 So, welcome back after the tea break, you must be happy that it is the last lecture, but I will make you happiness even more, because I am not going to give any lecture at all. So, I have converted this to a surprise tutorial, because I wanted to tell, no, no, I am not going to give you any exercise, I am going to do it, because I wanted to show you how a tutorial is conducted by me, when I take students and I do not make only them to do it, I do it myself and then ask them to do it. So, what we are going to do is tutorial on, and what I want to impress on you is that the tutorial has a very specific purpose, just as laboratory has a specific purpose. The tutorial has to bring in the mind of the student that connection between what you have been teaching or what is in the textbook to reality. So, reality is that some such circuits are actually used, required somewhere and so on, and I am teaching the theory of it and now you have to make this connection. So, making this connection is also the same purpose as in lab, that you have to make this connection that when you teach a common mode amplifier, the connection should be made that I want to make a audio amplifier, not a common mode amplifier. The laboratory experiment cannot be called common mode amplifier, because that is the same thing in the book, it has to be called a audio amplifier. Same way if I conducted tutorial, I cannot go on teaching the theory as the excuse or I can give you only numerical, I have to make that connection between what is start and what is the numerical and what are the realities. So, I have a few things and like yesterday you can also have access to this later, but I am going to show you something of interest. So, I am going to show you a few illusions. So, this lecture will have less jokes and few things like this to make that connection. So, you see there are some dots and I want you to look at it carefully, focus on that center cross point plus, I have to also do it myself from here, you know apply it again because you are not doing meditation, if you do a meditation it should not come back again, everything disappears. So, let me if you focus on your task, all the other paraphernalia goes away. So, that is the important focusing. So, what we are doing is to focus, no, that is the next one, no, only one dot moving on the circle, from here it is even better. So, the nearer you are to this, the easier it is to focus, the further you are and the mind plays games on it. Now, this one, what is the color of A is some gray, color of A is some gray, color of B is some gray or gray is in A and B same color, obviously not. So, I did this exercise, I went to Microsoft Pent and with a dropper, the dropper picked up this color and made a rectangle like this and filled that color. So, I took a dropper, I picked up the same color and I filled a rectangle. Now, you look in your A it looks perfect, look in your B also looks perfect, but A and B are still looking different, but they are the same color, so this is optical illusion. This is real man, the other one required meditation, concentration and all that, this does not require concentration, this is real color in A and B are same. Now, appearing different, why, because there is a contrast, there is a shadow of this green cylinder falling across and the eye has this perception that it only remembers the contrast, it does not remember anything else. So, if you look at a very bright object and close the eye, that bright object becomes a black object in your eye, you can try that at night. So, eye and the brain perceive only contrast, it does not perceive any real color. So, this is the syllabus of your university all uniform and you are all 20 people teaching the same syllabus in different, different colleges, but the people teach differently, professor A will teach in this way, so it appears bright, professor B same, professor C syllabus appears difficult, same syllabus is being taught, depends on who is teaching it appears difficult or it appears simple, the syllabus is same, I did not make any change, I just change the contrast and it looks different, I can go back one step to one, absolutely same. So, you please pick up your style of teaching, make the syllabus contained brighter, so it is in your interest and it is in everybody's interest that the student get interested in what you do and what is interesting is in India, you respect only the age, that is why you are all polite, saying that ok sir, sir, the same thing when I was 25 year old, I will show that you know you will hate the fellow, what is the young fellow think of himself, making a joke instead of giving a lecture he is telling any other thing. So, I think you all as soon as you become older and older, you will get more and more respect, which is natural, but the younger faculty when they face the first year class, you know they get all the you know shouting and laughing and banging table as soon as they turn towards the board. So, they got to be bright, they have got to make the course bright for the student to listen, we become older then it is ok, anyway you teach is fine, as long as you cover the course it is ok. So, this is the acute problem, so Laplace transform in circuit analysis, so I am going to take this quiz. So, one is how do you do apply Laplace transform to a differential equation, that is the fundamental, then you take a circuit, so no equations are given, now