welcome back to this ah lecture series on pulsewidth modulation for power electronic converters so we have been going through you know this is the fourteenth lecture we have been going through a number of modules here in the first module we looked at various kinds of power electronic converters different topologies dc dc converters and dc ac converters like two level multilevel and all those we saw there and after that we just had an overview of just the voltage source converter what we are looking at here as voltage source converter we just add that and after that we looked at the purpose of pulsewidth modulation certain introductory ideas about pulsewidth modulation why you require pulsewidth modulation basically we want to control the fundamental voltage and you also want to control the harmonic voltages such that they are not very high and their undesirable effects are kept as well as possible so we try to do that here and in this fourth module we have been looking at low switching frequency pwm methods for this now thats essentially what we have our voltage source inverter we are going to switch them and we are switching them at a frequency which is very low so right now we are taking only two switching angles per quarter cycle or the switching frequency is only about five times the fundamental frequency and with this we are trying to understand how the fundamental and harmonic voltages are related to the switching angles thats the whole burden of our ah you know the recent past few lectures so as i was mentioning we have been considering only two switching angles per quarter the waveform has quarter wave and half wave symmetries so what it has in essence it has about five switching cycles within a line cycle now so it is switches at five times the fundamental frequency as the fundamental frequency is fifty hertz this switching frequency would be two hundred fifty hertz now so you only have two angles alpha one and alpha two the other angle here being fixed at zero that would be one other switching at one eighty degrees now with only two angles we keeping the problem as simple as possible we are trying we have been trying to understand how the fundamental and the harmonic voltages are related to switching angles so that we can achieve the necessary fundamental voltage at the same time we can mitigate the harmonics and their ill effects so we were looking at how the fundamental voltage is vary this is how the vary that is along this line on the alpha towards is alpha plane any point on this line gives you zero fundamental voltage and any point on this line gives a point two per unit fundamental voltage where one per unit stands for the fundamental voltage equal to the square wave operation there are the six step operation of the inverter now so we have been doing all this through hand sketches we have been trying to calculate the values and sketch them by hand in the past few lectures because we want to get a good idea of whats really happening there now that we have gain certain amount of insight we can start using some computer programs for plotting them lets say you plot this constant view on contours now so what i have done here is i have plotted for different values here this is for v one is equal to zero so this entire line is v one is equal to zero you can choose any value of alpha to comma alpha one on that particular line the resultant would be a pwm waveform whose fundamental component is zero right then the next one that we have is point two on the next line that you have corresponds to point four per unit fundamental voltage then this is point six this is point eight and this is one point zero so your fundamental voltages are varying in this fashion now if you one point eight it in effect it means you can choose any of these points which point you choose that would depend on which harmonic you want to be lower or whatever you can so the first objective of controlling the fundamental voltage is easily attained from here so for v one is equal to point eight any of this on this line you can choose any point on the line which point you can decide based on the harmonic content etcetera now so which is the principle harmonic the fifth harmonic would be the principal harmonic in the third you know the ah three phase inverter if there are any third harmonics in the pole voltages they will get subtracted vro as some third harmonic vyo will have the same third harmonic and vro minus vyo when you are doing you know the line to line voltage that effectively gets cancelled out so we dont have to worry about triple in harmonics as we saw earlier we will only to worry about the fifth harmonic now and this is the specific problem of fifth harmonic elimination which again we discussed in the last class we want the fundamental voltage to be equal to the desired value the alpha one and alpha two should be such that the fundamental voltage is equal to the desired value at the same time the alpha one and alpha two have to be such that the fifth harmonic has got to be zero this is the problem of fifth harmonic elimination you have two equations

simultaneous equations but they are nonlinear transcendental equations you use a numerical iterative procedure to solve what we are trying to do is rather than just using a procedure we are trying to understand whats happening behind the scene thats what we are trying do here so we plotted all the v five is equal to zero contours last ten we tried plotting various values plugging in several numbers of alpha one for example into this equation and calculating the corresponding values of alpha two and we found how the contours are coming up to be and now the same contours are now plotted using a computer program so you know this is a v five equal to zero contour this is another contour that is along this line also you have v five equals zero this contour is nothing but the same fellow shifted up by seventy two degrees shifted up by seventy two degrees because seventy two degrees is three hundred and sixty by five that gives the periodicity for the you know fifth harmonic so the same contour when its shifted to the right appears here it appears here and its only a part of it it appears here because this is our area of interest why this line marks alpha one is zero we cannot have alpha one less than zero so alpha one is equal to zero is one boundary and alpha one alpha two equals ninety is another boundary again we have a constraint that alpha two has to be greater than alpha one this is alpha two equals alpha one and so we have to operate only in the top right angle here now and within the top right angle also you have this is zero per unit fundamental voltage this is one per unit fundamental voltage now and on this side you are going to you will still get some waveform you may get something like point four point five per unit that you might want but it will have a different phase as i was mentioning in a previous lecture now so we are essentially going to operate not even within the top triangle we are going to operate within these two blue lines marked here as zero per unit and one per unit now so these are the fifth harmonic contours that are of interest towards now so if you want to choose any value of fundamental voltage let us say i want to choose point five if i choose v one is equal to point five and a plot point five v one is equal to point five on this i see that it intersects is two points these are the solutions for the fifth harmonic elimination problem so along the line i can choose any point firstly to achieve the necessary fundamental voltage of