 we were looking into filters, we last time did something about Butterworth. Butterworth is a polynomial which is given by a function b square omega which is equal to 1 plus zeta square omega by omega 0 for 2n and it is called the order of Butterworth order of polynomial. Now the idea in filtering aids that since you are looking for a typical pass band, stop band, ripples, these functions which are mathematical functions, they can actually fit to the required desired response by proper choice of order of the functions. The Butterworth functions are also called maximally flat filters, essentially they mean their ripples are extremely small ok. The last time we are doing this something about this, I just thought I will repeat where I started and I did derive out the expression that a Butterworth function, normalized Butterworth function can be written as 20 log 1 upon epsilon square omega of omega zar where this epsilon or zeta whatever it is is a function of ripple frequencies. So obviously, larger m means larger attenuation. So, one can see from this transfer function if I want to attenuate sharp, larger the n number I create, larger will be fall in the from output to the higher value to the lower value and I already showed you as I increase the Butterworth polynomial number order n is equal to 1, 2, 3, 4, 5, 6, 7 the sharper falls starts ok. Now this essentially means if you are looking for a very great sharp filters, also you need a very low ripple, you may have to require 7 stages, n is the number of stages or sections, each n is a pole, each n will give a pole. So, each section will have one circuit which will give realize a pole and if you are larger number of poles those many sections will have to create. So, if I see for example, for 7 one can see it is a very sharp filtering has been done, this is for a low pass, similar thing can be done for high pass. So, a sharper fall essentially means larger value of the function you are creating or the n being larger and that is the problem. So, we and as I said last time larger the n value, larger is the issue of money and therefore, the cost wise it is better if your n number is related to be smaller. So, I will give an example to solve my problem, let us say this is a attenuation function, please remember hj 0 by h0 the attenuating function maximum value is 1 and it will start going down that means start attenuating higher and higher and you can see larger n will attenuate at lower frequencies means sharper bands ok and also you can see the flatness of the curves ok in the pass band. Ideally what is the low pass filter response I am looking, I am looking as if something like this is an ideal low pass filter response and I am trying to replicate it by using a polynomial function which fits closer to this. All that we are trying is trying to fit the function as close to that and doing so we have figured out that larger than I put closer I will come to the reality ok. What normally I said you other day there will be a ripple ok and then there is a some kind of a fall which is a transition band. So, I want to reduce transition band, I want to reduce the ripple, I want sharper falls. So, I require larger polynomial functions, larger number means as I said each n will give a as we shall show you the function, each n will give one pole larger the number of poles you create one can understand why larger will fall faster each is 20 dB, 20 dB, 20 dB say 7 poles 140 dB it will go down ok. So, the trick of the trade is how many sections you can tolerate by your money in your pocket ok. So, this is the better worth function and one can see from here if you are noted down I will just show you ok. Here the HS is the transfer function which is nothing but V0 by VNS which has to be a as by BS and if it has only poles or no zeros it is called all pole transfer function it can be 0 but we normally want to have all pole transfer functions. Then HS let us say there are no zeros. So, AS is never at except at omega equal to 0 it is a pole probably or 0 but otherwise it is H0 upon BS and the function which is called better worth polynomial is given by B square omega is 1 plus zeta square or epsilon square omega by omega 0 by 2n where omega 0 is the cutoff frequency at which you want first pole to occur ok. Absalon is a I would not say constant say variable or rather you may say for the particular polynomial say constant which is a function of ripple frequency. It is some function which has we will actually evaluate that value of how much it should be or normally they may give zeta is typically taken 1, zeta is typically taken 1 in the case of better worth polynomial sharper this zeta is taken 1 in better worth but the zeta will be taken something else in case of the other filters which is shall be shown ok. So, unless assumed unless not critically specified use always epsilon or zeta whatever you feel as equal to 1 unless stated otherwise ok. So, if this is the better worth function and what I am trying to do I want to fit this kind of function into the pole 0 functions and I will like to see whether it replicates the desired response. This is all that I do when I say I am designing a low pass filter or for that matter any filter. If I calculate hj omega function as I gave you there it is h0 upon 1 upon epsilon square omega by omega 0 to the power 2n at omega equal to omega 0 hj omega 0 is h square upon epsilon square and normally as I say epsilon is treated 1 most cases hj omega 0 square by h0 is half. So, if this epsilon if you want the value it is 10 to the power a by 10 minus 1 is the actual gain h0 I say it should have been h0 there. So, the maximum transmission a max occur at this value and therefore epsilon value I can calculate, but typically as I say I choose the value of 1. So, this function goes to a value of half a max is no no it is a hj 0 by h0 magnitude of that ok. It is a normalized gain function. So, I named it separately for that is that correct hj 0 