Lecture: 25

So, in today's session we will discuss about the second order response of the circuits means those circuits which contain inductor, capacitor and register all three components are there or we have a two inductors or two capacitor which cannot be clubbed to become a equivalent inductor or capacitor. So, in those cases how we will solve the circuit we will discuss in today's lecture. So, let us start the discussion about the second order response. So, let us first understand what do you mean by second order circuit. So, second order circuit is characterized by the second order differential equation. Now, we can also say that the circuit which contains two storage elements particularly inductors and capacitors then those are called as a second order circuit because their responses include differential equations that contains second order derivatives. So, second order circuits can typically series RLC circuit or parallel RLC circuit or may be the two storage elements of different type or same type. Same type means the elements can be either two inductors or two capacitors, but these elements cannot be represented by an equivalent single element. So, in those case you need to create the governing equations which are the second order in nature. Now, let us see the examples of such type of circuits. So, you will see that this series RLC circuit, this is second order circuit also the parallel RLC circuit is also second order circuit. Now, if you have a two inductors and you cannot simplify this circuit and represent as a single inductor then also it can be considered as a second order circuit. Similarly, in this case also if we have two capacitors, but cannot be clubbed to represent as an single equivalent capacitor of the circuit in that case also the circuit would be considered as second order circuit. Now, the first thing which we have to understand is that how we will find the initial and final values for these type of circuits. Now, it is easier to find the initial and final values for v and i because these are something which can be considered or you can derive it while just doing the inspection of the circuit for time t is equal to 0 or time t tends to infinity, but it is little bit difficult in finding the initial value for their derivatives that means, dv by dt or di by dt. So, we have to keep a two major points in mind while we determine the initial conditions. First is that polarity of voltage vt across capacitor and direction of current through the inductor we have to always keep in mind, then voltage across the capacitor cannot change abruptly. So, that is why we say that voltage at 0 minus instance equal to voltage at 0 plus instance in case of capacitor. Similarly, in case of inductor the inductor current cannot change abruptly therefore, i 0 minus would be equal to i 0 plus. So, thus in finding the initial conditions we first focus on those variables that cannot change abruptly that means, we will concentrate on voltage of the capacitor and inductor current flowing through the inductor. Now, let us first take the case of source v series RLC circuit. So, we will try to find out the solution of those circuits which are a series combination of RL and C without any source. So, we assume that there is some initial current flowing through the inductor and the value is i naught and capacitor is initially charged with some voltage v naught. So, these 2 would be considered as a source which provide current to the circuit. So, now, what we can write at time t is equal to 0 v naught v 0 can be written as 1 upon C integral minus infinity to 0 idt and that will be equal to v naught because this is our initial condition. Similarly, i 0 that is i at time t is equal to 0 will be i naught. Now, if you write a KVL around this particular loop where the we are assuming that current i is flowing in this particular loop. So, you can write the KVL equation for this loop what you will write R i plus L di by dt plus 1 upon C integral idt and i the integration ranges between minus infinity to t equal to 0. So, this equation you can eliminate the integration ah portion what you can do you can differentiate again with respect to time t. So, when you again differentiate this equation what you will get you will get d 2 i by dt 2 into L plus R di by dt plus 1 upon C i. So, if you rearrange the second ah derivative of the equation you will get the equation as shown in the slide that is d 2 i by dt 2 plus R by L di by dt plus i by L C equal to 0. So, this you got simply by differentiating this equation and rearranging the terms. So, rearranging means the L terms L term which you will see here will be in denominator in both of the components and then you will get the equation as shown in the slide. So, now, if you see the equation as second order derivative. So, that is why it is called as a second order differential equation. Now, if you are asked to solve such a second order differential equation you require two initial conditions. Now, we have we know that initial value of i is equal to i naught. Now, next what you need to do is the first derivative of initial value of i. So, you have to calculate the value of the d i by dt at time t is equal to 0 and the value of initial value of v across the capacitor we know. So, we can utilize that value. So, if you see the previous equation 1 upon C integral i dt at time t is equal to 0 is nothing, but v naught. So, what you can write for this equation you can simply write R at time t is equal to 0 R into i at time t is equal to 0 plus L d i by dt at time t is equal to 0 plus v naught is equal to 0. Now, you know that i at time t is equal to 0 is nothing, but i naught. So, if you rearrange you will get d i by dt at time t is equal to 0 is nothing, but minus 1 by L R i naught plus v naught. So, now, you have the initial value of i and initial value of d i by dt in your hand which you can utilize to solve the equation. If you ah see that ah the previous discussion ah what we did when we discuss about the ah the source free response of R L circuit and R C circuit we came to know that typically the solution of these kind of circuits is exponential. So, let us assume that the solution for this case that is the equation which we have is i is equal to A e to the power sigma t where A and this is sigma. So, A and sigma are constants which we have to determine. Now, what you will do? You will put the value