private property it's you welcome it's good to see you again we are here in the heart of Brineston once again and this is Devonshire Road one of the most exclusive roads in a very very exclusive area a stone throw away from the four-way CBD a nice up-and-coming CBD as well as sands and CBD so you know you're not gonna be sitting in traffic for too long but this home is something special about it it is bringing in the bush felt to Brineston so follow me in because I think it is time for me to show you this house so as we walk in to this property the first thing you see is this double wide entrance hall with this beautiful bamboo flooring now this house is 860 meters squared inside was 2,622 meters squared of ground space there is tons and tons of space here and as you can see walking through this beautiful home this home has been made to bring that game lodge vibe into Brineston and into this urban area as you can see on my right is this beautiful river stone feature wall and if we carry on just on my right of the year there is a mahogany fireplace that brings so much warmth into this house again double vaulted ceilings with this thatch mmm it smells like I'm in the Kruger now this house is retailing for just about 7 million rent and we're gonna talk more about how you can purchase this house very very shortly follow me and let me show you as this is the first lounge feet let me show you the second one so as you walk off this sitting room as soon as you come into the house you've come now to a smaller although nothing in this house is small by any standards but a smaller and more personal sitting room this could be a cigar room this could be your TV room this could be really anything you want of it and the first thing you've noticed these floor to halfway ceiling curved windows behind me and it's another architectural design throughout the house where they do these curved windows it gives you such a nice 180 degree view of your ground and in this case that beautiful beautiful blue pool but another thing that this lounge does is this lounge opens up into the first of the two dining areas a little bit of a more formal dining area and what you'll do is you'll come right up these stairs and you are leading into it immediately and it is placed just off the two sitting rooms and as you can see it's perfectly sized to seat six people again it's got some beautiful windows in here letting in a lot of natural light with this high ceiling and the chandelier that really does encapsulate that bush felt vibe again in the heart of Brynston it really is a place of relaxation more than something that you'd come in and feel like you're in a home it does feel like you're in a holiday yeah but follow me let me take you to the kitchen and as we come around the corner we immediately be led into the kitchen slash dining room nook area on the left beautiful little nook for you to sit down and enjoy breakfast with your family and on the right a entertainers dream kitchen this kitchen as you can see has got sandstone granite toppings as well as a four burner gas stove with two electric gas hubs and the immediately what you'll see is just how open and free-flowing this is where you can sit you and prepare a meal for your family and still be able to talk to everybody sitting in the dining area but another very very cool feature of this kitchen is just how much storage space it has so on my right you can see just how big the area for your fridges are your double door fridges will fit there very very easily and as we come in here ample ample ample storage space but another key feature of this kitchen is is our double oven here D5 and as you can see the Gemini Gourmet range so for all of you budding chefs out there this kitchen is already catering for you now as we move forward you'll see that this hallway over here leads off to a scullery that is where you'll be putting all your appliances your dish washes your tumble dryers your washing machines and they are out of the way you will see ample kitchen space and kitchen cupboard space on your right hand side as well as on your left hand side as well as a lot of space for your different appliances your dish washes your washing machines your tumble dryers they can all fit here and fit away from prying eyes another really cool feature about the scullery is your double kitchen sink so all your dishes after a hearty meal you can throw into the scullery put them away and they're out of sight until you can deal with them a little bit later but we've got a lot more to show you so let's go along off the left of the hallway you'll find yourself in the study the study has floor to ceiling windows letting in a lot of light and air which is exactly what you need during your early mornings and late nights working the study incorporates the same beautiful thatch letting in that aroma as the rest of the house does it also has bookshelves on both sides of it which you can take mighty good use of if you bring it in your novels and here we are at the end of the hallway but we are far from showing you everything that this house has to offer and right now i'm going to be taking you into the master bedroom so let's go take a look and here we are within the master suite and this suite has so many things that really do screen presidential and one of the first is those floor to ceiling windows 180 degree views of the grounds it really is such a presidential field standing here and looking at it it makes me think of the White House another really cool feature of this master suite is the backing take a look at that wallpaper they've really done a cool job of bringing in even though the bed is recessed bringing in that wall ample space for a king-size bed here you can see inside the recess i just want to lie back and kick it off for the rest of the day but that is not all within this master suite and we're going to be showing you the en suite bathroom and just what that has to offer as well let's go and leaving the master suite we have come in to the en suite bathroom to my left you will see the double vanities that should stop a lot of fighting in the morning and on my right you'll see a very luxurious jacuzzi bath as well as a double space shower and if you look very closely in that shower there's beautiful windows right in front of you so that you've got the view of this outside private garden and if you follow me outside you'll see with these super high walls it offers you a lot of privacy this is a little oasis a bubble away from the outside area but it's not the only thing you can do here and if i whip off this jacket i can very much go and enjoy this outdoor shower now we're going to take a look at what the upstairs has to offer but i'm going to need you for that come on let's go as you arrive at the top of the landing you'll be greeted by two different bedrooms one on the left and one on the right the left side bedroom is a guest room and the right side is a feature bedroom so let's take a look at our right hand bedroom throughout the tour i've spoken how this house brings so much of that lodge like living to brineston and nothing encapsulates that more than this bedroom it brings in a feel like you are at a safari with the way that they've incorporated the bed and the bathroom behind it and as you can see just to my left again that 180 degree views of the outside as well as the balcony let's go take a look at that now our left hand side room has you covered with windows and lighting as well as a full bathroom area to your right hand side so if you come with me on this side here you'll see the bathroom area leads all the way across the guest bedroom into a shower a full bath as well as a vanity suite there the bedroom is just in front of me the bed's in front of me it again brings in that lodge living it is so peaceful here it cannot be understated another really cool feature about this room is that window looks onto the garden the grounds here are huge and it's something that all of the architecture within this place takes advantage of and i love that design feature checkmate and let me tell you what else you need to check in this entertainment area the first on my right is a built-in jacuzzi which i'm sure must be tons of fun whenever you're hosting a brine whenever you're having friends over because this area is an entertainer's dream we've got a full on seating area let me just move that baby out the way full on seating area here for your friends we have got a lounge suite there for you want to have a little bit of a party which after covid i'm sure every single person wants to do after lockdown another thing a lot of people are going to want to do is take advantage of this built-in bar it's right there behind us another thing is just behind me you'll see that there is seating that is built in recessed off the wall underneath the windows that you can take advantage of and enjoy while relaxing out here and enjoying the incredible south african sunsets this area also provides spectacular views of the grounds in front of us as well as the pool area and i'm very very keen to show you more of that right now the house is surrounded by expansive grounds following this pathway leads you to an idyllic swing smartly placed away from the main house as you move up the hill you see the feature posts of the house holding up the second-story balcony moving off to the left we have another shaded sitting area the perfect place to enjoy a sunday cup of tea this is also where the main rooms outdoor shower is located some of the features of the house to consider is the massive double garage easily tall enough and wide enough to fit two cars or various size length and height also something to definitely keep in mind is the security measures the owners have taken there's 10 strand electric fence surrounding the property as well as an intercom system and a camera system another thing i'd definitely like to mention is the beautiful water features that you find throughout the house specifically as soon as you walk into it the water feature almost overwhelms the senses again bringing in such a nice holiday out in the bush felt vibe throughout the house so from myself chair de veiros and the whole team at private property we hope that you find your dream home thank you and goodbye come to a six bedroom six bathroom one thousand square meter mega mansion in brineston this is number 42 a saint ordly my name is chair de veiros and i'll be taking you all over this country showing you the most exclusive property south africa has to offer and today it's all about living presidential a perfect marriage of high ceilings and gloss the driveway and motorcade lead you to the front door and a palatial entrance hall through the entrance hall we pass two seating areas on the left and right as well as a cozy reception area and come to a centerpiece broad-iron staircase leading up to the second level one out of the five seating areas greets us directly off of the staircase with beautiful views of the grounds surrounding the house can be enjoyed from the second story balcony the first bedroom we're going to be looking at is located on the top floor sharing those green views off of not one but two personal balconies with an en suite bathroom that houses both a toilet and a shower fit for a king with three incorporated shower heads this bathroom is sure to please here we have the second bathroom also with its own personal balcony third upstairs bedroom is perfectly situated to take advantage of the light even during a rainy day like this and offers stunning views of the santa and