by 0 magnitude of that is 20 log of that is this ok. So, this epsilon is essentially something to relate with maximally flat conditions and maximally flat conditions means what the gain which I have already shown the figure as flat instead of ripple as flat as I get a response here that is what I say maximally flat because ideally what do I want sharp constant gain and fall. So, as close I choose epsilon I can get more and more flatness, but normally the maximum value which all filter people use is epsilon equal to 1 ok. I will give an example and I think that all things which you have worries will be very clear now ok. Here is an example which is very easy to understand. Let us say I want to design a low pass filter using better functions and we want an attenuation of 40 dB at frequency of omega equal to omega by omega 0 equal to 2. Is that clear? I am giving an attenuation at omega twice the omega omega there is the first cutoff. Is that point clear? Which value I am talking? If this is my omega 0 twice this frequency what is the value transfer function value goes down ok. That value I say is 40 dB down it should go by when the frequency doubles from the cutoff I want how much attenuation 40 dB attenuation I want ok and we as I say I use epsilon 1. So, if I write better function h j 0 by h square is equal to 1 plus omega 0 to the power n and since I am given h j 0 by h 0 is 1 upon 100. Is that ok? Why 1 upon 100? 40 dB is 1 upon 100 magnitude and that is equal to 0.01. If I substitute here in this function h j 0 h j omega upon h 0 square I put 10 to power minus 2 square 10 to power minus 4 equal to 1 upon omega by omega 0 is 2. So, 2 to the power 2 n evaluate what I am going evaluation what is that I am doing from this what the value I am calculating order of polynomial which will give me at twice the frequency from the cutoff the attenuation by 40 dB ok. So, I calculate n from this value n there this is equal to around 6.64 so roughly 7. So, if I have a 7 section polynomial or 7 or 7th order polynomial then what is the output of a filter will be that it will have a cutoff of omega 0 and at twice that frequency the attenuation will be 40 dB that means it will fall 40 dB down by double the frequency is that clear. This means so now you can see from here if I want even more attenuation what will happen let us say 60 dB this n will further increase 7 may become 12, 11, 10 whatever number it comes. So, larger the attenuation you are expecting at a smaller frequency shifts you can say omega equal to omega 0 itself that value occur it will be even n may be even more ok. At that point if I want very sharp at omega 0 itself then it will be even larger n number will appear in the real life and that means what is the problem in larger n. So, many poles you will have to realize so many realize each pole will be realized by one section as we say. So, so many sections will be required. So, you have a large low pass filter actually creating a relatively not great filter response because most cases you are not really looking for ideal low pass. If you are really looking for very ideal low pass yes all that we say 3 dB down if this starts falling it is fair enough. So, the problem start how much is that transition band you should allow and therefore that is where the next when it starts again because if you reduce do not reduce transition band then the next filtering frequency will also come immediately. So, we want to separate the pass band from their stop band. So, I want to as sharply I do it that is what I am trying to look at. Now, I have shown you one method of designing a filter by using a Butterworth polynomial ok. Now, what is the problem I said everything is fine repels are almost close to 0, but larger number of sections are required. Here is another function which does similar job and is called Sheba-Shev ok. So, here is another transfer function which polynomial which can create a transfer function which will also do attenuation the only problem with this transfer function is that it will have a ripple. The typical Sheba-Shev transfer function can be represented as and it has two possibilities when the frequency is less than omega 0 and the frequency is larger than omega 0 the transfer function has this kind of expression 1 upon epsilon square cos square this Cn is a name given to Sheba-Shev function. This is my Sheba-Shev function ok. This is cos square n bracket n cos minus omega by omega 0 is the Sheba-Shev function ok and this becomes can you tell why cos become cos h? e to the power j phi minus e to the power plus e to the power minus j phi by 2 is cos and if by 2j in case of signs ok and if you remove that j's then it will become cos h functions. So, whenever the function is larger than this according to this theory the function becomes cos h coins ok this is how the functions have been. This function has nothing to do with filters per se. This is a mathematical function which was used by filter theory by Mr. Sheba-Shev. This is not he only actually got this kind of function. So, how do you get new functions? You have a lot of mathematical functions available ok. You try to see which one comes closer to the kind of actual response you are looking for ok and then many a times in engineering what we do we put a fitting function on that. So, like for example, I told you give me any conic section I can always put any order of polynomial a 0 plus a 1 x plus a a to the some millions I can do but everything can be fitted I will get the coefficient. So, I need so many terms to evaluate so many coefficients to evaluate but the end of the day I can fit any kind of shape in that. Any random shape conic of course is the relatively periodic shapes but even any kind of shapes. This idea is therefore valid after you are looking for some kind of this response. So, you know initially it should be constant and then should fall ok or initially it should start rising and then become constant. So, we are looking