 I have been saying that either it could be a 120 degree adhesion or it could be a 6 step harmonic spectrum is the same. Now, if I have to reduce the thd, what do I do? Now, I am not going to, I will just show you a result based on that we will make conclusions and we will go ahead. If you are not convinced, I will show you. See, there are large number of pulses here and this is the current wave form, load current wave form. It is almost a sinusoid, almost a sinusoid, almost a sinusoid. See here, there are large number of pulses in one cycle and this is the current wave form, almost a sinusoid. I am assuming that load is inductive, load is inductive. So, if I have, if I turn on the device and allow the device to conduct for 180 degrees, there is a problem. Harmonics is 6 and plus or minus 1. Instead, if I have a large number of pulses in one cycle, in half a cycle, current wave form is going to be sinusoidal. How are we getting the sinusoidal? I will explain to you with the equivalent circuit. So, one conclusion that we can make is, if there are large number of pulses, my current wave form is almost a sinusoid. So, therefore, the question immediately arises, how to have this so many number of pulses? Who will tell us how, when to turn on and when to turn it off? For that, I need to have a strategy, what is known as a pulse width modulation strategy. There are various pulse width modulation strategies. Before doing that, I will explain to you using the equivalent circuit, how we can get a sinusoidal current from this pulse supply. One way is to write the Fourier series using the Bessel's function and get the harmonics series or the harmonic spectrum of this. It is mathematically involved. Second is from the equivalent circuit. I will just take the second method, which is relatively simple. See, I have a large number of pulses and these pulses are applied to an RL circuit. Load is inductive, highly inductive. What will happen? A DC is applied for a very short time. So, definitely current will increase. Current starts, whereas in a 180 degree, we turned on the switch and we allowed it to conduct for 180 degrees. So, current increase and try to saturate. Whereas here, whereas here, I am applying it for a very short duration. The circuit is inductive, RL circuit and we know that initial portion of this current waveform is linear. This is almost linear. So, this waveform or this current rise, I can approximate it to a straight line. See, after some time, I am turning it off, applying a 0 voltage. So, what will happen when I apply a 0 voltage here to an RL circuit? Current will decay. Current will slowly decay. Again, I will turn it on. Now, the width of the second pulse is slightly higher than the width of the first pulse. So, what will happen? Current will again increase. After some time, I will decrease. I will turn it off. Similarly, if there are large number of pulses, I have an approximately a sine wave. So, that is nothing but the pulse is modulation strategies. The technique is very simple. I have large number of pulses and because of the load being inductive, I can have almost a sinusoid. So, what I can infer? See, if I apply sinusoid voltage to this RL load, see, if I apply sinusoid voltage to this RL load, current is sinusoidal, may be lagging by some angle theta, perfect. At steady state, at steady state. If I apply a pulse waveform, my current is still a sinusoid. That is what I showed you and that is what I can infer from this diagram as well. There are large number of pulses, yet the current waveform is sinusoid. What can I conclude? I will repeat my question. I have an RL load, apply sinusoidal voltage, current is a sinusoid. That I can, all of us know. Now, instead of applying a sinusoid, I will apply a square wave. In the sense, these are all square waves. There are large number of pulses in one cycle, large number of pulses in one cycle. Current waveform is a sinusoid. What can I infer? One simple inference is current waveform drawn by this load is independent of the excitation. I will repeat. Current drawn is sinusoid for load is being, when the load is RL load, applied voltage is sinusoidal, current is also sinusoid. Current drawn is sinusoid when the voltage applied is non-sinusoidals. So, what conclusion can I make? The conclusion is that, looks like, this pulse waveform has a fundamental component, has a fundamental component of 40 hertz, which is sinusoid and has a very high harmonic, other harmonic components and for those harmonic components, my RL circuit does not respond or for those high frequency harmonic content, impedance offered by the RL circuit is so high. Therefore, the current, the circuit behaves as if it is an open circuit. I will repeat. My conclusion is that, this step waveform or