 lecture on natural response of 2nd order circuits continued. We have been discussing 2 cases that is capital D greater than 0, in this case there are no oscillations, the solution consists of the sum of 2 exponentials e to the minus s1 t and e to the minus s2 t where s1 and s2 are both real and negative alright e to the s1 t minus e to the s2 t a1 plus a2 this is the solution where s1 and s2 are both negative there are no oscillations, so the current starts from 0, attains a maximum and then goes to 0. On the other hand if D is less than 0 then we have oscillations we have we have V0 by L omega 0 e to the minus alpha t sin of omega 0 t alright. The border line between these 2 or the transition case which is which happens at D equal to 0 occurs obviously when r square by 4 L square is equal to 1 over LC alright this is a very special condition and the condition on the resistance r to satisfy this is called r critical, it is a critical value of resistance for which this shall be true alright. If this equality becomes an inequality there shall be either oscillations or no oscillations and therefore this is a critical case and this case is called the case of critical damping, critical damping that is if the damping is slightly less then there shall be what? There shall be oscillations if the damping is slightly less there shall be oscillations alpha is the damping coefficient if the damping is slightly more there should be no oscillations and therefore this case D greater than 0 is called an over damped case, over damped means no oscillations in the natural response and naturally D less than 0 should be called an under damped case alright under damped case and D equal to 0 is the so called critical damping case, critical damping. It determines the boundary between oscillations and no oscillations let us see what the solution to this case is our S 1 2 was equal to minus r by 2 L and since capital D equal to 0 there is nothing else plus minus 0 and therefore the 2 roots are now coincident coincident on if you take the sigma j omega S plane coincident where on the negative real axis there are 2 roots at this distance is equal to r by 2 L the 2 roots are no longer complex they are both negative real and they are coincident. So what is the solution under this condition? Obviously if you take our general solution if you recall e to the S 1 t plus a 2 e to the S 2 t and if S 1 is equal to S t obviously the 2 solutions are no longer independent of each other is not that right they are identical solutions in fact except for the constant a 1 and a 2 well if S 1 and S 2 are equal I can only write this as a 1 plus a 2 e to the S 1 t if S 1 is equal to S 2 and therefore this cannot represent the general solution. The general solution must have 2 arbitrary constants because it is a second order system it must have 2 arbitrary constants and the 2 solutions should be independent of each other this should be independent solution and as you know as you know the solution of differential equations the solution in this case shall be given by a 1 plus a 2 times t e to the power let us say S 1 or S 2 S 1 t all right this is the general form of the solution if under the this is the case for critical damping that is that is capital D equal to 0 or which means that R squared equal to well R would be equal to R square by 4 L squared R would be equal to twice square root of L by C is that correct all right. This value of resistance is called the critical value of resistance is R critical and under this condition this will be the general solution. Now if I apply the initial condition that is I 0 equal to 0 then obviously you see a 1 should be equal to 0 is that right if I put t equal to 0 here then obviously a 1 is equal to 0 and therefore my I of t shall be equal to a 2 t e to the power S 1 t the second boundary condition the second initial condition is that L d i d t L d i d t at t equal to 0 should be equal to v 0 should be equal to v 0 if I differentiate this then I get L a 2 t S 1 e to the S 1 t plus e to the S 1 t that should be equal to v 0 at t equal to 0 I have differentiated this this expression and I have to put t equal to 0 if I do that then what do I get as a 2 a 2 is v 0 by L no longer omega 0 comes all right a 2 is v 0 by L which means that my total solution is given by v 0 by L t e to the S 1 t where S 1 is equal to minus R by 2 L this is the solution to the equation when the 2 roots are coincident for example if L is equal to 1 Henry and C equal to 1 quarter Farad what should be the R critical twice square root of L by C can be 4 ohms in other words if R is greater than 4 then this case will be which one over under over damped if R is equal to 4 ohms of course this corresponds to critically damped case critically damped case and if R is less than 4 then this is the under damped case all right now let us look at this situation let us look at this situation in the S plane all right the complex plane and I will repeat I will repeat what I have been saying in terms of the complex plane you see if I have I have 3 situations one is capital D capital D greater than 0 capital D less than 0 and capital D equal to 0 all right and the roots of the characteristic equation let me show the axis sigma j omega j omega sigma sigma j omega there are 3 cases and you see in the 1st case D less than 0 there are 2 negative real roots minus S 1 and minus S 2 in the 2nd case that is the under damped case we have 2 complex conjugate roots where this distance this point is minus alpha and this point is j omega 0 this point is minus j omega 0 all right minus alpha plus minus