for some kind of a function. So, easiest function. So, essentially even an exponential function which is at my cos and cos hs are appearing because an exponential function trying to fit something ok. So, any mathematical function which tries to replicate the response you are looking for can be used but these two became very popular because these can be implemented on a circuit much easier way. Why only these two? Because they were very easily implementable on circuit performance which we have blocks like opams, resistive capacitors. Using this many of them can be easily implemented. Therefore, these two became extremely popular but I am not saying there cannot be any other function better or worse than that ok. There cannot be many other functions. These two became only popular. Some other way we will say ok there are other kinds of function also fits in and maybe fits in better at some other positive ok. But for this course let us take it that these are the only two filters. So, if I have this Shevashar filter and it is always said all Shevashar filters are all pole functions ok there are no zeros in that ok. However, they always cost you on ripple. The ripple of all Shevashar filter is larger compared to Butterworth. Butterworth is called maximally flat ok. In the Shevashar there is going to be a ripple. But as in Butterworth sharper fall for lower number of sections probably if you can get compared to Butterworth at the cost of ripple then you are doing something better one way other something you are losing other is that point clear. I can reduce either a ripple or I can reduce the sharper fall. So, if I say a very sharp fall then I know I am going to increase my ripples ok. If I want relatively lower falls or sharper lower sharpness then I may have reduced the ripple itself. That ripple is essentially given by what we call epsilon term and it is related to a ripple attenu whatever equivalent to in ripple gamma in dv's by this function epsilon square is 10 to the power gamma by 10 minus 1. This is the expression given by Shevashar no proofs needed epsilon square is 10 to the power gamma by 10 minus 1 minus is outside. So, we say if gamma is half dv if the ripple is half dv is that clear earlier in Butterworth how much epsilon or a gamma zeta value I chose one only. Now, I say depending on the gamma I can choose different epsilon values is that correct depending on gamma value I choose I can substitute gamma half dv, 3 4 dv, 1 dv whatever ripple you can tolerate you substitute here that though many dv's and evaluate epsilon. So, for two cases which is the most popular case in Shevashar filters one is half dv ripple the other is 1 dv ripple for which epsilon is 0.3493 for the first case and epsilon is 0.5089 this is just maths nothing great and these values are also not evaluated by me I am just giving from Sadrashmi's book ok. Hopefully their maths their maths is also correct because they are also copied from Shevashar's first paper. So, hopefully Shevashar was right ok but it is very simple maths so you can verify on a calculator in a few minutes time I have not done it this case. So, please I do not want to gouge on that but must be right yes for the same 40 dv let us say at the same frequency if I get Shevashar filter which has been lower than this then I have achieved my part at the cost of ripple is that point what I said I will get I will give the example I will use the same 40 dv fall at double the frequency ok and I say the sections which I get how many are there less than the Butterworth if they yes I have achieved but what cost I am paying this either half dv or 1 dv ripple I am through which in the case of Butterworth I have no ripple there. So, if you are really looking only flat bands ok then you require larger n to use Butterworth's I agree section is a one pole realization what section word I am using n is one pole 7 poles I have to realize ok you are you have a point but let us wait for it. So, is that value I can calculate epsilon for any dv ripple whichever you are asking is that point clear to you is that point clear if epsilon is larger denominator is larger the dv will be sharper falls is that correct. So, one dv as a ripple layer ok the ripple frequency omega c which is also related to 3db point call it omega c or omega 0 ok. So, no ripple frequency omega c and omega h sorry the cutoff frequency here please remember where is the ripple coming again well let us look at it this is my good band this is what I am going to get and at this frequency 3db point this is related how is that correct this is 3db point that is our actual bandwidth up to which passband exists ok and this is the ripple frequency. So, this ripple frequency is related by Shebesha function to this corner frequency by this expression as I said the function has when the I change that I have used the transfer function which I have given you I start increasing omega from say 0 onwards. So, initially because of the function of cost is varying function it starts rippling ok at omega equal to omega 0 or omega h whatever 3db point we say the transfer function net value suddenly starts falling because of the other n e to the to the power n values which I have given as it starts falling I say ripple at that corner whatever is the ripple available I want to make relationship with this corner frequency to the ripple frequency. Essentially ripple is also not universe one frequency term it is a multiple frequency term, but the fundamental of that ripple is chosen as a ripple frequency whatever is the ripple frequency at that point that I am related to the corner frequency by this expression which is from Shebesha theory. I cannot assume as I say I am not doing Shebesha function because it will take longer time for me your next course you should hit them ok. The cutoff frequency 3db cutoff frequency is related to the ripple frequency ripple is as I say ripple is not constant it is like this like this is kind of