fifty percent that is fifty percent of the square wave right now along this i have two different points at these two points i am guaranteed to have fifth harmonic to be zero we dont know what is a seventh harmonic we dont know what would be the eleventh harmonic we would know what is about thirteenth harmonic we dont know what would be the overall things now but one thing that we can certain is lets go back to the slide if you look at the slide here so you see this at this point now let me mark this point is a and let me mark this point b now we will see a little later that a and b are equally good as far as fifth harmonic is concerned both are eliminating fifth harmonic but you will find that b is better in terms of seventh harmonic after fifth harmonic the most dominant harmonic is actually seventh harmonic so its probably better to choose b as we will see later now firstly what we have to see is sometimes your equations are possible you know its you can get multiple solutions out of them you may start solving a particular problem and your friend may also solve the same equation you may come with one solution he may come with a different solution now why is that because this equation at at in certain ranges can give you multiple solutions and this is what we are trying to see now so what are the range is now at point five for example it gives two different solutions the fifth harmonic elimination the same fifth harmonic elimination problem if i consider a different fundamental voltage of point nine per unit gives only one solution it does not intersect with the other v five equal to zero contour on this side it doesnt intersect here it only intersects here and let me call this point as c so c point c is the only solution when it is point nine you have nothing else whatever is the seventh harmonic whatever the eleventh harmonic whatever the thirteenth harmonic at that point you just have to accept it if you say that i want v one is equal to point nine per unit and v five should be zero then this is the only solution for you when you are using only two switching angles per quarter cycle now so lets see that this you know there is this so some range in which you have multiple solutions and some range you have unique solution and there is also certain range where you have no solution at all for example you take v one is equal to one per unit and you take anything close to that something like point nine eight per unit that will be very close to this line and that may not intersect on this red contour also i mean it may not intersect with this also thats what we will see in the next one we will see the ranges now first let us say if i take alpha one is equal to zero what i am getting is for v one is corresponding values of alpha two i am getting a sixty and twelve degrees that is for alpha one is equal to zero this is twelve degrees this is sixty degrees and this is one of the curves now then on along this line you have alpha two is equal to ninety degrees so when

i use alpha two is equal to ninety degrees the corresponding values of alpha one are sixty and eighty four degrees so i go here so this is sixty degrees alpha one this is eighty four degrees now so this is what we have so along this curve let us see how fundamental voltage varies and along this curve how the fundamental voltage varies now if you are starting from here the fundamental voltage is zero as in move past in this curve you know if the fundamental voltage goes on increasing and at this point it reaches v one is equal to point eight per unit and i have plotted that point eight per unit line here if you fundamental voltage increases beyond point eight there is not going to be any solution out of this particular curve now so up to zero to point eight per unit this is zero here and this is point eight here you have one solution as we saw before and if you are looking at this curve your this is zero here and this is some point nine six per unit here so if you are moving starting from this point we moving all the way here the fundamental voltage goes on increasing the fifth harmonic is always zero the fundamental goes on increasing and you reach when you are when you are reaching when you are reaching here you you get something like point nine six per unit and beyond that point nine six per unit there is no solution now so zero to point eight you have two solutions and between point eight and point nine six we have one solution and beyond that you have no solutions so this is what we are coming up here so zero to point eight you have two solutions and this one and this sometime no solution now now lets take the counterpart of the seventh harmonic elimination problem and before that lets look at these you know the two options how you can look at them how do you really implement them so this stored waveform pwm it comes from them that is what you are trying to is let me just explain now so there are two solutions here now i am focusing on one of the solutions which are marked in green here so what i want to do is i would start from this point and i move along this curve and i have reached alpha sorry i have reached alpha two is equal to ninety and that alpha two is equal to ninety along alpha two equals ninety itself i move and go and reach the end point so what is that this curve along with the straight line is a locus of a point which starts from v one is equal to zero sorry this is v one is equal to zero and this is v one is equal to one per unit so this curve starts from there and goes there so this is we define a pwm technique if you could recall in a previous lecture a pwm technique can be defined as any contour i mean any locus of a point which starts from one of the points on v one is equal to zero contour or any v one is equal to v minimum contour and ends at v one is equal to one or v one is equal to v max thats all that we wanted to now so any line that you can draw you can just draw a straight line from here to here or you can call that a pwm technique you can draw some wavy line from here you can call this as a pwm technique now in this case we can call this as a pwm technique this green curve can be called as a pwm technique what does it essentially say it only gives the relationship between alpha one at alpha two here and v one varies as we move along the curve so along the locus of the point v one varies monotonically you start from here and you go on doing that even goes on increasing increasing increasing it reaches point eight here and from here it goes on again further increasing it goes to one now so along this locus it increases monotonically so if you want to implement a pwm method what you can do is you can just save this curve what is that curve it has some alpha one and alpha two defined at every value of alpha one and alpha two on this curve there is a v one now so let us say if you want to implement a pwm method you can save this as a look up table and hence the name stored waveform pwm when you want v one is equal point four then v one is equal to point four could be some point like this could be i am sorry v one is equal to point four could be a point like this now at that point you have some alpha one and alpha two what those values are you can pre compute them and you can store them in a look up table and you can output them and this is the idea of what is called as a stored waveform pwm and thats what we are trying to look at today so the selective harmonic elimination methods are one variety of stored waveform pwm now you can store this green color curve in a look up table and you can output it as required now alternatively what is not necessarily that this green curve you can also use this other curve you can also use this other curve now incidentally i have called this as pwm one that is pwm two thats ok we have been shifting here now so instead of using that curve what i am trying to do is i am using this green curve here and now this green curves if i move along this green curve it starts from zero and the fundamental voltage goes on increasing

