 Let me start in the so far we discussed the voltage source inverter. Power is coming from the solar side to the DC link and that power I can connect it to the grid, sorry that power I can feed it to the grid. If there is a spare capacity, this inverter can supply Q to the grid as well. It can supply both P as well as Q. Now, how to determine P? Now, let us see instead of doing the inverter theory, I will briefly discuss, I will briefly derive the expression for power. The vector representation of instantaneous three-phase quantities, what we again the definition of space vector itself, three current vectors I can generate as the one vector, what is known as a space vector. This is these are the in-phase component and this is the quadrature component. Similarly, therefore I alpha I beta, this is I 0, do not worry why exactly why 0, that is because if I know, see I can get I alpha I beta or the d axis and q axis components. See alpha d axis and q axis component or x axis and y axis components I can get from I a, I b, I c. See here, if I know I a, I b, I c I can get I alpha I beta. Now, if I know I alpha I beta, how can I get I a, I b, I c? Looks like this is a non-singular matrix. So, what do I need to do? I need to add a third row, need to have a square matrix. So, that is the reason we have added this third row, do not get into finer details. So, if I know I a, I b, I c I can get I alpha I beta. Similarly, voltages if I know V a, V b, V c I can get I can get e alpha and e beta or e x, e x and e y or e d or e q. So, these are the d and q axis components of current. These are d and q components of voltage. What is the expression for power in terms of, what is the expression for power in terms of this d and q components? See, we know that expression for instantaneous power is given by V a, I a plus V b, I b plus V c, I c. So, I will substitute for V a, I a, V b, I b, V c, I c in terms of alpha beta components and these alpha beta components I can get from these, the previous two matrices. And I will simplify expression for power, instantaneous power is given by 3 by 2 into e alpha, I alpha plus e beta, I beta. X axis component of voltage, x axis component of current, y axis component of voltage, y axis component of current. So, in vector form this is nothing but the dot product itself e alpha, alpha plus e beta, I beta. Similarly, I can derive the expression for q, the reactive power, the reactive power, that I can, what is reactive power? See, V s is the space vector, voltage space vector, I s is the current space vector. The expression for power is V s, I s, cos of the angle between, cos of the angle between V s and I s and reactive power is V s, I s into sin of the angle between of V s and I s, that is the expression for power and reactive power. And gone simplifying, I can get a very simple expression V s, I s into cos phi, that is what it is. And similarly, reactive power V s, I s into sin phi in terms of alpha beta components is e alpha, I beta minus e beta, I alpha. This is the expression for q, this is the expression for q, where e alpha is the d axis component of the voltage, I beta is the y axis component of the current, e beta is the y axis component of the voltage, I alpha is the x axis component of the current. So, it is possible to determine the q supplied by the inverter using these expressions, using this expression and p using this expression V s, I s into cos phi, little bit of simple theory. Now, change of reference frame, why are we doing this? There are few very good advantages, I will tell you the advantages. See, I have had 3 phase voltages here, at 3 phase voltages 50 hertz, what did we do? We converted them into alpha beta components, x axis and y axis components. If the frequency of I a, I b, I c is 50 hertz, the frequency of I alpha and I beta is also 50 hertz, if we just converted I a half of I b half of I c root 3 by 2 I b root 3 by 2 and I c. So, I am just multiplying by a constant term. So, there is no change in the frequency, frequency will remain the same. So, frequency of I a, I b, I c if it is 50 hertz, I alpha I beta also 50 hertz. If their frequency change in a very narrow band, that is what happens in the grid, I alpha I beta frequency also will change accordingly. Now, what I will do is, I will transform, so what did I do? 3 phase stationary frame, with respect to some stationary, remember it is 50 hertz. What I will do is, I will transfer this 3 phase to 2 phase alpha beta and from this 3 phase, 2 phase alpha beta frame 2, what is known as a frame which is rotating at, which is rotating at the synchronous speed. This is rotating at synchronous speed. At that time, what happens? This 50 hertz frame will appear as if it is stationary. So, I had 50 hertz current, 3 phase current, I converted into 2 phase 50 hertz current and if I transform, if I transform them to a rotating frame, which is rotating, a frame which is rotating at omega s, I get d c currents. So, this is why am I doing this? I will tell you. Now, first let us worry about how to convert or how to convert this 2 phase current, which are referred to the stationary frame to a frame which is rotating at omega s. See here, d s and q s are stationary or this I can call as x and y or alpha beta d s and d s, q s are