 Hello everyone welcome to lecture 19 of basic electrical circuits. In the previous lecture we were looking at how to calculate responses to a step of a first order circuit which could be an RC circuit or an RL circuit. In this lecture what we will do is we will look at what to do for a second order circuit that is second order RLC circuit. At the end of the previous lecture we considered two second order circuits and then wrote down the differential equations for them. In this lecture we will start from that point and calculate the responses and different parts of the response just like we did for the first order circuits. As I said we considered two circuits I will assume that we are interested in the capacitor voltage VC. Now this is not necessarily always the case it could be that I am interested in the current or the voltage across the resistance I can write down the differential equation in terms of any variable. Now I could write it in terms of VC I could write it in terms of IL or VR or anything else. Now if I write it in terms of VC I will get LC times second derivative of VC plus RC first derivative of VC plus VC equals VS. Now the same thing could be written in terms of some other variable let us say we choose to write it in terms of the inductor current. Now it turns out I will not show it here but even if we write it in terms of the inductor current we would get the left hand side of the differential equation to be exactly the same that is what I mean is here with VC as variable we will have LC second derivative plus RC times first derivative plus VC equals something dependent on the input VS ok that is my convention I always put the independent source on the right hand side. If I choose to write it in terms of IL then I will get exactly the same LC second derivative of IL plus RC first derivative of IL plus IL equals this will be some other function of VS ok. So I am not going to find that here but the point is if you take a given circuit and then identify any variable and write the differential equation for it the left hand side will remain the same ok meaning the differential equation the left hand side part or the homogenous part of the differential equation will remain exactly the same. Is this ok any questions about this? Now let me write it in terms of write it for a different second order circuit which is I have an input VS and I have not looked at what kind of input it is. So the differential equation we have written is quite general then depending on the type of input whether it is a step or a sinusoid we can go and do analysis further. Again I identify this VC as variable I believe that is what I did last time as well ok in that case I will get LC second derivative of VC plus L by R first derivative of VC plus VC equals L by R first derivative of VS ok. Now again I can take some other quantity as variable instead of VC I will take there appears to be some problem with this let me fix that ok. So I think we are back in action. So if I choose some other quantity as a variable let us say the inductor current IL again as before I will get the same equation on the left hand side ok and on the right hand side I will get some other function of VS ok. Now you see also that for this particular circuit and this particular circuit the left hand side looks different that is the middle term for instance is RC times first derivative in the left side case and L by R times the first derivative in the right side case ok. Now these are two sort of standard types of second order LC circuits ok. Now we will do many things with the solution we will see how to solve this for these things for step response etcetera and those will be general for any second order circuit. In fact they do not even have to be LC circuits. Now in this particular case we have chosen two different LC circuits and they will have a slightly different types of responses ok. Now what are the what is the difference between these two? I would like answers from the participants what is the difference between these two circuits? Yeah so there was an answer from Vikas that the first one is a series RLC and the second one is series parallel but it is actually parallel RLC what I mean by this is the following ok. With a nulled circuit that is you null the independent source that means that in this case we have only a we have only an independent voltage source ok. So we said VS equal to 0 if we had a circuit with current sources we would also said that to be equal to 0 ok. So once we null this what we get is R, L and C ok. We get a single loop of RLC in series for the circuit on the left side and for the circuit on the right side if you null this we will see RLC in parallel ok. Now I have drawn the R like this but it is very clear that R and L and C are in parallel with each other ok. So we get only two nodes with RLC in parallel with each other ok. So basically by looking at the nulled circuit we can tell whether it is a series RLC. So the circuit on the right side is known as sorry the circuit on the left side is known as a series RLC circuit and the circuit on the right side is known as a parallel RLC circuit ok. Now it depends on basically the nulled circuit ok. So now it does not matter where the source is for instance what I mean is. So let us say I had a circuit like this or a circuit like this if you null the source is these two look different from each other but when you null the source you will simply get R, L and C in series ok. So these are basically series RLC circuits. Similarly even in this case I could have sources in different places voltage sources in series with one of these branches or current source in parallel with something else ok. In all of those cases I will get R, L and C in parallel if I do get that it is a parallel RLC circuit ok. So there is a question on previously recorded videos I believe they should be online but