monotonically and it finally reaches point nine six roughly and at beyond this point fifth harmonic elimination is not possible but i want to still control the fundamental voltage therefore i move along this green line to the origin now so this green line is again a pwm technique and with this pwm technique the fifth harmonic is eliminated all the way from zero fundamental unit to point nine six per unit and the fifth harmonic is not eliminated between point nine six and one there is certain amount of fifth harmonic is always there now similarly with the other curve that if you had chosen this one fifth harmonic wouldnt be there between zero to point eight and it will be there now between point eight and one there will be certain amount of fifth harmonic certainly there now it is also possible for you to start with this curve and follow this curve up to some point and jump back from there to here with the same v one says at v one is equal to five six or point seven some intermediate point you can jump over here and you can follow this curve and you can move were there now thats also possible now so as i was mentioning pwm is a locus of a point on this alpha towards is alpha one plane and the locus starts from v one is equal to zero or v one is equal to minimum be minimum and ends up at v one is equal to one or v one is equal to maximum contour so this is what you can do so this is another example of a stored pwm waveform now these stored pwm waveforms are you know stored stored waveform methods are targeted at eliminating the fifth harmonic now so you can also think of the other method of eliminating seventh harmonic voltage now its a similar exercise but you will see that there are differences now so this is v one you want decide v one to be equal to your desired v star at the same time sorry v seven you want it eliminated instead of v five mathematically it is the same procedure two equations two unknowns so you go about solving it iteratively you have a good starting solution you can converge faster and you go about doing that now we are trying to see a graphical approach as we have been doing so now this v seven is equal to zero contours we plotted them again by hand in the last few classes now you can see that i have plotted them here now and many a times you know are the way i plot it would look like a semicircle and i kept on telling you its not exactly semicircle now the point we can very clearly see that point its not exactly a semicircle now its actually flat red the middles and its more curved at this edges now the four corners thats thats what you can see and you can see that between v five and v seven they are appearing more frequently because the space seeing you know the periodicity in case of v five equals zero was three hundred and sixty by five that is seventy two degrees here it is three hundred and sixty by seven that is something like fifty one point four degrees so they are they you find closer and you find the other two curves also appearing prominently in in the case of v five they appeared somewhere the corners now the the two curves are other two curves are also appearing more prominently here now so this is v seven is equal to zero contour now and any point here you will have v seven is equal to zero again on any point here you will have v seven is equal to zero and interestingly this v seven is equal to zero contour starts at a point here starts at a point here at this point the fundamental voltage is not equal to zero it again ends at a point where the fundamental voltage is not equal to one that is not surprising because when v one is equal to one its square wave operation and they fifth harmonic and seventh harmonic cannot be zero right so but it is interesting to note that there are certain v seven equals zero contours which do not start at v one is equal to zero they start at some other point this is something we could not see when we brought it v five equal to zero contours this is one reason why one would like to plot v one is equal to v five equal to zero some time we seven is equal to zero we eleven equal to zero we thirteen equal to zero etcetera these are exercises that will tell us you know ah will make us understand the relation between the switching angles and the harmonic voltage is better now you have this now on the other hand you have this contour now this point here v one is equal to zero at this point also v one is equal to zero so you start and if you start from this point and move along this contour here ok you start from here zero it goes on increasing it goes on increasing at some point it reaches a peak the v one becomes something like point eight seven or so and beyond that point it goes on reducing and at this point it becomes and on this side of the contour you still have v seven is equal to zero you get fundamental voltage but this negative what do you mean by that you will get certain fundamental voltage but the phase will be reverse of what you will get on this side that what is called many case now so if you see if you are considering v one is equal to zero there are two points available for you here and here these two points eliminate seventh harmonic for you well guarantee what a v one you want you take any v one which is small like point ah like zero two two percent or three percent you are going to have something very close here here so you will have two different solutions possible now correct so which are the solution you will choose in both the cases you can get the desired fundamental voltage in both the solution you can get v

seven is equal to zero but you might probably want to choose a solution where v five is also low may not be zero but at least low you will see that this point v five is zero and anything close to this point v five is fairly close to zero on the other hand if you take this point v five is at its maximum really and any point close to this v five is going to be very high we will see that little later so i would say that i would probably take a point here rather than take a point here when i wanted to seventh harmonic elimination if fifth harmonic is a you know criterion two so behind certain range now so trying to illustrate this with certain specific examples one is v one is equal to point four excuse me i want to take v one is equal to point four so v one is equal to point four is a curve that moves like this let me write point four on this so this is the point four curve i drawn it in black so it does not intersect in this side but it intersects with this curve at these two points now so there are two solutions possible so for certain range starting from zero to some number there are two solutions possible then i have been considering v one is equal to point eight and this is v one is equal to point eight i have shown plotted here if i take v one is equal to point eight in plot that contour it intersects this curve here and it intersects this contour at two different points so there are three different solutions available for you three different solutions available for you you have a choice to choose from great number of choices to choose from so one