j omega 0 on the other hand if I have the critically damped case then the 2 roots coincide on the negative real axis all right is it okay let us consider a 4th case a 4th case in which capital R equal to 0 can you tell me where the roots should be on the imaginary axis okay can you tell me the value 1 by LC in terms of omega n it is j omega n is that okay right now I want you to look at this figure carefully if the figure is too small on the monitor I can draw it again and last version can you can you see distinctly from the last pinch okay now look at this case D less than 0 this is omega 0 and this distance is alpha so what is the distance of the root from the origin omega not squared plus alpha square is in this simply equal to omega n square okay now omega n depends on L and C only is it not right omega n does not depend on R is that clear by definition omega n is the undamped natural frequency oscillation and undamped means R equal to 0 it is this omega n we are talking of on the other hand both alpha and omega 0 naturally depend on R is that clear because alpha depends on R alpha is equal to R by 2 L therefore omega 0 depends on R omega 0 square after all is omega n square minus alpha square is that clear I am trying to introduce a concept so please be with me what I have said is omega n is independent of R alpha as well as omega 0 depend on R and therefore if I take a circuit if I take a second order circuit C L and R and if I adjust R if I adjust R how do the roots change how do the roots change alright this is what I want to find out suppose I wish to draw the roots of the characteristic equation on the S plane then when R is equal to 0 when R is equal to 0 start from the 0 value it cannot go below 0 alright when R is equal to 0 where are the roots roots are on the imaginary axis at plus minus j omega n is that okay when R is slightly increased what happens to the roots they move to the left half plane this is the right half plane this is the left half plane they move to the left half plane such that the distance of the root distance of either root from the origin remains a constant and that constant is equal to omega n and therefore the movement of the roots shall naturally be on a semi circle where this distance is omega n alright is that clear as R is increased this point corresponds to R equal to 0 and as R is increased the roots the two roots this root moves in this direction and this roots this root moves in this direction such that they are always complex conjugates that they must remain on the circle alright now when the roots approach each other and meet on the negative real axis what is the situation the two roots are coincided and therefore capital R must have reached its critical value so this point shall correspond to R equal to twice square root of L by C where the roots become coincident roots are no longer complex they are real negative and coincident alright what happens after this they move on the real axis well what happens is you recall that the roots are minus R by 12 plus minus square root of R by 12 whole square minus 1 over L C now when capital D this point corresponds to capital D equal to 0 alright that means the roots are at minus R by 12 now when resistance increases further what happens is that this minus R by 12 is either added to a quantity less than it or subtracted that is a negative quantity which is less than it is added to it what does it mean it means that the roots one of the roots goes towards the origin and the other root goes towards infinity alright if capital R is increased beyond the critical value the two roots the case is that of over a damped case and the two roots move along the negative real axis the root follows remain negative real one increases in magnitude the other decreases in magnitude the one that decreases naturally goes towards the origin and the one that increases goes towards infinity alright this can also be seen from the analytical expression try to follow this very carefully when capital R tends to infinity when capital R tends to infinity one of them one of the roots should be 1 by L C naturally can be neglected then and therefore one of the roots should be minus R by 12 minus R by 12 and the other root would be minus R by 12 plus R by 12 so one of the roots goes to the origin the other goes to infinity is that clear okay so both from the analytical expression from explanation and from the figure you can see that the locus of the two roots is given by this semicircle and then two straight lines one going towards the origin the other going towards infinity and this picture this picture which gives the motion of the roots or the locus of the roots as one of the parameters capital R changes in the circuit this picture is called a root locus for obvious reason it describes the locus of the roots as one of the parameters that is capital R changes now if you argue what is so sacred about capital R why don't you keep R constant and very L we can again obtain another sketch which will not be exactly this it will be another sketch but you can draw it and I leave that to you as an exercise once again you can obtain a critical damping condition and under damped condition and an over damped condition you can also draw root locus for variation of capital C capital C can also vary and this would be a third kind of sketch now is this point clear the concept of root locus the concept of over damping under damping and critical damping alright usually in measuring instruments in measuring instruments which can be usually modeled as a second or a system what do you think we should