stationary. Whereas, d r and q r, they are rotating at omega s at the synchronous speed, continuously changing. In other words, theta s is changing with respect to time. So, in this to this reference frame, stationary reference frame, frequency of current is 50 hertz. Since this frame itself is rotating at synchronous speed, the currents will appear as if they are d c current, d c currents. How do I transform? It is very simple. I will resolve these 2 components along d s and q s. Now, see here, what is d r component? The d r component is d s into d s into cos theta s. Similarly, similarly q s into sin theta s, the d x is d r component. What is the q r component? The quadrature, I have to resolve this d r, this d s and q s vector along d r and q r. If it is confusing, what you do is, you resolve d r and q r along d s and q s and invert it. That is all. Simple. So, you will get this matrix cos theta s sin theta s minus sin theta s into cos theta s. So, if I know d s and q s, if I know d s and q s, I can convert into d r and q r if I know theta s. If I know how to get this theta s, we will find out. That is the crux of the problem. So, I will repeat. If I know, I can d s and q s are known. When I know I a, I b, I c, when I know I a, I b, I c, I can get d s and q s or I can get alpha s and beta s. Nothing is just a constant multiplication 1 minus of minus of root 3 by 2 and root minus root 3 by 2. I can get d s and q s. Now, if I know theta s, assume that we know theta s, I can get d r and q r. The d r and q r, nothing but a frame which is rotating at synchronous p or synchronously rotating reference frame, synchronously rotating reference frame. These currents will become d c. Now, all this time we found that we have discussed that inverter can supply active power, it can supply reactive power. Assume that inverter has to supply active power to the grid at unity power factor. The k v rating should be equal to kilowatt. You have chosen, you have a 10 kilo k b i inverter and power that is coming from the solar cell is also approximately 10 kilowatt. So, in other words, your inverter has to operate at unity power factor or you want to operate the inverter at unity power factor. V s and I s should be in phase. This V s and I s should be in phase. How do I make that angle phi s 0? This phi 0. How do I make this phi 0? I want to feed the power to the grid at unity power factor. Now, please, so many times I said assume, assume. All these assumptions that we made, we will try to find out. In the sense, are they valid assumptions or are this assuming? I told you that previous case here theta s is known. Assume that theta s is known. So, in other words, at any given instant d r and q r are known. This theta s is continuously changing. That frame is continuously rotating. So, what I will make is I will make this V s vector, V s vector I will align it along d r. What we will do is this V s vector is aligned along d r. In other words, V s vector does not have a q axis component. It has only a d r. For any power factor phi I s is making an angle, this angle with V s. I want to feed or the criteria here is power fed to the grid should be at unity power factor. In other words, this angle should be 0. What do I need to make? I need to make this q component 0. I need to make this q component 0. If I make q component 0, V s and I s are in phase. In other words, we are feeding power to the grid at unity power factor. In addition, if there is a spare capacity and if you want to supply q as well, then I have to have a finite angle between V s and I s. How to determine this finite angle? So, what are the various steps that are required now? V s and I s. So, I have to transform these space vectors to the synchronous irrotating reference frame. How do I to do that before? I have to convert it to a two phase stationary currents and these two stationary two phase stationary currents I can get from a three phase currents just by multiplying by constant terms. So, here is a complete procedure a complete approach. Transform all the variables to the synchronous irrotating reference frame. Fundamental components of V and I will become d c. See, when I transform them to the synchronous irrotating reference frame, the 50 hertz component, they become d c. If there are higher frequency components, if there are higher frequency components, they appear as if, how do they appear? It looks something like this. See, if your harmonic spectrum has a 50 hertz component and I say 11th, 13th, any frequency component no problem. If the reference frame is rotating at 50 hertz, this component will appear as if it is a d c and this 11th frequency component and 13th frequency component will appear as if it is a high frequency component on the d c. This is rotating there at 11th. So, 50 has become stationary. So, 11th may appear as if it is a appears as 500 hertz, 13 may appear as if it is a 600 hertz. So, what I will do? I will give this to a small low pass filter. I will give it a small low pass filter and I will get a constant d c. I will get a constant d c and this d c value corresponds to this 50 hertz component. Now, the question is how to get this value of theta s or in other words, how do I generate the reference? This is the question. This grid