fundamental frequency who score ripple bull thing or ripple frequency it has the largest component among all other frequency terms that frequency to the corner frequency correlation here ok. So let us take the same example I have a 1db ripple accepted n is 5 this is 1.03 relation saying don't know me ok what does this mean omega c and omega ripple frequency is 1.03 if the cutoff frequency and the frequency of the ripple is very close to each other what is the advantage you will get when you substitute in this the h function will start falling sharper at that point ok as close you come to the that means ripple is you know iska kitna yeh jojo frequency is mainly iska bhi frequency why iska kya matlab hai kiye asai hai actually there is a very little other frequency components going through. So it is almost becoming average value is very close to the maximally flat value average value that is what we are looking for the frequency of fundamental should be such that its average value is close to flat value is that correct uska amplitude itna kam ho ki wo flat value ke kari this is what shevesha is trying. So I calculate for 1db let us say same example is that clear what was the last example 40db down at twice the cutoff frequency we want when twice the frequency attenuation is 40db and now I am saying ripple is 1db 1db ke ripple ke liye epsilon ka value point pi 0 8 9 tha isme substitute karo aur yeh aapka shevesha hai function hai cn square by 2. Why this 2 omega by omega 0 is 2 I did this maths for you I have this shevesha function yaha likha hoi aapko phir se I substituted this value in this cn square is 10 to power 4 minus 1 upon this this I yaha se calculate kia cn 2 nikala why I am calculating I use this expression substitute attenuation and evaluate the value of cn square 2 is that correct aapko 40db jahi tha transfer function should reduce by 40db so 10 to power minus 4 this is 1 plus why a square kia mn a wonder root tha isle either square kadhyam expression so 1 plus epsilon square into the shevesha function square at 2 at omega by omega 0 equal to 2 so I went to evaluate this value from this I know this I know this I calculate cn square 2 to be 3.86 into 10 to power 4 that means cn the shevesha function at omega equal to twice omega 0 is 3 point under root and this into 10 to power 2 this function iski value a 1 196.5 and the shevesha function is cos h why I am using cos h omega greater than omega 0 this is cn 2 is 3.86 this is 196.5 equal to cos h n cos h inverse 2 so n becomes 4.53 is that clear is that correct 1db ka ripple except karia ifi half db ka karte toh kya hata thoda sa 5 ka bhaja 6 aata tha epsilon denominator mehna over point 3 jive value uski toh is ka madhna n section thoda sa number power over but it was saath ho gaya toh there is no point in then using shevesha filter is that clear toh shevesha kab use karte maal ka jab larger na bhaj sharper chahia rata 60db aur this tab jaroor shevesha use kare kiyon because corresponding Butterworth meh require 12 15 larger number shevesha meh require 8 9 10 so 4 5 section ka para kya so jab larger sharper false ab dekh nahi 60db 80db short then you better use shevesha filters if you are looking relatively 40db or something of that kind you may as well use Butterworth with no flat no ripple around so a thoda sa choice karna bata hai ki kitna akirati, kitna passband, kitna ripple accepted hai is hevesha say you can decide is that correct this is what the design of a filter is all about is that correct so sub se baile aapko kya batayenge we will just tell you that I want a attenuation at this frequency so much okay then you will use both functions is that correct and evaluate how much n that is giving how much n this is giving for both values of gamma also half db as well as 1db and then figure out which one you should choose I first take once I am given at this frequency this is the attenuation I am asking for you then you say okay I will use shevesha formula for both gamma equal to half db and gamma equal to 1db I will also use Butterworth function find for ripple 0 how much is n if n is much lower in Butterworth for the you want then you need not go for shevesha at all okay but let us say you are getting shevesha hai much smaller than the Butterworth for a even for a 1db ripple you better go for it is that clear so the choice of filled kind of function choose and kind of sections you can use is decided by what specifications filter is asking is that correct so it is the word clear why I said design because what will be specified to us so many db is down at this this is specified then we calculated what normally what we should do okay n is pi substitute get the value so okay it will give 43 db this is not what we have to do someone will tell this is the filter I want this should do like this then I should reverse we are going figure out what should I do should I use shevesha with half db ripple or 1db ripple or should I use Butterworth you have a choice so calculate with this kind of thing all three and choose whichever cost wise you feel is cheaper and acceptable this is what the design is all about so I this course I am trying to again and again bring to your notice that at the end of the day we are not analyzing things but to do this I must know the function themselves I must know how we are using them so I will analysis is only required to know what options I hold okay because at then I will be given by customer as we say all designs are custom designs what does that custom design means the customer specifies this is what I want you go to a shop and you make a choice I want this I do not want this so that is the choice a customer has manufacturer has no choice he has to provide he say okay you want this this cost if not you know it is even costlier a TV LCD 39,000 come be 65,000 come be LED 75,000 what you want money is accordingly okay if you tolerate repel okay that is the trick all designers follow please remember I my attempt in