after this lecture I will go and check if it is not online I will ask them to have it put up online as soon as possible ok. So the idea is to have every lecture online before the next lecture ok. So now I showed that this is the differential equation governing this circuit this is the differential equation governing that circuit but in general if you have a series RLC circuit you will get a differential equation of the form L, C first second derivative of V, C like I said this could be any other quantity you will get the same thing ok. The difference will be in the right hand side. Now like I said you could have the input as a voltage source or a current source or something else as long as the null circuit reduces to this the left hand side of the equation will be of this form for any variable. Similarly if the null circuit reduces to a parallel RLC circuit the left hand side of the differential equation will be of the form L, C second derivative of the variable I am taking V, C as an example but as I keep emphasizing it could be any other variable. L by R first derivative plus V, C equals this will be related to the independent source in some way. So once you are able to set up the left hand side the transient response or the natural response can be determined because for instance you can set the right hand side to 0 and that gives you a natural response with 0 input that is the 0 input response which is also the natural response for 0 input. Obviously with 0 input there is no force response force response is something that is proportional to the input with 0 input the force response will be 0 ok. Any questions about this I classified the second order RLC circuit into two types but the method of solving the second order circuit will be the same even if it is not RLC ok. Now RLC is an important practical case. So I have classified them into two types where one has series RLC in a single loop the other has RL and C in parallel with each other. If you have series RLC in a single loop the differential equation will be of this form where right hand side could be anything. If you have RL and C in parallel the differential equation will be of this form and the right hand side could be anything ok. Any questions? Now there is a question asking is it possible to calculate the time constant. Now we will do that a first order circuit has a single time constant and a second order circuit will have two time constants. In general an nth order circuit will have n time constants ok. So we can calculate that. Now it turns out that for higher order circuits that is order higher than one you can mention the time constants you can specify the time constants or you can specify alternatively the characteristic frequencies ok. Frequently that is what is done. So we will look at that ok. So if you recall let us go back to the first order circuit. This is again an example R C but it could be any first order circuit. The differential equation governing this with V C as variable was R C dV C dt plus V C equals V S and we could write down and we could write down V C for a step input. If V S is a step input that goes from 0 to V P ok. I could write V C as V P times 1 minus exponential minus T by R C ok plus the initial condition exponential minus T by R C. This we have done before ok and we call this the zero state response and this is the zero input response ok. And this can also be rewritten as V P plus V C of 0 minus V P exponential minus T by R C. This is specifically for a step input or a constant input ok. When I say a step it is really a piecewise constant input and we call this the steady state response and this the transient response. Alternatively this is also called the force response or this is called the natural response ok. Now for the second order circuit also we will calculate these things that is the steady state response or the force response and the transient response or the natural response ok. And at every step I will try to make an analogy with the first order case so that you can associate the results and understand the results clearly ok. Because I will not be deriving formally the solution to the second order differential equation. I will give you the solution and say how it is similar to what we have in the first order case. There is a question asking what is the transient response and natural response ok. The transient response and natural response are the same thing ok. They are just two different names for the same thing. Now if you look at the total response there will be a part in the first order case containing exponential minus T by R C ok. Now this exponential minus T by R C is the natural response of the first order circuit. So, the part that contains this is the transient response or the natural response ok. Now in a more complicated circuit there will be what are known as natural modes of the circuit ok which will be of similar form exponential minus T by something related to the circuit. So, the parts of the response that contain those things are known as transient response or natural response. And this part which depends only on the input which depends on the input is known as the steady state response ok. Is this clear? Basically for a first order case whatever contains exponential of minus T by the time constant of the circuit is the natural response. So, now first thing we observe is that for a first order circuit the steady state response or the force response is a constant for a constant input ok. What I mean is the input changes from 0 to V p that is input is basically V p after T equal to 0 ok. That gives a steady state response which is V p which is also a constant. Now it happens to be the same as this, but in general it will be some constant. And you would have seen earlier that this was true for any constant input case we discussed the step response of first order circuits extensively. We