 In the previous classes, we have been looking at how to calculate the response of an RC circuit to a unit step. Now, this method is applicable to any system described by a first order differential equation and excited by a step input. So, I think we have a good handle on this. Today, we will look at one particular case which needs extra care if we have to get the step response correctly. We said that the step response in an RC circuit and exactly the same thing will hold in an RL circuit as well in a first order RC RL circuit. What we need to know are, let us say we are calculating a particular quantity Vx. We could be calculating anything. We could be calculating some voltage or some current. It does not matter. So, we need to know value of Vx at t equals 0 plus that is just after the step is applied value of Vx at t equals infinity that is when the circuit reaches steady state and the time constant tau. Now, once you have all of this, the response is given by. So, this is a general expression that is applicable to any quantity in a first order RC or RL circuits. In our examples, we have been taking only an RC case that is with a capacitor, but the same thing applies to an inductor. If we have time, we will work out an example today. Now, how do we find the value of Vx at t equal to 0 plus? We do that by shorting all capacitors. We can short circuit all capacitors or if the capacitors have initial conditions, what we really do is replace capacitors by voltage sources whose value equals the initial condition. And how do you find that t equal to infinity? You simply open circuit all capacitors because if the voltages reaches a steady state, a constant value across all capacitors, then their currents will be 0. By the way, we are talking only about first order systems here, but this is the case for all RC or RL circuits. And in case of inductors, we have to short circuit all inductors and also it turns out just like you replace capacitors by voltage sources that represents the initial condition for t equal to 0 plus, you have to replace inductors by current sources. And finally, how do you find the time constant tau? You find the Thevenin equivalent resistance across the capacitor. So, this is the algorithm. Now, are there any questions about this? Any questions about these steps? How to follow them? How they come about? So, if not, let me go to a slightly more complicated example than what we looked at in the previous lecture. So, again this is V s and it goes from 0 to, jumps from 0 to V p. And I will have RC and let me call this C 1. So, V s jumps from 0 to V p and let us say I am interested in this voltage V c, the voltage across the capacitor C. So, what I would like to know is to how to go about finding this? As usual, I need to find out V c of 0 plus that is just after the step is applied, V c of infinity that is after an infinitely long time and the time constant tau. We have to find all these 3, but first let us start with V c of 0 plus. So, I would like the participants to give me an answer. What is the value of V c? Just after the step is applied that is V c of 0 plus. What is the value? The input V s jumps from 0 to V p. What do you think is the value? There is an answer that says it is V c of 0 the initial condition on the capacitor. Now, let us say the initial condition on the capacitor is 0 volts. Then what would be the answer? Let us say that this is initially discharged. So, now clearly the contradiction is obvious. Now, the problem is that there is a loop of voltage source and these 2 capacitors. Now, so far we were considering that the voltage across a capacitor will not change instantaneously. We know that that is the case if the current through the capacitor is finite. Now, if you consider just after the step is applied. So, this voltage here which is the sum of voltages across C 1 and C that changes from 0 to V p. So, when it changes from 0 to V p, if we say that the voltage across capacitor C does not change, the voltage across capacitor C 1 has to change. Alternatively, if we say that the voltage across C 1 does not change, then the voltage across C has to change. That is we have this loop of capacitor 2 capacitors and a voltage source. If the voltage source changes its value, the voltage across at least one of the capacitors has to change instantaneously. So, this is the problem. At least one capacitor has to change its voltage instantaneously. Now, it cannot be just one which changes because if the voltage across C 1 changes instantaneously, that means that there is infinite current flowing through it. But then we assume that capacitor C did not change instantaneously. So, then KCL at this node will be violated because you have finite currents through C and R, but infinite current through C 1. So, it cannot be that only one changes. In fact, both the old capacitors have to change their voltages instantaneously. So, while logically at least one have to change, it cannot be only one. In fact, the answer is that both C and C 1 have to change their voltages instantaneously. So, now that means that if they are changing instantaneously, infinite currents are flowing through both C 1 and C. Now, because infinite currents are flowing through them, whereas a finite current is flowing through R because the voltage across that is still finite, we can approximate the circuit without any loss of accuracy by removing the R altogether because the current