 Welcome to the fourth lecture in the course on engineering electromagnetic. We saw in the last lecture that in practice a transmission line would be terminated and therefore a wave propagating on the transmission line when it reaches the termination in general would partly be reflected and partly be transmitted. We saw the reflection coefficient, transmission coefficient, etcetera. And then we went on to consider some simple traveling wave situations. We continue these simple situations further and today what we consider first is a charged line connected to a resistor. The situation looks like this. We have a charged line and here we have connected a resistance R. The transmission line has a characteristic impedance Z naught. The line is lossless and is charged to a voltage V naught. So, that we may write here that the line is charged to a voltage V naught. How would the line be charged? Perhaps by connecting a voltage source V naught at one end in a situation similar to what we considered last time. Let the line length be L and let the line be open circuited at the end away from the end where the resistor R is connected. And then we say that at t equal to 0 R is connected let us say through a switch like this. What do we expect when the switch is put on? We expect that this voltage to which the line is charged V naught will act as a voltage source and this voltage will be shared between the resistance R and the characteristic impedance Z naught. As a result of which we may say that a voltage V R will develop here and a corresponding current through the resistance which may call I R will start flowing. Now, since the initial charged voltage V naught is going to be shared the voltage across the transmission line should drop. What will be the mechanism causing this drop? That mechanism will be a wave travelling to the on the transmission line away from the end at which the resistor is connected. And therefore, we say that a wave with associated voltage V plus and current I plus notionally indicated like this will start flowing at t equal to 0. Now, therefore to find out the values of V plus and V R what considerations are we going to make? See the line is charged to a voltage V naught we can consider it like a voltage source of voltage V naught at t equal to 0 this will be connected to two entities. What will be those two entities? The resistance R and the transmission line and depending on their relative values this voltage source will supply or share its voltage across these two entities. And therefore, we say that V R is going to be equal to V naught plus V plus when we write the relations you will find that we are consistent. This is the way we indicate the current in a wave travelling in this direction notionally. The actual direction will come from the actual magnitude and direction alright. Here also you would notice that we have written V R equal to V naught plus V plus knowing fully well that V R will have an overall voltage which is less than V naught which means that V plus will come out with a negative value. So, we should write a V naught equal to V R plus V plus. But this is quite consistent with the way we have been writing the total voltage for example across a load. We said V L is equal to V plus plus V minus. Similarly, this is the load in this instance and the total voltage across the load is equal to whatever was there earlier and whatever starts off at t equal to 0. It is a matter of notation we are being consistent with our notation. What will be the relation between I R and I plus? For the directions shown I R will be minus I plus. As long as these relations are written correctly consistently we will not be making a mistake. Now of course I R and I plus are not independent of V R and V plus. They are related through the resistance and the characteristic impedance. And therefore we can write I R is V R by R which should be equal to minus V plus by Z naught. Now we may number these equations 1, 2 and 3 and equations 1 and 3 involve two unknowns V R and V plus other quantities being known and therefore they can be solved. Very simple equations and when we do that what we get is V R is equal to V naught into R upon R plus Z naught and V plus is equal to minus V naught into Z naught upon R plus Z naught. Quite as expected V plus has come out negative assuming that V naught has a positive value. And you find that V plus and V R in the ratio of R and Z naught and they are sharing this notionally source voltage V naught in that ratio. When we have got these voltages we can also write the associated currents. Now instead of keeping things general for the sake of simplification we assume a particular value of R. We choose a particular simple value of R. Normally things will not be that simple but for illustration let us say that R is equal to Z naught. We will have V R equal to V naught by 2 and V plus equal to minus V naught by 2. So in a way such a voltage starts getting subtracted from the initial voltage on the transmission line as this wave starts travelling away from the end at which the resistor is connected. Effectively discharging the transmission line by half the amount. The corresponding currents can also be written now. I R is V R by R so that it is V naught by 2 R and I plus which is V plus by Z naught is equal to minus V naught by 2 Z naught. And when you consider the value of I plus with this sign you find that there is a current which is flowing in the counter clockwise direction. So this is what happens at T equal to 0 and at T greater than 0 this voltage wave starts propagating away from the resistor end and brings the charge, the voltage charge on the transmission line to half its value. Let us say that it will reach the other end at a time which is related to the velocity of the wave propagation on this transmission line. And we say that let V be equal to 1 by square root of L c so that the time taken in travelling from