the inductor okay. So capacitors and inductors are elements with memory okay so what we will do in this lecture is to try and analyze circuits which have inductors and capacitors in addition to resistors is this okay many questions so let me take a circuit that we are all familiar with a resistive divider okay now let's say I call this V s and the voltage is 0 okay what will be the output voltage please try to answer this what is the output voltage when I say output in this case I have defined the output to be this particular voltage okay what is that voltage going to be what will be the voltage here when the input voltage V s is 0 anybody what is the voltage going to be so this voltage clearly from the voltage divider formula is V s times r2 by r1 plus r2 okay and if V s is 0 this voltage is also 0 okay this is a simple this is a simple resistive divider let's say it jumps from 0 to some voltage V p okay what happens then the output voltage jumps from 0 to V p times r2 by r1 plus r2 okay so you can see that the variation of the output voltage with time the horizontal axis here is time this is time and here also it is time the variation of the output voltage with time is exactly the same as the variation of the input voltage with time with a scaling factor okay now let's say we do something we have V s and I replace r2 with a capacitor and I have R and C okay what happens in this case let's say initially the capacitor voltage is at 0 and the input is also 0 clearly in that case what happens is that the capacitor voltage is 0 and because V s is 0 no current flows through the resistor which means no current through the capacitor so if there is no current through the capacitor its voltage cannot change okay so in this case the voltage remains at 0 if the input is 0 now let's say the input jumps from 0 to V p okay what happens to the voltage please try to answer this let me call this V c V c when V s jumps from 0 to V p okay there is some of you attempt to answer this what is the voltage V c going to be when V s jumps from 0 to V p okay we have an answer it says that 1 over C integral I of t V p okay this is correct but the only issue we have now is that we don't even know what this I of t is okay so that also we have to solve for I mean we have done this for the simpler circuit before the current in this is V s by r1 plus r2 okay whereas here we have to find out and the current itself depends on the voltage of the capacitor okay so now what I would like to do is to write the equation for this circuit so that I can solve it I will do what I will do is this I will write kcl at this node so the current through this I will express everything in terms of the capacitor voltage V c so the current through this is C times dV c by dt we know that that is the current through a capacitor which has a voltage V c across it okay now that has to be equal to the current coming from the resistor so what is the current through the resistor in terms of V c please try to answer this okay that is correct so the capacitor current which is C times dV c by dt equals the current in the resistor which is V s minus V c divided by r okay what I will do is I will group all of the terms containing V c to the left hand side and I will also multiply both sides with r so here rc dV c by dt plus V c equals V s okay now this is a feature of circuits containing capacitors or inductors or both that the equation governing the circuit will be a differential equation okay whereas previously they could write if I still call this V c there is no capacitor here I will just call that voltage V c V c equals V s r2 by r1 plus r2 okay so it is an algebraic equation whereas in this case it is a differential equation if you have resistors or capacitors okay so we have to solve the differential equation to find the solution okay so let us try and do that so before that we will try and solve a couple of cases intuitively and then go on to finding the solution of the differential equation okay any questions so far so let me rewrite the circuit and the equation again okay this is the circuit and for arbitrary inputs it is not very easy to solve for these things so we will take specific kinds of input so initially in particular I will take an input that goes from 0 to V p at a particular time instant okay so this is t and let us say this is t equals 0 okay and we also know that the differential equation governing this is rc dV c by dt plus V c equals V s okay so now first let us forget the differential equation and try to sketch the output based on what we know about resistive capacitor and the voltage okay I already said that initially the voltage across the capacitor is 0 okay so if I try to plot V c versus time okay before t equal to 0 the voltage will be 0 so that is what I have assumed okay so this is because I say initial condition but I really mean 40 less than 0 okay on the capacitor now at the instant that at the instant that the V s steps from 0 to V p okay what happens to the capacitor will it change will the capacitor voltage change