calculated the steady state response by open circuiting all the capacitors. So, if we have a constant input the output will also be some constant ok. So, this is the first thing that we will use and it turns out the exactly the same thing is true for second order circuits also ok. So, let me first take the series RLC case and let me now consider this going from 0 to V p R L and C and I am looking at V c. So, V c of T will be of the form steady state response plus transient force response plus natural response ok. So, just like for a first order circuit for a constant input here you have to understand what is meant by constant the input changes from 0 to another constant. So, I am looking at what happens after T equal to 0 and that is a constant input the steady state response is also a constant. Now, I would like answers from the participants what is the steady state response of this circuit. Now, we have not evaluated the response in detail, but please look at the circuit and think about it the input is a constant at V p for T greater than 0. The steady state response I mean what happens after a very long time and all the quantities are constants. This will be the value of V c what is the steady state value of V c. So, now like I said the steady state quantities will be constant that is the voltage across the capacitor will be constant current through the inductor will be constant. Now, this is true even for arbitrarily complicated circuits. So, if the current through inductor is a constant then the voltage across the inductor is 0 if the voltage across the capacitor is a constant the current through the capacitor is 0. So, just like before for the just like for the first order case you evaluate steady state with constant inputs this is what is important constant inputs to evaluate this you short circuit the capacitors open circuit the sorry shorts I said it backwards sorry about that your short circuit the inductors open circuit the capacitors. So, if you do that for this particular circuit what will we get what is the value of steady state V c you please use this algorithm and find the steady state V c what is it going to be. So, clearly it will be the input source itself because we have V s which changes from 0 to V p and I have short circuited the inductor that is the inductor short circuited and the capacitor is open circuited. So, if you have V p here exactly the same thing will appear there. So, V c of infinity will be V p that is fine. So, a part of the response can be evaluated very easily. So, now let me give another example. So, that you clearly understand what is going on. So, let me take the other circuit let me say that V s again changes from 0 to V p and have RLC and in this case I am interested in the inductor current that is my variable of interest I know that I L will also be of the form of steady state response plus transient response. So, what is the steady state response of I L in this particular circuit? So, I got two different answers let us see which is correct R I will short circuit the inductor and open circuit the capacitor. So, if I have V p which is the constant voltage V s after t equal to 0 the current that flows this way would be V p by R. So, the steady state value of I L is clearly V p divided by R. So, this is how you can find the steady state response quite easy open circuit the capacitors short circuit the inductors and find the corresponding values. So, next thing we have to do is calculate the rest of it which is a little more complicated, but certainly possible. So, again I will make analogies to the first order circuit and go with it. So, now our goal is to find the transient response or the natural response. So, let me write it down for the first order case again now we know the exact solution for this, but let me do it in a slightly different way now the differential equation governing V c is R c d V c by d t plus V c equals something ok. Now, to evaluate the kind of natural response I do not need the right hand side I could also set this to 0 because even with 0 input I will have some natural response ok. Now, for a particular input you have to find the natural response that we will do by adjusting the initial condition ok I will explain exactly what I mean by that. Now, we know that the natural response for this differential equation will be of the form exponential sum exponential ok and let me call it exponential p times t ok let me call it exponential p times t remember we know exactly what the answer is it is exponential minus t by R c I am trying to get the same answer in a slightly different way ok. Now, we know that like for instance I could rewrite this and say that the time derivative is minus 1 by R c times V c this we did earlier what it means is that the time derivative is a scaled version of the function itself ok and we know that that property is true if the function is an exponential. If the function is an exponential the time derivative is also an exponential, but it could have some scaling factor ok. So, we know that exponential will satisfy this we just have to find the constant inside the exponential ok. Now, how do we find the constant now for this we have already found it, but I will do it in a more general way which will also be useful for second order circuits. So, what I do is I will assume that it is exponential of p times t ok. So, what will I get I will have R c the time derivative will be p times exponential of p times t plus V c which is exponential of p times t equals 0 and this exponential p t is common to the two parts and that will be that can cancel out ok. So, what I have will be R c times p plus 1 equal to 0 or p equals minus 1 by R c ok. So, this is an alternative way of finding the argument of the exponential ok. Now, we did it earlier we already know the answer now we know that this answer is correct because if I substitute this into