 Hello everyone, welcome to another lecture of basic electrical circuits. In the previous lecture we looked at second order systems and their transient response and so on. What we just started to do at the end of the lecture was to determine sinusoidal steady state response of linear circuits. Essentially what it means is that we want to determine the steady state solution to an input like cos omega t. Of course there can be a phase shift it could be cos omega t plus phi. Now it is possible to substitute that in the differential equation and substitute the known form of the solution which is some combination of cos and sin and work out the answers. But it turns out there is a lot easier way and that is what I was describing. I first showed it with the case of a resistor which seems rather trivial but its importance becomes very clear when we use that analysis for circuits containing capacitors and inductors. So that is what we are going to do in this lecture. If you have any questions about the previous lecture please go ahead and ask and I will discuss that before going on with the topics of this lecture. So let me start with our lecture today. What we mean by steady state response is that the transient response has already died out. Now that is a topic for another set of lectures in itself but you have to also verify whether the transient response dies out. Now in this course what we will assume is that for all the circuits that we consider with real R cell and C's the transient response indeed will die down if you wait long enough. So after that what you evaluate will be the steady state response. So that is why it is also useful to calculate this because the total response always consists of the transient and steady state response. After a certain period of time for a class of circuits known as stable circuits the transient response will die out. So the total response will be equal to the steady state response. That is why it is useful to calculate this. If the transient response was always dominant there is not much point calculating the steady state response but it turns out that the steady state response will dominate in a lot of practical cases after a long time. How long is long depends on the time constants of the circuit. So now let me go back to the case with the capacitor. So what I mean by this is response to some stimulus of the form cos omega t. So now I will assume that just like I did at the end of the previous lecture that the voltage across the capacitor is cos omega t. So let me call this V1. I1 then will be minus omega c sin omega t. Now let me apply sin omega t. I will get a current I2 equals c times the derivative of V2 which is omega c cos omega t. Now clearly these are linear components the capacitors here is the same capacitor. So now if I apply some superposition of V1 and V2 I will get the same superposition of I1 and I2. I hope this part is clear that I take some superposition of V1 and V2. What do I mean by that? I will take some alpha 1 times V1 plus alpha 2 times V2. Then the total response will be alpha 1 times I1 plus alpha 2 times I2. So let me do that and the particular linear combination I will take will be this multiplied by 1 and this multiplied by j. So V will be cos omega t plus j sin omega t and the response to that will be the same superposition. I have to multiply this by 1 and multiply this by j. So I would get omega c times minus sin omega t plus j cos omega t. This will be the total current I. So far I have not done anything profound. I just for a single capacitor given a voltage I can easily find the current by differentiating the voltage and now I have also taken some particular superposition of voltages in two cases and the currents in two cases. Now the important thing is that from basics of complex numbers we know what this is. This is equal to exponential j omega t. So essentially this superposition is like taking a capacitor C and applying a voltage V equals exponential j omega t to it. Now previously I first applied cos omega t and then sin omega t and consider that superposition. Now what I will do is I will take this exponential j omega t and try to calculate the current directly. Now this also is trivial in case of a capacitor. It is C times the time derivative of exponential of j omega t which is basically j omega c exponential j omega t. It does not look like I have done anything significant here. We calculated the response to exponential j omega t directly by differentiating and also by taking superposition of cos and sin omega t. Now but the most important thing which I pointed out last week as well is that in this case remember V was exponential j omega t. So the current for this particular value of V turns out to be j omega c times V. Now of course this is not a general rule. The general rule is that I is C times the derivative of the voltage. But if the voltage is an exponential j omega t then you clearly see that the current is proportional to voltage. So this is a very very important thing and in fact there is a whole body of work on linear circuit analysis based on this. Now we will not go into the depth of this but I will only point out that this exponential j omega t if you apply to any linear system the steady state response will consist of exponential j omega t times something. So the output will be proportional to the input in this particular case. This is true even if you have a differential equation like you have capacitors and inductors in a circuit. So you will have the circuit described by a differential equation not an algebraic equation. Even