through this is infinity, this is some finite value. If you add that to the infinite current, it does not change anything. So, we can approximately find the we can find the answer by removing the resistor altogether and having only C 1 and C because just as the step is applied, the current through this loop is infinite and this resistor contributes only a finite current. So, there is no loss of accuracy if you just remove the resistor, is this clear? So, then assuming that both of them change, we have to calculate the change in V C. So, let us assume again that it is C is initially discharged, C is initially discharged that is at t equal to 0 minus, we have 0 volts across this which also means that there is 0 volts across C 1 because before t equal to 0, V s is 0, the voltage across C is 0 and voltage across C 1 is 0. So, what I would like you to find out is the voltage V C that is the voltage across capacitor C when a unit step V p is when a step V p is applied. There is an answer saying it is V p, but it cannot be V p because if this changes by V p, it means that the voltage across C 1 does not change at all. That says that if there is no change in voltage across C 1, there will be no current here, but the voltage across C has changed abruptly. So, there is infinite current there. So, KCL will be violated at this node. So, it cannot be that only one of them changes, both of them have to change and they have to change by an appropriate amount. So, that KCL here is also satisfied. So, now because the currents are infinite, we cannot do useful calculation with currents. So, we will work with charges which are integral of currents, time integral of currents. So, now if you consider this as a node, what KCL says is this current I 1 has to be equal to the current I 2. Now of course, when capacitor voltages change abruptly, these currents will both be infinite and we will not be able to do useful calculations with that ok. So, this is C 1, this is C 2 and this is V s which changes from 0 to V p. Now the same KCL can be rewritten by saying that instead of currents I will say the amount of charge that flows that way Q 1 has to be equal to the amount of charge that flows this way which is Q 2, is this fine. So, now let us consider the circuit for t equals 0 minus that is before the step appears. So, we have 0 volts here and 0 volts there and 0 volts over there and I also will write the circuit for t equal to 0 plus that is just after the step is applied. So, there will be some voltage across C, some other voltage across C 1 by C 1 of 0 plus and this value will be equal to V p. So, now you see that in this case the charge on this plate is 0 and the charge on this plate in the second case is what is the charge on this particular plate in the second case at t equal to 0 plus it will be equal to V c 1 is defined with this polarity right. So, let me just take a capacitor here, if I have a capacitor C with a certain voltage V across it what it also means is that on this plate there is plus C times V charge and on this plate there is minus C times V. So, on this plate we have minus C times V c 1 of 0 plus that is by the definition of a capacitor. When a capacitor C has a voltage V there will be plus C times V on one plate minus C times V on the other plate. It will be plus C times V on the plate with the positive sign of the voltage and minus C times V on the plate with negative sign of the voltage. So, the charge on this plate has changed from 0 to minus C V c 1 of 0 plus. Similarly, let me write down the charge on the other capacitor. The charge on this plate originally before the step will be 0 and just after the step will be C times V c of 0 plus that is plus C times V c of 0 plus. So, now we can find these charges the way I have written it is charge coming out of C 1 has to be equal to charge going into C 2. We can also write it in a conventional Kirchhoff's law formulation where the total charge leaving some node equals 0. Either way it is the same in this case I will stick to the definitions I have written. So, what is the value of q 1 that is the amount of charge that is going from left to right in this wire just after the step. What will be the value of q 1 and what is the value of q 2 q 1 is what goes from here to there when it transitions from t equal to 0 minus to t equal to 0 plus. So, now the charge here was 0 and it changed to minus C V c 1 of 0 plus. So, obviously a charge of plus C V c 1 0 plus has gone from left to right. What has gone this way is the negative of this quantity. So, q 1 equals C V c 1 of 0 plus and q 2 this q 2 which is the current flow as the charge flowing into the capacitor C this is not C 2 that is C equals C times V c of 0 plus. Because originally there was no charge on this plate later there is a charge of C times V c of 0 plus. So, q 2 is C times V c of 0 plus. I think I wrote this wrongly this is this capacitor C 1. So, this is minus C 1 V c 1 of 0 plus and the charge on this plate this capacitor is C times V c of 0 plus. So, this is also C 1 not C. So, now C 1 V c 1 of 0 plus equals C V c of 0 plus. So, this comes from KCL, but reformulated in terms of charges. In addition to this we also have the KVL which has to be always satisfied of course, which says that V c 1 of 0 plus plus V c of 0 plus that is the sum of these two voltages has to be equal to V p. Now, from these two you should be able to calculate both V c 1 just after the step is