that one what I get will be that ok. So, now, we can use the same method for second order equations it turns out ok. So, this equation from which we get the value of p is known as the characteristic equation. Now, remember we got this by substituting V c equals exponential p t ok. Now, we do not have to do it every time we know that if you have a first derivative you will have p and if you have V c you will not have anything and exponential p t will be common to every term and that will cancel out ok. This will become even more clear when I do the second order case. So, all we have to do is wherever I see a first derivative I put p and wherever I see the function I just have 1 ok. So, then I will get the characteristic equation. Is this clear any questions and solving this what we get are characteristic frequency in this case frequency in a higher order circuit it will be frequencies and what is this? This is nothing but the reciprocal of the time constant or rather negative reciprocal of the time constant. Any questions about this? There is a question of how does this p come about? See I know that if the differential equation is of this type that is d V c by d t is proportional to V c basically this differential equation I know that an exponential will satisfy it because what this is saying is the differential coefficient the derivative is the same as the function with some scaling factor ok. Now, you know from basic differentiation that if you have an exponential its derivative will also be an exponential with some scaling factor. So, I know that the form of V c is exponential of time of course, there can be a constant here. So, that constant I have assumed to be p I have to find out what p is for this particular circuit, but I have assumed that the solution is of the form exponential p t and I substitute it here I get R c times p times exponential p t plus exponential p t equal to 0. So, this is a standard way of solving differential equations because I know exponential p t can be a solution I do not know the value of p I will substitute that and then find the value of p is this fine you also know that the value of p we found is minus 1 by R c if I substitute it there I will get exponential minus t by R c which is consistent with what I had earlier these things we had already determined earlier this is whatever appears in the exponential as t by something that is the time constant by definition because we have seen that if you have exponential t by tau the amount of time it takes for it to settle is related to this tau. So, now I will do the same for a second order circuit and V s changes from 0 to V p and I am interested in V c. So, now the differential equation for this is L c d square V t by d t square plus R c d V c d t plus V c equals something to find the natural response I will simply set the right hand side to 0 then I will get the form of the natural response the coefficient of the natural response I will determine from initial conditions. So, again in this case also exponential will satisfy it. So, I will assume that V c of t is exponential p t. So, please substitute this here and find the characteristic equation please try this out please assume that V c of t is exponential p t and find the characteristic equation. So, I have an answer here which is correct. So, all we have to do is substitute this what I have would be exponential p t in place of V c plus R c times p exponential p t because this will be that one this will be that one and if I differentiate it once more I will get p square. So, I will have L c p square exponential p t this corresponds to this one and the right hand side is 0 and by canceling this exponential p t I will get L c p square plus R c times p plus 1 equals 0. Alternatively, it could also be written as p square plus R by L times p plus 1 over L c equals 0. This is the characteristic equation I hope all of you are able to derive this yourselves. All we have to do is to substitute V c is exponential p t in this equation any questions. Now, from this we can solve for the characteristic frequencies or equivalently characteristic time constants, but as I mentioned earlier for higher orders than 1 it is common to speak of characteristic frequencies p rather than characteristic time constants. But if you want to call them time constants that is okay then the reciprocal of p will be the time constant. So, now clearly there will be two possible solutions for p and also the nature of solution will be different depending on the values of the coefficients. So, this is the characteristic equation of series RLC circuit as I have been emphasizing I have taken V c as an example of the quantity of interest, but you take a series RLC circuit by that I mean if you null the independent sources you get a series RLC and then you pick any variable of interest okay the inductor current capacitor voltage whatever it is then you will get the same characteristic equation okay because the left hand side of the differential equation will remain the same. So, for any series RLC circuit wherever you apply the source or whichever variable you calculate the differential equation for the characteristic equation will remain exactly the same okay and that is p square plus R by L times p plus 1 by L c equals 0. So, this has two solutions okay now there are the two solution would be in general I think all of you know the formula for solution of a quadratic equation it will be minus R by 2 L plus square root of so this is the solution in general okay now this can give you three different kinds of solutions. The two characteristic frequencies p 1 and p 2 could be real and distinct what is the condition for the characteristic frequencies to be real and distinct please give me the answer when will p 1 and p 2 be real and distinct. See you know that solutions to a quadratic equation can be I mean you will always have