this pulse waveform has a fundamental component, which is a sinusoid and that forces the sinusoidal current. In addition, it has high frequency components. Those high frequency components are not able to supply any current because the impedance offered by this inductive load is high. It makes sense as the frequency increases, impedance offered by the inductance also increases. So, what do I need to do is, now the thd of this waveform is definitely is very less compared to, compared to this waveform, the so-called 6 step waveform. So, I cannot allow the 180 degree conduction. So, therefore, I need to a large number of pulses. Now, who will tell me how to, when to turn on the device and when to turn it off? That I told you a PWM strategy. There are various number of PWM strategies. I will try to do only two of them. One is a sinusoidal PWM technique and second is what is known as the space vector PWM technique. Sinusoidal PWM technique, all of us know, what I need to do is, there is a sinusoidal modulating function, a sinusoidal modulating waveform, a sinusoidal modulating waveform and there is a high frequency carrier waveform. They are compared, intersection of these two will determine the turning instant of S 1 and S 4. So, in a three phase inverter, there are three sinusoids, which are displaced by 120 degrees and there is only one carrier. They are compared and intersection will determine the switching instant. Now, if I want to have a 50 hertz supply, the frequency of the modulating wave is 50 hertz, but then how will I determine the magnitude of this triangular waveform or this carrier? By the way, the magnitude of the carrier is fixed and frequency is also fixed. How do I select the frequency? At every instant, every intersection, S 1 is turned on and S 4 is turned off and vice versa. At any given time, one switch is on. So, if S 1 is turned on, S 4 is turned off. So, depending upon the power level and depending upon the switches, I need to select or need to select the frequency of this triangular waveform. Lower the power and if I have a faster devices, higher could be the switching frequency. Say for example, if I have a 5 k wave inverter, triangular frequency could be of the order of 10 kilo hertz or so. Whereas, if I have a 1 mega volt inverter, triangular frequency could be of the order of 550 hertz, not more than 11 times. That also again at a higher upper limit. So, depending upon the power, the switching frequency or the carrier frequency is changed. And if I see the harmonic spectrum, they look like there is a fundamental component of 50 hertz and the immediate dominant harmonic is approximately equal to frequency of the triangular wave minus the frequency of the sinusoidal waveform. I will repeat the harmonic spectrum is fundamental component, which is 50 hertz here. And the dominant harmonic is or immediate dominant harmonic is the frequency of the triangular wave minus 50 hertz. And next is frequency of the triangular wave plus 50 hertz. So, at those two bands, there is a dominant harmonic. That is about sinusoidal p-dolume technique. Next one I will do, which is a space vector p-dolume technique, which is very popular if you are feeding power to the grid. Generally, space vector p-dolume technique is used. I will quickly I will cover what exactly is the space vector p-dolume techniques. Relatively simple, very popular. I will quickly I will explain. If V a is equal to V m sin omega t, V b is equal to sin V c is equal to this, V a b is equal to this. I can represent this three vectors in a two dimensional space. Voltage space vector V s by definition is given by this. So, I can substitute for V a, V b, V c. I get and simplifying I will get V s 3 by 2 V m sin omega t minus j omega t, the space vector having 3 by 2 V m and it rotates in space at omega radians per second. So, if I know V s and if I know the theta at any given time, I can get its d and q axis component and if I can get d and q axis component, it is possible to get the value of V a, V b, V c. So, that is what I am doing here nothing. I am just resolving these three phase vectors along x axis and y axis here. V a, V c what is the x axis component of a, b and c, x axis component of V a is 1, V b is cos 120 that is minus half and V c is cos 240 is again minus half, V y is y axis component is 0, y axis component of sin 120 is root 3 by 2, sin 240 is minus root 3 by 2. So, if I simplify V x is equal to 3 by 2 V a and V y is equal to root 3 by 2 will be V b minus V c, just matrix multiplication quick. Now, let us see. Now, all of us know that V a 0, V b 0, V c 0 could be plus V d c by 2 or minus V d c by 2. So, and I can write using K v l, V a 0 is equal to V a n plus V a n o. V a n is equal to V a n is equal to V a n, V a n is equal to V a 0 