way as i said you can probably look at what the fifth harmonic is at those points there are three points available xy zee and you can see at which of the points v five is the lowest one you can choose or you can also look at certain other criteria as i will say talk about little later you can also look at pulsating torque thats also a factor there are other issues that you can consider and choose one of these three points now firstly we should be aware that options are available what are the options available and then evaluate the options for evaluating the options we should know what are the criteria that we would like to pose so one of them could be v five being zero one of them could be the weighted thd of voltage being low other one could be the pulsating torque lets say the sixth harmonic torque being low etcetera etcetera so now these three points are then now let me go to another value that is point nine five if i go to this v one is equal to point nine five i have intersection only here and i this v one is equal to point nine five does not intersect this contour it does not intersect this contour at all right yeah so there is only one solution now so initially at two solutions and some range you have three solutions and in some range you are back to only one solution and finally when you are very close to v one is equal to one there will no solution so seventh harmonic cannot be eliminated here so let us delineate these ranges clearly so if you start along this curve and you move along this what are the maximum values now for v one is equal to zero you have some values here alpha one is sixty in alpha two is ninety and here you have thirty six and seventy two degrees thats what i am written here thirty six and seventy two degrees now so along this curve if you move zero to point eight seven roughly it can give you a solution zero to point eight seven it can give you two solutions this side of the curve can give you one solution this side of the curve can give you another solution so this particular contour alone can give you two solutions in the range zero to point eight seven and if you look at this and you look at what is the fundamental voltage at this point the fundamental voltage at that point is alpha one is zero alpha two is forty two point eight six the fundamental voltage is something like point four six six so up to zero to point four six six this fellow does not give you any solution from point four six six all the way up to point nine seven eight it is point nine seven eight here it gives you one solution so this fellow gives you two solutions between zero to point ah point eight seven and this one gives you two solutions one solution between point five four seven to point nine eight are so so from between zero to point four six this is point four six so i am excuse me so this is point four six six between zero to point four six six is one range where you can get two solutions and then you have point eight seven this is point eight seven between point four six six and point eight seven you get three solutions now and finally you have point nine seven eight so at this point it is point nine seven eight and this black curve this black curves its a little very close to v one is equal to one per unit so that between point eight seven and point nine seven eight you can get one solution and when your v one is beyond point seven nine seven eighth seventh harmonic cannot be eliminated thats what is indicated here you have two solutions and you have three solutions and you have one and you can have you have no solutions no solution here right

so lets move on so you can once again look at stored pwm waveform there are several options available and i need a pwm method how do i do i have an inverter i have a motor to run let us say using a constant voltage per hertz volts per hertz control and i need a pulsewidth modulation method to do that so one thing that i can do is let me just choose a green kind of color here i can start from here this is v one is equal to zero and i can move along this curve i can move along this curve as long as the fundamental voltage keeps on raising maybe up to some point here so i would have started off with zero and here i will have something like point eight seven per unit now up to this point this contour promises me seventh harmonic eliminated if there will be no seventh harmonic and from here i can just move along a straight line here i am sorry i should have used a different color here just a moment yeah i am sorry about that so let me go back to the contour that i wanted to say so i start from zero i follow this contour i follow this contour and from here i can come like this an india i can come in like this now so we are able to see the green curve this is a pwm method this green curve is a pwm method i can save this as a look up table and output this alternatively let me use another color shall i say yellow now i can start from here i can go along this curve i can go along this curve i can come back and join this v one is equal to one so this yellow curve is another pwm technique this also guarantees me you know the necessary fundamental voltage and it guarantees me seventh harmonic would be zero starting from zero to something like point eight seven per unit be another it doesnt eliminate ok so this is another one so the green curve can be stored as a look up table or the yellow curve can be stored as a look up table now alternatively you can also consider this method let me use red itself here let i can start from here i can move along this curve or i can move along this what a red curve and go back here this red is another pwm method now so i am giving you three different examples of pwm methods and these are stored waveform methods you can stored them in look up table and you can output them and you can use them now so they all can give you fundamental voltage ranging from zero to one per unit but you know in in the case of this green and the yellow the seventh harmonic will be eliminated all the way from zero to point eight seven but if i use the red one it is not so it between zero to point four seven it is not eliminate it is eliminate only between point four seven but up to point nine seven or so i can also move along one of them and jump over to the other one and do this such things are possible i mean just trying to indicate some examples for you now as to what can be done now so these are stored waveform methods and you can go about doing and the green is probably the most you know preferred solution i can easily tell you why because the range is very good it starts from zero and goes all the way up point eight seven just like the yellow also does but in the case of yellow curve as you can see here fifth harmonic is very high so the seventh harmonic may be eliminated but the fifth harmonic is very high this could result in a very high pulsating torque as we will see a little later now just briefly look at that so this could be this could be very bad in terms of fifth harmonic being bad terms of the sixth harmonic pulsating torque being very high in the molten drive this could particularly bad now so you may not want to use it so you have a options you know one you have a options you should know how to way various options and go about doing them and these are all examples of stored pwm waveforms now based on seventh harmonic elimination so selective harmonic elimination is one kind of stored waveform it is not necessarily you can also do many other kinds of waveforms you can come to you know several other pwm methods now this v five equals zero on v seven equals zero gives you a very good idea if you plot them they give you some kind of good idea as to how you can go about designing your pwm method so now we have only two switching angles we want to see how best go about doing it so we have done this excess of plotting v five equal to zero and v seven is equal to zero now let me start off let me use some two different options one option let me call it as the green option and the other option let me call it as the yellow option what is this green option