two solutions but the two solutions could be real and different from each other the two solutions could be real and the same as each other and also the two solutions could be complex conjugates of each other. My question is when are the two roots when are the two characteristic frequencies real and distinct that is real and different from each other I think the question is not very clear or somehow you have not understood the question properly we will have two values of p 1 and two values of p okay that is p 1 and p 2 okay the two values of p are given by p 1 and p 2 by these two solutions minus r by 2L plus half of square root of this minus r by 2L minus half square root of this whole thing okay this is just the solution to this quadratic equation that is all okay now why am I calculating the solution to this if you recall here I have a first order equation in p I solve that for p and I will get the characteristic frequency I have to put this into exponential here I have a second order equation for p so I will get the characteristic frequencies two of them and I have to put them into two exponentials clearly there will be two exponentials which will satisfy this okay so please tell me when will the two values p 1 and p 2 be identical to each other okay clearly there are two values one when this is plus one when this is minus okay so my question is when is it that the values of p 1 and p 2 be identical to each other so I have a couple of answers one of the answers is when there is no resistor now that is not correct actually the correct answer is see why do we get two different solutions let me expand this what this means is when I have this plus and minus one of the solutions p 1 is minus r by 2L minus half square root of r by L square minus 4 by LC and the second solution is minus r by 2L plus half square root of r by L whole square plus sorry minus 4 by LC now when will these two be equal to each other obviously if this term you have a minus sign here and plus sign here so they will be different from each other unless this entire thing happens to be 0 the term under the square root happens to be 0 okay so if this part is 0 that is r by L square is 4 by LC you can simplify this I think that is the answer that was given by Suganya but I will leave it like this under this condition you will have two identical roots okay now if this quantity under the square root is positive then it will have a real square root right if this is positive we will have a real square root and the two solutions p 1 and p 2 will be real and different from each other if the quantity inside the square root is negative then the square root will consist of an imaginary number okay then this p 1 and p 2 will be complex numbers and there will be complex conjugates of each other is this part clear please let me know in summary there are two solutions to p p 1 and 2 which are minus r by 2L plus or minus that is the meaning of the symbol 1 by 2 r by L square minus 4 by LC now we will have real distinct solutions if r by L square is more than 4 by LC okay this also can be written as this can be rewritten in a couple of ways I could say 1 by r square root of L by C is less than half or I could also write it as r by 2 square root of C by L greater than 1 okay these are some quantities that we will define later okay so then there will be two real and distinct solutions which are minus r by 2L minus half square root r by L square minus 4 by LC and the other one will be minus r by 2L plus half square root r by L square minus 4 by LC okay now there will be two identical solutions when I say two identical solutions you will calculate p 1 from this plus sign p 2 from this minus sign those two values will be the same as each other if r by L square equals 4 by LC in that case p 1 and p 2 will be equal to minus r by 2L okay and there will be complex conjugate solutions okay if r by L square is less than 4 by LC remember in that case there is number inside the square root will be negative and the square root of a negative number is an imaginary number okay there is a comment here I guess it is not a question say second order non-linear equation can solve by Laplace transformation first of all this is not a non-linear equation this is a linear differential equation and it is only a linear differential equation that can be solved by Laplace transform now we have not even come to Laplace transforms and we will not do that in this course so we will solve the transient response using time domain techniques it is because it has two roots it is there are many possibilities but it is quite simple to do that in the time domain also and in fact understanding this is a prerequisite to a good understanding of Laplace transforms okay so now under these conditions what happens is that I have to calculate square root of r by L square minus 4 by LC when this number is less than 0 what is that what is square root of minus 1 please answer this question what is square root of minus 1 yeah it is denoted by the letter i or j which by definition is square root of minus 1 now in electrical engineering because we use i for current we always use j for square root of minus 1 okay so what is this this is basically square root of minus 1 times 4 by LC minus r by L square okay which is basically j square root of 4 by LC minus r by L square so in this last case when r by L square is smaller than 4 by LC okay the two roots will be p1 is going to be minus r by 2L plus j square root of 4 by LC minus r by L square times 1 by 2 and p2 will be minus r by 2L minus j square root of 4 by LC minus r by L square okay so we will have two roots which are complex and not only are they complex you see that they have the same real parts minus r by 2L and imaginary parts with opposite signs plus half times this minus half times that so these are complex conjugate complex conjugate pair roots okay so in