skyline off of its rear balcony another en suite bathroom complements the package the fourth bedroom this one also with an en suite bathroom houses a standalone bathtub this room is one of my favorites as on a clear day the views from its massive balcony are absolutely stunning welcome to the master suite these rooms as i'm sure you've noticed are exceptionally large fit for a house of the stature each with its own balcony the master bedroom steps it all up a notch with plenty of glass and sunlight as well as some strategically placed frosted glass allowing in light but still allowing you your privacy the master bathroom continues the theme of excess with a fully enclosed glass shower and two three head shower units this is the epitome of opulence the master bathroom also has a double vanity as well as a freestanding bathtub for those relaxing soaks on a rainy day like this one heading downstairs we were introduced to a chef's dream a closed-off kitchen with ample cupboards windows are plenty and countertop space to cater to even the most demanding of sunday dinners this kitchen also has a built-in bean-to-cup coffee machine as well as in a microwave oven with a state-of-the-art six burner gas stove and electric oven off the side of the kitchen you have a scullery nook that allows you to place appliances and dishes out of the way keeping the main kitchen clean and tidy outside of the kitchen we have a large dining room and lounge area with floor-to-ceiling windows and stackable glass doors this is an incredible area allowing you to feel like you're dining outside regardless of the weather off of the main entrance hall is a guest bathroom as well as some office space taking the main entrance hall stairs down leads you to the ground floor area currently undergoing renovations this area would make for a perfect cigar lounge with a soon-to-be wine cellar neighboring it to the left and soon to be home theater system coming in on the third room at the far end of the ground floor area you'll find the laundry room coming back upstairs we explore the two sitting rooms adjacent to the entrance hall both offering views of the garden area through the floor-to-ceiling windows as well as the guest bedroom located above the sixth car garage we have a second TV room located between the entrance hall and dining room area and leading straight onto the outside veranda this 1500 square meter brineston mansion has a large garden with a pool and covered patio removed from the main house that would make for the perfect area to sit and relax enjoying the views of the house that you've worked very hard for as well as possibly enjoying a tasty meal from the brine this house is surrounded by a 10-strand electric fence and also houses a guard house built into the wall with a window overlooking your outside driveway that driveway is guarded by a 10-foot tall iron gate and an intercom system that connects directly to your cell phone and there you have it folks mega mansion living in the heart of brineston this was very very fun for me and i cannot wait to see you all next time where we show you more luxurious properties but right now i've got a plane to catch goodbye welcome to the home shopper show my name is Chad Rivera and i'll be taking you all around SA showing you some of the most luxurious properties South Africa has to offer and today we are in Pretoria East soon to be South Africa's richest square mile situated right now in mainland Maine and right over there on the 15th floor we have a two bedroom two bathroom apartment going for 5.7 million that i cannot wait to show you so come with me welcome to the 15th floor the coolest and most impressive view you're going to see anywhere in mainland Maine this balcony is one of the two within this apartment and it has space for both a gas bri as well as some storage space behind me where you can even fit an aircon however let's take you to the crowning glory of the apartment and that is the master bedroom come with me and here we are within the master bedroom now the master bedroom has got a lot of space but also a lot of attention to detail has been paid throughout the room the first thing that i really like are these cupboards they've turned a wall that would generally just be a wall of cupboards into a feature wall these cupboards sliding on hinges both top to bottom with a safe in one of them and these attention to details i feel really does turn an apartment that just might be average into a beautifully crafted apartment there's also as i mentioned a lot of space here and a lot of space in bedrooms becomes very important when you start looking at what types of beds you want to put in them this year is a queen bed and there would be no problem fitting in a king with extra length however my favorite feature within this bedroom is the bathroom behind me which i'm sure you can already see so let's go and explore that a great deal of thought has been placed within the conservation of space within this apartment there's a glass wall dominating the shower separating it from the rest of the bathroom and allowing you to enjoy those 180 degree views of Pretoria which is incredible because Pretoria one of the few cities that still has remained as green as it is and as you can see behind me those yields rolling in the background must make for incredible showers as well as baths as you can see on the right hand side of me there's a floating bathtub hunscrow taps within all of the apartment and this nice floating basin with a very very big mirror and floating toilets really does lend an air or sophistication towards an otherwise ordinary apartment coming out of a master bedroom you'll find yourself in the kitchen this kitchen is dominated by marble countertops throughout it also has three very very thoughtful features that i absolutely love the first of those features is the cupboard that houses your washer dryer the second is the cupboard that houses your microwave my third favorite feature of this kitchen is the hunscrow tap that has an inbuilt shower head how easy would it be to clean up your dishes with this both of the sinks are melee as well as both of the appliances behind me and the oven over here moving forward we've also got a melee stove top which is touch sensitive so it's very very easy to keep clean not needing to go through the knobs the tiles throughout this unique apartment also act as a feature splashback behind the stove the apartment also has space for a double door fridge over here at the end of the kitchen we've got an inbuilt wood top breakfast nook you can have breakfast here while enjoying the incredible views outside and another very very cool feature very thoughtful from the developers is this over here you've got these cupboards here that i'm sure many of you would think can just be storage space but not within this apartment we've got a coffee nook that's built in over here plant your storage space at the bottom with everything a caffeine lover needs let's close this up we're not in use and look at the uninterrupted facade going into the living room area a very very cool point of this is everything you see here today everything you see in this video is all included with this apartment other than unfortunately the paintings but this living room area is essentially a place for you to enjoy the 180 degree views you get from this balcony or Pretoria the light and the air another very very cool feature within both of the bedrooms and this living room are right at the top of the sliding doors there are hatch windows that you can open up and letting a little bit more fresh air but that is not all that you get with this apartment so let's go take a look at the second bedroom and bathroom here we are within the second bedroom and ensuite bathroom and you'll see the theme of space conservation continues it's so easy within apartments of this size to feel small and that is never the case here within this bathroom there's a floor to ceiling glass enclosed shower as well as floating toilet and a massive double volume vanity moving in to the bedroom again plenty of light plenty of air it's shared across the apartment and it really is so easy to enjoy it here but there's no better place to enjoy it than your personal balcony you can sit on that balcony having your morning coffees using that coffee and milk and getting the most out of it this bedroom has also got plenty of space for a king-size bed and also has those featured sliding cupboards thank you very much for joining us viewing this stunning two bedroom two bathroom 5.7 million rent property on the 15th floor in the heart of mainland Maine and we'll see you next time on The Home Shopper Show proudly brought to you by Private Property


In the last class, I finished discussing about DC to DC converters. I briefly told you about the soft switching inverter, soft switching DC to DC converters as well. Soft switching generally is used for to reduce the switching stress, switching stress on the device. So, if I use soft switching technique, it is possible to increase the switching frequency. Therefore, the size of filter components comes down. Since I am doing soft switching, there are only conduction losses now. Your switching losses are practically eliminated by using a separate L and C, what is known as the resonant elements. So, there is no EMI has been taken care. Since there are losses of reduce, my entire cooling requirement also has come down, but then the complexity has increased. So far, we have discussed about DC to DC converters depending upon the power level and depending upon the voltage level. As of now, we are in a position to decide on a suitable circuit topology. Basic circuit topologies we have discussed. Let us see sometime later whether we need to do some modification in the basic circuit topologies. Now, we will see given an input given a DC voltage and if I want to convert it to AC, what are the possible ways of doing this? Well, initially we will do the basic inverters and maybe when you are doing the overall system integration, the modification that are generally made depending upon the power level, we will see. I said basically there are two types current source and voltage source. I also told that the literature says that the weakest link in voltage source inverter is the electrolytic capacitor. So, in order to improve the total lifespan of the inverter, we may have to find a solution to this electrolytic capacitor. Can we replace this by some other more reliable storage element? A basic, still an open ended question. Second is a current source has certain advantages here. Here the voltage is practically at this in the voltage source inverter, current can reverse, current can reverse instantaneously like a capacitor. So, basically I need, but then voltage at this point cannot change. Similarly, here current cannot change instantaneously, whereas voltage across the inductor can change instantaneously. So, I need to, I cannot use the same type of devices in current source and voltage source. There are two different types of devices. Here it is a bider, the switch should be able to carry current in both direction, but it need not block the voltage in both direction, whereas here it need not carry the current in both directions, but then it should be able to block the voltage in both directions. We started discussing about half bridge voltage source inverter, basic building block. I need total volt, I mean the total input voltage is V dc, but then at the output I have a square wave whose magnitude is V dc by 2 and minus V dc by 2 is a square