have as far as damping is concerned critically damped well why not under damped let us understand it takes quite a bit of time the needle goes on oscillating and then finally it settles to some value whereas if it is critically damped critically damped then it reaches the final value alright if it is over damped then it takes a very long time to reach a steady state and therefore this concepts are important the concept of root locus as you should see is extremely important arises in control systems arises in many other physical systems the other concept that is of extreme importance is that of impedance and this is time that we introduce this concept impedance I should first give a formal definition it shows the locus of the roots as one of the parameters is varying so that is okay that is obvious but I mean what what is its importance no what is its importance well as you shall see later if you are a mechanical engineer in control systems in any process control for example which can be modeled as a second order or third order or fourth order system suppose the roots one of the roots due to a disturbance in a parameter goes to the right half plane if the root locus includes the right half plane what will happen or the imaginary axis then what will happen suppose the parameter is such that the roots are on the imaginary axis then the instrument or the process that you are experimenting will never settle down it will go on oscillating is that right so the root locus shows the locus of the roots when we design should be such that the roots are in the left half plane for example it is not always true that all physical systems we require a critically damped case no sometimes a bit of oscillation is necessary why if you want a quick response well after a couple of oscillations it stays it settles so a little bit of damping a little bit of under damping is quite in order so the root locus shows where to locate your roots and if you know where to locate your then the system design is obvious all right so this is the purpose of the root locus root locus is an analytical tool for designing a system I have drawn the root locus of a second order system there can be third order fourth order usually natural systems practical systems are of are usually second or higher order systems and the root locus is essential for determining the design criteria where should you locate the roots in order to satisfy what you want in order to satisfy your requirements is that okay any any other question regarding root locus all right next we introduce the the concept of impedance the the introduction of the definition that I am going to give you now would be a bit heuristic a bit inaccurate a bit unsophisticated but this will serve our purpose at the present moment the definition of impedance is like this consider a two terminal network a network which is only two terminals all right and you consider a linear network all right a linear network then you know that all that you can do is to is to connect a voltage generator and measure the current the excitation would be a voltage generator the response would be a current or you could connect a current generator here and measure the voltage all right now if you if you if you are excitation either a current generator voltage generator is of the form of e to the s t and this network is a linear network contains resistance capacitance and inductance then you know the differential coefficient of e to the s t is also of the form of e to the s t the integral of use is also of the form e to the s t so if you apply a voltage generator here which is of the form e to the s t the current should also be of the form e to the s t isn't that right similarly if you apply current generator which is of the form of e to the s t the voltage developed across this shall also be of the form e to the s t and therefore the ratio the ratio of V by I shall be a constant but it will depend on the value of S and therefore this is denoted by capital Z of S and this is called the impedance. I must caution you, so let me first define it, the impedance of any 2 terminal, any 2 terminal network is the ratio of voltage to current provided the excitation either the voltage or the current, excitation and the bracket, the either the voltage or the current is of the form of e to the power S t, is that clear? It is not true for any arbitrary excitation, it must be exponential, only then the ratio of V to I is called an impedance. Let us take an example, is the definition clear? The exponential excitation is an integral part of this definition, if that part is taken out then it does not make sense, all right? For example if I take a resistance, let us say 2 terminal network appear resistance R, then you know V is equal to I R and if V is of the form of e to the S t, I is e to the S t divided by R, if I is of the form of e to the S t then V is e to the S t multiplied by R and V by I is equal to R, this is a constant independent of S, all right? This is a constant independent of S. On the other hand if I have an inductor L, V, L, I then you know V is equal to L, D, I, D, T and therefore if I is of the form of e to the S t then V by I shall be equal to S times L, so the impedance of an inductor of value L is simply equal to S times L, all right? What is the dimension of S? 1 by time, which means S is a frequency and S could be real or complex, all right? So in general small S is a complex frequency, real frequency, real quantity is a special case of a complex quantity, is not it right? So if you if you view life as complicated this is the most general