when I said t at t equal to 0 I mean at this instant will it suddenly change that's the question what do you think yeah so as Aarti answered the capacitor voltage cannot change suddenly now as I emphasized before this means that we are already assuming that the currents are finite okay so if the current is infinite the capacitor voltage can change suddenly but the current is finite and in this case we cannot have infinite current okay because if we do an infinite current through the capacitor it means that we also have infinite current through the resistor which in turn means an infinite voltage across the resistor okay so this capacitor voltage will not change will not jump at t equal to 0 okay now my question is what is the current through the capacitor at t equal to 0 what is the capacitor current at t equal to 0 so clearly it is V s by r or V p by r because the value of V s is V p so the capacitor current at t equal to 0 this is V p divided by r okay so that means that this current I see at t equal to 0 is V p by r now because the current is flowing through the capacitor its voltage tends to increase its voltage tends to change and from the direction of the current we know that V c is going to increase okay so what happens is it increases so I will show a little bit of increase like this now what is the slope of this increase going to be what will be the slope please try to find out because we know the current through the capacitor so we should know the slope of the voltage across the capacitor okay because the capacitor current is nothing but c times dv by dt dvc by dt so the current through the capacitor is basically related to the slope of the voltage across the capacitor okay so what is the slope of the capacitor voltage versus time so when I say slope of the capacitor voltage versus time obviously it should have a dimensions of volts over time volts per second something like that right so we know the capacitor current V p by r okay so we have to equate that we have to use that in the equation for the capacitor current and find out the slope of the capacitor voltage please try to calculate that so someone please try to answer this question so the current through the capacitor is V p by r which also in terms of the capacitor voltages c times dvc by dt okay so what I was looking for was dvc by dt the slope of the capacitor voltage okay so clearly that is equal to V p by rc okay so the slope at t equal to zero will be V p by rc okay so it increases a little bit I have shown it with this blue line over here now let's say after it increases a little bit let's say I come to this point what happens then to the current through the capacitor please try to answer this now initially at this point the current equals V p by r what will it be at the other point a little bit I mean for t a little bit more than zero the capacitor voltage increases so what happens to the current through the capacitor in this circuit somebody please attempt this current through the capacitor here what happens to the current through the capacitor so there is an answer that says slope less than one but keep in mind that here we are plotting voltage versus time so the slope is measured in volts per second okay so when we have a dimension quantity like that it doesn't make sense to say it is less than one okay so it has to be less than some quantity okay so another answer is that it is V p by r but is it true why is it V p by r what is the voltage across the resistor now the voltage across the resistor at this time instant what is it going to be is it V p is it less than V p is it more than V p exactly so the current through the capacitor is V p minus V c by r now when the capacitor voltage was zero the current was V p by r but when the capacitor voltage increases a little bit the current will decrease okay so the current here is smaller than the current there now what does that mean because the current is proportional to the slope of the capacitor voltage or the other way around the capacitor voltage the slope of the capacitor voltage is proportional to the current through it so the capacitor voltage will increase a little okay but after that the slope will reduce because the current has reduced now if you look at this point the voltage across the capacitor has increased further okay so that means the current through it is reduced further if the capacitor voltage increases the voltage across the resistor decreases because the capacitor voltage plus resistor voltage equals V p okay in this time in this range of time it equals V p which is a constant so as capacitor voltage increases the resistor voltage decreases so the slope reduces further okay so the slope will go on reducing okay now in reality it doesn't reduce in discrete steps like this continuously it will go on reducing because as soon as the capacitor voltage increases a little bit the slope will reduce then it increases a little further it will reduce and so on okay it turns out that the slope will go on reducing then finally it will asymptotically read some