this course all through was to bring to your notice that the designers how designers look at the theory which we do okay always measured one because Butterworth people believe that epsilon 1k over flatness creep creep 0 a gamma is almost 0 degree flatness it is related to ripple so gamma is higher which me epsilon is higher means you are closer to the kind of ripples you are looking okay no in butter so ripple magnitude if it is given then you say okay if maximally flat filters gives much lower ripple anyway isn't it and it still gives you lower section so why do you want to have a higher ripple with a lower you already got a lower value of the choice is yours you can still use a measure but if I can give equivalent sections or around the same section 5 or 6 I may actually go for 6 because then I say I don't have to worry about because if I am giving you 0 dB ripple which is anyway better than half dB ripple you are asking for whichever you are this so you cannot that is the maximum allowed to you below any means allowed is that correct so if you say that your 6 section and 5 section you have to make a choice you are better make a choice of 6 because you say okay ripple is now no more there is the choice is something like this if I say half dB ripple then I say okay it is roughly equal to the up that 10 to the power gain by 10 minus 1 you evaluate epsilon from there so that value will tell you what is the flatness on the gain function you want okay that flatness means ripple on that the flat a max to have a kid now scope or over reading Karna Chhatiya that is your value of epsilon you are choosing you should say 0 fine okay ripple 0 fine so epsilon is 1 for that so use one they are always related by same function okay is that okay okay here is I have just copied it Xerox together the this is the order of polynomial Butterworth polynomial 1, 2, 3, 4, 5, 6, 7, 8 this is normalized polynomial BNS if N1 the function which I am realizing in Butterworth is s plus 1 that is a pole s plus 1 if N is equal to 2 I am realizing s square plus root 2 s plus 1 okay if N is equal to 3 it is s plus 1 times s square plus s plus 1 if N is 4 s s 2 s square terms will appear 4 means 4 poles so s square plus s plus into s square plus s plus something so 4 poles is that these are number of poles 5 means s plus 1 into 2 s square bracketed term is that correct 5 poles so you can see whenever there is a odd number of poles there will be s plus 1 single pole s a square root 2 s plus 1 or s plus 1 actually pole make a difference a square expression such that the other 2 always will be a conjugate poles whereas this will be a real pole okay so what should I have to realize I must be able to realize the conjugate poles as well as a real pole so if I have a circuit which realizes a single pole which is real and also can replicate a s square s kind of terms then I am having a transformation so how many if this is the this third one so what is the kind of function I am saying 1 upon s plus 1 into s square plus root s plus 1 this is the kind of function I am having from the Butterworth is that correct each I should be able to realize now okay what is the method I said if you have a 3 functions 1 upon s plus 1 plus 1 upon a s plus b 1 upon s plus so what they can be 3 transfer function this into this into this so I realize first function and output of that I give it to the input of the next I have product of the 2 now output of I need to the third stage of the third field also so all that I do is each is now clearly what is the section section means realization of a single pole okay so 1 upon s plus 1 is 1 transfer function into 1 upon s plus something is second transfer function so I realize individually individually each pole is that clear and since I can realize a pole I had just shown you the theory earlier filter come a pole the one of the Rc 1 Rc say you are 2 R or 2 Rc say you always say I say functions 2 R or Rc say can make a function 2 R or 2 C say s square terms are such a take Rc say single pole are such a so we should be able to then realize any number of such sections in series of that is that that so n 7 when you understood 7 means there will be 7 sections of realizing individual poles is that correct that means there are so many series combinations are going on to realize that function and as he said the penalty essentially is the power larger the sections each will drop our okay okay so once you get either you are going for Shabesha or going for Butterworth you are decided this is the function you use and implement this is known that I will if it is a Shabesha function I will implement this if it is a Butterworth I will implement that is that clear is that clear so that is the I have started within a s upon bs is the trans function is all pole s is 0 constant so I am actually realizing that functions which is the trans function I said is that clear okay here up book may take a set us with my poor a table side if I am asking you to design something I will provide you this table is that clear I will you do not have to remember I will provide you this table only thing is the n you must get it and choose among them which function to realize is that correct so how to design a filter I repeat given the spec find which filter and which order and choose expression from here or there this is for half dv this is for 1 dv okay this is I say please do not have to remember anything this is just to show you s plus something you can see the functions are not very different here only thing it was s plus 1 here it is s plus 2.863 the next term is s square 1.425 s this is even derived from the Shibasha and better functions okay so please look into book or rather as I said just for the heck of it I just say you these are the functions I am going to realize this is called silent and key low pass section okay now this words section is clear how many poles this is going to realize each of them inverting and non-inverting two poles so