i would start from v one is equal to zero right what is a pwm method its a locus of a point which starts from this v one is equal to zero contour and ends up with v one is equal to you know one lets say so now i will start my green one one low pwm technique i will start from there and the another pwm technique which i would probably say the yellow one i will i can start from here what is the advantages of these two points at these two points you see that the fifth harmonic is zero you also see that the seventh harmonic is zero it is supposed to be it should be so because while this entire blue line guarantees you v one is equal to zero this blue line guarantees you v one is equal to zero these two are the only points on the blue line that guarantees that all the harmonics are also zero five v five v seven v eleven v thirteen etcetera are also zero because the waveform is a third harmonic square wave here and it is also third harmonic square wave was i told you in a previous lecture now so the fifth seventh eleventh thirteen harmonics do not exist so not only fifth and seventh which are shown here eleventh and thirteen also do not exist so both are equally good options now now let me say i want to slightly increase that v one is equal to zero is not a point at which you would operate we are trying to understand that v one equals zero mainly to understand the issues related to v one is equal to say two percent or three percent are small amount of voltage now now let me say let me go back to this i want to use something like v one is equal to like two percent or three percent now so if i were here what i would choose that curve can really be going somewhere like this now this is a small voltage now i can choose some point here alternatively i can choose some point here you see that it makes better sense to choose this point over this point let me choose the green point or the yellow point why is that so at the green point if i choose this green point i am both v five and v seven are kept reasonably low whereas if i choose this yellow point this may also give me the same fundamental voltage i am reasonably closer to v five equal to zero little farther to but i am far away from v seven is equal to zero now let me just finish it up this is the green yeah now let me say i just take a straight line moving like this i just take a straight line moving like this on the other hand with the yellow line let me just draw a line like this the green line is a pwm method and the yellow line is also pwm method as i have also mentioned before the green is good in certain respects because as the fundamental goes on increasing as the fundamental goes on increasing you are not straying away very far from v five equal zero contour or v seven is equal to zero contour you are not straying very far away so for a up to certain substantial value of fundamental voltage this curve moves along very close to v five equals zero and v seven guarantees you low values of v five and v seven so those are the two principal harmonics so the currents are going to be dominant you know like v five and v seven are more dominant or v eleven and v thirteen in terms of their currents v eleven and v thirteen phase higher harmonic reactances then v five and v seven do so if you can keep v five and v seven low whats the problem so you can consider a method like this now again what have what you can do is you can move on here if we consider the same yellow line what happens you are moving away an away from v seven goes in the other direction but your curve is moving in the other direction so your v seven starts increasing very quickly though v five may be reasonably low but its increasing so its going to lead to a substantial amount of v seven and also substantial amount of sixth harmonic pulsating torque so thats an issue there now so you can also consider certain variants so this is how you can look at when you draw all these curves you know what necessary and actually this point is a good point this point is a good point because v five and v seven are both equal to zero now so if you can look at a trajectory which passes through that point that is also fine and this point guarantees you ah zero pulsating torque because the pulsate is six harmonic torque is determined by the fifth harmonic voltage and the seventh harmonic voltage if they are both equal to zero sixth harmonic torque should be zero its actually a good point to operate in so you can look for any curve that goes through that goes close to this and things like this now so several possible contours you can draw infinite number of contours you can start form any point on v one is equal to zero and you can choose any point on v one is equal to one and from that point to this point you there are several contours again infinite number of contours it basically tells you that the infinite number of pwm techniques are possible but not all pwm techniques are equally good right how do you choose those good techniques now so not only a control

of fundamental voltage in terms of harmonics also thats be the burden of over thing now so if you look at v five and v seven just for example only two of them it makes it crystal clear that these are the only two points which are worth operating at when you are looking at v one is equal to zero when you are looking at v one is equal to some small number again you need to be close to one of these two points and it is better that you are close to this point rather than close to here because if you are close to here both your v five and v seven are guaranteed to be low on the other hand if you are here your v seven could be much higher this is now so it gives you an idea for to know on this know you can also plot v eleven and v thirteen and you can see you can get a better picture of what exactly happens now you generally see that in your here at low modulation indices when you are operating somewhere here it is very good it is something very good now so let me just plot v eleven v thirteen and such kind of curves now if i really plot them you know how we they would look most of the curves will actually go like this now those are loops i have indicated this is like v eleven is equal to zero v thirteen is equal to zero and go on and all the curves will move tangential at this particular point now so at that point all the harmonics are zero and you can see that if you are locus the pwm the locus that ten or the line that indicates the pwm technique itself moves tangentially to all the vn equal to zero contours so along the line the harmonic is not increasing very highly so thats one of the reasons so for at low modulation indices almost all the harmonics are very very close to zero here now as you will see in what is called as the minimization of weighted thd you would get many points normally close to this now that is if you are operating here your overall thd v five is guaranteed to be low v seven is guaranteed to be low v eleven is guaranteed to be low v thirteen is guaranteed to be low and overall the weighted thd is guaranteed to be