the circuit is given and from the circuit you have to write differential equation if necessary and then apply Laplace transform and then apply Laplace transform. The third one is the topic for today, I want us to realize the question phi a by using op-amps, that is the topic of today. So, this called long connection to op-amps, this is the filter and then I must make this filter using op-amps. So, I will not do the first two, the topic for today is how to make this filter, so this is the filter. So, s represents the Laplace transform variable and these are the numbers, whether you call it i of s or v of s is the same thing. So, my strategy always is that I have got this written out, so I always ask the students whether they are like the answers to model answers to this quizzes, they say yes sir. When I ask you teachers whether you need model answer to the quizzes, you will also say here, so that is human. So, I do the solve the question a like this, which I am just going to scroll now and I will actually do it with you, but I am little more tougher then I say b please do it yourself, I cannot give everything. So, like many good books now they give only answers to alternate problem, even the numerical answers are not given to the alternate, that means the student has to get that confidence that what he has done is correct. If he keeps on taking always the answer is correct, then in the exam all what is he supposed to do, naturally he has to ask to the nearest student whether his answer is correct or not, he has no choice and that is why equivalent to copying. So, for that also I have found a good way because here we can set our own paper. So, I give a circuit and ask them to do it by two methods, so they do not know how to ask anybody answer to both the answer by both methods have to be same. So, I think that is a wonderful tactic in the IIT we survive like that. So, they do not have to ask anybody, they have to ask themselves, the method they have confidence in the second method must get the answer which we have got with the first one. So, we are now not going to do B, I am going to give you as a take home later on you can do it. So, I am just going to solve this the first problem, but I do not need the use of this. So, when I begin my tutorial I first give all the theory. So, I give all this theory. So, this is the summary of previous lecture. So, we will skip it. So, this is the governing differential equation. So, this theory is for the Laplace transform. So, I am going to now have my problem was there, I do not remember the exact numbers, but I of s equal to 6 upon s. So, professor how far can you go? He only remembers 1, 2, 3 he cannot put anything more. So, some text book I even quite remarkably surprising data like one home register and one farad capacitor. I think the fellow who wrote the book does not have any idea how much one farad capacitance will be to occupy the whole room. So, like that is all 1, 2, 3. So, are we going to make a frequency of something 1 hertz or 2 hertz or even less than that, but this quite popular such examples are popular. So, we have this equation. So, this is a integral differential equation and then we can have the familiar first year form you know d, d, d etcetera. And then we know how to find the roots using the complex variable we substitute we get the formula which is in terms of s. Now, this is the most interesting part is that Laplace and Heaviside realize that e raise to s t is a unique and only function in the entire match known to date, date meaning that 150 years ago which maintains the form e raise to s t and the both differentiation integration and hence must be a solution to any differential or integral differential equation does not make any difference. And therefore, they immediately use this and came up with this particular form. Heaviside an English engineer for the first time use complex variable for impedances of inductances and capacitors and he generalize the impedance as inductive reactance of an inductor is s into l. And capacitive reactance of a capacitor is 1 over s c we know this now and he was the first fellow to also say that if you have got a s c circuit with a single frequency source in the circuit what we can do is just substitute s equal to j omega and everything is fine. And the j omega converts all your impedances into the corresponding real imaginary parts which is called Cartesian coordinates and the other one is called polar coordinates the magnitude and phase. So, the concept of phase and everything was very easy with s equal to j omega Laplace transform was developed and I do not want to go to the integrity of the 0 plus infinity infinity thing, but it is very easy for me just define because I assume that the student want to ultimately know what is the advantage of doing that. So, we quickly go down if you have a f of s the derivation just I skipped above is that the Laplace transform of a first derivative with time t is s of f s minus f of 0 which is the initial condition. Of course, you can apply the similar thing to the integral and the integral become f of s upon s and f inverse of this upon s f inverse is nothing but this integral. So, it is the charge on capacitor then it will be a charge integration if it is a current in inductor then it will be the initial current in the inductor. So, now we have a second order differential equation we will ultimately get this form the second