of i which you have assumed that is i is equal to A e to the power sigma t. You put the value of this current i in this equation that means, that you will differentiate it twice. So, you will get the first component. So, that becomes A into sigma square e to the power sigma t and then first derivative multiplied by R by L. So, you get A R by L sigma e to the power sigma t plus A by L C e to the power sigma t. So, this you will get simply by putting the values of i into this equation. Now, if you rearrange you can say that A e to the power sigma t you can take it out and in the parenthesis you will see the value which you get is sigma square plus R by L sigma plus 1 by L C equal to 0. Now, this solution of these type of equations ah if you see in in that case this factor that is A e to the power sigma t cannot be equal to 0 because this is our initial assumption that the value of A e to the power sigma t is nothing, but the value of current i. So, this cannot be 0. So, what we left with is sigma square plus R by L sigma plus 1 by L C would be equal to 0. So, we will say that the sigma square plus R by L sigma plus 1 by L C equal to 0 would be the governing equation for this and this is the quadratic equation and we say this type of equations as characteristic equation for the differential equation which we just saw. So, this will be considered as a characteristic equation of the differential equation. Now, if you see this is second order equation if you solve you will say that you there will be 2 roots of this equation. So, let us say these roots are sigma 1 and sigma 2 you can find out the value of these roots sigma 1 is equal to minus R by 2 L. So, how you will get if it is if you remember the equation is like A square plus B x plus C is equal to 0 then its root would be minus B plus minus root of B square minus 4 A C divided by 2 A. So, you know that when you want to find out the roots of a equation you can use this formula. So, because you have plus minus 2. So, one root would be related to plus second root would be related to minus. So, when you put these values you will get sigma 1 equal to this sigma 2 equal to this. So, now, you have these 2 roots in your hand then you can represent it in a more compact manner how you will represent. Let us say that the alpha is R by 2 L and omega naught is 1 upon root L C. So, if you assume these 2 factors and put the value of sigma 1 and sigma 2 in this particular form then you can simplify this as sigma 1 is equal to minus alpha plus root of alpha alpha square minus omega naught square and alpha 2 is equal to minus alpha minus root of alpha square minus omega naught square. Now, these 2 roots that is sigma 1 and sigma 2 are called natural frequencies and are measured in Neepers per second while omega naught is known as the resonant frequency or undamped natural frequency of the system which is expressed in radians per second alpha is the damping factor which is again expressed in Neepers per second. So, when you see this characteristic equation if you put the value of alpha and omega naught it will be simplified as sigma square plus 2 alpha sigma plus omega naught square is equal to 0. Now, this indicates that the there are 2 possible solutions. So, what we can say there would be 2 solutions like I 1 is equal to A 1 e to the power sigma 1 t I 2 is equal to A 2 e to the power sigma 2 t because we have 2 roots sigma 1 and sigma 2. So, we will have 2 solutions for the current that is I 1 and I 2 and now this equation that is our original equation d 2 y by d t 2 plus R by L d I by d t plus I upon L c equal to 0 this is a linear equation. So, any linear combination of the 2 distinct solutions with that is I 1 and I 2 is also a solution of this equation. So, for a complete solution of the above equation would therefore, require a linear combination of I 1 and I 2. So, what we can write we can write I t is nothing, but A 1 e to the power sigma 1 t plus A 2 e to the power sigma 2 t. So, this would be the total solution of the equation that is ah the second order equation ah related to our series RLC circuit. Now, here A 1 and A 2 are some constant which you can determine with the help of ah the I at time t is equal to 0 and d I by d t at time t is equal to 0 and these values we already have in our hand. So, what we can say that when you compare these ah this equation then you can say that the sigma 1 and sigma 2 which are the roots of the characteristic equation, we have 3 conditions or we say that there are 3 possible types of solutions related to the roots of sigma 1 and sigma 2. What are those solutions? One is when alpha is greater than omega naught. So, in that scenario we say that this is an over damped case, when alpha is equal to omega naught we say this is a critically damped case and if alpha is less than omega naught we have under damped case. Now, let us say the first case when the system is over damped means alpha is greater than omega naught, alpha is greater than omega naught means the value of C is greater than 4 L by R square. How we got? We just put the value of alpha and omega naught, alpha and omega naught we already have in our hand, we assumed that the values of alpha and omega naught are like this. So, if you put these value in this condition that is alpha greater than omega naught, you will come to know that this would be condition when the value of C is greater than 4 L by R square. So, in that case what will happen your both roots which are L sigma 1 and sigma 2 are negative and real. So, the response would be A 1 e to the power sigma 1 t plus A 2 e to the power sigma 2 t and the curve which you if you plot the value of current I t with respect to time t it will be like this as shown in the figure. So, which decays and approaches to 0 at as time t increases. So, this would be the ah type of response for the case when alpha is greater than omega naught that means, the system is over damped. Now, second case is the case when system is critically damped means alpha is equal to omega naught. So, what will happen in that case alpha is equal to omega naught means C is equal to 4 L by R square or we can say sigma 1 equal to sigma 2 is equal to minus R by 2 L. So, what will happen in that case you can simplify it because alpha is equal to omega naught when you will simplify you can say it as A 1 e to the power minus alpha t plus A 2 e to the power minus alpha t or you can