then the output will be proportional to input as long as the input is exponential j omega t. So now the point is that this the relationship looks similar to that of a resistor. For a resistor if you apply any V let us say V r the current through the resistor is always proportional to V r. I r is G times V r or 1 over r times V r. Now for a capacitor also if the voltage is exponential j omega t the similar type of relationship holds. This j omega t c is analogous to conductance of a resistor. So the point is now we have earlier had complicated circuits using resistors and we were able to analyze them. We have used nodal analysis, mesh analysis etc to set up the equations and we can invert some matrix and get the solution. Whereas it was a little more complicated for circuits with capacitors and inductors we ended up with differential equations. But now what I am saying is as long as the excitation is exponential j omega t the response will also be exponential j omega t times something. So the relationship looks similar to that of a resistor and every technique that we have used for calculating quantities in a resistive circuit can be used here. Any trick we have all the circuit's theorems we had for resistive circuits is also useful for these circuits. Is this clear or there any questions about this? This is the most important thing. The algebra I have showed so far is rather trivial but the important thing to realize is that for exponential j omega t the response is exponential j omega t times something. In other words the response is simply proportional to the stimulus. Any questions about this? Now this exponential j omega t is some fictitious input I created by superposing cos and sin. My actual input is cos omega t. Now I have to find the response to that. Now it turns out that if you calculate the response to exponential j omega t finding the response to cos omega t is very easy. This is because we generate this exponential by taking cos plus j times sin. The response also is the response to cos omega t plus j times the response to sin omega t. In this particular set of calculations and this is true of any other calculation involving real circuits and real inputs all the coefficients everything will be real. The only place where this j or square root of minus 1 appears is in this when you take the superposition. So each of the responses will be real numbers and the total response to exponential j omega t is response to cos plus j times response to sin. So the response to cos omega t can be simply obtained by taking the real part of the total response and this is what we are interested in. Now this itself is some real quantity that means that it has 0 imaginary part and this part also is a real quantities. So the response to cos omega t can be just obtained as real part of response to exponential j omega t. So now that is the convenient part of it. The response to exponential j omega t is easy to calculate because you get this proportionality between v and i for every component including capacitors and inductors. I have showed it for the capacitor I will quickly show it for the inductor also. So the analysis of any circuit is like analyzing a circuit with resistors and you do that and to calculate response to cos omega t you simply calculate real part of the response to exponential j omega t. This is much easier than applying cos omega t to your circuit and finding the response. So that is why this is extremely widely used and this analysis is known as sinusoidal steady state analysis and this will also lead to phasor analysis. Any questions? Now of course you do not always want to calculate response to cos omega t. What I mean is the stimulus may not always be just cos omega t. It may be cos omega t plus phi or it could be sin omega t which can also be expressed as cos omega t minus phi by 2 radians. So in general you may have a phase shift but that is very easy to accommodate here because if you have exponential j omega t plus phi that will be the response to cos omega t plus phi and plus j times response to sin omega t plus phi. So the response to cos omega t plus phi would simply be the real part of response to exponential j omega t plus phi. So we will see examples of this. Any questions about the principle or the basic idea of sinusoidal steady state analysis? So there is a question about how do we visualize a complex resistance? The term is impedance it is not called resistance anymore I will come to that. Now what is the visualization for it? What it really means is that in a resistor if you apply cos omega t as a voltage the current will be cos omega t. There will be no phase shift. I mean a capacitor or an inductor or in general a circuit containing capacitors and inductors. If the stimulus is cos omega t the response will be cos omega t plus some phase shift. So the complex impedance essentially has two numbers the magnitude and phase. So the magnitude tells you how much the magnitude of the sine wave is altered and the phase tells you how much the phase is altered. So you need two numbers whereas in case of resistive circuits you only need to tell how much the amplitude is scaled because the phase is not going to be changed at all. I hope that is clear. So let me quickly do it for an inductor. If I have I 1 of cos omega t plus phi I will take cos omega t plus phi directly in this case forced into an inductor. What will be the voltage across this? The inductor value is L. What is the value of V 1? Please let me know it would be L time derivative of I 1 which is minus omega L sin omega t plus phi that