 On impedance functions poles and zeros and their applications, we have already introduced the concept of impedance function as the ratio of voltage to current of a 2 terminal network whose excitation is exponential whose excitation is of the form e to the power s t the excitation could be either a voltage or a current and we have defined that under this condition z of s is the ratio of v by i with either v or i one of them is an excitation the other is a response of the form e to the power s t as an example if we have well one point that needs to be mentioned here and noticed by you is that when we speak of excitation as e to the s t we imply that the network under consideration when we say excitation e to the s t and the response is also e to the power s t we imply that the network under consideration is initially relaxed initially relaxed alright when we calculate the impedance function we do not take account of stored charges in inductors or stored energy in I am sorry stored charges in capacitors or stored energy in inductors alright so z of s when we say v is exponential i is also exponential well this is what it means now as an example if I take a simple let us say RLC circuit resistance inductance and capacitance a series circuit then the if i let us say suppose I excited by means of a current source i which is i 0 e to the power s t then you know that the voltage v across this shall be the sum of the voltages across R across L and across C and therefore v of v shall be equal to i 0 e to the power s t times R this is the voltage drop across the resistance plus L d i d t that is i 0 s e to the power s t plus 1 by C here is the question of initial relaxation we are not taking into into account any initial charges we assume that there are no charges stored therefore it is 1 by C integral integral i d t 0 to t alright and therefore this is simply this is simply equal to i 0 e to the power s t divided by s C in other words the impedance of the network between these 2 terminals is given simply by Z of s if I take the ratio v to i which is Z of s is equal to R plus SL plus 1 over SC as we shall see now the impedance concept helps us to determine the so called natural response of a circuit we shall see later that the impedance concept also helps us in determining the forced response or the steady state response as shall be demonstrated later in the last lecture at the last moment we mentioned that there is an associated concept of admittance y of s defined as the reciprocal of the impedance in other words the admittance of the series RLC circuit for example shall be 1 over R plus SL plus 1 over SC they are very simply related y of s can be defined as the ratio of the current to voltage when the excitation of the 2 terminal network is of the form of e to the power s t with the network initially relaxed that condition is important alright. Now let us take another example let us say we have a resistance and inductance and a capacitance these are the 2 terminals of the network the resistance inductance and capacitance so the impedance by looking at these terminals it is usually denoted by an arrow the impedance looking at these terminals is therefore given by R plus SL multiplied by 1 over SC combination rules are exactly those of resistances SL plus 1 over SC which I can write as I can simplify this as R plus s square LC plus 1 and here I shall have simply SL I can simplify this further Z of s is equal to R plus SL divided by s square LC plus 1 I can write this as s square LC plus 1 s square LC R plus SL plus R and this it is usual to write such expressions obviously these are these this expression is a rational function it is a ratio of polynomials 2 polynomials it is usual to write the polynomials with the highest power term coefficiented to unity that is the coefficient of the highest power term is reduced to unity obviously that is that can be done if I take LC R out from the numerator and LC out from the denominator and then I shall have R multiplied by s square plus s what shall I have here 1 over CR plus 1 over LC and here I shall have s square plus 1 over LC is the term rational function familiar to you okay a rational function is a ratio of polynomials and a polynomial is a finite series with integral powers integral positive powers alright a polynomial is a finite series with integral positive powers of the variable alright so I I see that this can be written as some constant some constant K which in this case is R constant means it is independent of s then a polynomial in s p of s divided by a polynomial in s q of s alright so the impedance function is in general of this form and you notice that the numerator and denominator both here are quadratics that is their second order polynomials and a second order polynomial shall have 2 roots alright both numerator and denominator have 2 roots and any polynomial can be written in terms of its roots as factors alright and this is one reason why the numerator here has been reduced to a form in which the coefficient of the highest power is unity so that I can write this in terms of its factors suppose well if you write the this particular rational function Z of s as equal to R s square plus s 1 by CR plus 1 over LC divided by s square plus 1 over LC you notice that the numerator roots numerator roots if I call them Z 1 and Z 2 small Z which I denote by a Z with a slash in the middle this is our distribution between cap Z and small Z the roots of the numerator function are given by minus 1 over 2 CR plus minus square root of 1 by 2 CR square minus 1 over LC is that okay alright this are the 2 roots of the numerator and let us say the denominator roots roots of the denominator polynomial are denoted by p 