low here at low modulation indices thats not the same when it comes to high modulation indice now let me say if i take this point here here obviously square wave so fifth cannot be zero seventh harmonic cannot be zero and similarly come close to higher modulations this is may not be the case now on the other hand if you consider anything here you consider this band here you consider any point here you are assured of v five as well as v seven being reasonably low and these are high modulation indices now if you plot v eleven equals zero v thirteen etcetera equal to zero also you will see that in this area you will see many of the vn equal zero contours passing here now so this essentially tells you that at high modulation indices if you operate closer to this origin you operate closer to this origin in this region you are guaranteed to get better results many of the harmonics could be lower at lower modulation indices its better that you operate closer to this particular point at higher modulation indices you operate in this range now so if i may have to give you something indicatively at lower modulation indices if i take this angle what is that this is sixty this is ninety both my alpha one and alpha two are within the range sixty to ninety degree if i choose my alpha one and alpha two be within the range sixty to ninety the lower order harmonics are likely to be much better at low modulation indices i mean and similarly if i am talking high modulation indices its better that operate somewhere in this range its better that i operate somewhere within this range that is zero to thirty degrees i have to make sure that both the switching angles are probably within zero to thirty degrees and closer to them now thats how you will get now right so this is basically to give you on this i would certainly encourage you to plot v eleven equal zero and v thirteen equal to zero contours etcetera you can certainly make use of you know way we were looking at v five equal to zero v seven equals zero it is also possible to plot for you roughly where it comes and you can also use a computer program to plot it precisely so you will see what happens in how exactly its going to come through and you see this where should now so there are many many pwm methods and the you know and its possible to when some of them will be better than the other and we are trying to get some idea as to how one could design better pwm methods now so if you are looking at this is only up to two switching angles now you want to go more than two switching angles because this two switching angle you must realize that we took it up only as a means to improve or understanding we just wanted to understand the relationship between the fundamental voltage and the harmonic voltages on one hand with the switching angles on the other hand so we we kept the problem simple by only two switching angles now so what is this two switching angle stuff we have been considering a pole voltage which is like this we have been considering pole voltage vro is like this now this is zero degree this is alpha one this is alpha two this is ninety degree this is what we have been considering now if you want to consider three switching angles here vro is switching from zero at zero it is switching from negative to positive it is common to consider the switching

the other way at zero degree vro would normally switch in the other direction and this could be your alpha one this could be your alpha two this is ninety degree alpha one alpha two i am sorry alpha one it goes back alpha two comes and then alpha three ninety degree its normal to use it the other way that is at the zero crossing your vro will be switching from positive to negative now so this is for three switching angle problem now so you have alpha one alpha two alpha three v one is related to alpha one alpha two alpha three vn is related to alpha one alpha two alpha three you can do all that you want to do the harmonic elimination so fifth harmonic and seventh harmonic two harmonics can be eliminated with controlling fundamental voltage now so with three switching angles two harmonics can be eliminated this is the nature of waveform will use now now let us say you want to use four switching angles so what you can probably do is you go like this so this is four switching angles per quarter alpha one alpha two alpha three alpha four and this is zero degree so generally for e one value for where n is the number of switchings per quarter and it is even value you choose this kind of a pwm waveform where vro switches from negative to positive and in general for the other side you will switch it the other way alpha one alpha two alpha three zero alpha one alpha two alpha three alpha four alpha five ninety degrees so for n is odd you are using this kind of waveforms and you can use for any value of n now so these are more and more switching angles now the exercises that we did for n is equal to the for two we can do similar exercises for n is equal to three we can plot here we could plot on for n is equal to two we could just plot it on a single plane here it has to be plot in a three dimensional space but still you can use your computer programs to plot those and get an idea of how v five equals zero and v seven equals zero contours are here how they would be you know instead of being a curve in this case sorry instead of being just a curve here it would become a surface here it would become a surface in the three dimensional space here and for all these you can just formulate the equations and you can go about solving them if you look at the selective harmonic elimination and you look at n switching angles so you can have n number of equations so v one is equal to v one star is they desired fundamental voltage this is one of your equation then you can go about eliminating n minus one harmonics capital n minus one harmonics where n is the number of switching angles per quarter you can eliminate seventh fifth harmonic seventh eleventh thirteenth and up to n minus one harmonics we can go about eliminating you can solve all these equations its capital n number of simultaneous equations these are all transcendental equations and you can solve them and you can come up with solutions now again you may come up with different solutions so sometimes it may give you same same set of equations you may come up with one solution your friend may come up with another solution because you know one reason could be that your starting solution and in his starting solutions very different so depending on numerical procedure and the starting solution that you take you would converge on to one or the other kind of solutions here now so we we are now better we we know that multiple solutions are possible and all such awareness that we suddenly have and with an idea of all this v five equals zero v seven equals zero where are the harmonics being zero its also possible for us to such as good values of starting solutions now ok so many a times if you want to eliminate all the harmonics you can also start let us say v one is equal to point five per unit is what you want to start you want to do so and all some n minus one harmonics have to be zero you can also start at a point where the waveform is effectively a third harmonic square wave or a ninth harmonic square wave or so on in that case all the harmonics are anyway guaranteed to be zero and the fundamental voltage is zero so once you know the solution for that you can calculate that is v one star is equal to zero the solution is very obvious and then you can calculate for v one star is equal to some small value point naught one point naught two and you can go about calculating so this can give you the solutions for various values of v one also ok so typically in the books you will find solutions for against v one now let me just write it down here now what you will find is against v one different values of alpha will be plotted here alpha one alpha two alpha three etcetera would be plotted against this now i am not doing this here you can certainly go about and doing this as an exercise and it is available in