is correct. And similarly if I have sin omega t plus phi the voltage V 2 would be L time derivative of I 2 which will be omega L cos omega t plus phi. So again if I form a new input which is I which is exponential j omega t plus phi this you can think of as being formed by I 1 times 1 plus I 2 times j. Voltage V will be L time derivative of I which is j omega L exponential j omega t plus phi. Now we can quickly verify whether what I said earlier is true this V is nothing but j omega L times cos omega t plus phi plus j sin omega t plus phi. And you can see the real part of this the real part of it would come from the product of j omega L and j sin omega t plus phi and it is nothing but minus omega L sin omega t plus phi. If you take the real part that is what we will get and also like I said the voltage will be j omega L times I when I is an exponential. So this works for inductors and capacitors and also here like I said this response to cos omega t is minus omega sin omega t and that is exactly the same as the real part of this one. So if you take the real part you will get minus omega c sin omega t. So in general if you want the response to cos omega t plus phi you find the response to exponential j omega t plus phi and take the real part. And the reason we do this seemingly round about thing is that the response to exponential j omega t plus phi is lot easier to calculate than for sin omega t or cos omega t. It does not involve any differentiation or integration or finding the or finding the solution to a differential equation. Now here when I did it I showed it by differentiation but you see that the voltage is just proportional to the current. So by finding the proportionality constant we will be able to solve it exactly as we did for resistors. So I hope that part is clear. So if for linear elements which is very important everything here uses linearity otherwise we could not have used superposition etcetera. If I is exponential j omega t we will be proportional to exponential j omega t. And this proportionality constant is in general a complex number. It is known as the impedance and it is in general a complex number. So if you take a resistance if I is exponential j omega t then V would be R times exponential j omega t. For C if I is exponential j omega t V would be 1 over j omega c exponential j omega t. This is the inverse of what we calculated. There we calculated i from V now we are calculating V from i as long as they are proportional it just becomes a reciprocal. And for l have j omega l exponential j omega t. So now the resistor of course was always proportional. So for a resistor V and i were always proportional and the proportionality constant was the resistance R. Now it is also proportional for C and l. So that is what I said and these numbers are nothing but the impedances of these components. The impedance of a resistor is nothing but R the resistance itself. The impedance of a capacitor is 1 over j omega c and the impedance of an inductor is j omega l. So instead of going from i to V we can also go from V to i. So let us say if V is exponential j omega t for a linear element the current i will be some number times exponential j omega t. And that number naturally is also a complex number and it is denoted by y and that is known as the admittance and it is analogous to conductance of a resistor. And if we evaluate for these three components V is exponential j omega t then i will be quite easy to calculate i 1 by R exponential j omega t, j omega c exponential j omega t and 1 by j omega l exponential j omega t. These numbers 1 by R j omega c and 1 by j omega l are the admittances of these components ok. Is this clear? So now because of this proportionality relationship between V and i the analysis of any circuit reduces to that of resistive circuits. Any questions about this? Let us take a very simple example. What I want to calculate is the response of this circuit R c to V p cos omega t ok. Now I could go ahead and solve the differential equation but the easier way which is what I have been advertising all along is instead of this. So in this case I will get some V c of t this is what I want to calculate. I will go about it in a somewhat roundabout way but which is actually easier. So let me make it to be with general cos omega t plus phi and here it will be exponential j omega t plus phi ok. Now the point is that because the excitation is exponential j omega t plus phi the voltages and currents everywhere will be of the form exponential j omega t plus phi times some number ok. So I showed this for capacitors and inductors but it is true of any linear circuit. So then what I do is because I have exponentials everywhere I can use the proportionality relationship between V and i ok. Now in case of resistors the relationship is the same as what it always was V equals i R and in case of capacitors I will denote the capacitor by this box which stands for some impedance and the impedance of the capacitor is 1 by j omega c ok. So this is and the impedance of resistor is just R ok. So now because everything is like a resistor in the sense that the voltages and currents are proportional to each other. We analyze this circuit just like we analyze resistive circuits. The only difference is that when we had purely resistive circuits all the constants were real. Now we also have complex numbers as constants ok and we will interpret that properly. So let me call this Zc the impedance of the capacitor. So what will be the voltage here Vc what is this going to be? What is the Vc going to be in this circuit when I