minus V a n o. Similarly, I write these equations, I will add them up V a n plus V b n plus V c n is equal to 0, substitute them here, I will get V a n o is equal to given by this equation. I will add them, I will get 3 n o is equal to V a 0 plus V b 0 plus V c 0. So, V a n o is equal to one third this. Now, substitute this value of V a n o in this equation, in this equation and in this equation and we can write V a n, V b n and V c n in matrix form, matrix form something like this. V a n, V b n, V c n is equal to V a 0, V 0, V c 0 2 minus 1, minus 1, minus 1, 2 minus 1, minus 1, minus 1, 2. Go back and simplify, it is simple. Do not get tristated, these are correct. If the, I told you there are 8 possible states of the inverter 0 0 0 to 1 1 1, these are null states V a n, V b n, V c n are 0. So, if V a n, V b n, V c n are 0, what will happen to, what will happen to x and y, V x and V y, V a n, V b n, V c n are 0. Therefore, V x and V y are 0, if V n direct V x and V y are 0 in the x y plane magnitude of the space vector is 0. And tan inverse of V y by V x is the position of the space vector, that means I am at the origin. So, for when all 3 switches are on and all 3 switches are off, magnitude of the space vector is 0 and V r at the origin. Now, what about the remaining 6 vectors? What I will do is a phase, a first leg upper is on and lower 2 are off, these are the phase voltages, sorry the pole voltages. I can substitute V a 0, V b 0, V c 0 in, please do not confuse, I know V a 0, V b 0, V c 0, I will substitute here, I can get V a n, V b n, V c n and if I know V a n, V b n, V c n, I can get x and y, V x and V y and if I can get V x and V y, I can get V s. So, this is the procedure that is all. So, for this 0 0 1, V a 0 is V dc by 2, V b 0 and V c 0 are minus V dc by 2. So, I got V a 0, V b 0, V c 0, quickly I write V a n, V b n, V c n and therefore, V x and V y. So, it so happens that V x is equal to V dc, V y is equal to 0. So, the magnitude of the space vector is V s and angle is 0, because y component is 0. So, if 0 0 1 corresponds to minus V dc, V dc, magnitude V dc 1 and angle 0 that means, I am along the x axis, along the x axis, 1 1 0 will be magnitude 1, angle 180 degree, because that is the complementary vector. I will repeat if 0 0 1 corresponds to 1 at an angle 0, 1 1 0 will be 1 angle 0. Now, second state I will take, I will take two devices are a phase and b phase, c phase is lower is on. So, this is the pole voltages, I know the pole voltages from those pole voltages, I can determine the phase voltages from that matrix and from this phase voltages, I can get x component and y component, x is half V dc and y is root 3 by 2 V dc. So, what is the magnitude V x square plus V y square, square root is 1, angle 60 degrees, because tan inverse of y by x, tan inverse of y by x is 60 degrees. So, the magnitude of the space vector is again 1 or V dc, similar to the previous case, previous two cases, but angle is 60 degrees. Therefore, for 0 1 1, magnitude is 1 at an angle 60 degrees, for 1 0 0, magnitude is 1, angle will be, where will I be? Tell me, it is 60 plus or minus 180. So, I will be in the third quadrant, third quadrant, it is a complementary of this vector. So, we have already found out the position of the space vector for four vectors, only two are remaining and what are they? For 1 1 0, it is V s and an angle 240, the last one is 0 1 0, B is on, A and C are off. Same procedure I can get or we get magnitude 1, angle 120 degrees, vector is moved by 120 degrees, the complementary will be 300, angle 300. So, see here these are the pole voltages for those three phase inverter, pole voltages for three phase inverter. This is A phase, B phase, C phase, A is on, B is off, B is off, there is nothing but 0 0 1 vector, A is on, B is off, B is 0, C is 0, that is 0 0 1, magnitude is 1, angle 0, we are here. Next is, A is on, B is on, C is off, A is on, B is on, C is off, lower. I said magnitude is 1, angle is 60 degrees, we are here. The next vector is only B is on, A and C are off, lower switches are on. Magnitude is 120, the magnitude is 1, angle is 120. Now, these three we determined the complementary we can immediately draw. If it is 0 0 1, it is 1 1 0, 0 1 1, it is 1 0 0, 0 1 0, it is 1 0 1. So, this is, so six active vectors, six active vectors occupy, occupy the vertices of an hexagon, hexagon. These are the active states, these are the active state and these are two or null states which are in the origin. So, six active vectors occupy six vertices of the hexagon. So, what we observe, from B 1 to B 2, from one position to another position, we go by just by turning on or off one device. What did we do here? We turned on B phase. See, we were at the origin, all upper switches are off 0 0 0, all upper switches