the resistor end to the other end is L by V which may write as capital T. Now let us see what happens at T equal to capital T when this wave that was initiated at T equal to 0 which is the other end which we have kept as an open circuit. At the open circuit no current can flow therefore the open circuit enforces a condition such that I minus plus I plus is equal to 0. And immediately we realize what is the origin of I minus I minus will correspond to a reflected wave. And in an attempt to satisfy the conditions along itself the open circuit will initiate a reflected wave with what V minus and I minus constituents they will be I minus equal to minus I plus and I plus is available to us so that I minus will have a value which is V naught by 2 z naught. How can we find the corresponding V minus for the reflected wave? We go using the relation V minus upon I minus equal to minus z naught. It is useful to start this way in a simple manner so that we do not go wrong. And therefore V minus is minus I minus z naught so that this also comes out to be minus V naught by 2. And therefore at T equal to T and greater instance of time there is a reflected wave with V minus equal to minus V naught by 2 and I minus equal to V naught by 2 z naught. With this value of I minus there is no current through the open circuit and with this value of V minus what happens? There is a reflected wave travelling from the open circuit end towards the resistor end and it wipes out the remaining charge on the transmission line. The total current on the transmission line will now be I plus plus I minus which of course will be 0. And the total voltage can be found out as in the following manner. We will do that in a short while. Now this kind of wave reaches the resistor end at T equal to 2T. Now at the resistor what is the voltage now? The total voltage using the same notation as we have been writing will be V naught plus V plus plus now V minus is also available. Knowing their values we see that this is equal to 0. And now this does not violate any condition in particular because it is not a source it is not a voltage source or a short circuit or an open circuit. And therefore the current which is V r by r is also equal to 0 and that is where the matter ends. Alternatively one could also look at it from this point of view that there is a voltage wave which hits this end. Now since r is equal to z naught it gets completely absorbed and there is no further reflection. So either way this is where the current and voltage will stop changing any further. And if we want to consider the current that has flown through the resistor as a function of time I r it is going to be of value V naught by 2 r at T equal to 0. And it continues to flow till time T equal to T. So such a current pulse of duration twice capital T seconds flows through the resistor in this case where a charged transmission line is connected to a resistor. Of course things are fairly simple in this case because of our choice of r equal to z naught otherwise you can see in a straight forward manner that this is not where the phenomena will stop. The charge on the transmission line would not be wiped out completely by the reflected wave and there will be some reflected wave which will start at T equal to twice capital T and so on. But applying similar considerations in a systematic manner one can find out what is the behavior for example for r less than z naught or r greater than z naught. So these were some very simple situations involving transmission lines. The next go on to consider a specific kind of time variation. What is that specific kind of time variation sinusoidal time variation that is the kind of signals that we utilize maximum in practice. Even if a signal is not exactly sinusoidal you know that through Fourier analysis it is possible to express it in terms of sinusoids of different frequencies and therefore the case of sinusoidal time variation is of great practical importance that is what we consider next. We go on to consider case of sinusoidal time variation that is we consider signals say a voltage signal which is a function of time and we put a curly bar over it to show that it is a sinusoidal time varying signals. So for example it could be V cosine of omega t omega being the radiant frequency and then it can be written as real of V times e to the power j omega t and then using phasor notation we may represent the actual time varying signal V in terms of simply a symbol V which in general could be complex and therefore would have a magnitude V naught and a phase. The origin of this phasor notation can be seen on the OHP and we say that this is the phasor V and when it is multiplied by the factor e to the power j omega t the tip of the phasor rotates on the circumference of a circle and when we consider the projection of this phasor with its tip at various points on the circumference on the real axis then the projection along the real axis has a sinusoidal or a cosine sinusoidal behavior. So that is very simply the basis of the phasor notation going back to the board therefore we can say that in phasor notation the actual time varying signal can be written as V naught magnitude e to the power j phi e to the power j omega t and we consider its real part. If we have a phasor how will we reconstruct the actual time varying signal that is what we are trying to show now and this is exactly V naught and then cosine of omega t plus phi the original time varying signal that we started. Therefore, whenever we have a phasor quantity if you want to recover the actual time varying quantity from this we will multiply by e to the power j omega t and take the real part at a working level that is the formula we could use. In phasor notation there is one great advantage the time derivative or the integral with respect to time has a very simple form that can be seen here in the following manner. We consider the time derivative