 This is the 13th lecture. We shall continue the discussion on that problem in one of the tutorial classes. 13th lecture today is 7th February and we discussed natural response of 2nd order circuit and if time permits, then we shall discuss these 2 terms poles and zeros. You remember in the 2nd order circuit that we had considered, what we had considered was we have a capacitance which is charged to which is charged to a voltage of V, that is a V0 and then a T equal to 0 is connected across an inductance L and the resistance R. The current that flows in the circuit is I. We are considering this particular LCR circuit. The initial condition is that Vc is 0 minus is V0 and then naturally as time increases, small i, small i at T equal to 0 that is when the capacitor is switched to this series combination, the current would be 0, current would be 0 and then the current would gradually increase and then decrease. After infinite amount of time it would decrease. Qualitatively this is the phenomenon but depending on the relative values of R, C and L, the way it does so would be quite different and we have discussed one of the cases but before that let me write the differential equation. The second equation was L d2i dt2 plus R di dt plus I by C equal to 0. How did we obtain this equation? We first obtained an integral differential equation that is Ri plus L di dt plus 1 by C integral I dt V0 minus that and then we differentiated this to obtain a homogeneous differential equation that is the right hand side is equal to 0 and then we argued that the solution to this equation shall be we shall we can try a solution of the form e to the st, some constant multiplied by this and this with this we ended up in a quadratic equation that is small s satisfies a quadratic equation of the form s squared plus s R by L plus 1 over L C equal to 0. If you substitute this in this equation and clear both sides of a e to the st then this is the equation that you get which clearly has 2 solutions and the 2 solutions are s12 is equal to minus R by 2 L plus minus square root of R by 2 L whole squared minus 1 over L C and if you if you denote the quantity under the square root sign as capital D then you know the capital D stands for discriminant and we had considered the case when discriminant is greater than 0 this leads to s1 not equal to s2 and both s1 and s2 are real and negative and therefore the general solution to the current function would be a1 e to the s1 t plus a2 e to the s2 t and we had taken a specific example to show how the current varies with time. The general solution we had evaluated a1 and a2 from the initial conditions and there are 2 initial conditions to bother about one of them was that i of 0 was equal to 0 i of 0 was equal to 0 and the other initial condition was that d i d t at 0 plus is equal to v0 by L these are the 2 initial conditions these initial conditions will apply to every situation whether capital D is greater than 0 whether capital D is equal to 0 less than 0 it does not matter it it satisfy this initial conditions apply to all conditions all parameter conditions that is values of R L and C and with this under the condition that d greater than 0 we had shown that the total solution to the current is given by v0 L s1 minus s2 e to the s1 t minus e to the s2 t and we had taken a special case and showed that the plot of the current would look like this that it would show a maximum and then it would fall exponentially with time okay this is one case now let us consider the other case in which d is less than 0 then we shall consider the junction juncture case namely d is equal to 0 that is the transition from d greater than 0 to d less than 0 or the other way round we shall consider the transition case at the last instance alright first let us consider we have considered capital D greater than 0 let us consider capital D less than 0 then s1 2 the 2 roots are minus R by 2 L plus minus square root of d is less than 0 that is R square by 4 L square is less than 1 over L C and therefore this quantity is negative I can write this is minus R by 2 L plus minus square root of I can take a minus 1 common here and then write a positive quantity 1 by L C minus R square by 4 L square and square root of minus 1 is J minus R by 2 L plus minus J in mathematics one writes an I but small I we use for current in electrical engineering and therefore electrical engineering we always use a J square root of minus 1 so J square root of 1 over L C minus R square by 4 L square this would be the nature of the roots and you see that unless capital R is 0 the roots are in general complex quantities this is a real quantity and this is a complex quantity let us consider let us call the real quantity as minus alpha that is minus R by 2 L and plus minus J let us call this quantity as omega 0 and omega 0 is a positive quantity we have taken plus minus here so the roots are in general complex minus alpha plus minus J omega 0 and any complex quantity can be represented in the complex plane where it is a 2 dimensional plane rectangular coordinates along the x axis we represent the real part along the y axis we represent the imaginary part and therefore that plane that complex plane we shall denote as the s plane as you see these values these values are values of s we had what is s s was the power