wave. It has its own fundamental component, given by 4 by pi into V dc by 2. Then I also found that though I am switching S1 and S2 are complementary, I need to turn off S1 and there has to be a finite time for which neither S1 and S2 are gated, but then even if I get S2, a finite time current may not flow through it because of the load power factor, for sometime current flows through d2. Now if I replace the load by an inductor, an ideal inductor, how does the current waveform look like? I said I replace this load by an ideal inductor when S1 is on, voltage applied to the load is V dc by 2 and it remains constant for t by 2 seconds. See I will repeat, I have connected an ideal inductor across AB that is the load. I close S1 for t by 2 seconds, voltage applied to the load is dc and it is equal to V dc by 2, a constant DC voltage is applied to an inductor therefore, current increases linearly, provided it does not saturate. Of course, we should not allow the inductor to saturate. After t by 2 seconds S1 is turned off and S2 is turned on, voltage applied to the load is minus V dc by 2 remains constant for t by 2 seconds. Therefore, current decreases linearly and at steady state, current waveform looks something like this, it is a triangular wave. So, this is t by 2, so this will be t by 4, for t by 4 duration seconds d1 conducts and after that S1 conducts for t 1 by 4 seconds and so on. So, if I see the compare the voltage and current waveform, see this is the voltage applied to the load and this is the current flowing, voltage applied is positive, current is negative. Therefore, the product that is nothing but the power is negative. So, source is absorbing power, in other words power is flowing back to the DC link. Similarly, here voltage positive, current positive, so product is positive, load is load, source is supplying power to the load, in this zone current positive voltage negative, so therefore, power is negative, source is absorbing power and the remaining last t by 2 seconds voltage negative, current negative, therefore power is positive, source is supplying power. So, the average power consumed by the load is 0, it makes sense because load is an ideal inductor. So, average power consumed by the inductor is 0. I have one, as of now let us neglect the devices, device losses as well, the so called the switching and the conduction losses. I will assume that the inverter is an ideal one. So, an ideal voltage source inverter feeding an ideal inductor, power, average power required or average power, average power input is 0. So, if the average power input is 0, so we can replace this battery by, battery by a capacitor, charge the capacitor once and forget about it, it will continue to work for ever. The sense I am neglecting the losses, there are no losses, loss system is lossless for sometime capacitor is supplying the energy and after sometime the remaining half cycle in the entire one capacitor receives power. So, capacitor voltage will change, when it is charging capacitor voltage will increase and when it is discharging capacitor voltage will fall. So, there is a going to be ripple in the DC link voltage very enough, but then I do not need to an active source. If I consider a non ideal system, in the sense this inductor has a finite resistance which is very small, devices have finite losses, again these losses are very small compared to the k v rating. So, the net losses they are finite. So, if I have a capacitor here in order to maintain the voltage constant, I need to have some input active power input to the DC link. So, this is a very important concept that is required when we connecting this inverter to the grid. If it is an, if the inverter is feeding or supplying power to an ideal inductor and if I consider ideal devices, so average power required is 0. If I take the non idealities into account, active power that is required to maintain or to sustain this DC link is very small compared to the k v a r supplied by the source. Why k v r? It is supplied by the source or supplying reactive power or supplying an inductor, v a is watts plus war, j war. So, you require a delta p to sustain or to address this or to account for the losses, you require delta p to account for losses while supplying q, supplying q. So, this power active power may be equal to delta p has to come either from the AC grid or I need to have an active source to supply this power. So, if I do not have the grid here and if I connect a load and connect an ideal inductor, then I need to have some sort of a either, I need to have either a battery or may be power input from as part of the power input from solar is being utilized to address this losses. Remember an inverter can supply any q, any q in the sense as long as it does not exceed its rating. To do that, it requires delta p, a very important result I will just write here inverter requires delta p to supply q. So, in a grid feeding inverter as up now we do not know how to connect it to the grid. Let us assume that we know how to connect the inverter to the grid. Having connected the inverter to the grid, there is no solar insulation in the sense in the night grid requires reactive power support. At that time inverter can supply q, to supply this q it requires a very small amount of power. So, this power can be drawn from the grid. So, at night there is power, so if there is no input from the sun grid requires reactive power support at that time inverter your existing setup or grid feeding inverter can supply q which can be equal to its kv rating of the inverter. But to do that it draws a very small amount of power from the grid that is equal to delta p. The control loop, the control structure we will see sometime either on Tuesday, the second half or on Wednesday. What are the limitations of off bridge? I said input voltage is total is V dc, but then output is V dc by 2. And may be only one device is conducting at a time, so may be losses are less. Now, see here off bridge inverter I have actually V dc divided into two parts, but then I get V dc by 2. So, has a fundamental component which is sinusoidal which is equal to 4 by pi into V dc by 2. Now, if you want a 230 volts RMS 4 by pi into V dc by 2 is 300. So, calculate what is V dc is required. Now, instead of using that can I use full bridge inverter. So, what happens at full bridge inverter? There are four devices, two are conducting at a time s 1 and s 2 on. Voltage applied to the load is V dc, turn off s 1 and s 2 turn on s 3 s 4. Voltage applied to the load is minus V dc, but then two devices are conducting at a time. It could be an issue, see two devices are conducting at a time. If the input voltage is say around 400 volts, device drop is of the order of 2 to 2.5 volts, one device conducting a two device conducting it may not make any difference. See I will repeat, the input DC link is high say is of the order of 400 volts. How did I get this magic figure 400 volts? I will tell you later. Right now we assume that it is 400 volts. So, if it is around 400 volts, device drop could be of the order of 2 to 2.5, one device conducting or two device conducting makes it may not make any difference. Now, instead if I have a 12 volts solar panel or set 24 volt solar panel, V dc by 2 is going to be 12, 12 minus 2 is going to be 10 or means or not 12 minus 4 is going to be 8 or if I have a 24 volts panel, 24 minus 4 is 20 is 1 over 6, really significant, really significant. So, it depends on the input voltage. Another limitation is output voltage is a square wave. So, if I take the 4 year, if I write the 4 year series, the harmonic spectrum is going to be fundamental. If T is 20 millisecond, I have a 50 yard supply, then I have third, fifth, seventh, ninth, all odd harmonics are present, all odd harmonics are present. So, if you require, if you want a sinusoidal supply and the harmonic content should be as low as possible, I need to do something else. I just cannot allow the device to conduct for 180 degrees. See, each device is conducting for 180 degrees. Having turned on, it is maintained in that state for 180 degrees. I will repeat, having turned on S 1 and S 2, they remain on for 180 degrees. So, limitation here is it has all odd harmonics, three phase inverter, single phase we discussed. Now, let us see a three phase inverter. We require three legs, three legs of an inverter. Where is that? Here is it, three legs, there are three legs of the inverter is an IPM. I require three such modules, three such modules and I do not know how many such devices are required because these are four amperes, this is 75 amperes. I do not know what could be the number of devices, what could be the size, whether will this such a small heat sink will do. I do not know, we need to find out. I have three legs are connected in this fashion. As of now, I have taken center point as a reference. It is not required for analysis I have taken. The rule here is at any given time, one switch is on. At any given time, one switch is on in each leg. So, that is the principle of operation in VSI. At any given time, one leg, one device is on. So, each leg, one device is on. So, therefore, there are three devices on. I can have any combination, no issues. By the way, here S 1 implies a control switch. In addition, there is a diode across it because this switch should be able to carry current in both direction, but then voltage in only one direction. It need not block the voltage into this. So, I need to have three phase supply. I will turn on S 1 after 120 degrees. I will turn on S 3 and after 120 degree S 5. And when S 1 is on, V A 0 is V A 0 is V dc by 2. When S 3 is on, it is V B 0 is V dc by 2. So, when S 5 is on, V C 0 is V dc by 2. Similarly, the complementary switch is on, voltage is minus V dc by 2. Here are the waveforms. V A 0 is a square of V B 0, each of magnitude V dc by 2 and V C 0. Though V C 0, S, see here, do not get confused here. S 5 is turned on after 240 degrees because after 120 S 3 and after 120 S 5 and that will remain on for 180. So, compared to the positive 0 crossing of S 1, this turns on after 240 and it is allowed to remain there for 180. So, 240 plus 180 is how much? 240 plus 180 that completes one cycle and 60 degrees still remaining. So, this is 60 degrees where S 5 is turns off and S 2 is turned on. So, this is one complete cycle, each of 60 degrees pi by 3, 2 pi by 3 and pi. So, 0 to pi by 3, see here, attaining given term 3 devices are on. 1, 6 is on in the lower half and S 5 upper device in C phase. What is V A 0? See this voltage, when the switches are on, see switches are on in a voltage source inverter, the so called pole voltages are known. Pole voltage means voltage of A or B or C with respect to either the midpoint of the DC link or with respect to the negative DC link, it is known. I told, I will let me repeat, if the devices are, if I know the conducting state of the device, if I know the conducting state of the device, I know the pole voltages. Pole voltage is nothing but voltage of A or B or C with respect to the negative of the DC link or center point of the DC link. So, had plotted the pole voltages. Now, how do I determine the line voltage? Line voltage is nothing but V A B. V A B is V A 0 minus V B 0, V A 0 minus V B 0, V A 0 is V DC by 2, V B 0 is minus V DC by 2. So, V A 0 minus V B 0 is V DC. So, for 2 pi by 3 radians, V A 0 is a square wave of magnitude V DC. From 2 pi by 3 to pi, V A 0 is also positive, V B 0 is also positive. So, V A 0 minus V B 0 is 0. So, there is a 0 voltage period in line to line voltage wave form and similarly, in the negative