applied and V c just after the step is applied. Please do that and let me know the value of V c of 0 plus. So, as usual we use KCL and KVL to determine everything, but in this case KCL is formulated in terms of charges because the capacitor voltages are changing instantaneously and the currents are infinite that is all. So, we have two equations two unknowns. So, we should be able to solve them. So, please calculate V c of 0 plus and let me know what it is that is exactly right. So, when a step of V p is applied to a capacitive divider you see that this looks somewhat like a resistive divider, but with capacitors with C 1 and C the voltage here at t equal to 0 plus will change also instantaneously and it changes from 0 to V p times C by sorry C 1 by C plus C 1. C 1 is this capacitor and C is that capacitor and you see that this formula is also reminiscent of the resistive divider formula except that the roles are interchange. In the case of resistors you get R by R plus R 1 here you get C 1 by C plus C 1 and just to complete the picture the voltage V c 1 will change from 0 to V p times C by C plus C 1. So, when you have a loop of voltage source and capacitors and you apply a step the voltages of all capacitors in the loop will change abruptly they have to ok. So, that I explained and if now you can also calculate how much they change by and all you need to do is to write down KCL in terms of charges instead of currents that is all ok. Any questions about this? Any questions on circuits with loops of capacitors and voltage sources? Now of course, we have not completely solved the problem yet. What we were trying to solve was this when V s a step of V p volts is applied to a circuit of this sort C 1 and C and we want to determine V c of t for t greater than 0 and we have the standard formula based on the value of value of V c at t equal to 0 plus and t equal to infinity. So, in order to use that formula we had to find V c of 0 plus V c of infinity and the time constant tau. Now this part we have determined already and that is equal to V p times C 1 by C plus C 1 ok. The next thing is V c of infinity what is the value of V c of infinity that is you apply a step wait for a long time for things to reach steady state what will be the value of V c of infinity please find out that is right. Clearly it will be equal to V p because if we open circuit both these capacitors we get a circuit which is the same as what we have seen earlier and very simple this voltage is V p and we have R and because no current is flowing through R obviously the voltage here also will be V p ok. So, V c of infinity is V p and what is the value of tau the time constant what is the value of tau what is the value of tau the algorithm I outlined earlier you null the input the time constant is the property of the circuit it has nothing to do with the input applied ok. Then what happens is I had earlier said that you find the resistance across the capacitor. Now in this case there are 2 capacitors ok, but please hold on for a second there is some problem here I have to restart the journal ok sorry about that interruption. So, we have to null the circuit and when I have the null circuit I have 2 capacitors C 1 and C ok, but as you can observe this C 1 and this C appear in parallel ok they are connected between the same 2 terminals they will be in parallel. In fact this will always be the case a first order circuit which is what we are considering means that there will be only 1 capacitor. Now there are apparently 2 capacitors what I mean by there is only 1 capacitor is that when you null the circuit all the capacitors can be combined into a single capacitor ok that is the meaning of a first order circuit in fact if you cannot do this it is not a first order circuit ok. So, now we have R C 1 and C or in other words R times R and C 1 plus C ok. So, now please tell me what the time constant will be ok I have already given you the null circuit. So, you should be able to find out the time constant. So, clearly it is the product of the resistance you see here and the capacitor that is equal to R times C 1 plus C ok. So, now we have determined all the quantities let us take a numerical example and please try to solve that because it is only when you solve it independently that you get a hang of what is going on and give me the expression for the voltage we see as a function of time ok. Before that if you have any questions regarding this please ask. So, let us say the input jumps from 0 to 5 volts and it is the same circuit I had earlier and I will say this is 2 nanofarads and this is 3 nanofarads and this is 1 kilo ohm ok and both are initially discharged both of these capacitors that is 0 charge just before the step is applied at t equal to 0 minus. Now, what I want is the expression for V C of t. So, please find this as usual you have to find the voltage V C at t equal to 0 plus just after the jump at t equal to infinity that is at that is the final value and finally the time constant. So, the value at t equal to 0 plus this is 2 nanofarad by 2 nanofarad plus 3 nanofarad times 5 volts which is equal to 2 volts and V C of infinity is found by open circuiting these capacitors. So, all of this 5 volts appears here. So, that is 5 volts itself and finally, the time constant tau we saw that it is R times C plus C 1 when I null this we have these 2 capacitors in parallel that makes for 5 nanofarads and 1 kilo ohm in