general a second order system will have three possibilities it will always have two characteristic frequencies the two characteristic frequencies will be identical if r by L whole square equals 4 by LC okay this is for a series RLC circuit and if r by L whole square is more than 4 by LC it will have two real roots p1 and p2 which are different from each other and if r by L whole square is less than 4 by LC it will have a complex conjugate pair of roots is this part clear so we have to calculate p1 and p2 which are characteristic frequencies from the characteristic equation then what do we do with it the solution can be exponential of p1t or exponential of p2t both these are possible okay okay now it turns out that in a linear system when you have a linear differential equation when there are two possible equations the general solution is a linear combination of the two equations okay so finally the let me write down the natural response okay the natural response will be of the form a1 exponential p1t plus a2 exponential p2t where p1 and 2 are characteristic frequencies okay or basically there are the roots of the characteristic equation and what is the characteristic equation all I have to do is to substitute p square where I have second derivative and p where I have first derivative and 1 where I have only vc equals 0 okay this is exactly what you will get if you substitute vc by exponential of p times t okay and now we have to find out a1 and a2 from initial conditions okay just to show what I mean I will go back to the first order circuit and do it that way okay let us say vs goes from 0 to vp and we have rc and I know that the steady state solution to vc is steady state response equals vp and the natural response is some constant a times exponential pt where p equals minus 1 by rc how did I get this I get this from the characteristic equation characteristic equation rc times p plus 1 equals 0 so the total response is the steady state response vp plus transient or natural response a exponential p times t okay there is a question which asks whether the characteristic frequency says something about the damping of the signal now we will talk about that later we did not introduce the term damping factor okay now it turns out that if the characteristic frequencies are both real and negative you will have an over damped system they are identical it is a critically damped system and otherwise it is an under damped system but we will come to that in the next lecture okay coming back to the first order system the total response is the steady state response plus transient response but we have this constant a that has to be determined how do we do that that we do by looking at the initial condition okay if you look at this total response and substitute t equal to 0 or 0 plus okay what do we get what is the value of this at t equal to 0 plus just all you have to do is a mathematical substitution so the question is the total response is of the form vp plus a times exponential p times t what is the value of this at t equal to 0 all you have to do is to substitute t equal to 0 in this and if you do that you will get vp plus a okay so if you have vp plus a at t equal to 0 so vc of 0 happens to be vp plus a okay so obviously then a has to be equal to v a has to be equal to vc of 0 minus vp okay and I can go back and substitute this into the equation so I will get the total response to be vp plus vc of 0 minus vp exponential p times t which is minus t by rc okay and this is exactly what we had obtained earlier using a different technique by completely solving the differential equation okay but the advantage of this method is that it can be extended to second order circuits as well okay so in general the way to solve it is as follows okay when we have constant or step inputs you first determine the steady state solution okay and this you do by open circuit the capacitors short circuit the inductors then you determine the natural or the characteristic frequencies for a second order system you will have two of them p1 and p2 so then this means that the natural response will be of the form a1 exponential p1 t plus a2 exponential p2 t okay and finally the total response will be steady state plus natural response and you will have to determine these two constants a1 and a2 which you do based on initial conditions okay so this I showed for a first order system exactly the same thing is true for a second order system okay this is clear any questions any questions about this in a second order circuit we will have a second order differential equation and two characteristic frequencies okay I see some question but it seems to be half typed I am not able to understand what the question is okay in case of first order there is only one characteristic frequency and it does not matter what the values are okay the characteristic frequency is always going to be negative it is minus 1 by rc okay for a passive first order rc circuit for an rl circuit it will be minus r by l okay so there are in case of second order there are three distinct classes of solutions because there are three possibilities for the characteristic frequencies in case of first order there is no such thing there is just one case and it will always be an exponential okay okay so in the next lecture what we will do is we will start from this point and see what these solutions look like and that will also take us to these different kinds of responses which are underdamped critically damped and overdamped okay I will not go through every detail of this but what I want to point out is for a second order system also you should be able to write down the natural response and even the complete response in case of a constant input that is a step input okay by evaluating the initial conditions correctly okay so now if there are no more questions we will stop here and continue next week.