half. So, line to line voltage wave form, I have pulse width of 120 degrees, the magnitude is V DC and 0 for 60 degrees. Now, if I know the voltage wave form, depending upon the type of load, I cannot draw the current wave form, no issues. Now, I have a 3 phase inverter, the load could be either delta connected or star connected. Now, star connected, if it is a star connected load, I need to know the phase voltage. If it is a delta connected load, I cannot determine the phase current if I know the phase voltage. So, I will repeat, if the load is delta connected, if I know the voltage wave form, line to line voltage wave form and if I know the type of load, it is possible to determine or it is possible to draw the load current wave form. I mean the phase current wave form, phase current not the line current, if I know 2 phase currents, I need to add them, wave forms I need to add to get the line wave form. Please do not say that root, phase current into root 3 is line current, no it is not valid. So, that is how it looks, constant voltage is being applied to VAB, depending upon the type of load, current wave form will change. If it is a RL load, voltage applied is constant DC. So, current will exponentially increase at steady state. If it starts from at t is equal to 0, if I energize, current has to start from 0. At steady state, it has a negative value, exponential increase, voltage applied to the load is 0. If the voltage applied to the load is 0 and if the load is RL, current will decay because applied voltage is 0. If the load is purely inductor, this will be linear in this zone, current will remain constant because voltage applied to the inductor is 0. So, current will remain constant and here current will decay. So, this is the current wave form IAB. I can draw similarly IAC, IA I can determine by adding these two wave forms. I have to add these two wave forms to get IA, this is how it looks. I will not discuss much time on this. The observations, there are six steps in each cycle. The number of steps in each cycle are six, hence the name six step inverter. V s 0, V b 0, V c 0 are pole voltages. Predominant harmonic, what is the predominant harmonic in the pole voltages? So, it is a square wave, definitely all order harmonic 3, 5, 7, 9, 11, everything. But then, line to line voltage is a square wave of 120 degrees duration. It can be proved that harmonic only 6 n plus r minus 1 harmonics will be present, 5, 7, 11, 13, so on. One way is to write the Fourier series or write the Fourier series for V s 0. We all know that this is 1, 3, 5, 7, 9, 11, everything is there. Here also 1, 3, 5, 7, 11 harmonics. Now, V AB is V s 0 minus V b 0 and then you get the harmonic spectrum. Third harmonic and third harmonic get cancelled. It can be proved. You just prove it. So, in the line to line voltage wave form, we cannot have third harmonic. It is a very popular or very important result that we have studied. So, predominant harmonic is 5th, 7th, 11th, 13th, 6 n 1 plus r minus 1. So, this is the V s 0. This is the V b 0 harmonic spectrum. This we can get by writing V b 0 and etcetera. So, the r m s value of the line to line voltage is square root of 2 by 3 into V d c. What is the r m s value of this? We will get as 2 by 3 into V d c of this wave form. And the fundamental component here is root 6 by pi into V d c, 0.78 V d c. Fundamental is 0.78 V d c, r m s value of the line to line, r m s value of the line to line voltage is 816 V d c. For star connected load, I need to determine the phase voltage. There are 2 ways to do it. One is depending upon the switching combination, draw the equivalent circuit and determine what is V a n, what is V b n, V c n. I mean, it is something like this. Say, SCA attaining one time 3 devices are on, SCA is on, this is on, lower switch of SB and lower switch of SC is on. I will take this example. See, attaining one time, one of them is on. I will take SCA is on, lower is lower of SB, lower of SC is on. So, when SCA is on point, what could be the equivalent circuit? What could be the equivalent circuit? When SCA is on, A gets connected to the positive DC bus. When the lower switch is on, B gets connected to the negative DC bus. Similarly, I have assumed that lower switch is on, C gets connected to the negative DC bus. So, how does the equivalent circuit look like? It looks something like this. I am not going to draw, explain to you each and every step. One is on. So, this is A, this is B, this is C, finally connected to positive, negative, this is V d c. Assuming the load to be balanced, assuming the load to be balanced. So, what will happen? I have the magnitude of and the phase angle of the load impedance is the same. So, I can prove that V a n is 2 by 3 V d c. V b n is equal to, V c n is equal to minus one third V d c. If you say, assume that this is R, R, R. So, combine this, this is going to be R by 2, this is R. So, voltage at this point comes to be one third, this comes to be two third. So, can I make a generalized rule? I am not going to derive for other combinations. See, I have three switches. They can take two values.  There are two possibilities. Under that condition, when all three of them are on, A gets B and C, they get connected to the positive d c bus. So, all three line terminals are at the same potential. So, therefore, source does not supply any power to the load. It is also known as load is freewheeling through the positive d c bus. Similarly, when the lower switches are on, these three points are connected to the negative d c bus. Power supplied by the source is 0, load freewheels through the negative d c bus. So, when A, B and C are at the same point, B A B line to line voltage, B A B is 0, B B C is 0, B C A is also 0. All three line to line voltages are 0, A B B C C A and therefore, you can prove that B A N is also 0, V B N is also 0, B C N is also 0. So, this is also known as, these two vectors also known as 0 voltage vectors, 0 voltage vectors. Why? Magnitude of the voltage applied is 0. All three terminals are connected, either they are connected to the positive d c bus or they are connected to the negative d c bus. The remaining six, what are they? One is on on top, two are on in the bottom or two are on in the top, one in the bottom. So, there are six such possibilities, six such possibilities. So, one such possibility is, I said I assumed A is, upper device A is on and lower switches of B and C are on. What is the magnitude of the voltage that is obtained? If one device is on, we found that this voltage is 2 by 3, 2 by 3 and when two devices are on, we found that this magnitude is one-third, one-third. Remember, if one device is on, magnitude is 2 by 3 and if two devices are on, magnitude is one-third. What is the sign? They could be either positive or negative. If only one upper switch is on, the corresponding phase voltage is positive. Similarly, if one lower device is on, the magnitude is minus, sorry the magnitude is 2 by 3 and sign is negative. Instead, two upper devices are on, magnitude is one-third, sign is positive. Here, if the two devices are on, magnitude is one-third, sign is negative. That is all. V B N and V C N are minus one-third V D C. Why these two devices are on? V A N is one-third, sorry V A N is two-third V D C. So, there are six active states and two null states. So, I can draw V A N, V B N, V B N, V A N. And V C N, how does the waveform look like? How does the waveform look like? See, I will go back, can draw from here also. In this zone, see here, V A 0 is positive, V C 0 is positive, but V B 0 is negative. So, therefore, V A N is plus one-third, plus one-third. Why upper two devices are on and V A is also positive, so plus one-third, whereas V B N is minus two-third. Now, what happens from pi by 3 to 2 pi by 3? Only once which S 1 is on, B in the second leg S 6 is on, that is the lower device and in the third leg S 2, which is again a lower device. There are two lower devices in the lower half are on, so therefore, V B N and V C N are minus one-third each and only one upper switch is on. So, therefore, V A N is two-third V D C. See in this zone, A and B, there are two devices are on. So, V A N is plus one-third V D C. You can draw the equivalent circuit and derive. So, the phase voltage waveform here is one-third V D C between 0 to pi by 3, pi by 3 to 2 pi by 3 is 2 by 3 V D C and 2 pi by 3 to pi it is again one-third V D C. So, again there are six steps, hence the name six step inverter. See the entire analysis is there. You can go back and read it is simple, one is zero vector vectors. This is equivalent circuit approach. This is how the waveform look like, one-third V D C, two-third V D C, one-third V D C hence the name six steps. The harmonic spectrum, you can prove that, I will just show you. The harmonic spectrum must be, again the phase voltage has 6 N plus R minus 1, harmonic can be proved. So, as of now, we learnt how to convert D C to a higher voltage D C, any voltage magnitude that I can get now. If I want to have a very high voltage, I need to use a transformer in the sense and depending upon the power rating, I can choose either a fly back forward or a push pull and there are other topologies also exist. Now, given a AC voltage, I can convert to AC. The output voltage waveform is a square wave. If it is a, the line to line voltage has 120 degree duration, whose magnitude is V D C or 60 degrees it is zero. Phase voltage waveform is one-third V D C and two-third V D C. The harmonic spectrum is approximately the same, 6 N plus R minus 1, harmonics are there. So, we are using it as a UPS here to feed, assuming that, now there are, you need to make two assumptions. What are they? I have solar panel over that is coming here. This is D C to D C. This is D C to D C. This is D C to AC. Now, I can, sorry in the sense depending if I want to have a 50 hertz supply, 50 hertz supply, the predominant harmonic is 6 N plus R minus 1 harmonics, nothing but 250 and 350, 550 and that is 11 and 13, 650 hertz. This is the predominant harmonics, assuming that you want a 50 hertz supply. If you want to have a variable frequency supply, then therefore, as the frequency of the fundamental changes, frequency of the predominant harmonic also changes. As of now, I will assume that it is a 50 hertz supply. Predominant harmonic is 250 and 350. I do not think we can directly connect this inverter to the grid, because harmonic magnitude of this fifth and harmonic is substantial. So, the total harmonic distortion T S D of the voltage waveform is very high. So, your, the entire black box may not be able to satisfy the given norms of I triple E and I E C. So, you may not be able to connect it to the grid. So, if I have to connect it to the grid, definitely I need to reduce the harmonic content. How do I reduce this harmonic content? That I will discuss in my next lecture for another 10 minutes or so, I will take up the questions. I will encourage you to come on chat mode, because it reduces the time. In this, I can directly read the question. I will repeat it instead of going into each and every center and repeating the question. Please feel free. War angle, over to you. Sir, the one leg of a 3 phase inverter. Both the switch are complementary. That means, if the upper switch is on, then the lower switch should be off. If the lower switch is off, then upper switch will be on. The module that you have given, where the three legs which we have shown, it is always that condition or by any means we can switch off both the switch, the upper switch as well as the lower switch, so that you can do any other modifications. So, I hope you follow my questions, sir. Now, I do follow in a voltage source inverter, each leg, the devices in each leg are complementary. If S 1 is on, S 4 is off or vice versa, they are complementary. They are complementary. Say that again? No, both the switches can be made off. Both the switches can be made off. No problem. You can turn it off. What is going to happen? See, if both the switches are off and if there is some current flowing this side, yes, it can flow. Are you with me? Some current can flow provided there is some current coming from the load side. In fact, it happens here, is not it here?  