several other references which i am going to mention now there all right ok so now if you move to the next point now this is about selective harmonic elimination that is not the only stored waveform quality what you can do is you can go about minimizing thd you can go about minimizing the thd the thd is the line current the thd in the line current is equal to the weighted thd w stands for weighted weighted total harmonic distortion in the voltage waveform now if you see that there is a factor vn what is this vn vn is the nth harmonic voltage the nth harmonic voltage sees a harmonic reactance which is n times that seen by the fundamental voltage so you have vn by n factor here and that is whole square so this gives you basically that the i have missed out a square root on this i am sorry so what do you normally have is you would have a square root on this you would have a so this vn by n vn by n is indicative of the nth harmonic current and the sigma vn by n the whole squared is indicative of sigma i n squared is the measure of sigma i n squared that is all the harmonic comes added together the total ripple correct so that is normalized with respect to the fundamental voltage now this is called the weighted thd of the voltage waveform which is equivalent to the current thd if you multiply the weighted thd by a factor which is roughly equal to the ah no load impedance divided by the block total impedance or the machine you will get your current thd roughly so now theoretically it it contains all the values of n but you will see that higher values of n know they you know the vn itself will be low and vn by n will be very low so they can be ignored so i have just shown like v five seven eleven thirteen you can probably stop here i mean you can consider more values i am just trying to tell you that n can be truncated at some point in approximately now you choose your alpha one and alpha two such that they they guarantee going to the same switching angles the alpha one and alpha two they guarantee the required fundamental voltage at the same time they minimize this weighted thd that is you look at the entire possible solution of entire set of all values of alpha one and alpha two which will give you the required v one star and you choose that particular value which gives you minimum amount of weighted thd that is what is called as offline optimal pwm for optimizing minimizing things so you can do an offline optimization you can calculate the optimal switching angles so for v one is equal to point one per unit what is my alpha one optimal an alpha two optimal v one is equal to point two per unit what should be my alpha one optimal and alpha two optimal and you can store all of them in a look up table so this is again a stored waveform pwm method so this waveforms will give you not just eliminating fifth harmonic and seventh harmonic they will not be eliminate but they will try to minimize the weighted thd fifth harmonic may not be zero seventh may not be zero but the overall weighted thd of the voltage waveform or the current thd or the thd of the current waveform will be very low so this is what is an offline optimal pwm method now so minimization of thd of line current or weighted thd of line voltage is what is done here alternatively you can also try in minimize pulsating torque how do you do by minimizing pulsating torque you will have to understand what causes pulsating torque what causes pulsating torque had the motor been operating with sinusoidal voltages there is going to be no pulsating torque at all its going to be steady torque only because the motor is operating with non sinusoidal voltages you are getting pulsating torque what are the effects of a non sinusoidal voltage the flux is non sinusoidal and the currents are also non sinusoidal so if you if you consider the flux you have a sinusoidal component you also have a non sinusoidal components again similarly consider the currents the motor currents they also have fundamental and non sinusoidal the fundamental flux on the fundamental current interact to produce the steady torque whereas the fundamental flux and the ripple current harmonic current produce the pulsating torque the same way the fundamental current and the harmonic fluxes interact to produce the pulsating torque also we you know harmonic fluxes and the harmonic currents can interact but they are negligible because harmonic fluxes fluxes will be negligible and harmonic currents will be negligible and their product would be much more you know negligible you dont have to worry about so what causes pulsating torque is one is the fundamental flux interacting with harmonic current and the other one is the fundamental current interacting with harmonic flux and analysis can show you that the fundamental flux interacting with harmonic currents is what is really higher so it is really cause of harmonic current now so now let me say i have these harmonic currents one is i five fifth harmonic current what does this fifth harmonic current do this fifth harmonic current would interact with the fundamental flux and the relative speeds would be six times the

fundamental frequency and this will end up producing a sixth harmonic pulsating torque the same way the seventh harmonic current will also do the same job the seventh harmonic current and the fundamental flux both revolve in the same direction they have a relative speed of six times the fundamental frequency so this i seven fellow also interacts the fundamental flux and gives you a produce a sixth harmonic torque so both i five and i seven are responsible for sixth harmonic torque so if you want t six to be zero ideally what you should do is make sure that i five and i seven are zero you want to make i five and i seven zero what you should do is v five and v seven both should be zero but it may not be possible particularly if you are considering only two switching angles its not possible so if you analyze the problem clearly you would see that t six depends on the difference between i five and i seven it actually depends on the difference between i five and i seven so if you can make sure the difference between i five and i seven is very low you can make sure that the difference between i five and i seven is very low then you can eliminate sixth harmonic pulsation now so let me just change this further now so what is i five i five is basically v five upon five and v v seven is basically v seven upon seven so if you if your fifth harmonic and seventh harmonic are such that v five by five and v seven by seven are close to one another then the six harmonic pulsating torque is reduced i am trying to simplify the arguments and giving a very simple picture now so with two switching angles itself it is possible for you to make sure that the fifth harmonic and the seventh harmonic are both close to zero you know then you can get this kind of a waveform that you here now lets go back to the v five equal to zero and v seven equals zero contour and there we will see this point now so here i want to point out to you