implement one upon s square plus something and that constant 1.425 r1 r2 c1 c2 values is that clear the constants which you are derived expression I mean you have got from functions had to be now implemented through r1 c1 r2 c2 values okay but that is normally taken higher than this you can see now your gain is now reduced so give it up you have a point this is 1 2 pole theory may n is equal to 2 is as a 2 poles so 1 s square sections is that correct n is equal to 5 means s 1 upon s plus 1 into s square a bracket or this s square a bracket 3 sections s square kale a 1 section yoh gaya na jar hai to doi section lagging or the punch wa to a s plus 1 ke liye be a section lagga wo dikhatao me haan wo casket karna is ka upot next ke upor as I section jo next ke liye banah rahein please take it this value just to give an idea let us take I am doing Butterworth and is equal to 4 of I so s plus 1 hai wo alak se banana hai 2 term a s square plus 0.618 s plus 1 ek yeh realize ho na hai 2 rha s square plus 1.618 s plus 1 yeh 2 rha realize ho na to yeh jo s square term wala yeh ek 7 keeka section jo abhi dikhatao abhi ekor section amko dena yeh 2 section lagging hai na paash ke liye is that correct each s square term can be realized by one section. So, you require 2 sections per 2 s square terms the third kaun saan hai real poll jahiye to uske liye ekor section lagga jo haan last me denge abhi dikhatao me is that clear. So, given the transfer function once you make a choice you can actually start realizing using 7 keys. So paash ke liye kitne section lagging hai 3, 6 ke liye 3 ne lagging is that clear abhi me ne pichhe war aapko ek jawab diya tha ki paach aur 6 ho toh 6 aah leh saptye ho ki section ke issaaf se 3 section ne lagging is that okay function wise is that clear and is the section is leh me wala tha ki me ne galak boljaya shayal n usse thoda tha confusion ho gaya s square ek section se aata in a 2 poles ekhi section se milte and n is the number of poles risk 5 hai toh paash pole hain, paat 2 pole ekhi section se manaya jaw saan is that clear. So, section wise I am sorry I made a mistake section wise s square term can be realized by one section another s square term ho gaya toh ekor section aur ek s plus type ka ho gaya toh ek 3rda section is that correct. So, 3 section I will show you an example abhi aap socho ki jaw aapka final transfer function plus a anaya maina khanaya waitha uske issaaf se sign laga uske okay, ek shayal inverting ho gaya toh 2ra norm inverting okay abhi me ne aapko bola ki I need conjugate pole toh s square ke maale thei but there may be a real pole requirement is that correct, a real pole requirement. So, here is a single real pole aapko kya vola ki RC ek toh load ho jayega nahi nahi RC is that the out this is the filter input is not loading input is coming here with high impedance on that so it is not getting loaded by this. If you say odd drive karna hai kahin toh maha bhi ek buffer daal nahi single pole mila ye one upon one plus RC hse nahi ek single pole mili aapko one upon RC par ek pole hai aapka. If inverting chhata hain toh normal R2 R1 laga hiye toh ek additional term aajaya achaya isko R2 by R1 ko hata nahi toh kya karna chhe hai equal kardo R2 is equal to R1 toh one ho jayega okay. If you want to gain fine if you do not want put R2 is equal to R1 so that will become one upon S by one upon R3 with a minus sign is that clear choice of R are yours so you can always get rid of R's. So, this is called creation of a real pole is that clear real single pole why it is called real again and again I am telling you okay. So, this finishes the filters band pass kese banayenge ek low pass banayenge uska omega zero nishit karenge ek high pass banayenge uska omega achaya iskaya hi pass banayenge toh kya karna chahiye R and C ko ulte karoge toh hi pass hojata is that kare replace R by C by R okay. Accept the feedback o R1 R2 ko C mat kar deng. If you see the function aabhi dekhha nahi kya figure meri. A R aur a C hai bas. Yeh bola nahi inko mat karna yeh feedback hai mo. Yeh C R, C R okay. So, you will get that high pass is that okay. So, I have now I can create a band pass band reject low pass high pass using Salon ki filters sections okay. Either using Shabesha coins or using this. If you see it is trying to do the same thing is that clear. So, there I did not name that is a Salon ki section but it is essentially a Salon ki section is that clear. Yes, aabh iska thoda tha analysis karo aur aapko milh sakta. Ek ko ashoon karte baaki sa proportion liye nikal. Okay, ek C ko lete hain uske proportion sabko nikal. Because over ratio me milen gya. Okay. So, this finishes filters. Ek aasai mujhe strike kiya thoda tha. Please take it. May I so dasa sadras me dekh rata toh. I figured out this is something which I did not tell you. So, I thought maybe quickly I will tell you something more interesting what they did for stability. So, that is why I named it revisit. Okay. What is the way I told you to find the stability of a amplifier ki aap AOL nikalo beta nikalo AOL beta bana aur loop gain plot karo. Okay. Aur loop gain me uska phase nikalo phase ke upar aur edhi wo gain margin ne ne positive hai gaya japki phase 180 degree cross ho gaya toh unstable hai aapko bataar. This is very interesting match has been done by Sadras Smith and I thought you can read that more detail. This is just to give his idea which is fantastic. So, I should have also this we do in other method, but I hear they think. This is the loop gain is A beta. Is that clear? How can I write A beta? A upon 1 by beta. A beta can be written as A upon 1 by beta. I take logs. So, 20 log A beta is nothing but loop gain in dv's which can be written as 20 log AOL minus 20 log 1 upon beta. So, abhi hum kaya kaya nahi ki you do not find 20 log A beta at all loop gain nikalo emat. Plot AOL versus omega bode plot for open loop amplifier. Is that point clear? Do not get A beta just plot bode for magnitude of AOL versus omega and phase