Assume that you do not turn on S 2. What will happen? Whatever the current that was flowing through the inductor, this was the direction of current, it will start flowing through D 2 and when this current becomes 0, conduction stops. Are you with me? So, if there is some current coming from the load side, yes, it can flow through the diode. If there is no current coming, that is it. Is that clear to you? Yes, actually why I was asking means suppose if I want to go for other topologies, higher topologies, higher levels. Higher levels also the principle is the same. In a 3 level inverter, higher level also has in a 3 level inverter, 2 devices are on. There is a rule to be followed, rule to be followed. Here, one device is on at any given time. If the both the devices are off, there is no power flow. It is only when if there is some current coming from the load side, it can come and charge the DC link, that is all. In fact, that is being used sometimes. Suppose it is like this, I have, say for example, I have a voltage, a 3 phase voltage source inverter. This is line A, this is B and C and I connect a capacitor here. I will not give any pulses to the active switches connected to AC supply, connected to AC supply. I have an uncontrolled bridge charging the capacitor. I have an uncontrolled bridge charging the capacitor. So, we will see sometime later. This is being used initially to charge the DC link capacitor in a voltage source inverter, feeding power to the grid. Is that okay? If I do not switch on, I have an uncontrolled bridge, 3 phase uncontrolled bridge. So, there is a question in the sense inverter, the question I will read out. Sir, you said we can apply inverter for supplying reactive power, whether it depends upon our load or on the inverter can direct to supply Q. I have a 3 phase source supplying power to the load. As of now, I will, load is, I will just assume that, as of now I will assume that it is a linear load. Power factor is less, not equal to 1, linear. After sometime we will see it can have a non-linear load as well. So, if, when I am saying linear load, current drawn is sinusoidal. If the input is sinusoidal, current drawn will be sinusoidal, but then power factor is not equal to 1. Under this condition, your voltage source inverter, how you do not ask me right now, can supply this Q. Here is, if the load is, sorry, SL is power load plus or minus j QL, as of now I will assume that inductive load plus QL. This inverter can supply QL, this inverter can supply QL and thus QL should be less than or equal to, I will repeat this QL should be less than or equal to S inverter, K V rating of the inverter, K V rating of the inverter. So, Q supplied by the inverter is, could be as high as the K V rating of the inverter. To do that, a small amount of delta P is required, to do that a small amount of delta P is required. The delta P is to account for losses in the inductor and the losses taking place in this inverter, because whenever I switch the device, there are losses. The circuit theory base, for this I will explain to you. Right now, I proved that inverter can supply any Q, Q could be as high as its inverter, its K V rating. To do that, it requires delta P. Suppose I do not have a load here, I have just three phase transmission line and if it requires Q, yes this inverter can supply Q, that Q is equal to or less than or equal to inverter, it is inverter, it is V rating and so and this inverter will draw a small amount of delta P from the source. So, question is you do not need to have a load. Now, load could be grid or load could be some whatever. The question is what are the advantages of three phase inverter over single phase inverter? Now, how do I answer? What is the advantage of single phase supply over a three phase supply? That is all. If you want a three phase inverter, three phase supply, you need to have a three phase inverter and if you want to have a single phase supply, you need to have a single phase inverter, that is as simple as that. So, when do I go for a three phase? For low power application, I use single phase supply. For as the power rating increases, I go in for a three phase supply, go to the market and find out, can you get, can you get say more than three HP pump, irrigation pump, a single phase three phase, a single phase three HP irrigation pump, you may not get. So, you have to go in for a three phase supply. So, if you want a three, if you have a three phase load, I need to have a three phase supply. What are the reason that makes electrolytic capacitor unreliable? Come on, that is a device physics or whatever the capacitor construction or design aspect, you go and read the construction of a capacitor, why they fail? What is the reason that makes an electrolytic capacitor unreliable? How to generate a sine wave from the output from the inverter? Do not worry, I will discuss right now. How to generate a sine wave from the inverter? I will discuss right now. Sir, out of two modes namely 120 degree and 180 degree mode inverter, which is good for star and delta connected load? Let us not talk about 120 and 180 degree, this is just a basic block. If you have to sell this, sell your inverter in the international market or in the market, neither your 120 degree will work, neither 120 nor 180 degree will work, you have to do something else. In the next class, we will discuss after this. Please, 120 and 180, they are simple to understand. Since they are simple to understand, I did this and I told you the limitations. Now, we will discuss what are the limitations. I told you the harmonic content. So, if it is a 50 hertz supply, I have a 250 hertz and 350 hertz harmonic. So, you may not be able to connect your load. If you connect your load, there will be harmonic torques that will be generated, which I am not going to discuss. You may not be able to feed it to the grid either. So, you have to do something else. What is that? We will see in my next lecture. What is the advantage of using split power supply in 3 phase inverter? Who is using the split power supply? I did tell you that you do not need to use center point of the DC link as a reference. In fact, it is not there. It is not available. I can use one single capacitor. It is just for convenience sake that I have used a center point. Of course, by the way, I do not know in what context you are asking this question. I can use this if I have a split capacitor topology and if I connect the center point to the neutral of the source, there are certain advantages. Please, KJSM, I do not know in what context are you using. In what context you ask this question? If your question is based on this slide, let me tell you one thing you do not require this point at all, do not require this connection. In fact, in a conventional 3 phase voltage source inverter, this point is not available. But then there are certain topologies. I can use this point and if I connect it to the neutral, there are certain advantages. What are the advantages? I will discuss it. I hope you meant that. You can confirm through chat. Now, 36, I will take two more questions. KJSM asks, sir, I was asking about the split power supply. Where is the split power supply? We asked the question about split power supply. Where is the split power supply? I know that you already answered. No, no, I have not answered in the sense. Wait a minute. I have only one part here I have answered. Assuming that the question is based on that photograph. But then there are circuit topologies wherein we connect the center point of the DC link to the neutral. When you do that, there are lot of other advantages. What are they we will see? Now, which question is that? Did you ask the first part or second part? Now, what? Yes, first part, first part you do not require. What are we doing still in DC? How much efficient to apply a multilevel inverter rating? How much efficient to apply a multilevel inverter for obtaining a sine wave inverter? Now, why multilevel inverter? We are still in 2 level. We will see multilevel sometime later. Multilevel generally used for high power application. What are the advantages we will see? First we will try to understand 2 level of p-dolume technique and we will see 3 level or multilevel later. For DC to DC conversion, how to select whether to go by buck, boost, buck, boost, fly back? Tell us formula. Sir, I am not going to tell you the formula for DC to DC conversion. How to select whether to go for buck, boost, buck, boost, push, pull, fly? Kindly tell us formula related to converter selection. No, sir, depending upon your power level, depending upon the input output voltages, you have to select one of them or you have to modify them. I do not want to tell you and nor I will tell you which one you should use. You choose fine, fine. You take one topology and if it is wrong, what price will you pay? Fine, you will learn. You will gain some experience. That is more important. So, I am not going to tell you which one you should use. It depends on the power level as well as the input and output voltage ratios. Sir, suppose the inverter is connected to the grid, then on what basis we can select the coupling inductor value? Wait a minute. Let us connect the, first we will connect the inverter to the grid and we will see later. We have not connected the inverter to the grid so far. What could be the value of the T H D in ideal sine wave inverter? Hey, Bhagawan, what should be the value of T H D in an ideal sine wave? What do you mean by T H D? T H D is total harmonic distortion and you say in an ideal inverter, ideal sine wave. I think there is a inner contradiction, I guess. Total harmonic distortion and ideal sine. Anyway, wait a minute. This class, what is A S, A S, E B? The value of inductor, value of capacitor, we will see. Let us connect the inverter first to the grid. Then we will find out what should be the value of L.