of this quantity del by d t del by del t of V t the actual time varying quantity that is going to be minus omega V naught sin of omega t plus phi which we could cast in this form by rewriting it as real of j omega V e to the power j omega t. We are just rewriting it in a different form and therefore, we can generalize that the time derivative of a sinusoidal time varying quantity in terms of the phasor V is simply j omega V. Many times the phasor quantities are represented say in text books in different type phase to make it very clear that it is a phasor quantity that facility we do not have here, but from the context it will usually be clear that we have a phasor quantity in mind yes please if you have a question the time derivative of the sinusoidal time varying quantity V t and through this very simple consideration we have seen that the two are equivalent of this phasor if you want to recover out of this phasor if you want to recover the actual time varying quantity that will be simply its time derivative. Now a phasor is a complex quantity and actual time varying quantity is not a complex quantity but this complex notation makes representation mathematically very simple and therefore we use it. In particular the time derivative in a similar manner the time integral they have a very simple form in phasor notation. So using this phasor notation we now consider the behavior of voltage and current on a transmission line. We have seen that on a transmission line the voltage and current satisfy an equation which we named as wave equation and then replacing the general voltage and current in this by the sinusoidal time varying voltage and current if we rewrite the wave equation we will have del 2 V by del z square equal to l c and then del 2 V by del t square. Earlier the wave equation that we derived was a general one applicable to all kinds of time variations since we did not consider any particular time variation and therefore it should be applicable to sinusoidal time varying signals as well and making that substitution this is the form that we will get for the wave equation. Now we use the fact that in phasor notation the time derivative is equivalent to a factor j omega and therefore we rewrite this wave equation phasor notation in which case it will become del 2 V by del z square equal to minus omega square l c times V since each del by del t factor is equivalent to a factor j omega as we have seen in phasor notation. The lower equation now is using phasor notation V is a phasor quantity and in that the time derivatives can be replaced simply by the factor j omega. We may write this as minus beta squared times V where beta is omega times square root of l c. Beta is an important quantity we will identify its nature in a short while. Written in this form one can write the solutions for V the general solution without much difficulty it will be V naught plus e to the power minus j beta z that will be one kind of solution when we substitute this back we find that the equation is satisfied and equally well the other kind of solution would be V naught minus e to the power plus j beta z. Now from our experience with voltage and current on a transmission line we expect that these two terms in the general solution should be representing waves propagating in opposite directions. This one should be representing wave propagating in the positive z direction and the other term a wave propagating in the negative z direction. However it is not obvious from this form the kind of functions the kind of arguments that we identified for the functions which could be solutions of wave equations. Apparently that kind of form is not here what I want to bring to your attention is that that form is there it is implicit in this it is not obvious because we are using the phasor notation. Sir why we are used V naught plus e minus j beta z why not V naught minus e minus j beta z. Because from our experience we are anticipating that this term would represent waves propagating in the positive z direction. So its amplitude we have kept as V naught plus we have used some knowledge. Now let us justify this assumption that the first term would represent a wave propagating in the positive z direction and similarly the second term how can we do that all we have to do is to rewrite this in a form in which the time variation is explicitly there that is we need to write this as V which is now a function of both z and t it is certainly a function of z it is also a function of time but in phasor notation that time variation is implicit. Now we want to bring it out explicitly to clarify this what do we do the formula that we stated we multiply the phasor of interest by e to the power j omega t and take the real part. So we do that we consider the real part of V naught plus e to the power j omega t minus beta z plus V naught minus e to the power j omega t plus beta z the operation is very straight forward that will give us V naught plus cosine of omega t minus beta z plus V naught minus cosine of omega t plus beta z and we manipulate this in a very simple manner we write this as V naught plus and then cosine of we take out minus beta as common. So that it is z minus omega by beta t and similarly the second term V naught minus cosine of beta into z plus omega by beta into t and now I am sure you can recognize the kind of argument that is associated with propagating waves they are here this is simply of the form z minus V t and this is simply of the form z plus V t and immediately we can say that the velocity of wave propagation here is omega by beta and using the expression for beta it is simply 1 by square root of L c quite consistent quite the same as we saw for signals where no particular time variation was specified. We have already made out the behavior of functions of this kind of arguments this a function of this kind of argument represents a wave propagating in the positive z