are off, we are at the origin. Now, I will turn on only phase A. See, somewhere we are here, at this point at pi by 3, I turn on phase A, magnitude is 1, angle 0. If I wait for one sixth of a cycle, I will repeat, if I wait for one sixth of a cycle and at the end of one sixth of a cycle, if I turn on B phase also, I, the vector moves by 60 degrees to this place. See here, 0 0 1 and at this point B phase, also turned on. We will allow the vector to wait for 60 degrees in that place or once, sorry we will allow the vector to wait, remain there for one sixth of a cycle, turn off phase A, the upper switch vector. When you turn, when you do that, it goes, it moves by another 60 degrees. See here, 0 0 1, phase A turned on, phase B turned on, phase A turned off here, phase C turned on, phase B turned off, phase A turned on, phase C turned off. I mean, upper switches are turned off. I am just talking about upper switches. So, what exactly are we doing? Every one sixth of a cycle, we are changing the conduction state of one switch and if I do that, I will move or the space vector moves by 60 degrees. Space vector fibrous change in conduction state of switch moves by 60 degrees and it completes one cycle in six steps. So, this is nothing but six step operation. This is nothing but six step operation. The predominant harmonic is again fifth and seventh, but then we need to have a very low THD. Therefore, we have to eliminate the lower order harmonics. Therefore, we cannot remain idle for one sixth of a cycle at one particular state. See, I will repeat, we have to reduce the THD. Therefore, we require large number of pulses. We cannot allow the space vector to remain idle for one sixth of a cycle at one particular position. So, now, how do we have pulse width modulation or how do we have large number of pulses here? Because I can move from, see the origin I came here by turning on or of only one switch. Immediately, when I turn on another switch, I will go here. It moves by 60 degrees from here to here by another 60 degrees. I have only this choice, no other choice, but then using these choices, I have to construct a smooth sinusoid. I cannot have six step operation. What do we do? What do we do? We need to have a pure sinusoid. By the way, if I have three pure sinusoids, pure sinusoids, what is the trajectory? What is the space vector? How does the space vector look like? What is the locus of the space vector v s? Here, I have a hexagon because the line voltages are pole voltages are of 180 degree duration and they are square wave. Therefore, I get a hexagon, but then I want the output voltage to be sinusoidal. Now, if I get a sinusoid, if I have 3 pi sinusoid, what is the locus of the space vector? The locus of the space vector is a circle. Circle, it rotates at a uniform of the omega. Now, how to realize these sinusoids? It is a question that I am asking. You need to listen to me carefully for another just for 10 minutes. That should be sufficient. See, here are observation. Active vectors are pi radians apart. Describe a hexagon boundary. Six active sector occupy six vectors of hexagon, 0 vectors at the origin. This what it is? Modulation index. If the output voltages are pure sinusoid, I can write v s is equal to m into e to the power j omega t. m is the modulation index and output voltage of the inverter is proportional to m. Output voltage of the inverter is proportional to m provided we keep the DC link voltage constant. I will repeat. Output voltage of the inverter is proportional to the modulation index m if I keep the DC link voltage constant. What is m in sinusoidal PWM technique? m in sinusoidal PWM technique is magnitude of the reference sinusoid or the magnitude of the modulating wave divided by the magnitude of the triangular wave. If I want to have a constant 230 volt 50 hertz supply, I need to keep the magnitude of the sine wave constant. Also, I need to keep the magnitude of the triangular wave constant. If I want to have a variable output voltage, I need to change m or I need to change the modulating the magnitude of the modulating wave. We do not do anything to the magnitude of the carrier wave. It is maintained constant. So, m is output frequency. It is 50 hertz in radians per second. Of course, locus is a circle. So, ideal trajectory of V s should be a circle and should rotate at uniform angular velocity should be achieved by a PWM process. So, in other words, see what I am saying. I may be saying a sine wave. How do I digitize the sine wave? Or if I digitize, how does it look like? There are large number of steps. That is nothing but a digitizing a sine wave. What happens? Suppose, if I sample this wave form and I keep it hold it for some time, what is known as the T s time. If I observe in Oslo scope, how