of the exponent divided by t so these are the values of s s 1 2 are also the roots of the characteristic equation roots of the characteristic equation and these roots can be represented on the so called argend diagram if you are not acquainted with the name it does not matter but what we do is the real part of s we consider an s plane the real part of s is represented on the x axis and we call it the sigma axis and the imaginary part of s is represented on the y axis the orthogonal axis and we call the omega axis or the j omega axis this is the usual convention for electrical engineers that we show the imaginary value explicitly sigma plus j omega for example this is the 0 0 point so our s in general therefore is sigma plus j omega now if we have let us say minus 1 plus minus plus j 2 then we shall have we shall go minus 1 in this direction and we go 2 units perpendicular to it in the upward direction and our point will be somewhere here now in our case the roots are s 1 2 is equal to minus alpha plus minus j omega 0 so what we do is we go minus alpha to the left let us say this is the point minus alpha then we go up and down by a distance of omega 0 so we have one of the roots here and we show this by a circle where this distance is j omega 0 and we go down by the same distance and put a circle here this distance is minus j omega 0 so these 2 points these 2 points painted green inside are the 2 roots of the characteristic equation and they can be represented in the complex plane this plane is called a complex plane because we are representing complex quantities on this plane or the s plane this circuit an RLC series circuit RLC in which V c 0 minus is equal to some voltage V 0 the current in the circuit if this is the direction of the current I of 0 minus is equal to 0 and by continuity of capacitor voltage and inductor current we know that this should be true the other initial condition is that L d I d t at t equal to 0 plus should be equal to V 0 and with this initial conditions we have solved the circuit for the case when d the discriminant is greater than 0 if you recall the characteristic equation of the circuit characteristic equation was s square plus yes R by L s plus 1 over L c equal to 0 which has 2 roots s 1 2 equal to minus R by 2 L plus minus square root of R by 2 L whole squared minus 1 over L c and it is this quantity under the square root sign which we have called capital D for d greater than 0 we saw that the roots are real distinct and negative and we saw that the complete solution could be worked out in terms of the initial conditions as if you recall d greater than 0 the current was given by V 0 by L s 1 minus s 2 e to the s 1 t minus e to the s 2 t is that correct I hope that is correct yes and we are trying to see what happens if d is less than 0 we are good that if d is less than 0 then s 1 2 can be written as minus alpha plus minus j omega 0 where alpha is equal to R by 2 L and omega 0 squared is equal to 1 over L c minus R by 2 L whole squared and as you can see R by 2 L is simply alpha so I can write omega 0 squared as minus alpha squared and this quantity 1 by L c let us use another symbol that is omega m squared let us use omega m squared as 1 by L c then you notice that omega m is simply equal to omega 0 when alpha equals 0 that means if there is no resistance in the circuit if there is no resistance in the circuit then omega 0 is simply equal to omega n if there is no resistance in the circuit where are the roots of the characteristic equation they are purely imaginary plus minus j square root of 1 by L c which means that the roots shall be plus minus j omega n and therefore omega n has a physical significance now under this condition that d is less than 0 we showed last time that the current solution shall be of the form a 1 e to the power minus alpha plus j omega 0 t plus a 2 e to the power minus alpha minus j omega 0 t and we are good that if I take e to the minus alpha t out e to the minus alpha t out then the rest of the quantity inside the brackets could be put in terms of cosine and sine and the cosine and sine could then be combined into this particular form sine of omega 0 t plus a theta where you see the initial constants I did this steps last time okay now you see the initial constants a 1 and a 2 have now been replaced by modified constants c and theta which have to be evaluated from the initial conditions let us look at let us look at the initial conditions i of 0 plus is equal to 0 this means if I put t equal to 0 here then you see c sine theta should be equal to 0 which means that theta should be equal to 0 alright and therefore my solution is i of t is equal to c e to the minus alpha t sine of omega 0 t once again we know that l d i d t the other equation is that l d i d t at t equal to 0 plus is equal to v 0 if I substitute i of t equal to this here and put t equal to 0 then what do you think I will get I will simply get l times I should use some other constant here I should not use a c why c i have used for capacitance so let us use a c 1 alright let us use a c 1 then if I do this if I substitute here the differentiation and put t equal to 0 can I skip the algebra and write simply that c 1 will be equal to v 0 divided by l that is what it shall be alright which means is it this okay if you want to do the algebra