1 2 small p 1 2 then you see these are purely imaginary that is it would be plus minus J 1 over square root of LC the roots of the numerator can be either real or complex can they be imaginary yes they can also be purely imaginary but if that is so purely imaginary under what condition capital R capital R is infinity did I make a mistake somewhere no this is okay that means purely imaginary okay if it is purely imaginary then you see it is identically equal to 1 but you see the function would be identically equal to 1 because this term will drop out capital R infinity means s square plus 1 by LC and s square plus 1 by LC so you are left with only capital R this becomes a degenerate case let us not worry about it what I am saying is that the numerator roots can be either real or complex alright the denominator roots are purely imaginary alright suppose suppose one suppose this quantity the quantity under the square root sign suppose this is greater than 0 then obviously both the roots shall be Z 1 2 shall be real and negative alright now these are indicated on the complex plane or the s plane like this Sigma J Omega and the 2 the 2 roots of the numerator are denoted by circles suppose they are negative then there will be 1 will be here and 1 will be here they are indicated by small circles and the roots of the denominator are indicated by crosses they shall be on the imaginary axis purely imaginary this is the s plane on which the roots of the numerator and denominator are indicated are plotted you notice that the original expression Z of s can now be written in terms of its roots like this s minus Z 1 times s minus Z 2 and s minus p 1 multiplied by s minus p 2 alright the original expression can be written like this and this is the reason why the highest power coefficient was reduced to unity alright and it is obvious that when s takes the value either Z 1 or Z 2 the expression becomes a 0 Z of s becomes 0 when s takes a value Z 1 s equals to Z 1 Z of s becomes equal to 0 when s takes a value Z 2 Z of s becomes a 0 and therefore Z 1 and Z 2 are called the zeros of the impedance function Z 1 and Z 2 are called zeros of the impedance function and that is why we use the symbol small Z small Z for zeros on the other hand when s takes a value p 1 or p 2 the impedance function becomes infinite it blows up alright such values of s for which the impedance function becomes infinity are called the poles of the impedance function p 1 and p 2 are poles and therefore this diagram that we have drawn here is called a pole 0 sketch or pole 0 diagram pole 0 diagram alright let us take let us take another case is there any question on this definition of poles and zeros are let me repeat if you write Z of s as equal to k times in general it is not necessary that the degrees be 2 they could they could be higher than 2 it is not necessary that the degrees of the numerator and denominator be equal it is not necessary alright for example if you take let us say the simple RLC network alright 2 terminal network then you know the impedance is is R plus s L plus 1 over s C and this can be written as 1 over s C s square L C plus s C R plus 1 which I can write as L L C taken out and C taken out from here s then we have s squared plus C R by L C s R by L plus 1 over L C and you notice that the roots that the numerator is a degree 2 and the denominator is of degree 1 alright so the 2 degrees can be different can be different now the numerator has therefore 2 roots the denominator has only 1 root where are these roots the denominator root where is it it is at the origin so there is a there is a single pole at the origin and if the roots of the numerator are let us say complex then we shall have 2 zeros like this you also notice that if the roots are complex then they are complex conjugates that means if I have minus a plus j b as one of the roots then the other root must be minus a minus j b now why is that why is this constrained because the coefficients are all real alright and therefore it is a property of real polynomial real coefficient polynomials that if there is a complex roots its conjugate must also be a root now you notice that the point that I was making is that the numerator and denominator degrees p and q degrees need not necessarily be the same they may be different and therefore the number of poles and the number of zeros may be different here for example at s equal to 0 the impedance function blows up it becomes infinity that is it allows no current to flow s equal to 0 means what e to the st s equal to 0 means a constant which means a dc and obviously this is a corroborated by physical considerations that the that the capacitor does not allow a dc to pass alright okay then the 2 zeros here the 2 zeros here may be complex may be real and negative alright and at a 0 at a 0 the function becomes 0 which means that which means that the the the the total circuit acts as a short circuit for that particular value of s not for any other value of s we were taking we are going to take another example let us take that example before we generalize this let us take an example let us say we have a resistance r 1 a resistance r 2 and let us say a capacitor c the impedance function between these 2 points obviously is given by r 1 plus r 2 divided by s c r 2 plus 1 does it sound magical no what I have done is r 2 1 over s c divided by r 2 plus 1 over s c and this is what I have simplified to this okay I avoided that algebra so I can write this as s c r 2 plus 1 s c r 1 r 2 plus r 1 plus r 2 and now I must reduce it to that form that is the coefficients of the highest power should be 1 and so I take c r 1 r 2 out and c r 2 out therefore I get r 1 and denominator I get s plus 1 over c