corresponding. Is that clear? Then I want to subtract. This is that 20 log M is bode plot of that. Beta is constant in this case. So, 20 log 1 upon beta is a constant value. Ab beta jaro nikalo. Ota ab nikalo gaya nahi. So, 20 log 1 upon beta ka ek line banao jaha par. Kitne db hain ho? This is our without this this is our AOL transfer function. Okay. Normal open loop ka poles hain jitne bhi 3, 4 jitne bhi hum. Jaha par mein hain ek line lagai aisi which has the value of 20 log 1 by beta. Okay. Iski jobi value hain yaha laga di main hain. Respect line. This is our AOL. This is 20 log 1 by beta. This is 20 log A. The difference kya hain? Loop gain. Yaha par kitne value uski? Loop gain ki value kitne hain? Zero db difference hain hain? Kitne value? One? Nahi to bhi hum beta hain? Phase kaha nikalo hain? Chapki loop gain one hain? Faiz dekh rahit hain hain? Normal method bhi main hain kya bola A beta versus omega plot korot jaha wo one huta hain? Faiz dekh ho. Ab main hain kaag hi hain to loop gain ki value one hain hain? Iska phase dhundo kitne hain? Phase diagram par. Jadi hain 180 se kum hain to stable hain? Cross kar gaya. Minus 220 hain. Minus 185 ho gaya. To unstable. To jaisahi apne beta fix ki aap, aap beta yaha alagalag line bana sakte hain? Is that point clear? For different feedbacks, beta values, I can have different straight lines which means the cutoff point will change. Is that correct? If the cutoff point change, yaha kya stable lagta hain kya? Sambhya yaha 160 db aaya 160 degree minus aaya. To stable hain? Stable hain? Jadi aap niche jayinge. Sambhya yaha aah rakhinge. To ho sakta yaha par jo pole ho gaha 180 cross kar gaya ho gaha. So, if the 1 upon 20 log 1 upon beta 50 se sambhya 50 db hain, ye 100 db hain to 50 db jam niche aaya aap, to phase kitna kar gaya cross 180 cross kar gaya. To ish feedback kele system unstable ho gaya. Ish feedback kele system unstable ho gaya. To choice of beta can be actually predefined, kitna beta hu de sakta yaha aap ki so that remain system remain stable. Is that correct? AOL. No, no, no, plot the phase of AOL. Jahi main batana jo normal amplifier ka bode plot karo. Is that correct? All that I am saying what is this point? Loop gain 1, par phase to aaya uska jo niche hain. Jadi ho 180 cross kar raha hain minus 180 se abhao ho raha hain to system unstable ho gaha hain. Jadi uske niche rahe raha hain to system stable. So, a given transfer function jayi aap open loop ka equivalent banayinge uska hain bode plot kar bide. Beta ka value plot karia ho niche dekhye phase kitna hain. Stable, unstable wohin pata raha. Is that clear? If this is 1, this is 0. A beta 1 is 20 log that is 0 that means these must be equal. At this point both are equal. Okay, given feedback network, sir amplifier, first make AOL jo to aapko bataya hain ki open loop with loading chahi aapko. Wo AOL s nikah lo aap. Usko bode plot bana uska. Aap jo feedback dekhye aap beta network ka wo beta nikah lo. 20 log 1 upon beta ki value nikah lo. Jo AOL ka transfer function ka bode plot tha usme ye 20 log 1 upon beta ki ek line laga ho. Wo toh constant hain na, beta constant hain. Wo jaha bhi intersect kar rahe hain. That means this value is equal to this value. That means loop gain is 1, okay. So this is the point where loop gain is 1. And we said in stability wherever loop gain becomes 1, find the phase. Okay, that is phase margin should be plus 180 se pahin niche rahe na chahi aapko phase plot pe toh was stable hain. So that 20 log 1 beta is lower value now. That is feedback is larger let us say per se. Aap yaha toh 2-3 pole hoonge kahin bhi hoonge. Toh ye point aage aapke pass. Jaha pe these are equal. That is loop gain is 1. Jaha pe phase do not do. Jadi wo 180 cross kar chukah hain toh system unstable hoonge. Kaha par ye 0 margin hoonga. You will have to find when they are equal. Find the value of such value of beta where the phase is 180 degree. At that point deko samjo ye 180 ka point de samjo yaha pe. Sabye yaha pe. Ye 180 aata phase. Toh ye value jo hain, iska ye 20 log 1 upon beta, se jo beta value aega that is the minimum that is the minimum beta you must use. So that the atleast stability start hoonge. Usse upar jaana aapko. Usse niche rahe gaye toh unstable hoonge. This method has been provided as an alternate method in Cedra Smith book. Alternate dikhayon hoonge isle main hain pada bhi aapke. So aap ek baar dekh lo because this is much easier in my opinion. But if you are using my earlier technique of finding loop gain and plotting is nothing wrong. Anyway aadha kaam do karee rahe, aav bhi nikal rahe, beta bhi nikal rahe, product bhi nikal satne uska phase bhi nikal satne. But this is easier way of doing. Is that clear? This is what they have suggested and I thought you should know. This next analog block which is required by each and every one of us in our system is oscillators. When I once I said give you some joke that whenever I design an amplifier, it oscillates. And whenever I design an oscillate, it amplifies or rather attenuates. Oki. O filter jesa kaam karne. So iska mada oscillator ka criteria kuch hoonan jaya ho mahi par wa exact oscillate kar jaya. So let us say what is that criteria. This is our standard feedback network. The only difference now I am saying compared to the earlier feedback network. Kya farag dikhara hume aapar? Anyone? X is input and X is output. That is why. Ye kya dikhaya mainin? Beta is, pahele kya liya tha amplifier ke sath? Beta is constant. Oki. Now I say beta can be a function of frequency. What does that mean? Beta mein ka network kya ho sakta aap? Ele frequency term haane toh? Or inductance kuch bhi ho sakta hain? So now you realize that you have seen oscillators mein tank circuit lagate L and C. Ye R and C. This is exactly what it means. The beta network has now frequency dependence. All amplifiers we avoid that. We want to abhi dikhaya ne ek figure beta constant hain. Nyaa beta frequency dependent. We already said loop gain TS or in