value okay it will read some value and after that the voltage does not change much okay so the slope becomes almost zero now what is this voltage what is the voltage it almost reaches what is the voltage at which it stops changing further please try to answer this clearly if the capacitor voltage is V p the voltage here is also V p and the resistance has zero volts across it so that means there is no current flowing through it okay so the capacitor voltage is V p and there is no current through the capacitor so that means that the capacitor voltage will not change at all okay so this would be equal to V p okay so I hope this intuitive description of how an RC circuit responds to a step is clear if not please ask your questions and I will clarify it further what happens is let's say you start from a discharged capacitor that is a capacitor which has no charge and in this circuit at t equal to zero there will be no infinite currents so the capacitor voltage will not change instantaneously but at t equal to zero because the input voltage increased to V p the current will increase okay so the current through the capacitor increases and sorry the current through the capacitor will be non-zero it will be V p by r so that means that the voltage across the capacitor increases as the voltage across the capacitor increases the voltage across the resistor reduces and that reduces the current and the slope of the capacitor voltage okay and this happens continuously the capacitor voltage will increase little by by little and the current through it will reduce little by little reducing the slope of increase. Finally, the slope of the capacitor voltage almost becomes zero and you can kind of say that the capacitor voltage reaches some value and that value is Vp. That is because the capacitor voltage will stop changing when the current through this is zero. If the current through this is zero, current through the resistor is zero, the voltage drop across the resistor is zero. So, that means that here also you have Vp. So, please think about this and ask me any questions you may have. The slope is Vp by R at k equal to zero. So, gradually reduces as Vc increases and Vc syntotically reaches Vp. What is meant by asymptotically is that it gets very very close to Vp, but does not quite get there. It gets there only at t equal to infinity. So, finally, when Vc equals Vp, current through the resistor and the capacitor is zero and Vc does not change anymore. So, any questions about this? Behaviour of an RC circuit for a step input, this kind of input which jumps from one value to another value is known as a step input. Now, let us go the other way around just for completing the calculation. So, let us say initially the capacitor voltage Vs is at Vp and the capacitor voltage Vc is also at Vp for t less than zero. So, imagine that we take the capacitor and charge it up to Vp. So, previously we charged it almost to Vp. So, we charge it up to Vp and then after some time we bring the input Vs down to zero. So, this is exactly the opposite of what we had earlier. Instead of going from zero to Vp, we go from Vp to zero. So, now if you look at the differential equation that governs this circuit, it is RC dVc by dt plus Vc equals Vs and this is nothing but zero for t greater than zero. So, we have a differential equation with zero on the right hand side. That is the differential equation that governs this particular circuit in this condition. So, again I put Vc versus time. So, like I said initially Vc is at a value Vp and also for convenience I will redraw the circuit with Vs equal to zero. With Vs equal to zero it is a short circuit and I have this C and R and the voltage across this is Vp at t equal to zero. Now, what will be the current through the capacitor? So, let us say I call this as Ic and because Vs equals zero I have shown it as a short circuit. So, if the capacitor current is flowing like this I have used the passive sign convention. What is the current Ic? So, there is an answer Vp by R, but please mind the signs because I have used passive sign convention for Ic. So, flowing from top to bottom that is right. First of all the voltage across the capacitor and the resistor they are the same in this circuit they are in parallel. So, this voltage with this polarity is Vp at t equal to zero which means that a current Vp by R flows in that direction. So, this Ic which is the current flowing from top to bottom in the capacitor is minus Vp by R. So, what happens to the capacitor voltage? After t equal to zero capacitor voltage we see for t greater than zero. What happens now? Will it increase or decrease or remain the same? So, that is right it is going to decrease. So, it will decrease and the slope is again given by the capacitor current. Minus Vp by R is the capacitor current C dVc by dt. So, dVc by dt is nothing but minus Vp by Rc. So, slope is minus Vp by Rc. So, there was a question here that was whether after charging a capacitor to some voltage can we use it as a voltage source? Yes we can in fact it is used