 So, there are some topics like Bode diagram for example, that is there in the syllabus, but we are we do not have any lectures on that. And so, I thought we will just go through this and there are also some other things like RC circuits, RLT circuits which are not explicitly in these lectures, but it may be useful to at least go through some of those as well if there is time. So, let us start with Bode plots and the sequential circuits we will just postpone to tomorrow. So, if you are talking about Bode plot, first of all what is decibel? That is the first thing that the students need to know about. The unit decibel is deci means one ten and it is used to represent, deci means ten. It is used to represent quantities on a logarithmic scale. And while logarithmic scale, there is a good reason for that. Because of the log scale, dB is convenient for representing numbers that vary in a wide range. So, that is why we use this unit, bell or decibel. And how wide is this range that we will see soon. And the other reason, very important reason is that this log scaling roughly corresponds to human perception of sound and light. So, we actually are able to see or hear other things which vary very widely in intensity. And that is why this log scaling actually helps to quantify the sound. What is another advantage of a log scale? That it has lost multiplication and division to be replaced by plus and minus. It is much simpler that way. If you want to multiply 4.2 and 3.7, instead of multiplying, if you have the log of those available, you can just add those and then take the anti-log. So, sometimes it is much easier to understand. So, little history, it is always good to do some little historical details to students because they know, that way they know what happened then. So, the unit bell was actually developed in the 1920s by bell laboratories, engineers. And why did they do this? Because they wanted to quantify the attenuation of an audio signal over one mile of cable. In those days, we did not have mobile phones and wireless and so on. And all signals used to be sent on cables. And so, they wanted to know exactly how much a signal attenuates in one mile of cable. And that is why this unit was invented by Santis Stubbell in the 1920s. So, this is also a very important detail to know. Interesting facts, this is just on the side. So, Alexander Graham Bell, who invented the telephone in 1876, all of you know that, could never talk to his wife on the phone because she was there. That is one interesting fact. Imagine that today, if wife and husband cannot talk to each other on the mobile, the mobile companies will just close down. Bell considered the telephone an intrusion and refused to put one in his office. That is another interesting fact. So, Bell invented this telephone, but he thought it was a nuisance and he never wanted to put one in his office. Anyway, these are just on the side. This is not really technical. Interesting to know for students. So, what happened is this unit called bell, B-E-L. It turned out to be too large in practice. And that is why we have this unit called one tenth of a bell or decibel. So, a dB is a unit that describes the quantity on log scale with respect to a reference quantity. Now, this is very important and we see a lot of confusion about this amongst students sometimes. So, whenever we talk about dB, it always has a reference. That is extremely important. So, x, suppose we are talking about some quantity x. In dB, it will be 10 times log of that x divided by some reference x. So, that reference is extremely important. Without that, there is no meaning to this quantity. And why are we multiplying by this 10? That is because this decibel is one tenth of a bell. So, the bell is log of x times x. So, if you have so many bells, you will have 10 times so many decibels and that is why we need to multiply by 10. So, all of these things I am saying because students, if they are just given the formula 20 log V 2 by V etcetera, they do not often appreciate what is why how it came about and they do not really apply it properly. For example, if you have two powers, one is 20 watts and one is 1 watt and you want to know and if your reference is 1 watt that is this p-ref here, then this power P 1 in dB will be given by 10 times log of 20 watts divided by 1 watt and that turns out to be 13 dB as an example. Now, this is a very important point and it must be made very clear to the students. For voltages or currents, the ratio of squares is taken and in fact, there was also a question on this on Moodle. Why 0.707 about it had to do with the filter response, why 0.707 for the half weight and so on. So, it is all of these things are connected and this is because if you have a voltage applied across a resistor, then the power in that resistor is V squared by R or the power in that resistor absorbed by the resistor is I squared times R. Now, if your R is fixed and if you go on apply different voltages, then the powers will go as basically V squared or I squared and that is. So, let us see what it implies. For example, if V 1 is 1.2 volts and V ref is 1 millivolt, then the power that corresponds to 1 millivolt is 1 millivolt squared and power that corresponds to V 1 is 1.2 volts squared. So, now when you actually take the ratio of powers, you need to take 1.2 by 1 millivolt volt squared and that is how that 10 log of that becomes 20 log of the ratio of the voltages. Now, this dBm