situation, if you take the imaginary part out life becomes simple, all right? So S is a complex frequency and the impedance of a inductor Z L of S is equal to S times L. If I take a capacitor C, you know the current I equal to C D V D T and if V is of the form of e to the S t then Z of S is given by 1 over S times C, this is the impedance of a capacitor Z C S, all right? And once the impedance is defined for inductor and capacitor you can treat inductors and capacitors exactly like resistances, all right? The impedance of a resistance is R, the impedance of an inductor is S times L, the impedance of a capacitor is 1 by S C and if you have let us say a series combination of R, L and C then your V shall be equal to drop across R plus drop across L plus drop across C and if you take the voltage or the current of the form e to the S t you can show that Z of S here of R, L and C in series is simply given by R plus S L plus 1 over S C that is this quantities can now be added, all right? Similarly suppose I have a circuit like this R, L and C, suppose I have another combination, this is not a series combination, which is a series connection of R with a parallel connection of L and C, if this is V and this is I and one of them is exponential then the impedance of the circuit can be written the manipulation will be exactly like that of resistances, here an impedance S L is in parallel with an impedance of 1 by S C and therefore it is R plus S L times 1 by S C R 1 R 2 divided by R 1 plus R 2, now instead of R you say Z the impedance, so Z 1, Z 2 divided by Z 1 plus Z 2 that is S L plus 1 over S C which is equal to R plus S L divided by S square L C plus 1, now you ask me what is the use, why should we do that? Well if we do this, if we define an impedance, one of the advantages is that all kinds of circuits can now be treated in exactly the same manner that you have treated a resistive circuit, for example the Thevenin's theorem with this definition of impedance, now Thevenin's theorem is applicable to R and C circuits also, Thevenin's nortons and we can also do circuit analysis without writing a differential equation, once the impedance concept is introduced D D T shall be replaced by S and integral shall be replaced by 1 over S, E to the S T D T integral is E to the S T over S, D D T of E to the S T is S times E to the S T alright, let us take a specific example before we close this class, yes, please note. Sir for considering impedance has the applied voltage has to be the form of E to the S T. Exponential, if we consider the charging of the capacitor, sir we do not apply a voltage of 4 meter power S T, but still the voltage across our capacitor can be given as E to the S T, so can we say like can we define an impedance for it. That is a generational V to the S T, generational exponential reform, that is a question is completely different, this question is that of combining networks, you see if you simply want to charge a capacitor, now here E to the S T does not come into effect because your voltage, your source is a constant, it is not of the form E to the S, E to the S T with S equal to 0 and therefore you cannot introduce impedance concepts here alright, impedance concept will come only when the source is an exponential, let us take one example, let us take the same circuit that we have been talking of, let us say R L and C, now please please pay your attention to this, we are doing something new alright, suppose R equal to 2 M, L equals to 1 Henry and C equals to 1 quarter Farad alright and this V, this V is given as 6 E to the minus 2 T, is not it of the form of exponential, it is an exponential with S equal to minus 2, so if I wish to find this current let us say I for example, then what I do is, I first find out Z of S, Z of S as you have seen already is R plus S L divided by S squared L C plus 1, now substitute the values, Z of minus 2 alright would be equal to 2 plus, S is minus 2 multiplied by 1 divided by S squared is minus 2 squared times 1 times 1 quarter plus 1, which means Z of minus 2 equal to how much, 1 alright, Z of minus 2 is equal to 1, so what do you think this current shall be then, V by I is Z of S, so it will also be 6 E to the power minus 2 T, is that clear, is it minus, no it is plus, I shall be equal to V divided by Z of minus 2 and therefore it is 6 E to the minus 2 T, what do you think this voltage shall be V 1, it will be V minus I R, so what would this be, V is, so this would be minus 6 E to the minus 2 T, I know this voltage, if I know this voltage, do I know this current I L, I L shall be equal to this voltage divided by S L, where S shall be equal to minus 2, do I know this current I C, yes it would be C d V d T, C d V d T that is the differentiation of this, so I know all currents and voltages in this circuit, you see the excitation was exponential, I did not have to write a second order differential equation, normally yeah, you wanted to ask a question, normally in a circuit like this I have to write a differential equation and then solve it, that is write a 1 E to the S 1 T plus it, I did not have to do anything like that because I recognized that the excitation is exponential and if the excitation is exponential I can work purely in terms of impedances alright, I will just 5 minutes, I will just introduce another term to you, it is admittance, I told you impedance is the ratio of voltage to current, the reciprocal of an impedance is the admittance, admittance is 1 over impedance, sometimes it is easier to work in terms of admittance rather than impedance and it is with this that we conclude the class today.