frequency does change for a very narrow band. How do I generate this reference waveform? See, even if I use a sinusoidal PWM technique, I have to generate 3 sinusoids whose frequency is same as that of the supply. We are discussing now how to synchronize the inverter to the grid. Now, I said we have to use the pulse width modulation technique. It could be a sinusoidal PWM technique or it could be a space-wetter PWM technique. Sinusoidal PWM technique, I need to generate 3 sinusoids. The output voltage of the inverter is proportional to proportional to m into v d c. I will assume that v d c remains constant. Therefore, output voltage is proportional to m. Now, see the problem. Power transferred to the grid is v 1 v 2 divided by x into sin delta. v 1 is the grid voltage. v 2 is the output voltage of the inverter. v 1 frequency does change over a very narrow band. As this frequency changes, this frequency also should change. So, in other words, both of them should be in synchronism plus this delta. What is this delta? Delta is the angle between v 1 and v 2. In other words, if v 2 is supplying power, I am assuming that v 2 is supplying power, v 2 should lead v 1 by an angle delta. I can measure this angle only when these two frequencies are same. Frequency of these two sinusoidal is the same. I just cannot generate 50 at sinusoidal because frequency of the grid also changes over a very narrow band. So, as this frequency changes, I need to change the frequency as well of v 2 and v 2 frequency is nothing but the frequency of the modulating wave in a sinusoidal p-dolm technique or the frequency of the space vector v s. How do I generate? How do I generate the modulating wave for a sinusoidal p-dolm technique? There are two methods. One is the hardware method and the software method. Hardware method is basically outdated, but those who are good in hardware they can do, but then there is another software method as well. See here, frequency over a narrow band. How will you do it? P L L through hardware, phase locked loop through hardware. What I will do is, this is a source. What I will do is, I have three sinusoid from the source using a zero crossing detector. A zero crossing detector, I will convert this is a sinusoid voltage of the source to a square wave. So, as if the frequency of the sinusoid is changing, the frequency of the square wave also will change. In addition, I will digitize a sine wave into n number of parts. n could be anything, either 2 1's, 2 5 6's or 5 1 2 or 1 0 2 4 or 2 0 4 8 k whatever, 2 k 3 k whatever. If it is 10 0 2 4 parts, in other words what I am saying is this entire period is divided into 2 4 parts. So, the resolution is going to be 1 0 into 2 4 parts. I will divide them, sorry I am in the white board is it not. I will divide this entire period into 1 0 2 4 parts and I will store it in EEPROM. I will store it in EEPROM. In this EEPROM, there are 1 0 2 4 steps of sine wave. Now, I have to address that EEPROM. How do I address? To address, I need to have a counter. Counter requires a clock. How do I generate this clock frequency? So, if I address this 1 0 2 4 parts, if I address this 1 0 2 4 parts in one cycle, I can get a sinusoid, I can get a sinusoid whose frequency is same as that of the supply voltage. See, I have digitized a sine wave and stored in an EEPROM. Output of the EEPROM is connected to a DAC. This EEPROM has to be addressed. That I will use through a counter. Here I am using a 10 bit counter because it is a 1 0 2 4 1 k EEPROM. If I divide this EEPROM into 2 56 parts, then I need an 8 bit counter. Counter requires a clock. This I will generate using a PLL. What is a PLL? At any given time, these 2 waveforms are in synchronization. Frequency of this is the same. What I will convert this is a sine wave using a step down transformer. Convert it to a square wave. I get a square wave here. The phase comparator, a voltage controlled oscillator and a divide by n counter. At any given time, these 2 waveforms are in synchronism. I am using PLL as a multiplier. I am using a PLL as a multiplier. I will divide this parts into 1 0 2 4. This entire 360 into 1 0 2 4 parts. That I can do it using a PLL. PLL as a multiplier. Here at this point, I have 50 or no f 1 into 50. Why f 1? f 1 is the frequency of the supply. That waveform is divided by n. Here I have f itself. I can generate a sinusoid. I can generate a sinusoid whose frequency is same as that of the supply. I will use this as a modulating wave. For power transfer, angle between this sinusoid and this sinusoid should be delta. Say for example, angle between these 2 sinusoid should be 5 degrees. What do I do? At this 0 crossing, I need to address a location which is equal to sine 5, sine 5 degrees. I will repeat. Angle between these 2 should be 5 degrees. I know the 0 crossing of this. At the 0 crossing, I need to address a from here whose location, to a location wherein sine 5 degrees address and I will go on increasing from there. Therefore, I have generated a sinusoid which leads the source voltage by an angle 5 degrees. That 5 angles happen to be delta. If delta is around 10 degrees, yes, I need to address at the 0 crossing to a value which has in which I have stored sine 10 degrees. So, this is how I generate 3 modulating