have exponential j omega t plus phi? Like I said you can analyze this exactly as you would analyze resistive circuits and that is because V and i are proportional to each other ok. I am sure you would be it would be very easy for you to analyze some circuit like this where I had R and then some R1 I asked you for the voltage here and this circuit is exactly the same except instead of R1 we have this impedance Zc ok. So please tell me the value of Vc what would be the value of Vc? So clearly you can use the voltage divider formula and this Vc is going to be Zc by Zc plus R times the applied voltage which is Vp exponential j omega t plus phi ok which can be further written as 1 by j omega c R plus 1 by j omega c Vp exponential j omega t plus phi which can be further simplified R plus 1 by j omega c. So 1 by 1 plus j omega c R Vp exponential j omega t plus phi ok this is ok. Now if I take the real part of this I will get Vc of t I will leave that to you to calculate it will be some sinusoid ok but if you take the real part you will get the response to Vc ok. Is this part clear? But it is not very difficult to do so let me do it and show it to you. So Vc of t in response to cos omega t plus phi is real part of ok. So in general for all these analysis you need to brush up your manipulation of complex numbers it is rather simple but you still need to know to use them fluently. So we will have first rationalize this part ok. So when you take the real part you will get one term from the multiplication of these two and another one from multiplication of those two ok and I will write out the answer right away which is that Vp times cross omega t plus omega C R sin omega t divided by 1 plus omega square C square R square ok. So this is what it is and you can always reduce this to single sinusoid with some phase shift. Is this ok? Any questions about any of this? In fact we will find a simpler way to write down Vc of t just in a moment. So the easier way to do that is response to exponential j omega t we saw was 1 by 1 plus j omega C R Vp exponential j omega t plus phi ok and in general it will be some number times exponential j omega t ok. So in this case I also have exponential j phi ok by the way when I say j omega t plus phi it j is multiplying phi as well ok. So everywhere I wrote j omega t plus phi without brackets but what it meant was that it is j times omega t plus phi ok the real part of that is cos of omega t plus phi that is what I had meant ok. So now this there will be some complex number multiplying this exponential j omega t which also it is easy if you write it in polar form ok that is let us say this number is rather let me do it only for this part of it can be written in the magnitude and phase form the magnitude of this is 1 over 1 plus omega C R square the square root of that and the phase of that is exponential some j theta where theta is minus tan inverse omega C R. So in general whatever complex numbers you have you write it in polar form ok. So if I call this complex number as h ok it can be written as the absolute value of h or the magnitude of h times exponential j angle of h ok. So once you have that the response becomes magnitude of h exponential j times angle of h times v p times exponential j omega t plus phi which of course can be written as magnitude of h times v p and exponential j omega t plus phi plus angle of h ok. So you can clearly see the role of the magnitude and phase earlier there was a question from Charmille about how do you interpret a complex impedance. Now in this case it is not an impedance but any complex number multiplying exponential j omega t should be interpreted like this ok. So the magnitude scales the amplitude of the sinusoid and the phase shifts the phase of the sinusoid because if I write it in this form taking the real part is quite trivial right. If I take the real part what do I get I will get a sinusoid whose amplitude is absolute value of h or magnitude of h times v p times cosine of omega t plus phi plus angle of h ok. So this modifies the magnitude and this modifies the phase ok. So and this is true for single elements also if you have let us say the voltage across an inductor being j omega l times the current through the inductor there is a modification of magnitude by omega l and modification of phase by 90 degrees ok. So you can see that this can be written as omega l exponential j phi by 2. So this v l becomes omega l times exponential j omega t plus phi by 2 and let me just for clarity say that the peak value of the current is I p then we will have I p here also and here. So you can see that the magnitude of the impedance omega l modifies the magnitude of the voltage or the amplitude of the voltage ok. So and then the phase of the impedance which is pi by 2 modifies the phase of the sinusoid ok. So if I take the real part of this I would get omega l I p which is the amplitude of the sinusoid of the voltage times exponential j omega t plus pi by 2 which is minus sin omega t ok. So I hope this is clear if not you just do a calculation with a couple of inductors and capacitors and it will become very clear ok. So if there are any questions on these I will take them now and now you see that everything it becomes as easy as analyzing a resistive circuit ok. Only thing is you have to handle complex numbers that is all and that is not very difficult. But what you should not lose out in all this algebra is the connection to reality which is that you are trying to find response to cos omega t plus phi that you do by finding the response to exponential j omega t plus phi