All right. Good evening, everyone. It is six o'clock. I will call the 14th regular Common Council meeting to order. Will the clerk state the court of the evening? Thank you, Mr. Mayor. Happiness is an inside job. All right. Will the clerk call the roll? All the person Ackley. All the person Decker. Here. All the person Feldy. Excused. All the person Flicky Paneski. Here. All the person Heidemann. Here. All the person Mitchell. Here. All the person Porello. Here. All the person Ramey. Here. All the person Rust. Present. All the person Salazar. Here. There are nine present. All right. Thank you. Next we'll do the Pledge of Allegiance. I pledge allegiance to the flag of the United States of America and to the Republic for which it stands, one nation under God, indivisible, with liberty and justice for all. All right. Item number three is approval of the minutes from our last meeting. Alder Decker. Thank you, Mayor. I move to approve the minutes from the 13th regular council meeting held on October 2nd, 2023. Second. We've been seconded. Any discussion on the previous minutes? Seeing none. All those in favor of approving the minutes. State aye. Aye. Any objection? The minutes are approved. Next is mayoral appointments city attorney. Mayor has two appointments for your confirmation. Melissa Parra and Aubrey Lockwood to be considered for appointment to the mayor's international committee. Thank you and those will lay over. City Clerk, anyone for a public forum this evening? No one this evening. Thank you. All right. Just one item of recognition for tonight for mayor's announcements. I'll keep it brief since we will be going and discussing our capital improvements budget. I just want to recognize that one of our firefighters, Tyler Meyer, passed away a few days ago in his services for his funeral where yesterday. Tyler's been with the fire department since December of 2007 and this would have been his 16th year as a member of the city fire department as well. So our thoughts and prayers are with him and his family as well as the men and women that worked alongside of him in the fire department as well too. So I just wanted to recognize his services that he gave to the city and I'll call on chief Montiano for additional comments as well. Thank you, mayor. Thank you for the support. It hit the crews pretty hard but Tyler was one of those individuals that nobody ever had anything bad to say about him. He was just one of those guys that would make you smile and laugh no matter what he did. You could not be mad at that individual and he truly would give you the shirt off his back so he will be sorely missed. It was a phenomenal service but I really appreciate the acknowledgement and the support so thank you. Thank you chief. All right with that we will proceed into our consent agenda items 8 through 23. Alder Decker. Thank you mayor. I move to receive and file all ROs receive all RCs and adopt all resolutions and ordinances. Second. Moved and seconded. Anyone have any discussion on any of the items in the consent agenda? Alder Flicky Panesky. Thank you. Item number 8 is this the alley that is north of the Lutheran church which is north of the Meade Library? It says section 128 of the old plaque book and I don't know what 127 of the original plaque book. Chuck you want to do that? Yes. Or Ryan Sazma? I guess Q did. Either one. The answer is yes. The answer is yes. Okay yeah it's part of the part of an ongoing thank you sought finance and personnel that there's been some discussion about transferring that parking lot and this is all part of that process and working with the school and church as well as with the theater. Okay it does not include the north south section of that alley? It does not.  Just when I looked at the map I only saw east west. Okay. But it is only the part that goes east west. Okay thank you. Other problem? Yeah so I had the same question about which one was that because I wasn't able to understand it and then also I didn't hear you see the attorney I'm sorry for that actually I couldn't hear the back and forth so in addition to understand where that alley is I also would like to understand what the issue is with that alley I don't know I didn't understand it. So there is an ongoing discussion to sell that lot to the church and school. That process though is including the theater because the theater has a vested interest they've got some loading docks back there but this this is all part of that of that process. Thank you. Any other discussion on that consent agenda? Alder Flicky Paneski? Thank you very much. Items 16 and 17 and they both involve getting new traffic lights at two separate intersections in town and from what I could gather the billing went up 65% for one of them and 105% for another the cost without any change in the in the design. So can somebody tell me why we're spending 165% more for two intersections? Right city engineer cue then yes those extra costs are strictly materials went up that much in two years material costs for for the traffic signals and the controllers and everything else yes those uh yeah those those contracts are are ran through the Wisconsin DOT so yeah that's that's that's the cost the nice thing about this program is it's a 90 10 split even with these increases it's 90% DOT funding or federal funding in 10% city. So the dollar amount we see it increased from 37 to 62 and then the other one increased from 45 to 93 thousands. Yes. We only pay 10% of the 93 and 10% of the 62. You got it. That keeps me happier thank you. Like I said it's strictly material cost it's not labor it's not design it's materials. Yeah all right. Anyone else for items on the consent agenda? All right seeing no more queues. This will be up Alder Flicky Podesky. Thank you. Item 23 it is asking me to vote on licenses already granted. So I have a difficult time voting on something that's already been done second only to voting on suspending rules so can somebody explain to me why we vote on something that's already been accomplished? You're not good. So you are receiving a report of officer that indicates that she has issued those licenses that's all you're doing. Okay thank you. All right any other items on the consent agenda? All right this will be a roll call vote. Nine eyes. All right those items are approved. Next report of officers item 24 RO number 59 23 24 by the City Clerk submitting a communication from the Wisconsin Department of Administration regarding the Berkey Company annexation. Alder Mitchell. Thank you Mayor I move to receive the RO and file. Moved and seconded discussion on this item? Seeing none. This is a voice vote all those in favor of accepting state in filing state aye. Aye. Any objection? All right that item is approved. Item 25 RO number 61 23 24 by the City Planning Commission to whom was referred General ordinance number 24 23 24 by other person Mitchell and RO number 48 23 24 by the City Clerk submitting petition for annexing territory from the town of Wilson into the city of Sheboygan, Wisconsin. Alder Mitchell. Thank you Mayor I move to receive the RO, file the application and adopt the ordinance. Moved and seconded. Any discussion on this item? Seeing none. This is a roll call vote. Nine eyes. That item is approved. Item 26 27 28 29 will be referred to the respective committees. Next item 30 is under resolutions number 78 23 24 by older persons Decker and Russ authorizing the director of engineering and public works to grant permission to Lake Shore Cap to temporary maintain a campsite on property owned and maintained by the city and to maintain a warming fire on such property during an overnight event beginning November 4th. Alder Decker. I ask to suspend the rules. Any objection to suspension? There's been an objection. We will proceed into a voice vote any all those in favor of suspending the rules state aye. Aye. Any objection? All right in the opinion of the chairs the eyes have it the rules are suspended. Alder Decker please proceed with your motion. Thank you Mayor I move to adopt the resolution. Second. Moved and seconded. Any discussion on this resolution? Alder Flicky Panesky. Thank you. I am not so much opposed to doing the activity. I am opposed to the fact that those kinds of things have been planned previously and it could have come to us in a proper time frame according to my estimation. So that's why I've objected and why I will vote no. All right thank you. Any other items for discussion on this item? That sounded weird. Okay seeing no more queues. This is a roll call vote to approve the resolution. Eight eyes one no. All right that item is approved. Next item 31 resolution number 79 23 24 by older person Salazar and Feldy authorizing the chief of the police to take necessary actions to receive the 2024 Wisconsin justice system improvement beat patrol grant. Alder Salazar. I ask to suspend the rules. Any objection to suspension? I object. All right there's been an objection. All those in favor of suspending the rules state aye. Aye. Any opposed? Aye. All right in the opinion the chairs the eyes have it. Alder Salazar please proceed with your motion. I move to adopt the resolution. Second moved and seconded. Any discussion on the resolution? All right seeing none this is a roll call vote. All right that item is approved. Next is item 32 resolution number 77 23 24 by older person Salazar and Feldy adopting the weights and measures device license schedule fee. Alder Salazar. I ask to suspend the rules. Any objection to suspension? I object. All right there's been an objection. All those in favor of suspending the rules state aye. Aye. Anyone opposed? Aye. All right in the opinion the chairs the eyes have it. Alder Salazar please proceed. I move to adopt the resolution. Is there a second? Second moved and seconded. Any discussion on the item? Seeing none this is a roll call vote. I move to adopt the resolution.


Stakeholders in industrial safety, led by the International Human Rights Commission, have raised the alarm over the increase in dangers of installation of liquefied petroleum gas, LPG, otherwise known as cooking gas plants, within the residential areas. Ambassador-at-Large and Head of Diplomatic Mission, IHRC Nigeria, Duru and Zikaya, who made the stakeholders who addressed the stakeholders at the interactive session of the project's safety and collaboration with relevant stakeholders in Abuja, made a reference to lives that have been lost over the years due to gas explosion, challenging the stakeholders to support the new campaign to enlighten the public and forstall any future incident or minimize the impact. We have witnessed and have heard a lot of cases. We are people's life, we are cutshot, something we can ordinarily avoid, something that we can just ordinarily avoid. Anything we are going to do to make sure that we bring to an end the incident or the avoidable death occasioned by this gas explosion, we are going to do it. We are going to use all the astronauts in our disposition to ensure that we get it right this time. They, however, suggested that strict measures should be adopted by the relevant government agencies to strengthen monitoring and licensing of apparatus of the LPG facilities in order to protect the ecosystem. The stakeholders said they are concerned about the lives that would be lost in properties that will be destroyed in case of any mishandling of the LPG facilities which could result in explosion. We recognize that the existence of LPG facilities in residential areas valuates national environmental guidelines and poses significant risk to the well-being of our communities. Through this project, we aim to engage state governments, create awareness amongst stakeholders, and collaborate with organizations and individuals who share our commitment to safety by implementing enforceable regulations, promoting sensitization initiatives, and advocating for the relocation of gas plants from residential areas. We can reduce the risk posed by these facilities and protect our communities. This camping gas, you are not supposed to use it inside your home. You are supposed to use it for camping because when you are using it for camping outside, even if there is any leakage, it will diffuse into the atmosphere. But because it is small and because of people do not have enough money to buy gas, why sometimes they bring it into the home and use it inside the house. And when you are using it inside the house, once there is a leakage, because it's within a small environment, compacted area that maybe does not have a deviation, then there will be a tendency of accidents and that's what we are behind.