parallel with that. So, 5 nanofarad times 1 kilo ohm gives a time constant of 5 microseconds and from this from what we know earlier we can write out the expression for V C of t which is the final value plus initial minus final value times exponential minus t by tau which is 5 microseconds which is to say it is 5 volts minus 3 volts exponential minus t by 5 microseconds. So, it is interesting to I hope the technique is clear if it is not please ask me and I will explain again. Now, let me copy this circuit and here in the second case what I will do is I will remove this capacitor and have a single capacitor of value 5 nanofarad. In both cases I will assume that all capacitors are discharged for t equal to 0 minus. So, there is a question how to find the initial value I will repeat that. Now, what is the time constant of this circuit? The circuit on the right side. So, clearly this is 5 microseconds because when I null this circuit I have 1 kilo ohm in parallel with 5 nanofarad. In fact, it looks similar to this when it is null that is the reason I chose the value tau is 5 microseconds for both of these. Now, in this case V C of t we determined to be 5 volts minus 3 volts exponential minus t by tau and in this case I will write out the expression. If you have forgotten please go back to the previous lectures notes and you should be able to find it and anyway you should know how to solve this one it will be 5 volts minus 5 volts exponential minus t by tau ok. Let me put that in green. Now, what I will do is I will plot everything on the same graph versus t ok. This is V C in both circuits and let us say the input I plot here it jumps from 0 to 5 volts ok. In the circuit on the left side what happens is that at t equal to 0 there will be a jump ok. Because of this voltage source and this capacitor loop the output will jump and it will jump up by 2 volts ok. So, let me put this in red it will be 0 it will jump up to 2 volts ok. This we have determined earlier we saw that V C of 0 plus is 2 volts and then after that it will go towards 5 volts with a time constant of 5 microseconds. In the circuit on the right side there will not be any jump in the output voltage ok. Because here the currents will be finite because we have a resistor and no capacitor here. So, the output there is no jump at t equal to 0. So, the output starts from 0 and it remains at 0 immediately after the input step and it will go towards 5 volts also with a time constant of 5 microseconds ok. So, that is the difference between these 2 circuits they are related in that they have the same time constant, but because of this capacitor here the output will jump whereas in that case there will be no jump ok. So, we have studied extensively the algorithm for writing down the solution to a first order RC circuit in case of a step input ok. Now of course, it is not good to simply mug up the algorithm you should understand where the algorithm came from that we discussed earlier ok. When we have a first order RC circuit for t equal to infinity when all the voltages are constant you can open circuit the capacitors that is how you find the final voltage. For initial voltage just after the step you can short circuit the capacitors and find it. And in those cases where you have loops of voltage source and capacitors you cannot use this algorithm because you cannot short circuit both of these. If you short circuit both these capacitors then you have a short across a voltage source so that is not possible. In that case you have to find the output voltage by applying KCL and KVL ok. And this KCL and KVL instead of using charges instead of using currents you have to use charges because the currents are infinite and charges are finite ok. So, that is how you find the initial voltage that is at t equal to 0 plus and finally the time constant is found by nulling the circuit and finding the resistance across the capacitor. When you null a first order circuit you will always end up with a single resistor and a single capacitor effectively ok. Now, regarding finding the initial voltage in this case I will say finding Vc of 0 plus but it applies to any quantity just after the jump it does not have to be a capacitor voltage ok. Now, if there are no voltage source capacitor loops then it is very simple short circuit the capacitors and find the voltages and if there are voltage source and capacitor loops what you have to do is apply KCL in terms of charges and KVL and find the voltages ok. And in this case the resistors in the loop can be neglected that is if you have a capacitor in parallel with a resistor let me not say resistors in the loop but resistors in parallel with capacitors can be neglected ok. So, I showed the example earlier the input jumps by VP. So, if I write the circuits at t equal to 0 plus C 1 and C and t equal to 0 minus that is just before applying the voltage. So, this is 0 volts and I have assumed 0 initial conditions ok. So, this is 0 volts and that is 0 volts and here it is VP voltage across this is Vc of 0 plus and Vc 1 of 0 plus ok. Then we have to find the charge that is flowing this way and the charge that is flowing that way and equate the 2. To do that we have to find the initial charges here it is 0 there also it is 0 and here it is plus C Vc of 0 plus and here it is minus C 1 Vc 1 of 0 plus that is the charge. So, that means that this Q 1 is basically the difference between the initial and final charge