I will gladly do it there is no problem okay I am glad you caught that this will be v 0 divided by l omega that is correct because a differentiation of this shall bring omega 0 alright so c omega 0 cosine omega 0 t multiplied by e to the minus l t alpha t put t equal to 0 this omega 0 shall be there thank you very much okay and therefore and therefore my current solution would be total solution is v 0 by l omega 0 sine of omega 0 t times e to the minus alpha t that factor shall be there alright now suppose sine of omega 0 t was not there if this was not there then you see dependence on time is exponentially decaying because alpha is capital R if you recall alpha is capital R divided by 2 l it is a positive quantity and therefore e to the minus alpha t decays with time the presence of sine omega 0 t means that it is the product of an exponentially decaying curve and a sine curve therefore what it means is that if I plot I of t versus t the envelope of the curve shall exponentially decay whereas inside there shall be a sinusoid if alpha was 0 if this was not there then it would be a non decaying sinusoid on the other hand if alpha t is there as time increases naturally the amplitude of the sine wave gradually decreases it decreases exponentially and therefore the figure that I shall get it shall start from here the figure that I shall get would be like this it shall start from here and then gradually the sine wave as time proceeds the maxima and minima shall gradually decrease it is not exactly a periodic wave because after one period it does not repeat the amplitude goes down so this is a decaying decaying sinusoid how does it decay the the rule for decay is exponential so exponentially decaying sinusoid it is as if it is as if the sinusoid is being damped with time it is as if the sinusoid is damped with time and therefore this situation is called damped oscillation okay there is an oscillation there is an oscillation but the oscillation amplitude gradually decays so it is a damped oscillation now if alpha was equal to 0 then you see the oscillations would have been undamped that is the amplitude would have remained constant whatever the time is and therefore therefore you now get a physical significance of the quantity Omega 0 Omega 0 is obviously the damped frequency of oscillation or frequency of oscillation under the damped condition and you see Omega n then is the frequency of oscillation under undamped condition and Omega n therefore is called the natural frequency of oscillation natural frequency of oscillation that is if if the damping was equal to 0 then the oscillation well undamped natural frequency undamped natural frequency of oscillation the word natural comes because it is not a forced oscillation it is an oscillation because of the initial energy in the system the system has been left alone and therefore it shows its natural behaviour so Omega n has the interpretation of undamped natural frequency of oscillation and Omega 0 is the damped natural frequency of oscillation alright damped natural frequency of oscillation and alpha is the damping coefficient alpha is the damping coefficient what is the unit of alpha 1 over t that is correct and you know that Omega 0 squared is Omega n squared minus Alpha squared alright so all the 3 quantities Omega 0 Omega n and Alpha they have a physical significance and this is the significance the the envelope decays according to this is V 0 by L Omega 0 e to the power minus Alpha t we can take an example at this point this is the general solution we can take an example the condition is that R squared by 4 L squared should be equal should be less than 1 by L c this is the condition under which oscillations take place and the oscillations shall be damped oscillations now 1 there are various combinations of R L and C which satisfy this but one of them let us say C is 117 farad we have taken a peculiar value so as to get nice numbers and capital R equal to let us say 2 ohms then the roots are S 1 2 is equal to you can see minus R by 2 L so minus 1 plus minus now R squared by 4 L squared is given by 4 and 1 by L c is 17 R squared by 4 L squared is 1 that is right so 1 definitely is less than 17 and therefore see that the roots are complex and 17 minus 1 is 16 square root of 16 is 4 so minus 1 plus minus J 4 and from since we have found out the general solution already we can write down the current solution as what will be the constant V 0 V 0 divided by L omega 0 L is 1 Henry what is omega 0 4 and therefore V 0 by 4 e to the power minus what is alpha 1 so it is minus T now alpha is 1 the roots are minus alpha plus minus J omega 0 alright so alpha is equal to 1 then sin of S 4 T this is the solution alright if you know the general solution for any given combination you can always find this out now we have considered 2 cases 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 S 1 T and e to the minus S 2 T where S 1 and S 2 are both real and negative alright e to the S 1 T minus e to the S 2 T A 1 plus A 2 this is the solution where S 1 and S 2 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 V 0 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 the 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 a damping coefficient if the damping is slightly mode 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.