order differential is s square i s s of i 0 and i dash of 0 that means this the initial value of the first derivative same way first s of s s of i s and first initial value of the normal value. And similarly the integral over the earlier part the interesting part is if you transfer various terms on the right hand side. So, we have f of s any initial condition was 0 this would disappear if the initial condition of the first derivative was 0 it will disappear again this and if all integral prior to 0 was again 0 then this is also disappear. So, it will be i s will be equal to f of s divided by this this is what is our usual relationship if all initial conditions are 0. So, ultimately you get a form which is i of s equal to a numerator and denominator polynomials the only condition applicable to remember is that the order of the denominator polynomial polynomial has to be more than the numerator polynomial. Because if it is less then we will get constant term if they are equal we get a constant term if it is less then we will get terms in s and so on. So, if it is equal we get a constant and this numerator denominator, but generally we are given polynomial in the numerator is having higher order. So, let us see because students will like to know such things suppose I have got a unit step function and the symbol is u of t u of t 0 for all values less than or equal to t up to what we call t 0 minus and is equal to 1 for t greater than 0 including 0 plus. Now, how do you show 0 plus to a student I have no choice, but to draw a graph like this and the graph has to be drawn in this way which is a little bit of cheating because that is not correct. This is that peculiar small difference between the Laplace transform whether value at 0 of current should be taken 0 or 1 just to differentiate that at 0 plus the value is 1 at 0 minus the value 0. So, whenever you have any confusion you must use 0 plus. So, the Laplace transform when you integrate also it is from 0 plus to infinity. So, there is very minor detail and the diagram has to be drawn like that to indicate that 0 plus becomes 1. So, at 0 or 0 minus it is 0. So, you do this and you get it as 1 of an s that is well known. So, if you have got a input which is in time domain a step function then I can actually use that as a Laplace transform 1 by s. Exponentialization it is 1 upon s plus alpha. So, Laplace transform e raise to minus alpha t and alpha is this constant associated with this negative exponent and become plus. If it is e raise to plus alpha then it will be s minus alpha alpha is already the sign is already included. So, when this is minus alpha it is become s plus alpha. If it is a ramp function then it is 1 upon s square it is all done by integration by part for sinusoidal slightly more lengthy, but we get it here function of sign of omega t we get omega upon s square plus omega square and for cos of omega t we get s upon s square plus omega square. But we will be intelligent enough not ever required to use this by using the substitution s equal to j omega. So, this is the table and in IIT we allow all this thing for student to take to the exam hall because we do not believe in mugging up. So, this is called a formula sheet and then they can carry it inside and we give two marks for a very good formula sheet. There are many words for that, but I like the formula sheet as the best one all others have a implication of copying and other things, but since it is allowed because otherwise they keep asking me as a paper setter. So, it is 1 upon s plus alpha s minus alpha and things like that. So, I remove their numbers by allowing them the sheet and therefore, we now have such a circuit I of s is simply this and then P in of s. Now, we put whatever is the V in if it is a unit step function then we put 1 upon s if it is anything else we put appropriately that. So, you see this one has the making of one term having s in the numerator and 2 poles in the denominator and if V in of s is a unit step function then I would have got another s here. So, it would have been like this with another s here. So, this particular circuit has got that kind of form. So, I am only interested in demonstrating the form I am not going to solve this I am going to solve only the filter application with op-amp that is our target, but this is the background I have to give. Now, there are 2, 3 possibilities one of the possibility is that I convert this since your teachers I am telling you, but students do not know that I can convert this to something called a upon s b upon s plus 2 c upon s plus 3. This is called partial fraction that means I can find this is an equivalent and this being identically equal to this I would have got all the constant a b c. If I put identity then a b c can be found, but heavy side was a very good friend of us engineers he gave a very interesting and very simple procedure this n of s as I have told you already includes the f of s that is to the circuit whatever is the input you are given is Laplace transform also included in the numerator remember that v in s was there. So, that includes it and this one totally depends only on the circuit it does not depend on that. So, whatever is the circuit that you connect to the op-amp it will depend on those impedances. So, order of d is usually higher than that of, but if it is equal we have to separate the constant and then have the remaining part as order here is