does it look like? Wave form is sampled at a particular value, instantaneous value and it is held constant for whatever the T s time. T s time is known as the sampling time. It looks something like this. So, during T s reference vector, T s reference vector is assumed to remain stationary at that particular time, particular point. In other words, the vector is rotating at a uniform frequency or radians per second. In the sampling time, that vector is supposed to remain constant. As time increases, the position of the vector changes. But, in that particular time, sampling time, it remains stationary. If the vector is remain stationary at a particular point, at that particular point, I will find out what is the x axis value of that vector and what is the y axis value of that vector. See, for example, what I will do is, this is a trajectory of the output voltage wave form which is a circle provided the voltage is a sinusoid. For T s time, T s is a sampling time, I will assume that this vector remains stationary at theta. After T s, this vector will rotate by another angle delta theta. This is after time delta t. Then again, it will remain stationary for some time. So, if I know theta and of course, this is nothing but a modulation index itself, I can get the x axis component and I can get the y axis component. I can get V x and V pi. So, what will I do after knowing V x and V y? I will go back to the inverter. I will repeat. If I assume that now I have three phase sinusoids, those three phase sinusoids whose magnitude is proportional to m, they rotate at constant omega. At a particular time, I know theta and this vector is stationary for time T s, what is one as a sampling time. So, since I know theta and if I know the modulation index, I can get V x and V y. Now, can I get this V x and V y by using the inverter switches? The question is, can I get V x and V y by using the inverter switches that is all? Assume that we are in the sector 1, sector 1 in the first 60 degrees and the vector, I want the output voltage whose value is proportional to V s star and is making angle theta. I told you that I can resolve along x and y axis, V x and V i, V x and V y are thus their values. Now, I have to get this V x and V y using this inverter switches. See here, when I, when we, when the applied vector, when all the three switches are off, we are at the origin. The moment phase A be moved here, this vector has only x axis component. There is no y axis component. So, I will apply this vector for some seconds, T 1. Do not ask me, what is the relationship between T 1 and T s? After some time, after T 1 seconds, I will apply V 2. I will turn on B phase. The vector rotates by 60 degrees and this vector will be there for T 2 seconds. See, when I apply V 2, it has a x axis component and a y axis component. What is the x axis component? V 2 cos 60 and y axis component is V 2 sin 60. So, what is the total x axis component? The total x axis component here is, or the so called volt second is V 1 into T 1. Similar plus V 2 into T 2 into cos 60. It is a net x axis component of the volt second. Y volt second, I will tell you. What is the y axis component? It is 0. y axis component of V 2 is V 2 into T 2 into sin 60. So, the total x axis component here is V 2 into cos 60. So, volt second balance is, what is the volt second balance? Volt second balance is nothing but the flux. Volt second is nothing but the flux. So, sigma x axis of R H is V dc into T 1. Why V dc? It is V 1. V 1 is V dc is again V 2 cos 60 into T 2. Sigma x of L H s, what is sigma L of H s? It is V x itself. See here, V s into T s because voltage vector V s is supposed to remain there for T s seconds and x axis component is V s into T s into cos theta. V s into T s into cos theta is the x axis component. That should be equal to V 1 T 1 into cos 0 V 2 into T 2 into cos 60. That is the x axis component of these two vectors. Similarly, y axis component of the space vector is V s into T s into sin theta. That should be equal to V 2 into T 2 into cos, sorry, into sin 60. That is all. So, simplify it. You will get a very simple expression, very simple expression, very simple expression. This is T 1. T 2 is C T 2. T 2 is given by T c. T c is nothing but T s by 2. A is nothing but the modulation index. Sin 60 is known. Sin theta, sin theta, theta is the position of the space vector, position of the space vector, sin theta. At some T, it was at this position. After some time, it moves by an angle delta theta. So, I can get an expression for T 2, time for which T 2 is on, sorry, time for which this vector is applied, time for which this applied, applied is T 2, time for which V 1 is applied is T 1. So, expression for T 1 and T 2 are here. They are very simple expressions. I will stop here.