right. If this is 0 and this is minus C 1 Vc 1 of 0 plus Q 1 has to be C 1 Vc 1 of 0 plus Q 2 is C times Vc of 0 plus. So, we have to equate these 2 Q 1 equals Q 2 and also we know that Vc 1 of 0 plus voltage across this plus Vc of 0 plus the voltage across that equals VP from these 2 you can find Vc 1 of 0 plus and Vc of 0 plus ok. So, if there are any questions about this please let me know. I very quickly outlined how to find the initial voltage in a case where there are loops of voltage sources and capacitors. Clearly you cannot short capacitors because if you short both capacitors you will end up with a short circuit across the voltage source VP ok. Now, the charge across C 1 is not negative it just goes from the definition of voltages ok. So, there is no meaning to saying charge across C 1 is negative. If you have a capacitor C with a voltage V with this polarity the charge on this plate is C times V plus C times V the charge on this other plate is minus C times V ok. So, that is what happens in any capacitor. Here I have a capacitor C 1 with a voltage V C 1 defined in this polarity. So, the charge on this plate will be plus C 1 times V C 1 of 0 plus and charge on this right side plate will be minus C 1 times V C 1 of 0 plus ok. Similarly, here the charge on this plate will be minus C V C of 0 plus ok. Any other questions ok. Now, I will not discuss circuits with inductors in great detail, but let me outline briefly what happens ok. You can compare that to this RC circuit ok and in both cases I will take V as jumping from 0 to VP. Now, the current through this clearly is V R divided by R and the same current flows through the inductor and the voltage across the inductor is L times DIL by DT where I L is the inductor current and V S the applied voltage is the sum of V L and V R ok. So, L times DIL by DT which is L times DVR by DT times 1 by R ok this is equal to L DIL by DT plus V R equals V S ok. So, L by R times DVR by DT plus V R equals V S and in this case I will not derive it again, but we had done it earlier RC DVC by DT plus VC equals V S ok. So, now in the RC case this was the time constant in the LR case this turns out to be the time constant ok. I said earlier that a first order RC circuit if you null the sources will always be reduced to a single capacitor and a single resistor across each other ok, single capacitor and resistor and parallel. Similarly, a first order RL circuit will always reduce to a single inductor and a single resistor and parallel ok this always happens and once you do that the time constant will be equal to L by R where L is that single inductor and R is that single resistor. Now in the circuit there can be more than one inductor right what I mean is I could have something like this, but when I null the source what I have is that and you can see that this L 1 and L 2 appear in series this is equal into L 1 plus L 2 and R and parallel. So, what will be the time constant of this circuit? The time constant is the inductance by resistance and it is equal to L 1 plus L 2 divided by R. I think somebody answered saying L 1 plus L 2 times R it should be divided by R ok. And because the differential equation looks the same for RC and RL circuits the solution also looks the same. Let me rewrite it here first order RL circuit in this case we if you want to find a voltage V x as a function of time for t greater than 0 and I am assuming that there is a step input at t equal to 0 ok. V x of t will be V x of 0 plus sorry V x of infinity plus V x of 0 plus minus V x of infinity times exponential minus t by tau. So, you have to find V x of infinity for this you short circuit all inductors this is analogous to open circuiting all capacitors. This is because when all the quantities have reached steady state which is a constant ok. We know that the voltage across an inductor is the time derivative of the current through the inductor times the inductance. If I L is a constant ok V L has to be 0 and that means that you replace the inductor by a short circuit at t equal to infinity. V x of 0 plus what did we do? We short circuited all the capacitors to find the quantity just after the jump. So, in this case we open circuit all inductors and finally, the time constant tau is given by L by R where L is the total inductance you see it could be a combination of more than 1 in the nulled circuit and R is the total resistance. Again it could be the equivalent resistance of some complicated circuit across L ok. So, the algorithm is exactly the same except that the time constant is of the form L by R ok. Hopefully with this information you will be able to solve for step response in any R L circuit as well ok. Any questions about this? Any questions about way to find step responses in an RC or R L circuit? Now just like when we had loops of voltage sources and capacitors, if we have current sources and inductors in parallel current sources and inductors connected to the same two nodes, then you have to have a special case. You can think about that special case yourself I am not going to discuss it here ok, but it will be analogous to having loops of voltage sources and capacitors. There is a question what does the total inductance in nulled circuit? I am not clear of what this question means, but for instance in this case there are two inductors, but they appear in series in the nulled circuit. So, obviously the total inductance is the sum of inductances. Now