Okay, yeah, thank you. So this is my last lecture in this series and I think the last lecture of the entire subject this year, and I'm going to talk about the computation. And this really should be understood to apply to algebraic number fields because that is where the real interest lies, but to begin with, let me remind you of the situation that you may be familiar with over the rational numbers. So suppose if I take P of prime and a integer that is not divisible by P. Then, we have the Legendre symbol, the Legendre symbol A on P, and that is defined by two properties. The first is that its value is either one or minus one. And secondly, what you know P. It is congruent to a to the power P minus one over two. If you square this right hand side, then by the most little theorem you get one regular P. So this is a square root of one and then the field will be, you have two square roots of one, and that is what the agenda symbol is. And this is also called a quadratic symbol, because if a is congruent to a square model of P, then the value is one. And if a is not congruent to a square model of P, then the value is minus one. So from its very definition, it is perfectly straightforward to see that there is a polynomial time algorithm that on input A and P computes the symbol by repeated squareings and multiplications model of P. The immediate results should also be computed more than a P to avoid getting numbers that are too large. And in fact, there is a faster algorithm, of course, still polynomial time to do this. And that relies on a generalization that is the Jacobi symbol, and the Jacobi symbol is defined for a and B. G, B is positive and odd, it is composed of such primes P and A and B are required to be called prime. And then the Jacobi symbol is the same as the Legendre symbol, if B is a prime number, and in general, you extend the definition of symbol by multiplicativity in the lower argument. So this is a product that ranges over all of primes and then you take A and B, the Legendre symbol, and you raise it to the power, the number of factors P that occur in B, which of course, for most primes will be zero. And it is good to know, and clear from the definition that this depends only on the residue class of A modulo B, because this one only depends on the value of A modulo P. If you fix B, then you can reduce A modulo B without changing the value of the symbol, you can replace A by its remainder upon division by B. Then your A is, once your A is less than B, then you can often make use of a remarkable theorem that is due to Gauss, which is the reciprocity law, which initially was stated for two prime numbers, but it is very easy to extend it to the Jacobi symbol. And that is as follows, if A and B are both positive and all, and they are still called prime, then there is a relation between the two symbols, A on B and B on A, and that relation is that they are equal, except if both A and B are three modulo four, in which case they are actually different. And this is something that you can apply by first reducing A modulo B, then A is the smaller one, if A is also odd, you can turn the symbol around, in case you are trying to compute it, you have to keep track of this sign. If A is even then there are a few special, a few special laws, the first is that two on B is equal to one, if B is plus or minus one, mod eight, and it is minus one if, in other case when B is plus or minus three modulo eight, and I did allow A to be negative here, so you also want to know what the minus one is, and that just depends on B modulo four. So here you see that you have enough material for computing the symbol, essentially by means of the Euclidean algorithm, you do have to pay a little special attention to parity issues, as I pointed out. If you look in the notes, collection notes written by Dan van Gent, you will see that those can be taken care of without very much effort, so that the computation of the symbol, even when the denominator is a prime number can be done as fast as the Euclidean algorithm, which is a bit faster than an entire differentiation. And of course, it should be pointed out that there is one way not to compute it which is from its definition, because that would require the factorization of B into prime numbers, which is typically in accessible at the current state of mathematics. What I'm going to do in this lecture is that I will tell you how this Jacobi symbol can be extended to algebraic number fields that turns out to be pretty straightforward, except that there will be a difference. There is already this A modulo B in the case of algebraic number fields, this B, so to speak the denominator, this B will not be an element, it will be an ideal of the number field. Of course it may be a principle ideal, but that is certainly not the only interesting case so there's already a certain asymmetry, and you really before you have a reciprocity law like this, then you have to restrict the principle of the number fields so that there is already something of a problem when you try to compute that generalized symbol in a similar manner. And of course there's also the problem that the Euclidean algorithm that we have been using here for integers that will in general be in applicable to rings of algebraic integers. So that is the challenge of the lecture, how to compute the Jacobi symbol in polynomial time. Let me first give you the definitions. I will start with replacing Z by a ring that I call R, which is an order in a number field. That means a subring of the ring of integers. The ring of integers itself, as I pointed out earlier this week, that is really an object that in the context of algorithms, better be avoided, because it is very hard to compute without something like factoring. It is not only difficult to compute, it is also difficult to recognize it is hard to say whether a given order is actually equal to the ring of integers of its field of fractions. If this is just surprising then I will, I should suggest to you just to think a little bit about the case that your number field is just a quadratic field of degree two over Q, already in that case, maximal orders are hard to compute. Okay, so not necessarily maximal. We have in here an ideal, and I call it be. This is the role of the be a moment ago, and I was in the case of integers I had a requirement that be was positive value is an ideal so the sign makes no difference. But it was also the oddness. So what I want is that this ideal has the property that if I look at our money will be that that is all. It will be a finite number, as long as I avoid the zero ideal. So in this ring arm of the no be the number two is inferno. And then I also take an element any in our and just as for the integers, I like a to be called prime to be. And that is to say that are a the ideal generated by a is called prime to be in the sense that together, they generate the unit idea, and that is equivalent to saying that this are a plus B contains the unit element. And that this element a, when I view it as the element in the residue class ring the coset of a mobile be actually belongs to the unit group so this last condition is equivalent to saying that a mobile be a mobile be belongs to the unit group of this finite ring. It has an inverse model B.  It is. Again, either one or minus one. B is a prime number. I say prime number, but I mean, if it is a prime ideal, so that our mod B is finite field, then it should be congruent to a to the power, have the number of units of the residue class field. So that is what you know be, and that will be what you know be plus one or minus one for the same reason as before. So this is if B is prime. And then, in general, for general be this a on B is a product of the prime ideals, and these are the prime ideals of with our more P. So those are the prime ideals of it's the residue characteristic is greater than two. This symbol is a W it stands for with. And then I have here able P. And then I have to face a slight problem, because here I have to write if I want to imitate the situation or the integers I have to write here the number of factors P in B and if R is a maximal order, then you have unique factorization into prime ideals. And that makes perfect sense. But in general, I have to define what I mean by this experiment LP be it is the P length of B. And let me tell you how this P length can be defined so as to generalize the situation of the full ring of integers. And then it's done by writing down a sequence of ideals be zero be one. Until let's say BT, which is B, and these ideals. It should be a chain pair wise difference or strictly descending. But it should not be possible to refine it so that means that there is never any ideal in between to successive be ice. So what I am effectively doing here is that I write down a composition series for the R module R module B. And then if you look at one of these successive closures be I minus one model of BI, then that will be a simple module over R. It has exactly two sub modules, the zero module and the whole thing, because otherwise there would be something intermediate in my chain. So that means that for each, I, there will be a prime ideal P a maximum ideal, such that as our modules be I minus one what you know be I is isomorphic to our more P, I should have said there is a P, the unique P, which is the unilater of the such that this module is isomorphic to our more P, and then LP of B is the number of I, for which this, let's call this P PI, for which P is PI. So that is the number of times that P occurs in this chain. And that clearly generalizes the multiplicity of P and B, in case my R is indeed dedicated domain. So if I see that if B is prime, then the P length of B is equal to B is equal to one, and all for all other P that P length will be zero. So this clearly generalizes the Jacobi symbol that I defined for Z. And now the question is, can we determine this symbol in polynomial time. The answer is yes. So let me write down the theorem here, there is a polynomial time algorithm PTA, that means polynomial time algorithm that with input. The input is just all you need for making the symbol meaningful. So first you input R and inputting R, that means that you input R as an a Boolean group, as I explained yesterday. And it is just a free Boolean group, so you just have to specify the rank. And secondly, you have to specify the multiplication, which is a map from R tensor R to R, which you can specify in the way I explained yesterday, and you specify in general morphisms between finally generated B groups. And with B it is the same, B is an ideal, and you simply specify it as an additive subgroup of R, and you are doing that only for the case that B is actually an ideal. So I'm just going to satisfy my conditions non zero and an old index, and also a, all of these as before, compute the symbol a on B. We have to do this. While avoiding the problems that I pointed out the fact that the reciprocity law is less than ideal. And the fact that there is no Euclidean algorithm. And that is one of the things that you encounter more often in algorithmic algebraic number theory that you run into perfectly natural question. In this case, computing a symbol a symbol that plays a role, for example in classical theory, but all of the theorems of classical theory are not enough to provide a polynomial time algorithm. And what you have to do here is to develop your own theorems in order to make this possible. And those theorems, I will next proceed to formulate and those theorems, they are about another symbol, and that is a symbol that is defined in the following situation. Notice that B is a finite abelian group as a G means finite abelian group. And in the application to this algorithm that is going to be the additive group of our model of being different type for being. So my beta. That is the second thing that you need to know that is an automorphism will be. And then automorphism. Well, if I have this, the billion group are more be that it is actually a ring, and a is a unit in there, and multiplication by that unit gives you an additive automorphism of our mobile. And that is the way you have to think of beta. And then we have a beta comma B, that is a symbol that was surprisingly generalizes the Jacobi symbol, it takes values in one, and in minus one, and it is defined to be the sign. The of beta view. As a permutation of the underlying set of B. So in other words, you have this automorphism group of B. It consists of commutations of the underlying set. So it lies in the symmetric group of B, which has order and factorial if N is the order of B. And then you have the sign map from elementary algebra. And if you compose this, these two maps the inclusion and the sign map, then you get this symbol beta B, which already shows that if you fix capital B, then this is multiplicative in beta. So that is the case for the Jacobi symbol, which I didn't point out, but if I fix my B, then this function of a is really a group of morphism from the unit group of our will be to plus or minus one. Okay, so that is the sign. And then we have two more theorems. If this is theorem one, then we have zero to which is the one that I suggested already. That is a B that you get it. This is symbol. So here you will have the automorphism of the additive group of our will be, which is given by multiplication by a, and here you have that's right in our B plus our B plus is just the additive group of our will be. And if you compute this sign in that situation, then the theorem asserts that what you get is the Jacobi symbol that is theorem two, and I will suggest much of the proof in the present lecture. And then we have another theorem, and that is following, there is a polynomial time algorithm with input B and beta and B is specified to the algorithm in the way I explained yesterday with generators and relations. The data likewise, you just told what beta does to the generator. And the output is of course, the output is my symbol. If you combine these two theorems, then you immediately see that my first year follows because this finite a billion group is easy to compute it is a cocoon of the inclusion map from B and R. And also this map is easy to compute because we know how to do multiplication in our. So, these are the two theorems that I want to sketch the proof of in this lecture, and then our problem will be solved so that I can use this part of the backward or other purposes. Yeah, it turns out that it is best first to prove theorem three in the sense that if you have enough materials for doing that, then the proof of theorem to drops out almost immediately. And in fact, you can say that both of these theorems, ultimately follow from one important lemma. Let me make a remark about this, however, since I should admit that I am not going to prove the entire theorem, I am going