waves. This modulating wave, the frequency of this modulating wave is the same as that of the supply frequency. As the frequency changes, the frequency of the sinusoid also changes and depending upon the delta, who decides delta, I will tell you. At the 0 crossing, depending upon the delta, at the 0 crossing, I need to address that location in a NEPROM. Therefore, as of now, we have found a way to generate the modulating waves for a sinusoidal p-dol-M technique whose frequency is same as that of the supply. If the frequency of the supply changes, modulating wave form also will change and depending upon the delta, I can generate this sinusoidal p-dol-M. So, the answer to Baramati is, I think as of now, we have tried to generate, we have managed to generate this source VAB whose frequency is same as that of source voltage V1. How did we achieve? That we achieved by generating a modulating wave whose frequency is determined by the supply frequency. The frequency of V2 is determined by the frequency of the modulating wave and we have generated the modulating wave using the 0 crossing output of the source voltage and we have divided this part into 1024 parts using a PLL phase locked loop. At any given time, at any given time, the frequency of those both the signals is the same. So, we have generated three phase sinusoids whose frequency is same as that of the modulating wave and if I know delta, that I can incorporate here. So, that should answer the so-called, so that should answer the question which is being asked repeatedly as to how to connect this inverter to the grid, software approach. What is known as a harmonic oscillator? I will quickly do this harmonic oscillator. Very simple. I do not know whether I will be able to, so this is the hardware method of generating using a software method. Will I be able to do it today? I do not know, but the philosophy is the same. Harmonic oscillator is the x dot is equal to omega into y and y dot is equal to minus omega into x, where omega is the instantaneous frequency of the harmonic oscillator. Under this condition, what is x and y? x is sin omega t and y is cos omega t. x dot is equal to y and y dot is equal to minus omega x. Now, I have two differential equation, I will solve them. What is this? So, if x is equal to sin omega t, what is x dot? x dot is equal to omega into cos omega t. Cos omega t, it is happen to be y. So, here is that x dot is equal to omega into y. y is cos omega t. See, what is y? y dot, y dot is equal to omega into minus sin omega t, but sin omega t happen to be x of t. So, therefore y dot is equal to minus omega into x. See, here is an implementation. y dot integrated, I will get y. y into omega is x dot. x dot integrated, you will get x of t. This is nothing but x dot, n plus 1 over n omega t. So, now comes the control loop. Now, I am going to how to determine what, how much p to supply and how much q to supply. I will go back to this block diagram. The power is coming from the solar side, from the solar panel. MPPT is the only one knows the power. It draws the maximum power, dumps it to the DC link and the inverter has to supply the power to the grid. In other words, inverter controller, I will repeat, it is the inverter controller has to choose the suitable value of delta. I am assuming that both modulation index also I am keeping constant and VDC also has to be regulated. If V1 and V2 remain constant, power output depends on, power output is just proportional to sin delta, where delta is the angle between V1 and V2. So, if I know delta, see if I know delta, I can generate the modulating wave whose frequency is same as that of the supply and the phase angle difference between V1 and V2 is delta. See, if I know delta, I can generate, I can incorporate this delta in my control circuit and that control circuit will generate a sinusoid which is phase displaced by that angle with respect to the supply. But the question is now how to determine delta? Who will tell me or how the controller will know to choose the exact value of delta? Now, assume that controller does not know to choose the appropriate value of delta, what will happen? Either delta could be higher than the required or it could be less than the required. If it is higher than required, inverter is supplying, inverter is supplying, trying to supply more power than what it is coming from the solar panel. I will repeat, the controller has chosen a wrong value of delta and that delta happens to be higher than the required. So, under that condition, inverter is trying to supply more power than what it is coming from or the power is coming from the DC, from the solar side. Under that condition, what happens? There is, we have one more storage stage that capacitor. It is trying to supply power because at any given time, there should be a power balance. If the output power is higher than the input power, in other words, inverter is trying to supply more power than what it is, what the power coming from the DC side. Capacitor starts supplying power. Under that condition, capacitor voltage starts falling. Instead, if the power supplied is less than the power output from the solar, the remaining power has to go to the capacitor. If the power supplied by the inverter is less than power output