and taking the real part. If you take the real part of that you will get some sinusoidal wave form which is the response to cos omega t plus phi ok. Any questions ok? So just to summarize and this implies analysis similar to analysis of resistive circuits and a response to cos omega t plus phi basically can be obtained by taking the real part of response to exponential j omega t plus phi and finally we should not forget why we did all of this where this is applicable and this is only for sinusoidal steady state. That means that you apply sinusoid which is cos omega t plus phi and then you find the steady state which is after the transients die out ok. I hope this part is clear. Now from here we it turns out we do not even have to write exponential j omega t everywhere because it appears everywhere in the in all the expressions. So we will omit that and the resulting stuff is what is known as phasor analysis ok. So let me take this RC circuit again and if I have I will just use exponential j omega t but whatever I say we will also apply when I have plus phi over there and I will just write out the solutions. You can verify these things yourselves here we saw the voltage was 1 over 1 plus j omega c r v p exponential j omega t and here it turns out it will be j omega c r by 1 plus j omega c r exponential j omega t ok. Now do not worry about the details what I want to point out is that if you have a loop ok where the input to the circuit. So I am assuming that we have a linear circuit and the inputs are of the form all inputs must be of the form exponential j omega t that is at the same frequency ok same omega if you have multiple inputs it is ok but they should all have the same frequency ok. So if you have multiple sinusoidal inputs of the same frequency the inputs all will be of this form ok there may be some multiplying constant but they will all be of this form ok. So then if you look at any loop every voltage across elements in a loop will have exponential j omega t right here I have illustrated this with exponential j omega t here there and there. Now this should be pretty clear because I said that any response in a linear circuit will be proportional to exponential j omega t if the input is exponential j omega t. So you take a loop and you have only applied exponential j omega t's so every branch voltage would consist only of exponential j omega t times something ok. So everything will be like this so the point of this is that what does KVL say Kirchhoff's voltage law it says that sum of voltages around the loop equals 0 ok. So let me call these voltage drops let us say V 1, V 2 in that polarity and V 3 ok V 1 plus V 2 plus V 3 is 0 in this loop this is the statement of KVL. Now I know that V 1 is something times exponential j omega t and V 2 is something else times exponential j omega t and V 3 is yet another thing times exponential j omega t and the sum is 0. Now because exponential j omega t is a common factor to all of these I can omit them and do my calculations only with these numbers that is I do not need exponential j omega t at all what I need are these numbers this, this and this. In fact I omitted VP here VP should be here as well ok. So in general these numbers multiplying exponential j omega t will be some complex numbers I will denote that by V 1 bar V 2 bar and V 3 bar V 1 V 2 V 3 bar will be some complex numbers and I can omit this multiplying factor exponential j omega t and write KVL only in terms of these complex multipliers of exponential j omega t. Now it is very clear that exactly the same holds for KCL as well because all currents will have exponential j omega t and KCL says that the sum of all currents entering a node or leaving a node is 0. Now because all currents contain exponential j omega t I can simply rewrite the currents in terms of the sum of the complex multiplying factors ok. So KVL which says V 1 of t plus V 2 of t plus I will take just 3 voltages equals 0 will reduce to some complex number V 1 exponential j omega t plus some complex number V 2 exponential j omega t plus V 3 exponential j omega t equals 0. Now this is common so I get rid of that I can simply write KVL in terms of the complex multipliers. Remember V 1 is not the voltage V 1 times exponential j omega t is the voltage but that appears as common everywhere so I will ignore that all together and I will write V 1 plus V 2 plus V 3 to be 0. Again the algebra is quite simple and I think you can learn it very easily but it is also important to figure out what is the meaning of this ok what do these complex numbers mean and it turns out that this is a very widely used analysis and these complex numbers which are multiplying exponential j omega t are known as phasors ok ok. And similarly just like this exponential j omega t is common to all voltages in the loop it is also common to all currents at a node and the statement of KCL which is I 1 of t plus I 2 of t plus I 3 of t equal to 0 becomes I 1 exponential j omega t plus I 2 exponential j omega t plus I 3 exponential j omega t equals 0 from which I will again remove the factors exponential j omega t. So, I 1 plus I 2 plus I 3 which are complex numbers are 0 and this I 1 bar I 2 bar I 3 bar are the phasors of currents ok. So, now what happens is that everything just looks like DC circuit analysis containing DC analysis of circuit containing only resistors ok. But once you see let us say a phasor I 1 to figure out what the voltages should take real part of I 1 times exponential j omega t ok. So, I can write KCL and KVL in terms of phasors basically omit exponential j omega