is an important unit and that means dB with respect to 1 millivolt as a reference and that is what is shown over here. So, very often students do not really make this distinction, they always write dB and they do not really know what that is. Now, very often the students will see the voltage gain of an amplifier described in dB and then I might just wonder why what is the reference over here. So, the reference here actually is the input voltage. So, it is the ratio of the output voltage to the input voltage and that we are multiplied by 20 and why 20 because for the same reason as this one described here. So, the V i the input voltage here actually serves as the reference voltage, very important to point it out. So, here is a simple example, maybe the font is not appropriate, but I would say that you just go through this later. It says that the input voltage is given as 2.5 millivolt and the output and the gain is given in dB and you are asked to find Bo the output voltage in dBm as well as in millivolt. So, you can go through this example later and see how it is done. So, there are two methods of doing it, one you convert the gain into a dimensionless number, second you use the gain as it is in dBm and convert your input to a dB value. Both of these are fine and they will of course be the same. Now, although this audio normally is not covered in a basic electronic score, it is good to know what audio, how audio things are, how audio intensity is actually characterised. So, and then that dB is sort of it essentially is the same, but it has different connotations because then it has to do with sound intensity and not electrical power. So, let us just take a quick look at this. So, when sound intensity is specified in dB, the reference pressure is P ref, that is the reference pressure that is 20 micro Pascal's. And that is why is that taken as the reference pressure because that is roughly over the hearing threshold. We cannot hear things which are waves, sound waves which have less pressure than that. Now, if the pressure corresponding to the sound being measured is P, we say that it is 20 times log of P by P ref dB. So, remember this is like voltage, the square of this is actually the power and that is why we have a 20 here as well. Now, although this is not in the electrical engineering domain, this is very interesting to know and it gives students a feel for this decibel unit as used in audio. So, some interesting numbers, if we have a sound proof number, so what is 0 dB exactly? So, that corresponds to this scenario, let us say you have a sound proof of room and you have a mosquito in that same room about 3 meters away from you. And you can just barely hear that and that is why it is that corresponds to 0 dB or that corresponds to a sound wave which is 20 micro Pascal's, which has a pressure of 20 micro Pascal's. So, that is a mosquito, a whisper is 20 dB, normal conversation is 60 to 70 dB, noisy factory is 90 to 100 dB and of course, there are strict laws which are not always followed about all of these things. Loud thunder is 110 dB, loudest sound that human ear can tolerate is 120 dB and if you have sound or above this level, your ear drum might actually get damaged and that is what happens if there is an explosion and so on. Okay, windows break, if you have glass windows and if there is an explosion, the sound waves can be so strong that of course you have to exceed this limit and that is the time when windows break. So, if you see old pictures of Hiroshima, Nagasaki, you will see many buildings without the windows, that is what happened at that time. Okay, all of that is on the side. Now, Bode plots, what is the Bode plot? So, there is an input, there is an output and if you cannot see the soldier, read it out for you. So, there is a V i of S where S is j omega, there is a V o of S which is V of j omega and there is an H which is a transfer function of S and what we are interested in is the transfer function H of S which is V o of S divided by V i of S and so this is, Bode plots are basically used to describe these transfer functions. For example, H of S is k over 1 plus H times tau. So, this is an example of a transfer function. We have of course more complicated transfer function. Now, remember that H of j omega is a complex number and therefore, a Bode plot will require both the magnitude and the phase of H of j omega and we, a Bode plot will require this information, the magnitude information versus omega and the phase information versus omega, both of these, only then the description is complete. So, Bode gave simple rules which allow construction of the above Bode plots in an approximate or asymptotic manner and that is why these plots are called Bode plots. So, here is a simple transfer function. Now, instead of just describing Bode plots theoretically, it is good to always take real example like an R C circuit and show all, where all this comes from. So, for this one you can actually derive V o over V i expression that turns out to be 1 over 1 plus H R C or 1 over 1 plus j omega by omega naught, where omega naught is 1 over R C. So, this circuit behaves like a low pass filter, we all know that. So, if omega is very small H tends to 1 and if omega is very large as much larger than omega 0, then what happens is that this 1 becomes small and this magnitude goes simply as 1 over omega. So, the magnitude and phase of H of j omega exactly are given by these expressions here