from the solar panel, the difference in these two powers has to go to the capacitor. Under that condition, capacitor voltage starts increasing. If there is a perfect balance between input and output, capacitor voltage will remain constant. So, what I will do is, I am not bothered about what is the output of the maximum power tracker. I will try to regulate the DC link voltage that is all. If I am able to do that, I have achieved the perfect power balance. In other words, the inverter is supplying whatever the power that is coming from the DC, from the solar panel to the grid. Of course, I am neglecting the losses or if I take the losses into account, power transferred to the grid is power coming from the solar minus the losses that are taking place. So, if I am able to regulate the DC link voltage, if I am able to regulate the DC link voltage, I have achieved the perfect power balance and I have achieved or the controller has chosen the right value of delta. So, if I know the right value of delta using the hardware PLL, using the hardware PLL, we are able to generate a modulating wave which is displaced by an angle delta and whose frequency is same as that of the supply. As the degree of frequency changes, supply frequency also will change. So, as of now, the two burning questions in the sense repeatedly being asked, how do I synchronize? Yes, using a hardware PLL, we are able to do it. How to use a software PLL, we will see. I will not be able to cover today, maybe tomorrow. How to supply P? That I have answered. Now, assuming that there is a spare capacity and inverter can supply Q, how to do that? Over at night, there is no insulation at all. There is no active power input from the solar. How do I do that? I will take the second case first and then we will discuss both P and Q together. At night, there is no active power input. There is no active power input. We know that inverter can supply any Q. Any Q as long as it is less than the required. To do that, this inverter has to draw some power from the grid. Now, the source is supplying a delta P, a small amount to the inverter. Now, the modulating wave has to lag the source voltage by an angle delta, because the direction of power flow has changed. In other words, now V1 is here, V2 and this delta. Delta is very small just to account for losses. What is the Q to be supplied by the inverter? Who will tell what should be the Q? There are two ways. One is if it is just supplying Q to the load, I will assume that there is a separate controller, which will tell, which informs this controller of the inverter to supply the required amount. Or if it has to supply Q to the grid, how to determine the exact value, how to determine that value of that Q? By the way, how to determine the required value of Q to the grid? How to determine the Q required by the grid? How do I determine? In the previous case, we regulated the DC link voltage. If I am able to regulate the DC link voltage, I have achieved perfect balance and I am transferring the required power to the grid. Now, how to supply the required power, required Q to the grid? What I will do is, now I will take the grid voltage or the voltage at the coupling point. I will take the voltage at the coupling point and take it as a reference. This is the grid voltage as the reference. This is the actual grid voltage plus minus delta. Sorry, this is the error. I will give it to a PI regulator. What is the output of the PI regulator? What it will be? It is the Q required, Q required. Output of the PI regulator is going to be the Q regulator. Why? That is because at steady state error is 0. At steady state error is 0 of the PI regulator because I am using a PI regulator. If error is 0, it implies that the actual grid voltage is actual grid voltage is equal to the reference. But if there is an error, there is a mismatch between the required value and the actual value. If this grid voltage is higher than the reference value, Q supplied by the inverter is more than the required. If the grid voltage is less than the required, the Q supplied by the inverter is less than is required. Here I am making assumption that inverter has the capacity supply the entire Q. So, using this loop, I will take grid voltage as the reference. Sorry, the required grid voltage to be maintained at a particular value. This is the reference voltage. Error, PI regulator, this will give Q required by the grid. In the previous case, I will keep the DC link sum reference, actual DC link plus minus error, a PI regulator and delta. This is the DC link voltage that has to be maintained at a particular reference value. Actual DC link that is measured. Error is processed by a PI regulator. Output of the PI regulator will determine delta. Why? Why it is delta? That is because at steady state, this error has to be 0 and that can happen only when there is a perfect power balance. In other words, error is 0 and error implies actual DC link voltage is equal to the reference. So, the inverter is supplying, inverter is supplying the correct amount of power to the grid. Whatever that is coming from the solar, it is transferring to the grid. So, I will stop here. I will take a few questions. The software PLL, I will discuss it in my next lecture. That is tomorrow and