if we are talking about a first order circuit, when you null the source you should have a single inductor ok. There may be separate inductors in this, but when you null it you should be able to combine many of them into a single one ok. So, that is always the case in a first order circuit. If this is not possible it is not a first order circuit. Everything we discussed so far applies only to first order circuits ok. Then we can go to so far we have considered circuits that have a resistor and a capacitor or a resistor and an inductor. We have not considered cases where we have both resistor and capacitor and inductor ok that is we have capacitors and inductors in the same circuit. So, now we will consider that and they will be known as second order circuits because they will lead to second order differential equations. So, first let me take a case where I have a voltage source V s and this is applied to a series combination of R, L and C and let me write everything with this capacitor voltage V c as the variable ok. So, if the capacitor voltage V c is the variable we know that all these three components are series. So, there is a certain current that is flowing through the loop. What will be I of t in terms of the capacitor voltage V c? Please give me the answer. What is I of t in terms of the capacitor voltage V c? I of t would be C dVc by dt because this I is nothing but the current flowing through the capacitor as well. So, it is C times dVc by dt. The voltage across the resistor V R, V R is I times R and that is Rc dVc by dt and the voltage across the inductor V L is L times di by dt which is Lc times the second derivative of V c ok. So, we can form this a differential equation V s or let me put V s on the right hand side. We know that the voltage across the resistor plus voltage across the inductor plus voltage across the capacitor equals V s. So, Lc second derivative of V c plus Rc first derivative of V c plus V c equals V s ok. So, this circuit with a series RLC obeys a second order differential equation that is given by this. Now, in general if you arrange the RLC in a slightly different way it will also obey a second order differential equation ok. Now, we can of course solve this and find what the solution is. We do not have time to go through detailed derivations of the solutions of this. So, I am just going to give you the solutions in certain particular cases ok. So, before we go there let me do it for a slightly different circuit or let me let me have R L and C arranged slightly differently and let me take the inductor current as the variable or rather maybe just like before I will take V c as the variable. So, in this circuit I know that there is certain I L and certain I C and this current equals I L plus I C ok. I C is nothing but C times d V c by d t and also you see that the voltage V c here is across the inductor as well ok. So, the voltage V c is related to the inductor current by L times d I L by d t or I L would be 1 over L integral V c d t that is because the capacitor and inductor are in parallel and finally, this current I R equals I L plus I C which is equal to 1 over L integral V c d t plus C d V c by d t. Also we know that the current through the resistor is the voltage across the resistor divided by R. So, V R equals V s minus V c divided by R sorry not V R I R equals V s minus V c divided by R or V s minus V c divided by R equals this whole thing 1 over L integral V c d t plus C times d V c by d t. Now, this has integral and differentials in the same equation we would like to have only a differential equation where everything is expressed as time derivatives ok. So, what I will do is I will differentiate the entire expression once more. So, I will end up getting I will also move R to the other side that is this R can be moved here. So, if I differentiate everything I will get R by L times V c because the derivative on the integral cancel plus this is R c R c second derivative of V c. So, if I rearrange this I will get R c R by L V c equals d V s d t. The reason I did this is to put the source on the right hand side and the unknown V c on the left hand side and I will rearrange this further I will multiply the whole thing by L by R I will get L c this plus L by R that one plus V c equals L by R d V s by d t ok. So, I am going to just derive the equations the whole idea was to show that they will be in the same similar form for both the circuits. So, let me put these side by side. So, this is the differential equation governing the circuit and if I take the other circuit I will have this other differential equation ok. The main point I wanted to make here is that the two are similar to each other mainly especially the left hand side it is of the same form you get a second order differential equation the coefficients are different. So, that means that the exact properties will be different but both of them will give you second order differential equation. Now, it turns out that this is also true if you find some other arrangement of R L and c and also you have current sources in the circuit and so on. So, what we will do in the next lecture is to get an idea of what the response of this circuit looks like ok. So, we will stop here today in the next lecture we will take up take up the second order circuits if you have any questions I will be happy to answer them now ok. If there are no questions we will close here I will see you on Thursday.