t ok. So, which means that to find some voltage v k of t you should take real part of the phasor of v k times exponential j omega t and this omega is something that you need to know. The very important thing is that even if you have multiple sources all of them should be at the same omega that is when you will have all the voltages and currents being proportional to just exponential j omega t ok. If you have omega 1 and omega 2 you will have exponential j omega 1 t and exponential j omega 2 t in different proportions and you will not be able to cancel them. If at all you find a situation where you do have multiple frequencies let us say omega 1 and omega 2 because these are linear circuits what you can do is first you disable all the sources of omega 2 and do the analysis of only the sources at the frequency omega 1 and then you disable all the sources at omega 1 and analyze the circuit for frequency omega 2 and then you add up the results separately after finding out the real part of phasor times exponential j omega t ok this is fine. So, again I will show this with a very simple example first of all the element relationships these we have already worked out, but I will show it just for the sake of clarity a resistor if it has some voltage v r exponential j omega t and a current i r exponential sorry v r bar exponential j omega t and i r bar exponential j omega t we omit exponential j omega t and simply write i r and v r and the relationship of course is the familiar relationship of the resistor v r by i r equals r and if you have a capacitor ok will have phasors i c and v c the phasors will be related by 1 over j omega c and I will have the inductor current and the inductor voltage which are phasors times exponential j omega t the relationship between these two with only phasors would be v l by i l equals the impedance of the inductor which is j omega l ok. So, we can omit this exponential j omega t from everywhere do our analysis and only when you need the expression for the signal in the time domain as a function of time you need to take this phasor times exponential j omega t and take the real part ok. Any questions about this? So, let us take a simple example. So, let me take the series RLC circuit and I want to find the response to v p cos omega t in steady state ok. So, now if I want v p cos omega t what is the phasor corresponding to this? The phasor is such that so let me call this v s of t lower case v s and the phasor of v s of t is such that real part of v s exponential j omega t is v p cos omega t and it is pretty clear what v s is v s must be equal to v p ok. So, if you have a source cos omega t v p cos omega t the phasor corresponding to that is v p alternatively if you had v p sin omega t the phasor would be what would be the phasor corresponding to v p sin omega t? What is the phasor? The phasor multiplied by exponential j omega t should give you the voltage. I worked it out for v p cos omega t if this has to be v p sin omega t what should be the phasor v s, v s bar please try and answer this question. So, clearly this will be minus j v p. In general if you need the phasor of v p cos omega t plus phi the phasor corresponding to that would be v p exponential j phi. So, this is what we will use as the source ok. So, I will take this v p cos omega t. So, the source itself would be represented by a phasor v p and the resistance will remain as the resistance and the inductance basically is an impedance j omega l and the capacitance is an impedance minus it is 1 by j omega c which can also be thought of as minus j by omega c. So, now each of these has some voltages across it which is represented by phasors v r, v l and v c and there will be a there is a single loop and there is a certain current i flowing through the loop. When I say a current i is i is the phasor and the actual current is exponential is i bar times sorry real part of i bar times exponential j omega t ok. So, I would be v s by r plus j omega l plus 1 over j omega c it is just v s divided by the total impedance and v c would be v s times 1 over j omega c r plus j omega l plus 1 over j omega c and so on ok. So, you will be able to find all of these things very easily and you see that this circuit is just as complicated to analyze as a single loop of 3 components, 3 resistors in series ok. So, that is all that is there to it this is how you do phasor analysis, but please do not forget the connection to reality which is that if you want the voltage across the capacitor you take real part of the phasor v c times exponential j omega t ok. Any questions these also can be found ok. There is a reason I took this second order circuit we will take up exactly what these things do to the input signal in later lectures ok. And just to for completeness I will give you another example ok. In this case let me take v s of t to be v p cos omega t plus phi and this time I will take a parallel RLC circuit as I mentioned earlier if you null the input source if LCR are in parallel it is a parallel RLC circuit ok. So, in terms of phasors I would have to replace v s by its phasor which is v p exponential j phi R will remain as an impedance R, L will be j omega L and C will be 1 by j omega C ok. So, now similarly everything can be calculated ok. So, for instance if you want the voltage across the inductor which is also the same as the voltage across the capacitor v c what do we have it is v s times the parallel combination of these two impedances ok. The parallel combination of these two impedances is j omega L parallel 1 over j omega C which is j omega L times 1 