r 2 and in the numerator I get s plus r 1 plus r 2 divided by c r 1 r 2 is that okay alright we can write this in a in a more elegant form yes you could not see the previous one okay s plus r 1 plus r 2 divided by c r 1 r 2 and s plus 1 by c r 2 now you notice that r 1 r 2 divided by r 1 plus r 2 is the equivalent resistance of r 1 and r 2 in parallel and therefore I can write it in a in a somewhat more decent form as r 1 s plus 1 over c r 2 s plus 1 over c r 1 parallel r 2 let us call this r 1 s minus p 1 there is only one pole s minus z 1 that is only 1 0 then z 1 is equal to minus 1 over c r 1 parallel r 2 and p 1 is equal to minus 1 over c r 2 so both the roots the roots of the numerator and denominator all of them are real and negative and therefore the pole 0 sketch shall be like this sigma g omega now if I go from 0 to the left what shall I meet first the pole or the 0 it has to be a unique answer the pole so the cross and then the 0 this would be minus 1 over c r 2 and this would be minus 1 over c r 1 parallel r 2 or I say this is minus z minus p 1 and this is minus z 1 this is the pole 0 sketch that is what I do not thank you this is p 1 and this is the one absolutely right okay now listen to me carefully I am going to make what I defined here as poles and zeros were purely mathematical convenience we said z of s is a is a rational function so I find out the roots of the numerator roots of the denominator I call the roots of the numerator as zeros and the roots of the denominator as poles alright so what is the physical interpretation of a pole or a 0 one of the interpretations is that at a pole the impedance becomes open circuit at a pole the impedance become infinity it becomes an open circuit so it does not allow a current to flow agreed on the other hand at a 0 the impedance becomes a short circuit so it does not allow any voltage to be dropped agreed they are completely dual of each other we can make the same kind of interpretation with regard to an admittance after all admittance zeros are impedance poles is not clear because one is the reciprocal of the other so I can make the same kind of interpretation now another physical interpretation can be like this it is the same thing but we will state in different terms and this is what will lead to application of the pole 0 concept in finding the response natural response of circuits you see v equals to z i provided v or i is exponential alright this is the definition now you will notice that if z equal to 0 if z equal to 0 at some value of s then v equals to 0 irrespective of i is not that right okay is that clear this is a simple mathematical statement that if z equals to 0 whatever i is it does not matter the product shall be 0 so v equal to 0 irrespective of i which means that a current can exist without a voltage agreed a current can exist without an external voltage at this value of s a current can exist without an external voltage which means well if for a current to flow there must be there must be charges available current after all is the flow of charge which means that if a current has to flow well it must be due to internally stored energy agreed and therefore 0 is in some way related to the natural response of a circuit the 0 is in some way related to the natural natural response what response current response natural current response of a circuit and precisely precisely if you think if you pondered a little you see that if z of s1 some value of s if z of s1 is equal to 0 then s1 is 1 of the roots for one of the functions occurring in the exponential of the natural response which means that if z of s1 is equal to 0 then the natural response natural current response of the circuit should include a term of the form i1 e to the s1 t I shall illustrate this in a moment but let the concept soak in the concept is that we start from this relation and we say that if z is 0 at some value of s then that s must have some relation to the natural response of a circuit and we have seen in the examples that you have worked out that the natural response is of the form of e to the st natural response for example for a second order circuit it was a1 e to the s1 t plus a2 e to the s2 t how did that e to the st come because we converted all equations into a homogenous differential equation and we sought the solution in the form of e to the st agreed and therefore the impedance concept therefore gives us the values of s which shall occur in the natural current response of a circuit let me illustrate this yes how z equal to 0 okay this is what I had explained if z equals to 0 then from this relation we see that v is equal to 0 irrespective of i in other words even without the presence of a voltage a current can flow now without any external voltage a current can flow if it is the natural response if it is if there is internally stored energy and therefore conceptually it appears that this must be related to the natural current response and s the value of s at which the function becomes 0 should occur in the natural current response of a circuit and s occurs as a power of e all right s and t the product occurs as a power of e and what is what is the dimension of s1 it is a frequency and therefore s1 the value of s at which the impedance is 0 is called the natural frequency of current response natural frequency of current response and in that sense in that sense the zeros identify the natural frequencies of current response the zeros of an impedance function identify the natural current response of a circuit yes the current shall be infinity the current shall be infinity but then we have to generate the voltage of the form e to the power s1 t okay those questions we shall take up later conceptually a 0 