like that in some cases, but of course you have to remember that a capacitor is not a voltage source. As you draw current from it its voltage is going to change. So, after some point the voltage changes so much that you cannot use it anymore. So, let us say you have you want a 5 volt source you can charge a capacitor to 5 volts and use it, but as you draw current from it the voltage will start decreasing and at some point it may go so below 5 volts so far below 5 volts that it may be useless. But in fact this is used if you want a source for a very short time where you are not drawing so much charge or so much current from the capacitor you can use that. In fact in many electronic gadgets when you want to change batteries during the time you take to change the battery there will be no source. So, some of those gadgets may use a very large capacitor. So, in the time that you take to change the battery the capacitor will be acting like a voltage source and supplying power, but of course it cannot do it indefinitely. So, if you leave it without batteries for a long time it will just discharge and die out. So, the slope is minus Vp by RC. Let us say we come to this point. So, what happens to the slope after that? What happens to the current through the capacitor and what happens to the slope of the voltage across the capacitor? The voltage across the capacitor is decreasing, but after it decreases a little bit what happens to the slope of the voltage? Now clearly the voltage across the resistor is also VC and as VC reduces the voltage across the resistor reduces which means that the current through it reduces and if the current reduces the slope of the voltage also reduces. So, it does that and then does that and that and that and so on. So, finally it turns out that it will asymptotically reach 0 because the input is 0 for T greater than 0 and if this capacitor voltage goes all the way to 0 then there will be no current through the resistor and that means that the capacitor voltage cannot change at all which in turn means that the capacitor has reached its steady state. So, this asymptotically goes to 0. Is this clear? Any questions about this? So, now we have looked at the behavior of the capacitor voltage for two cases when the input jumps from 0 to Vp or from Vp to 0. Qualitatively it is the same, it starts changing with a certain slope the capacitor voltage and the slope goes on reducing. The magnitude of the slope goes on reducing. So, it starts of steeply and then becomes less and less steep and finally asymptotically reaches some value. When you change the input from 0 to Vp that value is Vp. When you change the input from Vp to 0 that value is 0. This is fine. So, key things to remember are that are the values of the slopes and so on. So, this slope is minus Vp by Rc. So, it starts from Vp and if it went like a straight line with that slope it would cut the time axis at the equal to Rc. But of course it does not do that because the moment the voltage reduces a little the current also reduces. So, the slope reduces and so on. So, now what do you think this shape is? We have done it intuitively and graphically but we have to also be able to write a formula, an expression for Vc as a function of time. So, what is that function? Any guesses for what that is? So, let us see. Let me take the second case where the input Vs was 0 after t equal to 0. So, for t greater than 0 this is the model with Vs set to 0. Now, let us see what shape it is. Somebody answered parabolic. It turns out it is not a parabola. The differential equation governing this is Rc dVc by dt plus Vc equals 0. Now, I will rewrite this as dVc by dt is minus 1 by Rc times Vc. So, what this equation is saying is that the derivative dVc by dt is proportional to the function itself. So, the derivative is proportional to the function. So, what is the function that gives you a derivative which is essentially the same function? It may be a scaled version of that function, but it is the same function. So, for instance let me do it with some other variables. I could have something like this. If I have dx by dt equal to x, so in this case also the derivative is proportional to the function. In fact, I have made it exactly equal to the function. Now, you know how to differentiate functions. So, what is the function that you can recall that has the derivative which is the same function? Please try to answer this. Derivative is proportional to the function or maybe even equal to the function. What is the function that meets this criterion? So, if you want you can think of the table of derivatives that you would have studied when you studied calculus. So, think of a function whose derivative is the same function. So, clearly an exponential satisfies this because the derivative of the exponential is also an exponential. So, now, let us say I do not want the derivative to be equal to the function, but only proportional to the function. So, I also know that if I had a scaling factor here exponential at and I differentiate it, I get a times