higher. So, now we have got if this is s is a single one then we have got n plus root n plus 1 root s 1 s 2 s n. So, he said that it can always be represented as a sum of fractions a 0 a 1 a 2 a n by s n and he could easily give this procedure and this is one of the simplest. If I want to find the constant which is belonging to a of k say second all I have to do is to multiply by s plus s 2 this entire formula that I have here entire formula I just multiply by s plus s 2 that s 2 term in the numerator denominator gets cancelled leaving everything behind and s equal to minus 2 s 2 will give me the constant a 2. So, this is that a k is nothing, but multiply the entire i of s by s plus k and substitute for all other terms a value s equal to s minus k minus s k and therefore, it will be very easy for us to find this constant. So, one such example is that this constant is this one it does not matter. So, now we have got so many terms and then we do this we can actually use this as a function. So, if we multiply by s then we will get the constant a 0 and then we have to substitute by s equal to 0. So, then 1 divided by s 1 into s 2 to s n simple procedure having got this a 0 a 1 a n for this demonstration I do not have to do anything then I take the inverse Laplace term because this is in terms of the Laplace constant or variable then getting the time function. So, I get the time function a 0 into u of t a 1 into e raise to minus s 1 t etcetera. So, I have very carefully avoided any sinusoidal terms coming in because I want to only use the sinusoidal terms as s equal to j omega not unnecessarily complicate this. So, I think it is a good practice not to complicate inverse Laplace transform with normal substitution of s upon s square plus omega square for cosine omega t and omega upon s square plus omega square for sin omega t because it is unnecessarily more complicated. Then there are two additional things if you want to convert any function to the time domain then if I want all these. So, there is a initial value theorem. So, limit t tending to 0 if I want to find for the f of t, but I am not given f of t I am given f of s. So, from a given f of s I will how I will find the initial condition time t tending to 0. So, all I have to do is to do this left hand side. So, set limit s tending to infinity of a function called s into f s and that will give me the initial value exactly similar procedure. If I want to find the time function f of t when t tends to infinity same as s f of s when s tends to 0 it gives me the time equal to infinity. So, therefore very simple examples I have worked out, but we will not need to do that we will only need to do what is coming in our example for the. So, at the end of lecture I would have done such a circuit and I actually find applying loop analysis here r plus s l only here and like that and I just put the values and I actually find this for calculating I 1 or I 2. So, I get I 1 of s is a polynomial here can you see that polynomial here. So, I 1 of s is a polynomial like that l is a constant s plus 2 upon s s plus 3 s plus 1 this is a real circuit giving me a function in Laplace transform. What is the advantage if I go back to circuit now I have got the freedom I can give a time function like this or I can even give a sinusoidal function because my function is independent of that only my numerator will change depending on what I give as a input. So, I have already used numerator as 12 of ut in that function, but if I use a sinusoidal function I could have easily exactly got that function with the sinusoidal one which anyway I get it by putting s equal to j omega. So, I think this example with a real circuit puts the student that they can not only derive the transfer functions and then also solve it using the Laplace transform. You can also do it by taking the partial fractions to find a 0 a 1 a 2 once you find this constants then it is a very simple matter to invert it. Secondly, I always ask the student to check this initial value of this by using the initial value theorem and final value of this by final value and therefore, everything a student has this problem I told you in the exam hall he does not know whether he has got the right answer. So, this is the formula you do this from a real function you can find the initial value put time equal to 0 put time equal to infinity you get a final value. You can get the same from the original s function by applying the theorem and both match then you are happy. So, all you have to make a student happy and in the process you have checked whether he knows both the method. So, both ways also he gets. So, even if the examiner does not ask him to apply the initial value theorem and final value theorem we should encourage them because this is the best way to check the answer it tells. So, this is the habit we have to grow into the student that they have to know if there are alternate methods to check the actually. So, this is the tutorial prerequisite for conducting our. So, let us say answer to Q 5. So, here is the I of s and this includes V in of s and etcetera and then we can find this. Now, I am giving you a practice for doing this when I actually want to implement using op-amp I will use this procedure. So, this procedure is 6 upon s s plus 2 and s plus 3 a 0 by s a 1 by s 2 a 2 s plus 3. So, heavy side procedure a 0 is nothing, but multiply this entire thing by s. So, s and s get cancelled