that say lets say this point excuse me so if you say this point as i pointed out to you t six is equal to zero there t six is equal to zero at that point why is t six is equal to zero because both v five and v seven are equal to zero and therefore i five and i seven are equal to zero and therefore sixth harmonic torque is zero now let us say you consider a point something close to that you consider a point something close to this point v five is not zero and v seven is also not zero but it is quite possible that the v five and v seven amplitudes you have to search around that point you have to you its possible for it to find a point where v five and v seven are such that v five by five is equal to v seven by seven and also v five and v seven are the same phase if you kind of come up with such kind of a point then you can say that the sixth harmonic torque gets eliminated there also similarly this point you have a sixth harmonic torque is zero why because v five and v seven are both equal to zero and somewhere on this line you may come somewhere you know you will get a set of points here along these points that you indicating the v five and v seven are such that v five by five and v seven by seven are roughly equal v five by five is roughly equal to v seven by seven you can get a set of points now this is what would be sixth harmonic torque being zero now so you can look at selective elimination of harmonic torque and you can calculate the switching angles and you can save them as a look up table that is also stored we found pwm method instead of considering particular pulsating torque i am sorry instead of considering sixth harmonic torque you can also consider the sum of all the torques sixth twelfth everything you can consider the rms torque or the peak torque and you can try to minimize you come up with switching angles alpha one optimal alpha two optimal alpha three optimal etcetera that would minimize your pulsating torque the peak value of the pulsating torque or the rms value of pulsating torque whatever it could be are specifically sixth harmonic torque or twelfth harmonic torque that also can be called as an optimized pwm offline optimist your calculating all these offline when you are storing them as a look up table and using a during an implementation now and you can also attempt to minimize the motor losses these harmonics are causing certain amount of motor losses increase copper losses and there is also increased core losses you can if you understand the relationship between the switching angles and the motor losses if we can express this mathematically then its possible for you to come up with alpha one alpha two optimal values which will give you apart from the necessary v one which can also minimize the motor losses now you should know how to express motor losses in terms of switching angles you can also look at the total losses which is the inverter plus motor losses now so if you look at the inverter plus motor losses you should once again be able to express the total loss as a function of the switching angles then you can minimize you can treat this as an optimization problem and minimize and you can calculate your alpha one optimal alpha two optimal alpha three optimal at every possible values of v one like one percent two percent three percent four percent save it as a look up table and use it during the waveform now so thats what we have about offline optimal pwm here i am going to suggest some good papers

which has seminal work and there are also other papers in this area but i am just going to suggest a few of them for you to go and read and so that you get a better understanding of this so these are some of the earliest papers on harmonic elimination method now there is part one and part two which particularly dealt with harmonic elimination and voltage control methods now these are published in the years nineteen seventy three nineteen seventy four by patel and hoft so these are among the oldest there are still older paper here in nineteen sixty four etcetera but i am just giving you this paper this is one of the references now so this is all you also have this book by professor holmes and professor lipo on pulse width modulation for power converters principles in practice this summarizes you know many of the selective harmonic elimination methods the work that have been done over the years has been consolidated you will find lot of several graphs here which are very useful for you to understand you know what happens and things like this now so this i would suggest this as an addition reading to now then about the optimal pwm but minimization of thd this is a very good paper this is one of the first papers in this topic they calculate the switching angles in this paper buja and indri calculate the switching angles such that the thd is minimized or the weighted thd of the voltage is minimized this was published in the i triple e transactions is an industry applications in the year nineteen seventy seven so now you continue with that not only minimizing thd you can also minimize the motors pulsating torque of the torque ripple and the speed ripple so this paper by zach martinez and keplinger keplinger and seiser in the year nineteen eighty five on the same i triple e transactions and industry applications deals with that how do you calculate a switching angles such that the pulsating torque is low thats what you know i i just gave you an indication this paper discusses all the details there and similarly this is efficiency optimal pwm for ac drive so if they are trying to reduce the overall losses in the motor drive so the the switching angles the the losses are expressed ah are the efficiency of the motor drive is expressed as a function of switching angles and you try to optimize and get this so these are certain very useful references that i would like to suggest so all these are examples of the stored waveform pwm method now this is perhaps the last lecture that we will have an low frequency pwm methods in the low frequency we took it up essentially to understand what we are trying to you know get an essence of how fundamental voltage and harmonic voltages are affected by switching angles and so on so forth now we are practically done with low voltage ah things and in generally you know we will dealt with selective harmonic elimination and always optimal pwm you save them as a pwm waveform do that now from the next lecture onwards we will be looking at a situation where the switching frequency is much higher than the fundamental frequency we will say probably the fundamental frequency is fifty hertz you may look at switching frequency of five thousand hertz or so which is hundred times the fundamental frequency so the way you analyze and all that could be very different when you are doing there now so thats what we going to look at now so most of the modern methods are really like that ah unless you go to very high power levels etcetera you will do this here so in the modern context there are certain low frequency cases where when the high when the power level is very high your switching frequency is low therefore the ratio of switching frequencies the fundamental frequencies low and the other situation is the switching frequency is high but the modulation frequency is also high in case of high speed motors thats again a situation where the low frequency methods we studied will be helpful next lecture onwards we will be focusing on the high frequency methods and thank you for your interest and hope that you will continue to follow this lectures thank you very much