at a 0 of an impedance function the function becomes 0 yes are they clear yes in calculating the impedance or just just to understand your question in calculating the impedance we say it is the ratio of voltage to current and this voltage or current is an external excitation when we apply that all elements must be relaxed because otherwise you see the current response would not be purely e to the s t it might have a constant component also all right so what we are doing is in calculating the impedance is it is a it one has effect on the other in calculating the impedance we assume the circuit to be relaxed once we calculate that then the parameters of the impedance are this constant K I am sorry let me write it again this constant K and the coefficients of P and Q these are the parameters of the impedance alternatively we said that P and Q let them be factored out then these factors are the roots what is their physical interpretation this is what we are looking into the physical interpretation is that these the roots of P of s identify the natural frequencies of current response identify it has nothing to do with whether store charge is stored or not if you wish to find the natural response of a circuit s 1 s I am sorry z 1 z 2 z 3 and so on which are the roots of P of s shall show their teeth shall show their face let me take a very simple example which which might which might convince you that it is so let us take the familiar RLC circuit well the impedance is z of s is r plus s l plus 1 over s c which is equal to s c s square l c plus s c r plus 1 all right now which is which is we wrote this as l multiplied by s square plus what was the r by l s r by l plus 1 over l c divided by s now you put this equal to 0 which means that the numerator that is s square plus s r by l plus 1 over l c equal to 0 that you see that this is precisely the characteristic equation of the circuit if you write the differential equation and try a solution of e to the s t this is what we have got earlier and it is the roots of this s 1 2 in terms of which we express the natural response as e to the s 1 t plus a 2 e to the s 2 t agreed so s 1 and s 2 are indeed the natural frequencies of the current response agreed this simple example demonstrates that what we are saying is correct let us take let us take an example suppose I have a resistance let us say 2 ohms another resistance 4 and the capacitance of value 1 quarter third we assume the circuit to be initially relaxed we are calculating the impedance g of s obviously this is 2 plus 4 multiplied by s the impedance of 4 by s that is correct divided by 4 plus 4 by s I simplify this 2 plus 16 divided by 4 s plus 4 is that okay a multiplier s so 16 divided by 4 s plus 4 which I can write as 2 plus s plus 1 and here 4 this I can write as s plus 1 that we make it into a rational function 2 s plus 2 plus 4 so 2 s plus 6 and I write this as 2 s plus 3 divided by s plus 1 all right therefore the pole 0 diagram is sigma g omega the pole is at minus 1 minus 1 and the 0 is at minus 3 no I am sorry this should be a circle the 0 is at minus 3 all right now what we are saying is we have identified the 0 and if we wish to find the current response natural current response of the circuit then e to the minus 3 t shall appear in the natural current response let us see now how can you obtain a current without having an initially stored energy therefore let us say that in this circuit 2 4 let us say that in this circuit 1 quarter farad let us say that v sub c v sub c 0 minus let us say is equal to v 0 suppose v sub c 0 minus is equal to v 0 and then what we do is between these 2 points we have measured the impedance we put a switch on at t equal to 0 and I wish to find the current through this short circuit you see remember that v equal to z i what did we argue we argued that if z equal to 0 then i can exist without an applied voltage okay it is between these 2 points that we considered v and i we considered v between these 2 points so let me make v equal to 0 what does it mean it means that I short circuit this and let us say this is the current i then without doing any differential equation I can say that i must be of the form sum i 1 e to the power minus 3 t alright to find i 1 we have to apply an initial condition what is the initial condition i at t equal to 0 at t equal to 0 the current i now I needed initial condition on i i of 0 plus after the switch is closed obviously is v 0 I did not say what is v 0 minus v 0 by 2 that is correct because the voltage from here to here becomes v 0 current is in this direction assumed so it must be minus v 0 by 2 which means that this is equal to i 1 and therefore i is equal to minus v 0 by 2 e to the power minus 3 t I have not solved any differential equation I have simply found out the zeros of the impedance between these 2 points and then what I have done is I have shorted them shorted the 2 points found out the current through this and I suspect that e to the minus 3 t well I am convinced that e to the minus 3 t shall occur in the natural response natural current response will come to voltage a little later natural current response and therefore this must be the total solution for example instead of the voltage instead of this current suppose I have to find out v c t then I can write it immediately as the initial voltage is v 0 so it will be v 0 e to the power minus 3 t or I can find out the current through the capacitor i sub c well i sub c t would be 1 quarter c times d v c d t is that clear okay let us let us consider the other the other case in which we have to interpret the poles physical interpretation of the poles physical interpretation of zeros is let me repeat that you measure the impedance