substitute s equal to 0 in this. So, you get 6 substitute s equal to 0 you get 2 into 3 1 then a 1 multiply entire thing by s plus 2. So, you are left with 6 upon s plus 3 with s equal to minus 2 this we have said to 0. Therefore, it now remains inside this minus 2 and minus 2 plus 3 that is plus 1 therefore, it is minus 3 and plus 2. So, this is the way we are going to get the constant. Now, I have to do this exercise with you because op-amp tutorial I am not going to show you here, but I am going to show you on this one. So, are you with me up to now? What are the constants we already got? So, now I go to my good method called create satellite problems which are very simple to difficult. This is the difficult problem that is our target. So, the simple problem is first simple problem I of s equal to say 6 upon s plus 2. So, make a single filter and we can easily check as omega goes on increasing this reduces. So, it is a low pass filter and there is a cutoff frequency here at omega not equal to 2. So, can we plot the Bode plot for this just by visual thing? So, this is let us take 1 like 1 or 10. So, this is my output I of s is 10 log 10 of. So, now I get this is 10 dB and then this is 1 dB 0 dB sorry 0 dB and this minus 10. So, at omega not equal to 2 I am going to get my point at which it is going to drop 10 dB per decade. So, 10 log 10 I is correct or should it be 20 log 10? When is it correct? 10 log 10 power. So, what is the log 10 to the base 10? 1 and 20. So, if I go from here with 20 then it will drop by 20 dB. So, it drops 20 dB per decade. So, this is the Bode plot and this is the frequency at which it starts moving and then there is a exact 3 dB difference So, this is also called a 3 dB point exactly half power, but we do not want to go to that detail. Now, I want to make this using the op-amp. So, let us revise our op-amp basics what is this call it Z 1 V in. So, V in of t. So, it can be any signal that we give and we do not have any positive feedback only negative feedback. We are not making any oscillator or multivibrator. So, we have this function and we are very familiar that this function we only use R and C very popular R and C. Why are inductors never use this kind of circuits? I do not know anybody know? Well theory is very similar why only everybody take R and C not inductors. Secondly, we cannot integrate that easily inside of inside a VLSI or LSI very difficult to integrate a 3 dimensional thing which has about a magnetic field involved with that. But even if it is a external component they are very bulky and then get affected by any other magnetic component being nearby and so on. So, we do not want to do that. So, here what is the function 20 dB? So, let us take S plus 1 is absolutely right. We can just to show this when S is very very small I should take this as 1 and what I had written is correct. If I take is that 2 then S very small it would have become half of that both should now change to 1 and 10. I am telling you from this as S is very very small correct I get only 10. So, everything on the left of that is exactly 10. So, log 10 of 10 is 1 and 20 log 10 of 10 is 20. So, 20 dB on the left of it as S become very large right as S become very large say become 10 then I neglect this because that is J 10. So, I neglect this then it become 1 at this equal to 10 this become 1 and log of 1 is 0 that become 0 dB which is nothing but 20 dB down 20 dB per decade it will keep going like that it will cross here and there is no 0 like that this is the continuous minus 20 dB minus 40 dB etcetera it keeps going this is my low pass all rejecting all high presence. So, if I take 1 then I am happy because everything can be now explained. Now, we take the components some value C some value R. So, what is the best way I can actually find is equivalent. I can find the equivalent as a parallel component y of f this is S c the admittance of this and this is 1 by R right. So, we can now invert it R of f is R upon S c R plus 1 and for this simple circuit I will choose this as a simple resistor R 1. So, now my gain should be minus Z f by Z 1 right the effective gain if a is more than 10 S to 5 is. So, the gain of this V o of t. So, V o of t now I am writing in the terms of V o of s should be equal to minus Z f upon R f in this case I have only used R 1. So, minus R upon S c R upon R 1 this is R f minus R f upon R 1 S c R plus 1. So, now I compare this with this all I have to say is c and c R should be equal to 1 multiply by correct or we can write transfer function as by definition. So, I say by definition is V o is equal to this. So, if I want V o of s then I have to multiply it with the input transfer function I mean input in Laplace form. So, there are many a variety that if this numbers are 2 3 5 like that then there is a convenient thing to multiply and divide by c R then you write s plus 1 over c R and then we say 1 over c R should become 1 plus c R should become 3 depending on what we need. But I am quite comfortable with carrying it like this and we can convert everything in this particular form for this circuit. So, for c R equal to 1 I will get this and I can make this ratio you remember this goes to 0 I am left with 1 therefore R f upon R 1 should be 10. So, that is the advantage of actually carrying this s c R together because s c R does not then depend on the frequency the frequency goes down to 0 the c R also goes down with it. Otherwise