to prove only the half that is of interest to us. And we will assume in the proof and in the algorithm that the cardinality of B is all the cardinality of B that is a finite integer positive. Either odd or even in our application, it is always all and therefore I will restrict what I have to say about theorem three to the case of finite a billion groups of order. The theorem is also true then be has even or, but then the algorithm is completely different. Actually, it's fair to say that it is easier, but it is of no interest to the to go be simple and you can find the details of the case that B is even the carnival B is even in the lecture notes, maybe just in the exercises. The main lemma, so suppose that I have a short exact sequence of finite a billion groups, a short exact sequence of finite a billion groups of all or and suppose that I have three automorphisms. One for each of these groups. So, and alpha beta gamma are automorphisms of a B and C respectively, such that they are compatible with each other, which means that this diagram that I am drawing in which I repeat my original sequence. Once, and here I have those alpha beta and gamma, such that this diagram is commuted. So that is really equivalent to saying that I have an automorphism beta of B as I had it before. It would be such that a man I view it as a subgroup of B is exactly mapped to itself, and that will then induce a unique map from C to itself, which will automatically be an automorphism as well. And the lemma states that this symbol that we are talking about is multiplicative in such exact sequences, and that is the key, both to what we need to prove to get the Google algorithm, and to the fact that theorem to is valid. This lemma is completely a role with auto hypothesis that the groups of order than there will be different lemmas that are valid, but there is not in all cases, the situation to express the middle sign in terms of the other data. Okay, so let me give you at least a sketch of the proof of this lemma, because it is central to everything that I will have to say. Let me draw a picture of my group here I have my group B, and here I have C, B, and there is C, and here we have the unit element of C. So the kernel that is this group that is lying here, maybe I should find some color here. So that is, that is the group A. That is a subgroup of B. What I do for the purposes of my proof is that I pick a section. So for each element of C, I pick one element in the fiber above it, but you have to keep in mind that this is only a set theoretic sex and the image well the image is this piece. The image is S of C. And if you want to leave this element you end up there as of this element is the one over there. This is good for us because now I get a coordinate system on my group B for every B in B, I can pick two coordinates in a unique way, a coordinate little a capital A and little C capital C. And then the property should be that B is, let's say, as of C plus little a. Yeah, so if be lies here. Then you map it down to see, then this SC is lying above it in the lower row, and then a will be here. So B is the sum. You have to add to bees, then you have to add two of those SCs, which is a disaster, because S is not a group of both. But now what I do is I look at to. I look at three maps from B to itself three permutations, and the first one sense this element of B which I write in my coordinates, I just apply my automorphism to a. And it goes to as of C. Plus. Yeah, I have this automorphism beta I apply it to a so that's the same as applying alpha. So that is alpha of a that is not quite what beta is doing, because I have not been touching as of C. And my second map, it does touch as of C and leave the a long and how does it touch as of C. Well, I apply my automorphism at the bottom. So I apply gamma is called gamma to see and then I take a section. I add a itself the a is left alone, and I have here this gamma of C, and it is clear so what is happening in the picture is that the first map is just on this on this first column on a is just alpha, and on each of the other columns of alpha if you just translate it so this is just as a commutation. It is just as many copies of alpha as there are elements in C. So this is a commutation and the sign that is that is the sign of alpha raised to the power to order of C because I repeat that commutation. Once for every fire. But she is all so I can cross it out. And likewise, this communication is just what gamma does in every role. So the sign here. The sign of gamma C raised to the power the cardinality of a but a has also order. So rather than crossing it out. I don't even write it down. If you now compose these maps you first do the first and then you do the second, then you do not quite get your beta. But it turns out that what you do get is something that lies in the right fiber. That differs from beta simply by translation so the last map. It sends SC plus a and now I get my piece of paper, because this is a very complicated computation and I don't want to make a mistake. So I take beta, and I applied to the S of gamma inverse C. This is what's happening. And then I still add a. So this gamma inverse C that lies somewhere in C. If I applied as to it, then I lift it so that this is an element of be that maps to gamma inverse C. If I apply beta to it, then I get see again so this actually lies in the fiber overseas. So this lies in the fiber overseas. So that means that what you are doing with this communication is that in each fiber, you translate by the difference between these two elements. In the fiber over C. This is translation by an element of a namely this element minus that element if I give this element the name star, and it is by star minus S of C. I don't really care which element it is with this element of a and you see a is a group of order. So, this element by which I translate as order. So we need fiber. This is a permutation of order and permutation of order. They are even. So this is even communication. The purpose is of computing science. It is completely irrelevant. And then you sit down as I have been doing this afternoon, and then you discover that if you apply these maps, one after the other first the top one next to middle one then the bottom one. The compose map is exactly the same as beta. That is something that anybody can check for himself and I did it three times so I'm pretty sure it is true. So the sign of this permutation beta is the product of the science. This one is even so that gives me exactly what I wanted to prove. This is at least a sketch of the proof of my main. Okay. This group of earth has to be sacrificed, however, since we are now going to use this. Remember to, well, let's first prove this theorem to so use the proof of your to the states that alpha B is the sign. This is a on the sorry sign of multiplication by a on this ring are will be. And if B is prime, if B is a primary deal. Then this arm of B is finite field. And if I, it has a unit group that a belongs to a will be belongs to the unit group and the unit group of a finite field is cyclic. So let's suppose for the moment that a generates that unit group so able be generates the multiplicative group of that field, which is the entire field without the zero. Right now all elements here as powers of a, then you see that the map multiplication by a is one long cycle of length. The originality of this group, which is the order of a, and one cycle, which is just the zero. So, in this situation, you can easily compute the sign of the map, because be as odd norm. So, this group has even order there's an element of two in there minus one. So it is a cycle of even length and therefore it is all, and therefore this multiplication by a is minus one. And if a is a generator B, then it is certainly not a square will be so that is also a on B. So in that special case that B is prime and a generates the group, we have a perfectly straightforward proof of theorem two. And if I now replace my a by powers well both of them are for fixed be our group of morphisms as a function of a, and if to group of morphisms agree on a generator of the group. So that means that the case that B is prime, the proof it. And in the general case, well we go to our composition series, and you see that you have be zero be one be two, and you end up with BT, which is being. Each of these quotients as an arm or you is isomorphic to our windows on prime PI. As we saw before, and that is isomorphism of our modules. So that means that this isomorphism carries multiplication by a to multiplication by a, and therefore multiplication by a on this quotient. There's a sign so X goes to a X on this group be I minus one B I, that will be the same as what happens on the right hand side, which is as we just proved my. A on PI. And now you simply apply the main lemma about tea times. If you know what the sign is in the first layer, and in the second layer, and you combine it in the combined two layers, and you keep going. And if you apply it every step is main lemma. So in the first case, be one will be to will be my capital a be not will be to will be my capital B, and be not what will be one will be my capital C, and then you keep going, and you discover that the sign of X goes on the entire quotient is simply the product over I, of these reshondo symbols, and that is by definition. So, so that means that this thing here X goes to a X R B is a product over I, of a PI, and that is by definition simply over the course. So that is the proof of theorem to show. So these Jacobi symbols are ultimately no more than science of permutations that is shopping, how much structure one can throw away here these permutations, the proof also may use of permutations that will not group water morphisms at all. So we are really making use of the freedom that we have. The main lemma is also going to be used in the proof of theorem three and theorem three is in a sense not really a theorem is an algorithm and it asserts the existence of an algorithm and the algorithm. So we use the beta B, then B and beta are given as input to the algorithm in the manner that I explained yesterday. And that makes use of another lemma, which follows from the main lemma and that is the following lemma. So, suppose that I have two integers, T and N, which are T is non negative and N is positive. Then what I will do in this lemma is that I will look at a particularly beautiful be, namely a be that is of exponent and it's killed by N. So it's a module over Z and Z and it will be a free module of rank T. So, then for B equal to Z and Z to the power T, and this power T that is supposed to be a repeated direct sum. So if N is a prime number, then this will be a vector space over Z and Z of dimension T, one has, well, for all beta in the automorphism group of B, there is an expression for design. And it goes as follows. If you have an endomorphism and also for endomorphism, not just automorphism, if you have an endomorphism of a free module over a commutative ring, then that one endomorphism has a determinant and that is a multiplicative function of the endomorphism. If you compose two endomorphisms, then the determinants multiply and that determinant that lies in Z model. If beta is an automorphism, then this will be a unit modulo N, because beta has an inverse and the determinant of the inverse is the inverse of the determinant. So this is a number mod N and the statement is that all you have to do to compute the sign is that you take the Jacobi symbol of the determinant over N. You can see that while previously we were determining Jacobi symbols through signs, we are now computing signs through Jacobi symbols. There is an important difference. This is not the generalized Jacobi symbol. This is just the plain classical Jacobi symbol that I started with my lecture with just for integers. If you believe this lemma, then at least for simple-minded groups of this sort, you have reduced the computation of the sign to the computation, first of all, the determinant, which is somewhere I believe in the notes in an exercise, I guess.  And the determinant in my case is actually in Z model and Z star. So how, so let me briefly indicate how you prove this. First of all, you filter this group using the devices of N and that filtration you attack with the main lemma, and you discover that if you want to prove this, then by the main lemma, it reduces immediately to the case that N is a prime number, an old prime number. Oh, I forgot to say that N was old. And if N is a prime number, well, then you like to filter your vector space with a set of invariant subspaces so that you can again apply the main lemma. But not every automorphism allows such a filtration, but it is very easy to see that the betas which do allow such a filtration, generate the group so that this formula follows without any further ado. So let me take this lemma for granted. And what you now do in general is that you apply a theorem from last time to write your group as the direct sum of cyclic groups with the property that these numbers. These orders of these cyclic groups divide each other in succession. And if you then look at B modulo NMB and you do that also for the others. So here you have a filtration of the group. And one is the exponent of the group. So this is zero. Then you will see that each of the successive quotients is acted upon by my beta, my automorphism. And each of the quotients will be free of a certain rank. Here the rank is M, there is M minus one, M minus two, we go down to one over a ring of which the order C mod NZ is all here. This NM here is the quotient of the two successive ends. The last one is N1 divided by N2. So you see that you can use this lemma to roll up the entire group and multiply the signs so as to get the algorithm that theorem three loops to. Prove the theorem that I mentioned at the beginning. It is certainly not the end of the story since if you know a little more algebraic number theory, then you will know that there are also higher power residue symbols. So one more of what is happening because there are no higher signs on symmetry groups. Well, that is true. And nevertheless the whole thing generalizes. And one way to find out what you have to do there is just sign up for the summer school next year chances are that I will tell you how to do it and maybe you can even figure it out yourself in the meantime you have a whole year for it. Thank you for your attention. Thank you, Hendrik for that wonderful lecture. Are there questions. Well, Hendrik, I was going to ask about the power residue symbols but you say I have to wait till next year so I'll wait till next year. Okay, sure. Well, we can maybe communicate in private about this don't get away. Okay, I don't see any questions so let's thank Hendrik for all of his lectures this week was a great pleasure.