exponential at. So, clearly in this case if we see is as an exponential, this function will be satisfied. Also, I want a particular scaling factor. So, you see that this equation is of the same form as this. So, the derivative of something equals a constant times the same function. So, what is the function that will satisfy this equation? It is Vc of t of the form of exponential minus t by Rc. So, clearly now, I also can have a scaling factor here. So, let us say V0 because that will cancel out. If Vc of t is V0 exponential minus t by Rc, please give me the value of dVc by dt What is dVc by dt? If Vc is V0 exponential minus t by Rc, I use these expressions to find out exactly what goes inside the exponential. If it is exponential at, the multiplying factor here is a. I want the multiplying factor to be minus 1 by Rc. So, a has to be minus 1 by Rc. So, Vc of t is V0 exponential minus t by Rc. So, just differentiate this and tell me what the value is. Clearly this is minus V0 by Rc exponential minus t by Rc. So, you can see by inspection that this will satisfy this equation because the derivative is minus 1 by Rc. So, this part of it times the rest of it which is the function itself. So, the solution to Rc dVc by dt plus Vc equal to 0. The solution to this is Vc of t equals V0 exponential minus t by Rc and V0 can be anything as far as the differential equation is concerned. So, any value of V0, this differential equation will be satisfied. But in our circuit, we will need a particular value of V0. So, how will we find this? How will we find the value of V0? So, there is another answer which says V0 by Rc exponential minus t by Rc. I think you have omitted a minus sign it should be minus V0 by Rc exponential minus t by Rc. So, now we know that the shape of this blue curve here, the shape of this blue curve is of the form of V0 exponential minus t by Rc. So, this is this exponential describes the response of these types of circuits. And also we know that at t equal to 0, it starts from Vp. What is the value of this function at t equal to 0? The value of this function at t equal to 0 is if I substitute at t equal to 0 in this exponential, I will get V0. But I know from the circuit that it is going to be Vp. So, for this to correctly describe the blue curve here, this V0 has to be Vp. So, this curve is Vp exponential minus t by Rc. So, if initially the Kepler's curve had a voltage Vp and the input also had a voltage Vp, then the current through the capacitor, current through the resistor would be 0, current through the capacitor would be 0 and it would stay at Vp. But if Vp jumps, Vs jumps from Vp to 0, then the output will slowly discharge all the way to 0 towards 0. Now, we have also determined that function. The function is Vp exponential minus t by Rc. And we have intuitively determined the shape that it has some slope initially and then it goes on decreasing. And that is exactly what the exponential does. Please keep in mind that this exponential is the exponential of t, but with a minus sign inside. So, if we have an exponential with a plus sign, it will blow up. It will go on increasing with time. Here exponential of minus t by Rc, it will go on decreasing with time. So, any questions about this? The capacitor voltage across this will follow an exponential. I will redraw the input and output. Vs changes from Vp to 0, it jumps. So, this is Vs of t. And Vc of t, we will assume initially it was at Vp and it slowly goes to 0 like that. And this is Vc of t and it equals Vp exponential minus t by Rc. And the differential equation governing this is Rc dVc by dt plus Vc equals 0. And the solution is Vc of t is Vp exponential minus t by Rc. So, by now you should be comfortable with these results. If you have any questions at all, let me know. I will clarify that further. Any questions about this? Now, in this circuit, in this condition, the condition we are evaluating is with 0 input. So, if you look at this differential equation, it has only Vc, it does not have anything corresponding to the input. The input is 0. So, such a differential equation where the input or the right hand side is 0, I will always try to group the variables on the left hand side and the constant inputs on the right hand side. Here Vc is the variable and this kind of an equation which has 0 on the right hand side and only terms with Vc on the left hand side, this is known as a homogeneous equation. So, if you have taken some mathematics courses on differential equations, you will be familiar with this term. But in this course, whatever we need of differential equations, we will study them ourselves. And this comes because we have 0 input is known as the 0 input response. And this term which appears inside the exponential Rc, this is known as the time constant. It also appears here as coefficient of dVc by dt, but we have to be careful. The coefficient of Vc has to be 1. In that case, the coefficient of dVc by dt will be the time constant. So, it is called time constant because clearly Rc has dimensions of time. We have exponential