simply say A 1 plus A 2 is nothing, but A 3. So, the solution will become A 3 e to the power minus alpha t. Now, this cannot be a solution because the two initial conditions which we saw that is I at time t is equal to 0 and d I by d t at time t is equal to 0 cannot be satisfied with single constant A 3. So, that why it happened because our initial assumption of exponential solution is incorrect for this particular case of critical damping. So, now, what we will do let us rewrite the equation that is d 2 I by d t 2 plus R by L d I by d t plus I by L C equal to 0 where we will say alpha is equal to omega naught is equal to R by 2 L. So, when we will put this value our the our equation will become d 2 I by d t 2 plus 2 alpha d I by d t plus alpha square I or you can rearrange and say that d by d t over bracket d I by d t plus alpha I plus alpha over bracket d I by d t plus alpha I is equal to 0. That means, if you can rearrange it what you can do you can separate out 2 alpha d I by d t as alpha d I by d t plus alpha d I by d t and when you will rearrange you will get this equation. Now, let us assume that the d I by d t plus alpha I that is which is there in the parenthesis is some function f. So, what you can write? So, when you put the value of f here. So, this equation will become d f by d t plus alpha f equal to 0. Now, the solution of this because this is a first order differential equation and we have seen these kind of equations when we discuss the series R C or R L circuits in case of source free response. So, you can simply write that the solution of this equation would be f is equal to A 1 e to the power minus alpha t where A 1 is a constant. Now, if you put the value then d I by d t plus alpha I would be nothing, but A 1 e to the power minus alpha t. So, if you see this equation you have assumed f is equal to this value of f you have got as A 1 e to the power minus alpha t you put this value again in this equation. So, you will get d I by d t plus alpha I is equal to A 1 e to the power minus alpha t when you will rearrange you will get this equation. Now, if you see the left hand side of this e to the power alpha t d I by d t plus e to the power alpha t alpha I is equal to A 1 that is nothing, but d by d t of e to the power alpha t into I is equal to A 1. So, this equation you can simply rearrange and it will become d by d t whole of e to the power alpha t and I. Now, next what you have to do you have to simply integrate the equation on both sides when you will integrate what you will get you will get the value of I is nothing, but A 1 t plus A 2 e to the power minus alpha t. So, you have to just simply integrate it and you will get another constant A 2 because when you integrate you get one integration constant also. So, when you integrate both sides A 1 t you will get plus A 2 you will get and you will get e to the power alpha t into I. So, what you get finally, I is nothing, but A 1 t plus A 2 into e to the power minus alpha t. Now, the natural response of the critically damped circuit is the sum of 2 terms the negative exponential and negative exponential multiplied by a linear term. So, in that case what will be the response if you plot the variation of current I with respect to time t the response would be like shown in the figure. Now, if you see at time t is equal to 1 by alpha what will happen. So, at time is t is equal to 1 by alpha if you put this value in I t is equal to t into e to the power minus alpha t. So, the value will be maximum at t is equal to 1 by alpha and the value of maximum is e to the power minus 1 by alpha. So, that means, that after one time constant the current will reach to its peak value and then it will slowly decay to 0. So, this kind of response you will see when the system is critically damped. Now, the third scenario is that the value when alpha is less than omega naught. So, when alpha is less than omega naught means C is less than 4 L by R square. So, the roots which you have seen that is sigma 1 and sigma 2 minus alpha plus alpha square minus omega naught square in this case if alpha is less than omega naught what you can do you can rearrange these two ah the values and you can simply write since omega naught is greater than alpha you can rearrange the roots of a characteristic equation that is sigma 1 and sigma 2 as minus alpha plus root of minus of omega naught square minus alpha square. So, when you will say when you will see the root of minus 1 you can simply represent with the complex variable j. So, what you can say you can say that sigma is nothing, but minus alpha plus j omega d and simply sigma 2 also you can write minus alpha minus j omega d what is omega d omega d is nothing, but under root of omega square omega naught square minus alpha square. Now, since omega is greater than alpha this factor will always be positive. So, that is why we have taken minus 1 out. So, this omega d is nothing, but damping frequency and omega naught is the undamped natural frequency. So, when you will put the value of sigma 1 and sigma 2 in the in the solution which we have got that was I t is equal to A 1 e to the power minus ah sigma 1 t plus A 2 e to the power minus sigma 2 t. So, you will get the response of I in terms of t then you know that the Euler identity that e to the power j theta you can simply represent as cos theta plus j sin theta and e to the power minus j theta can be represented as cos theta minus j sin theta. So, what you can now write if you put the value I t will be ah like as shown in this expression. Now, if you replace constant A 1 plus A 2 and j A 1 minus A 2 with B 1 and B 2. So, this will be simplified as the expression for current I with respect to time t. Now, if you see this equation ah the expression for I t you will say that the the system will have cosine terms and sin terms that is sinusoidal term you will see plus exponential terms. So, how the response will look like in that case if you see the response for current it would be exponentially decaying plus sinusoidally varying also because you have sin cos terms in the equation. So, the response have time constant of 1 by alpha and the period will be t is equal to 2 pi by omega d. So, this you can see from the figure. So, with this we can close our today's session. We will continue our discussion in the last lecture and we will see when we apply step input to the RLC circuit how it will behave. Thank you .