over j omega C divided by j omega L plus 1 by j omega C which can also be written as 1 minus omega square LC by taking j omega C to the top ok. So, I will have j omega L 1 minus omega square LC j omega L 1 minus omega square LC plus R ok. It is just an impedance divider instead of a resistive divider we have parallel combination of these two impedances and that divided by the total impedance which is R plus the parallel combination ok. So, let me just complete this before ending the lecture which I will take R to the numerator and we will get something like. Now, again what does this mean there is a complex number multiplying v s which will modify the amplitude and phase of the sinusoid. If v s is a sinusoid v c will be a sinusoid at the same frequency and its amplitude and phase will be modified how much will they be modified by that will depend on the amplitude and phase of this number the magnitude and phase of this complex number ok. So, I hope that part is clear that brings us to the end of this lecture. So, the summaries as follows if you apply exponential j omega T to a linear circuit every voltage and current will also be of the form exponential j omega T. This gives us a lot of simplification in analysis. Now, exponential j omega T itself is first of all a fictitious input in the lab we would apply sin omega T cos omega T etcetera, but the response to cos omega T can be obtained from the response to exponential j omega T by taking the real part ok. This is because all the coefficients all the component values everywhere are real and I showed by superposition that you can think of exponential j omega T as combination of cos omega T and sin omega T ok. So, that is number 1. Secondly, because for every element voltage and current will have exponential j omega T the voltage becomes proportional to current just like that for a resistor. So, even though you have inductors and capacitors those circuits can be analyzed in a way that we use for resistive circuits which you agree was very easy because we did not have any differential equations. The only caveat here is that this applies only when the input is of the form exponential j omega T that is for sinusoidal inputs and in steady state because the transient response has to die out and that we cannot calculate using this method ok, but all the usual circuits that we consider in this course will have transient responses that die out over time. And finally, because this exponential j omega T is common to everything all voltages and currents we start writing Kirchhoff's current loss and voltage loss by omitting the exponential j omega T all together. And we will consider only the complex multipliers for each of these voltages and currents and those are known as phasors. And this analysis using phasors is known as phasor analysis and it is very very convenient to use whenever you have sinusoidal inputs to a linear system ok. So, any questions about this I will answer now ok, there is a question that says do phasors have a relationship to Laplace Fourier transform ok. Now the Laplace transform analysis of systems is derived from a similar reasoning that if you have exponential st given as an input to a linear system the response will also be of the form exponential st times something ok. This is true of any linear differential equation if you have a linear differential equation with an input of exponential st the solution will also be exponential st times something the steady state solution ok, so that is the relationship. Now what is the exact relationship if you take Laplace transforms if you do Laplace transform analysis of a circuit and substitute s equal to j omega you will get this essentially this phasor analysis ok. Now the Laplace transform and also Fourier transform analysis can also be used for signals which are not sinusoids ok the way we have presented it is only for sinusoids. So, if you reduce the if you use the Laplace transform for analysis of sinusoidal signals essentially what you get is the phasor analysis. So, you can think of this as a subset of Laplace or Fourier transform. Now Laplace transform is used for circuit analysis the Fourier transform in particular is mainly used for signal analysis there is another thing which we have not touched at all now we have showed that you can analyze circuits with sinusoidal inputs using phasor analysis. There is also the Fourier transform which says that any signal can be thought of as a combination of sinusoids ok, so like a square wave or a triangular wave can be thought of as a combination of sinusoids. So, what the Laplace transform and Fourier transform analysis do is to combine these two ideas that the response of exponential j omega t is exponential j omega t and any signal can be decomposed as some combination of exponential j omega t with this basically you can analyze any circuit any linear circuit with any input ok using Laplace transform and also may be less conveniently using Fourier transform ok. So, the spacer analysis is a sort of sub case of Laplace or Fourier transforms when you restrict the input to be single frequency sinusoids ok I hope that answers the question ok. So, then we can break now it turns out that for all the institutions which were subscribed to the online lecture the exams have already started. So, this will be the last online lecture, but we will continue to post the lectures by recording it offline and we will also have the assignments and you will have about 3 weeks or a months to complete them. So, please watch out the web page for these details they will be up shortly ok.