minus t by something. So, this has to have dimensions of time so that we have exponential and inside that some dimensionless constant, dimensionless number. So, this is called time constant. Any questions about this? So, like I said before, the initial slope is Vp by exponential of Rc. Now, at t equals Rc, what is the value of Vc? What is the value of Vc at t equal to Rc? Please try to calculate this. Many of you were able to answer this. It is Vp times exponential of minus 1 or Vp divided by exponential of 1 or numerical value is 0.3679 times Vp. So, if I mark one time constant over here, this is t equals Rc. Maybe it is not to scale. I should redraw this a little better. So, that is what happens. And here the value is 0.3679 Vp. Now, if I go to two time constant, it will be 0.3679 square and so on. So, it will be approximately 0.1353 and it will rapidly reduce. Now, it turns out that after approximately 5 time constants or 4.6 time constants, it goes down to 0.01 Vp. t equals 4.6 Rc. We approximate this to 5 Rc. The point here is that it will never completely discharge to 0 because if you look at this function Vp exponential by Rc, it will never become equal to 0. It becomes equal to 0 only when t tends to infinity. So, in a finite time it never goes to 0 but it becomes arbitrarily small. Now, how small you want it to be depends entirely on the context. Sometimes you may want it to be 0.1 percent of the initial value that is 0.01 Vp. Sometimes you may want it to be 0.1 percent or even smaller. But this is a good number to remember. For the voltage to reduce to 1 percent of the initial value, you have to wait for 4.6 time constants. So, many times the time constant is denoted by the letter tau and for this circuit it is equal to Rc. And in 4.6 time constants, it reduces to 0.01 Vp. Any questions about this? So, when the input steps from some value to 0, we know how to solve it. We have the homogeneous equation and by solving the homogeneous equation, by solving the homogeneous equation we get the output to be an exponential and this is the 0 input response also meaning we have calculated this response for the period when the input is 0. So, now let us get back to the other case where we have R and C, VS and the differential equation governing this is Rc dVc by dt plus Vc equals VS. The right hand side is no longer 0. So, again I will not take arbitrary functions for VS, but I will take something that jumps from 0 to Vp at t equal to 0. So, this red stuff is VS of t. So, for t greater than 0, the right hand side of this is a constant equal to Vp. So, now we already intuitively evaluated that it starts with a slope of Vp by Rc and it will do that. It will do something like that. So, now we also have to find a function that corresponds to this. So, any idea of what this function might be? What might be this function? So, there are a couple of answers. One of the answers is Vp 1 minus exponential minus t by Rc that is correct. In fact, we will derive it properly. Also there is an answer saying positive exponential and not sure what is meant there. If you mean that the argument of the exponential will be positive that is not correct because if the argument of the exponential is positive, you will get a curve that is like this. So, it will go on increasing, it will blow up. At t equal to infinity, it will become infinity. That is not the kind of function we have as we have determined from intuition. So, in terms of that we can very easily derive this because we know the result to the previous one when the right hand side is 0 for the homogeneous equation. In a very easy way, we can reduce this to the homogeneous equation. So, there are standard tricks and calculus that all of you are familiar with I think. So, let me redefine a Vc 1 which is Vc minus Vp. Now please tell me. So, Vp is a constant. So, keep this in mind. So, what will be dVc 1 by dt? What is this going to be? If Vc 1 is Vc minus Vp, what is dVc 1? What is dVc 1 by dt? So, please tell me what is dVc 1 by dt if Vc 1 is Vc minus Vp. So, clearly if Vp is a constant dVc 1 by dt is the same as dVc by dt. So, what I will do is, I will rewrite this equation in terms of the new variable Vc 1. So, first of all dVc by dt is the same as dVc 1 by dt and if you see this I have Vc minus Vp equal to 0 if I take Vp to the left side. So, that is nothing but Vc 1 equal to 0. So, here Vc 1 in terms of Vc 1, this is a homogeneous equation with the right hand side to be 0. Vc 1 is just a new variable I have defined. So, while solving problems you do this all the time. You define some variable in terms of other variables and substitute. In this case I have defined Vc 1 to be a variable Vc minus Vp. What I want to calculate for is Vc. So, instead of that first I will calculate Vc 1 and from that I will calculate Vc. So, if I do that and write the differential equation in terms of Vc 1, I will get a homogeneous equation. So, Vc 1 would be of the form V naught exponential minus 3 by Rc. So, Vc if I turn around this now Vc is Vc 1 plus Vp. So, Vc will be