from tomorrow onwards, I will also start the modification that we need to make in already discussed DC to DC converters and the inverters. Any questions? Baramati, are you, is it clear how to synchronize the inverter to the grid? If it is not clear, now I will explain to you again. Feel free. There are so many questions being asked. How to synchronize, how to, yesterday there was a question from COEP. How to, how the inverter can supply both P and Q? I have answered, if it is, if it is not clear, I will explain to you again. Yeah, Baramati back. Yeah. Sir, question is once again. Space vector PWM. Sir, in case of that space vector PWM, if we open all switches for the required dead time, will it do? In the sense, I do not know. You can either apply 0, 0, 0 or 1, 1, 1 for 0 voltage vector. But another condition is at any given time, only one switch should be turned on or off. This see, by the way, this is a very basic space vector PWM technique. There are a lot of modifications that have been done. I am discussing the very basic PWM technique. It says at any given time, only one switch should be turned on or off. Now, at t is equal to 0, you have to start. So, you have to start from the origin. All of them are off. Then I have to apply 0, 0, 1, then 1, 1, 0, 1, 1 and come back. So, you may not be able to start directly from 1, 1, 1 and come here. So, it all depends on your previous state. You can either make it all 0 or make it all 1 provided. You need to know the previous state. If the previous state is 1, 1, 0, 1, 1, then you can do 1, 1, 1. Previously, see, at any given time, you are at the origin, it could be either 0, 0, 0 or 1, 1, 1. Now, if you are 0, 0, 0, 0, do not you think you have to start from here, go there and come back. So, opening all the switches will not do. No, no, no. I will not say that. How can I possibly say that opening all this will not do? What is your initial state? Have you turned? See, have you turned off all the devices? So, have you turned off all the upper devices or have you turned on all the upper devices? No, that you need to know. What is your control algorithm? Assuming that all of them are off, you have 0, 0, 0. Next logical step is 0, 0, 1. If all of them are 1, then I think you have to go for 0, 0, 0, 0, 1, 1 and come back. I told you T 1 can be applied any time, T 2 can be applied any time, but then total sum of and they can be divided as well. So, if you have disabled the inverter or in the initial state is 1, 1, 1, all of them you have turned on, you have to come from 0, 1, 1 and 0, 0, 1. It is not only the whether to apply 0, 0, 1, 1, 1, you need to know the conducting state of the, you need to know the previous state as well. Is that answers your question? What exactly is exactly your doubt? Why are you, what is your main doubt? Sir, in case of implementing the dead band. Implementing the dead band. Dead band has nothing to do with the PWM technique. Dead band has nothing to do with the PWM technique. It is a, that you require because the voltage source inverter. Dead band, if it is IGBT, you provide around 3 to 4 microseconds that you provide. You provide 3 to 4 microseconds of dead time. Do not bother about whether you are in 0, 0, 0 or 1, 1, 1. You have a dead band of 3 to 4 microseconds. You want to make it, you want to complete, eliminate dead band. I would not suggest that to do. Please do not do that. No, I would not be eliminating. I want to implement dead band. So, that is why I am asking whether it is possible. How are you, how are you implementing the dead time in a hardware? How are you implementing this dead time through hardware? Software, DSP, processor. Then, yes, you have to, you may have to add up. In the sense, is there a control? Why are you increasing the software overheads? See the complexity that is involved. Why cannot you, of course, I am not saying that it is not possible to incorporate these 3 microseconds, whether the DSP has the time to do all this. You have to incorporate 3 microseconds, whether the DSP has time or not. I do not want to get into those issues either through hardware or software. You have to, you have to, you have to and I do not think it is depend on 0, 0, 0 or 1, 1, 1. It is a switching instant of switches belong to one leg and I am not able to understand exact point. You send me an email, I will try to answer that. One second, how to choose the gain? No, no, no. How to choose the gain of the controller? You have to do the system modeling, simulate and you have to choose the gain. The question is how to choose the gains of the PI controller? You have to do the system modeling, then you have to get the transfer function. Now, you have to choose, you can choose the values of gain. Once you have the transfer function, yes, by doing the stability studies, you can choose. For that, you need to do the modeling and simulation. Apply low possibility to alumina harmonics for high power rating. Will it affect our efficiency or not? A, B, B, C, what is the question? I am not able to understand. What efficiency are you talking about? Sir, in previous section, we are seeing ability to eliminate the harmonics. We are applying a low pass filter. It will eliminate our harmonics for 500 hertz and 600 hertz frequency. But in solar