between 2 terminals alright if you short circuit these 2 terminals that is if you make the volt applied voltage equal to 0 and there is initial stored energy somewhere in the circuit then the current response shall contain the frequencies that are identified by the zeros of the impedance function alright let us see the other case namely the poles for the poles of an impedance function poles of g of s we work in terms of admittance instead of impedance these are zeros of capital y of s capital y conventionally is the symbol for admittance alright and v equal to z i can be written as i equal to y v and we argue the same thing that is if y s 1 equal to 0 value of s at which at which the admittance is 0 then s 1 must be a pole of z alright 0 of y is a pole of z because one is the reciprocal of the other if y s 1 is equal to 0 then i equal to 0 and v is arbitrary any view do there is no restriction on v which means that without a current excitation a voltage can exist alright without a current excitation a voltage can exist so this value of s 1 must be related to the natural natural voltage response of the circuit natural voltage response of the circuit in other words the natural the voltage response must be of the form must must have a component v 1 e to the power s 1 t the voltage response v of t must have this particular component alright let us take another example to illustrate this point that is interpretation of poles i repeat zeros are associated with the natural current response and poles are associated with natural voltage response these are new concepts and i want i would go very slow and i want these concepts to soak in because in later in later classes we will simply assume we will not explain why we are doing this we will simply find out the poles zeros and say this is the voltage response this is the current response and so on let us take an example to illustrate this point that is interpretation of poles the same example we take r 1 equal to 2 ohms please follow this carefully r 2 is 4 ohms and c is let us say 1 quarter farad and we have a situation like this let us say we have a a switch now what we have to do is without a current excitation voltage can exist that means we will have to open the switch a current source has internal resistance of infinity and that we will have to open it okay so this is open that t equal to 0 and there is a source there is a source let us say current source here okay between these 2 points let us say a and b the impedance we have already found out g of s is 2 times s plus 3 divided by s plus 1 or y of s is half s plus 1 divided by s plus 3 so the 0 of y is at s equal to minus 1 now when this current source is here when the current source is here the capacitor accumulates a voltage alright the capacitor accumulates a voltage because the current passes like this and it charges the capacitor alright suppose the at t equal to 0 minus v c is 0 minus suppose the capacitor accumulates a voltage of v 0 now the switch is thrown open what you have to find out is v a b for t greater than equal to 0 plus obviously obviously this would be it must contain a term e to the minus t and some constant that constant obviously by inspection is v 0 alright because when you open it there is no drop in this so it must be the voltage across r 2 which is the same as the voltage across c this is point clear now let me generalize now let you generalize what you do is you take a 2 terminal network and between these 2 points between these 2 points you measure or you calculate the impedance or admittance let the impedance be of the form k p of s divided by q of s which I can write as k I can factor this out you remember how k was taken k is taken out such that p of s has the highest power coefficient as unity q of s is the highest power coefficient as unity alright then suppose the degree of p be equal to m this is how we write the degree of a polynomial just like temperature okay suppose the degree of q is equal to n then m and n need not necessarily be the same they can be different if it is so then the polynomial of degree m can always be written as the product are you acquainted with this symbol okay the product of factors like this s minus z i where i goes from 1 to m similarly in the denominator we shall have product of factors like s minus p i where i goes from 1 to n okay next if you wish to find out if you wish to find out the current flowing across the flowing through the short circuit that is if you short circuit these 2 terminals and if you want to find out the natural natural current response then the natural current response i of t shall be of the form a i e to the power z i t where i equals to 1 to m is that clear due to internal stored energy if you wish to find out the current when these 2 terminals are short circuited then the current response shall be of this form yes that we shall come back come a little later if there are repeated roots then we shall have to modify in the same manner that we did for differential equations that is instead of let us say z 1 and z 2 are equal then instead of taking a 1 e to the z 1 t and a 2 e to the z 2 t we take a 1 plus a 2 t multiplied by e to the power z 1 it is a same kind of modification on the other hand if these terminals are left open then the natural voltage response across the terminals shall be given by v t equal to summation b i e to the power p i t that is the poles shall now show their tip the poles determine the natural frequencies of voltage response where i equals to 1 to n alright and then these constants have to be evaluated from initial conditions we shall consider several examples next time to illustrate this point and that is the that is where we complete today