the c n R are outside and then they get also remain here and then where to take care of that. The design of such a thing will be that R 1 arbitrary should be at least 10 kilo ohm why we say that because we do not want the currents in this circuit to be very very small or very very big. So, 1 k is ok 100 k is ok. So, therefore I have a range like 1 k to 100 k I can either begin this with arbitrary suppose I do this I know 10 k is a reasonable range then R f is not arbitrary R f is then going to be exactly 10 times that. So, I have already derived that it should give me a ratio of 10 once R f is decided in this manner then C f suppose this is R f and C f then R f into C f should be equal to 1. So, I can find C f. So, C f should be 1 upon so we will get that value. So, this is the process what we need is 3 values and what we have is only 2 values 1 is the gain and other is this number is taken 1, but sometime it can be 2 etcetera. So, we have only values given, but we have to determine 3 unknowns. So, the design requires 3 parameters, but we have been given only 2 conditions. So, one is the DC gain and one is the frequency determining and one of them we have to take arbitrary. So, I like the student who can very proudly write that I have taken R 1 as arbitrary some students like 1 micro farad as arbitrary any one of them can be arbitrary and then we can derive the remaining 2 from that. So, I think it is just a matter of convenience what can be taken as arbitrary. So, let me write I will rub this out. So, 1 upon 10 raise to 5 will be what 10 micro farad right the product has to be 1 this product has to be 1 we have said no. So, it will be 10 micro farad. Now, we do this suppose we have to do all this I must therefore, for my convenient approach take this 2 out make this 0.5 right 2 outside I have not changed this 1 right 2 multiplying 0.5 s plus 1 is the same as s plus 2 correct I take 3 out and this is 1 upon 3 this become 1 and then I get this 6 out it goes there and makes it 1. I have now this 1 upon s plus 0.5 s plus 1.3 recurring s 1. So, now, I have got this in a convenient form. Now, I can split this into 3 as a product. So, this is my method 1 this is what I call by induction once I have shown you how to do 1 the rest will follow method 1 is that I will get this 1 upon s of course, this cannot do that because this has come because I have already input at the input in that I have to get this I of s, but if I have the transfer function this 1 by s would not have come. So, the transfer function this and for the input being step function I would have got this 1 by s. So, 1 by s we have ideal inverter of course, ideal integrator, but I am not of the ideal integrator. So, this normally which comes here because I have converted the transfer function to the output and I think for all discussion we are discussing of the voltage. So, I now convert function to exactly 2 parts 0.5 s plus 1 into 0.3 recurring s plus 1. So, now, I have got this 1 by s. So, I can make this using 2 op-amps in tandem and the form is same form is identical. So, this is my v in or v 1 of t then I get here v intermediate of t and then I give this to another resistor I take both this arbitrary 10 kilo ohm and I can easily determine the value because this is the one I have to get in here function I have to obtain here. So, this is called doing it in tandem this just split this into as a product, but I can also do by method 2 which is convert this to partial fraction then what I have to do is I have to make this one we know how to find a. So, we will not bother about it. So, now I have to make the function is of this type. So, it is the same my solution is same representing this we take this arbitrary then we find this. So, I get this function we can call it v 2 of s and this is v 3 of s and the two together will give me v out of s. So, I have another function I think the rate should be better than that. So, this is that v 2 of s is v 3 of s and now I have to get them in a addition mode. So, this is the summation mode and then I go to. So, I implement this two and I put it in the summation mode and of course, I have to take care of the number of inversions. So, necessary we can always do non inverting unity gain. So, if they were more than two terms I should get rid of this if I want transfer function. So, for a transfer function like this I will get that even if you actually have s and then you allow me to use a integrator it is ok, but normally I prefer one without the integrator because that one by s should come because of my input being a unit step function. No, this is only a going continuously down. So, if you have a single fellow then 20 dB down if you have got a two fellows it will go 40 dB down there will be two cut up points. So, there will be two cut up point one here then 20 dB down then next point onward it will go 40 dB down 20 dB per decade at the second cut up point it will go to 40 dB per decade. So, this is w naught this can be some other w 1 and w 2 this is giving w 1 and w 2. So, here is here is 1 by 2 1 by 0.5 is 2 and the other one is 1 by 3, but normally we will be having it sufficiently apart. So, that we can draw the border clock. So, it goes like this. So, this is just a low pass filter with more stop band yeah I mean sometimes there is a requirement that at the double that frequency it should be 50 dB down. So, to get that you know you have to put this more stages.