some book LS, AS into beta S. Hence if 1 plus A beta is 0 or A beta is minus 1, transfer function ke value kipne hogi? Infinite. If A beta plus 1 is 0, new denominator is 0, transfer function has a value of infinite. Is that clear? What does that mean in amplifier? V0 by Vn is infinite. What does that mean? Without having input Vn, I can create V0. Is that clear? V0 by 0 is infinite. Is that correct? So without having an input, I have an output now. Is that clear? By using a feedback theory which I said. Is that point clear? Without having any input, what is gain V0 by Vn? If I say gain is infinite which means without input, I have output. Is that clear? This term without having actual input provided, I actually can get constant value of V0 is called oscillator. The condition will lead to oscillations. Is the principle clear to you? So what we are saying? I pass some input to A which becomes X0, beta of that I return it to Xi and if it is such that this adds to this, then increase X0. Afterward, if I, it is certain value even if I remove Xi, it will still start feeding some input output. When that occurs, then we say we are having a oscillations. Because initially that means initially kahi to b Xi hona is that clear? What this X0 will be initially? No, actually in an oscillator, you never give any input. So how does it start? Noise. So a little bit of noise can start oscillations and it will grow to a standard frequency oscillations. Is that correct? So that is why no real inputs are actually required to start oscillations. As soon as you switch on the power supply, a pulse happens. You do not have to do anything. So there is a person called Mr. Barkhausan. He suggested a criteria which will give us unified oscillations. Now what did I say? Loop gain should be minus 1. So at any given frequency say F1, the magnitude of this trans function is 2 pi J F1 is equal to 1 magnitude and what is its phase? Because of its minus 1 value minus 180 degree. Is that correct? Minus 180 degree. So what Mr. Barkhausan says that at any frequency if the magnitude of this loop gain is unity and its phase is 180 degree then you will have oscillations. This is one possible condition. Let us say A is a 180 degree phase out input to output gain 180 degree. Minus GM, 180 degree phase has also come here. 180 degree has also come to you. So how much total phase has happened? 360 or 0. Is that correct? So the return signal you got, how much it has sent? In phase to the input. Is that correct? So you got a little input. It went to a little output. Return it to the in phase and then to the input. Okay. It increased again. You got more feedback from it. You got more input. Remove the input. There is no noise. In a little while it will go somewhere in a condition where the maximum beta that can be allowed, the maximum output you can get is stable. Is that correct? Now this is what oscillator is trying to say. This criteria says the magnitude of loop gain at a given frequency if it is 1, unity and its corresponding phase is minus 180 degree then oscillator, oscillation and F1 can be seen. This kind of alternative system is possible. If you take the transfer function here and you say its real value is minus 1 and imaginary value is 0. Even then this condition is satisfied. Is that clear to you? Please take it. Maths may both are possible. Magnitude 180 degree or 1 whole supply. Yeah, in transfer function imaginary value is 0. That is how much is the phase? 0 phase. Fair enough. Okay. But the imaginary part is 0 and the magnitude is 1 which is also is you have a problem situation in which system may become unstable or oscillate. This is what essentially Barkhausen suggests. And real life, how do I achieve that? That is to say oscillation conditions will be, phase shift through amplifier and feedback network must be such that it is either 0 or 360 degree. That is the feedback return is in phase with the input and the magnitude of A beta is 1 then you have a condition of oscillation called Barkhausen criteria. Exactly. A beta is okay but there is a return okay. Assumption is that amplifier gives you 180 degree. Is that correct? Yes. So this criteria is called Barkhausen criteria. So we have to adjust the circuit that you have to make. Output to return input should be how much is the phase? 180 degree. If you give an amplifier 180 or 180 feedback then the input is back to 0 degree phase. Okay. A amplifier is non-inverting. This is minus. If you bring this RC network in phase component 180 from here, 180 from here total essentially saying feedback is 180, this is giving 180. Return path is 0 degree. 0 degree. Amplifier of 180, feedback network of 180, total 360 or 0 whatever you call it. Okay. As I the condition you have adjusted, it is called phase shift oscillator. Why it is called phase shift? 180 phase is shifted by another 180 to create an oscillatory conditions. So in RC constant code, time constant. Okay. Is that clear? We will come back to it again. I just want to show what is Barkhausen's criteria. Okay. How is it different from this case? This condition is start of instability in amplifier. Is that point clear? This was the point. A beta equal to minus 1. Is that now point clear? Why sometimes amplifier do not be sure like amplifiers and oscillators do not because their phase may not become exactly 180. Then it will not. It will slightly grow down, die over. So, oscillator will not oscillate or amplify. Amplifier oscillators or growing functions are damping. That is what the stability theory suggests. That you must get your phase and magnitude. Exact. Okay. Is that clear to you? This is what essentially Barkhausen suggested based on this. We will have some three, four, five kinds of this. Then we will start moving from after few oscillator I will not go all of them next time I will start few of them and then I will start for you most important part in this A to D and D to A converters because the world is digital thank you.