direction and therefore anticipating this we had written the amplitude factor in front of this as V naught plus and the amplitude factor in front of this as V naught minus. Coming back to this form we say that beta z is the phase shift in sinusoidally time varying signals phase is clearly defined. So now we can identify that beta z is the phase shift and therefore beta is the phase shift per unit length or simply beta is called the phase shift constant with units of radians per meter one can also introduce the concept of wave length here wave length represented by lambda is the distance over which the phase change is by 2 pi radians. So that lambda times beta or beta times lambda is equal to 2 pi and it gives us lambda which is the wave length equal to 2 pi by beta or alternatively beta is 2 pi by lambda where lambda is the wave length of course in meters. I can show on the OHP the wave propagation for a single sinusoid and see its behavior this is the single sinusoid that is the kind of excitation we have considered and we consider its plot at different values of z and this is the way the plot will appear it will appear to have moved forward progressively on the time axis alternatively we could consider these snapshots of the same single sinusoid at different time instants and once again it would appear to have moved at different values of z it will have progressed along the z direction. So in a simple manner that is the way a single sinusoid would behave and as we have also demonstrated in the laboratory for a sinusoidal time varying signal the spatial variation and the temporal variation they are both sinusoidal which is what has come out nicely in this picture. If you have any questions up to this point we could try out those the question is that why have we written beta lambda equal to 2 pi actually we are trying to introduce the concept of wave length and we say that wavelength is the distance over which a wave changes in phase by 2 pi radians overall phase change is beta times a distance travel the distance travel for a wavelength is 1 wavelength so beta lambda equal to 2 pi that kind of distance will be wavelength by our definition so that is all what that we are doing. Any other questions okay then we go on to the next topic which also is of a very interesting nature first we make some space here we are familiar with low frequency circuits and we have no difficulty in saying or finding out the impedance that is connected in a low frequency circuit alright now what kind of impedance behavior do we expect on our transmission lines so for that purpose we take a typical transmission line the way we have been drawing these a transmission line with a load impedance ZL connected at Z equal to 0 let it be a transmission line of characteristic impedance Z0 and length small n the question that we pose for ourselves is at a plane which is at Z equal to minus L what is the input impedance C if we just extend our low frequency concepts there is really no problem the input impedance should be just the load impedance and these are just connecting wires but on a transmission line where time delay effects are present and have been incorporated into analysis it will be very interesting to see what is the input impedance at a distance L from the load impedance yes yes no at low frequencies you do not have the concept of Z0 so you just treat these as connecting wires and no matter how long they are as long as our low frequency approximations are satisfied the impedance here is the impedance C input impedance C here so this takes us to what we call the concept of impedance transformation and we find out this impedance in terms of the ratio of voltage and current at the plane of interest the procedure is going to be fairly simple we write V as V0 plus e to the power minus j beta Z plus V0 minus e to the power plus j beta Z according to the general solution that we just wrote down similarly the current can be written as I0 plus e to the power minus j beta Z plus I0 minus e to the power j beta Z. Where we consider that this wave is travelling in the positive direction usually we call this as the incident wave and depending on the termination there will be some reflected wave which will be initiated travelling in the negative z direction that is the usual framework that we have been considering. Now I0 plus and I0 minus are related to V0 plus and V0 minus and using that relation we see that it is 1 by Z0 V0 plus e to the power minus j beta Z minus V0 minus e to the power plus j beta Z. Now what is the input impedance it is going to be the total voltage divided by the total current at the plane of interest. So the total voltage and total current at different values of Z are already available we just evaluate these at Z equal to minus L and take the ratio that will be the quantity of interest and that becomes Z0 times V0 plus e to the power j beta L plus V0 minus e to the power minus j beta L since we are substituting Z equal to minus L and in the denominator we get V0 plus e to the power j beta L minus V0 minus e to the power minus j beta L which we process or manipulate so that it comes out in a nice interesting form. We can divide by V0 plus throughout and recognizing that V0 minus upon V0 plus is equal to rho the ratio of the reflected wave amplitude to the incident wave amplitude at the load that is precisely what V0 minus and V0 plus are. So using this you see that Zn is equal to Z0 and then we have here e to the power j beta L plus rho e to the power minus j beta L upon e to the power j beta L minus rho times e to the power minus j beta L since we are running out of time today we shall continue this consideration next time. Yes of course. See we have seen that these solutions arise out of the wave equation which holds good for the sinusoidal signal. I think we will stop here today and consider this either in the tutorial or in the next video. Thank you. Bye. Bye.