of the form Vp plus V naught exponential minus 3 by Rc. So, this is the general form of the solution to this equation which has Vp on the right side. So, this is not a homogeneous equation and the solution to this will have Vp. It will also have some terms like exponential minus 3 by Rc which was part of the 0 input response. It has this constant V naught which we have to adjust based on the initial conditions. So, what is V naught and Vp? First of all Vp is the size of the step right Vp the input jumps from 0 to Vp. So, that is Vp and V naught is just a constant. So, previously here I said that Vc will be some V naught times exponential minus 3 by Rc that will satisfy the differential equation and we calculated the particular value of V naught by looking at the actual solution at t equal to 0. So, V naught happened to be Vp in the previous case ok. In this case again we have to evaluate the value of V naught we will do that ok. So, now we have to evaluate the value of V naught what we will do is I know that Vc of t is Vp plus V naught exponential minus 3 by Rc and if I substitute t equal to 0 what will I get? What will I get if I substitute t equal to 0 in this function Vc of t? So, clearly exponential of 0 is 1. So, the answer is Vp plus V naught ok. Now what is the value of the function? Actual value of the function at t equal to 0 it is 0 right that we know initially the capacitor has 0 volts and just after the input step the capacitor still has a value of 0 volts. So, this should be equal to 0 which means that V naught equals minus Vp ok it was clear. So, there will be arbitrary constants in the solution to a differential equation because the differential equation in the general form could be satisfied for any value of this constant. But by looking at the initial condition you will be able to find the particular value that satisfies it for that case ok. So, in this case this V naught equals minus Vp. So, the actual solution to Vc is Vp minus V naught exponential minus t by Rc or minus Vp sorry Vp minus Vp exponential minus t by Rc. Also frequently this is written in this form Vp 1 minus exponential minus t by Rc ok. So, if you look at the solution here at t equal to 0 both are Vp and this cancels out or inside this this exponential will be 1 at t equal to 0 and you have 0 and as t becomes infinity this exponential part becomes 0. So, Vc of t will become Vp that is what we see from our intuition and it starts from Vp I sorry starts from 0 and then exponentially approaches this Vp ok. Is this clear any questions? So, ok. So, we have looked at the Rc circuit it is governed by a differential equation. We have solved for it intuitively we know what it means ok that is very important actually. We of course, taken a special case of only constant inputs. When I say constant it jumps from one constant value to another constant value ok. Now, if it does that what happens to the capacitor voltage and we will see later that this happens to every variable in the circuit is it undergoes some change with some exponential argument and the argument inside will be minus t by Rc or minus t by some time constant ok. Now, we solved it for 0 input that is you have some initial condition and the input then jumps to 0. We also solved it for you have 0 initial condition and it jumps to some value ok. So, let me just tabulate the results. The circuit we are looking at is this Rc and the variable I am trying to calculate is this I took two kinds of inputs one where the input jumps from some Vp to 0 ok and in this case the capacitor voltage Vc will do something like that ok and the other case was where the input jumped from 0 to Vp and the output it turns out thus something like that ok. This is Vc of t Vc of t. So, this corresponds to basically a 0 input and in this case we have a 0 initial condition on the capacitor voltage ok. Now, the capacitor voltage is also known as a state a capacitor can hold some voltage. So, that is known as the state of the capacitor. So, this is also known as 0 initial state or 0 state ok. So, the differential equation governing the 0 input circuit is that the important thing here is that you have terms containing Vc and its derivatives and the right hand side is 0 ok. So, this is known as the homogeneous equation and in the other case the differential equation is Rc dVc by dt plus Vc equals Vs and we are talking about constant values of Vs or Vp ok. And the solution to the 0 input case it turned out was Vp exponential minus 3 by Rc ok and the solution to this case is Vp minus Vp exponential minus 3 by Rc ok. So, we will look at these expressions and classify them into different parts ok. It turns out that this part is called the force response this is the natural response the entire thing is the 0 state response in this particular case and this is the 0 input response. We will look at all of those things in the next lecture. So, if you have any questions now I will clarify them any questions about any of this Rc circuits and differential equations ok. I will see you on Thursday ok.