panel, we are getting only low power. So, for that low power device, if we are applying a filter, it will take its own reactive losses. So, then how can we manage this? No, no, please. Now, you are totally, which low pass filter you are talking about? No, you misunderstood. I told you, the question to you is how to get this synchronous rotating reference frame. Once you get the synchronous rotating reference frame, I told you, all the components of, all the components or the fundamental component will become DC, rest will appear as high frequency AC. If I want to see, I have to generate a modulating wave or whatever reference wave, which is a sinusoid. I need to generate three pure sinusoids. How do I generate three pure sinusoids? How do I generate three pure sinusoids? If your waveform has harmonics, how do I generate? If the waveform has harmonics, how do I generate a pure sinusoid? If your waveform has harmonics, how do I generate pure sinusoid? Assume that grid… So, it is not possible if it contains harmonics. If has harmonics, fundamental component will appear as DC, all the harmonic components will appear high frequency. Now, this DC component corresponds to the fundamental frequency. I want only the fundamental component. I want to know only the fundamental component. How do I do? I have to put a low pass filter in the control circuit, not in the power circuit, in the control circuit, not in the power circuit, not in the power circuit. In the control circuit, I have to put a low pass filter. Efficiency, what efficiency? In the sense, losses of this resistor, this may be, this I am doing in a control circuit, not in power circuit. Sir, one more questions. Sir, in DC to DC converter, the voltage gets boosted up and the current gets dropped. There is a decrease in current. So, how do we deal that current when we are connecting it to an inverter, sir? Ma'am, it is a power balance. Power balance, do not confuse about voltage current. There is a solar panel, there is a solar panel. You have boosted, there is some voltage, but there is some current coming. So, energy is being stored in the capacitor. It has to go to the inverter and have to feed it to the grid. It is a power that should be balanced. P in should be equal to P out, assuming losses or neglect in the losses, neglect in the losses. How do I achieve? If I am controlling this voltage, see some reference, say 800 volts I want to keep. So, 800 volts is the reference and this is the actual DC link voltage. I will measure this. This is the actual plus, minus, error, P i, this is delta. At steady state, error is 0. So, at steady state, if error is, that can happen only when V actually is regulated at 800. So, under that condition, whatever the P, so capacitor is not supplying any power, so whatever the power that is coming is going out. So, I have achieved the power balance. I am not saying that converter is, if, so power balance is P out is nothing but V 1, V 2 divided by X into sin delta. I have chosen the right value of delta. If there is a mismatch of something, that will reflect in the actual DC link voltage. If there is a mismatch of power, DC link voltage has to change. If it is not changing, there is a power balance. Now, why worry when the voltage is high, currents are low, when the currents are low, when the currents is high, voltage is low? You talk about that. Is that okay? What are the parameters to be considered while selection of inductor in the bidirectional converter? One parameter is, one parameter is the harmonic current generated by the inverter. If the inverter, if the grid voltage is assumed to be sinusoidal, we cannot generate, we cannot generate an ideal sine wave from the inverter. We cannot generate. The inverter has its, the output voltage wave form of the inverter has its own harmonic spectrum. That is given by the switching frequency. This is fundamental and depending upon the p-dolium strategy, you will have a predominant harmonic. This is the equivalent circuit for the fundamental frequency. Now, I have to find out what is the current supplied to the grid by this harmonic. What is the harmonic equivalent circuit? It is, there is no harmonic voltage in the grid. So, it is 0. So, this is the magnitude of the harmonic. This is the impedance offered by this inductor for this harmonic. So, I n is equal to V n divided by n into omega 1 into x, sorry, into l. So, finally, you know what should be the THD that is given by the standard. So, one of the ways is to determine. So, that will determine your value of l. If your switching frequency is low, depending upon the p-dolium strategies, this magnitude is determined, magnitude gets fixed. It is approximately DC link voltage itself. It depends on a p-dolium strategy. Depending upon the p-dolium strategy, you will know the frequency of the predominant harmonic. Therefore, you know n as well as V n. The standards will tell you what should be the total THD. So, once I know THD, the various harmonic components also will be determined and therefore, l. That is one of the ways. Other way, when you do the current control, there are other methods also. Other parameters also will determine the value of, will affect the value of l that I will discuss later. Maybe, I think time is running out.