and at the Bode plot would be approximate representation of these two quantities. So, we are generally interested in the large variation in omega, several orders and it is effect on mod of H as well as angle of H and this is very important. We are interested in the variation of H over a very large frequency range and that could be several orders that is we are interested from 10 hertz to 1 mega hertz for example, that is several orders in frequency and that is why as we will see that is why we use log scale. But, if you just show the students from Bode plot and just show the log axis and without explaining why log is required and so on, then the students can actually not get much out of that. The magnitude H varies by orders of magnitude as well and that is also very clear. For example, here it goes as 1 over square root of 1 plus omega by omega not squared. So, imagine omega equal to 1 hertz and omega equal to 10 is to 6, the magnitude definitely is going to vary by orders of magnitude. On the other end, the phase varies in a limited scale. So, in this example, the phase can go from tan inverse of 0 to minus tan inverse of infinity which is pi by 2. So, it can only go from 0 to minus pi by 2. So, the phase actually varies in a limited range whereas the magnitude can vary in a wide range. So, here as I was saying it is important to actually describe why we need to use log scale. So, it is the same example as before for which we have already the transfer function and here is the magnitude of that transfer function. Now, if I use the linear scale from we are interested in the variation from let us say up to 10 raise to 6 hertz. So, this is 10 raise to 6 and if you do not if you cannot see that it has do not worry you can just look up the slides later and this is also on linear scale. So, this is 0 and this is 1 and the plot is the magnitude of h versus frequency. Now, if I just take both of these on a linear scale, the entire plot gets confined to this small region here and it is basically it looks like 0 all over here and we really cannot resolve the plot very well. Suppose I ask what is the value at 10 raise to 2 for example, 10 raise to 2 would be here, what is the value of 10 raise to 3 that would also be very close to this point and so on. So, therefore, the resolution is not correct. So, what we can do is to make the magnitude of axis logarithmic. So, now we have a better resolution. So, now I can say this is 10 raise to minus 3, this is 10 raise to minus 2 and so on which I could not do in this plot, but even that is not enough because all the small frequencies all small frequencies like 10 raise to 2, 10 raise to 3, 10 raise to 4 even they will all be concentrated on this axis. So, therefore, it is very important to also use the x axis as a logarithm only then you get a reasonable resolution. So, now with this and this is actually what a body plot will look like except that this is a exact and that will be a plot. So, here what we have is the frequencies on the dialogue scale. So, this corresponds to 10 raise to 0, this corresponds to 10 raise to 1, 10 raise to 2 and so on. We also have the mod of h on the log scale, this is 10 raise to 0, this 10 raise to minus 1, 10 raise to minus 2 and so on. Now, if somebody ask you what is the magnitude of 10 raise to 1, I can see what it is, what is the magnitude of 10 raise to 3, I can read it out, what is the magnitude of 10 raise to 4, I can read that out and so on or any number in between and that is why it is very important to show this give the students some idea of what goes wrong if you do not use logarithmic axis, where you often students have this confusion in mind, whether a logarithmic axis is the same as the d b, it is actually the same and these two figures, they both show a mod of h, this is mod of h in log scale like 10 raise to 0 here, 10 raise to minus 1, 10 raise to minus 2 and so on and because the decibel is already has log in it, the shape of this will actually remain the same. So, this point is you can say this 10 raise to 0 or it is the same as 0 d b, this point is 10 raise to minus 1, the same as minus 20 d b and so on. So, therefore, these two plots will have essentially identical nature, no difference and so it is good to explain this as well. All right, what happens if I change, if I use radians per second or hertz, very little difference except that this, the plot will just get a little shifted because the frequency in radians per second is 2 pi times frequency in hertz and so again that is important point out because students do have this confusion for time. So, this plot now shows the phase of the transfer function for the same example and if I have to use a linear frequency axis, I really cannot make out the information at low frequencies and therefore, once again we use the log scale for frequency and we use and this case the phase does not really change very widely, it goes from 0 to minus 90 degrees and therefore, you can continue to use the linear scale on the y axis and you can get a very good resolution. I can read off any, read off the phase at any given frequency using this plot. So, therefore, the frequency axis or the phase axis is linear, unlike the magnitude axis. So, one important now finally after all this introduction, we finally come to body plot and how do we construct body plot? So, for example, you might have a general body plot like a function like this k times some 0s divided by some 4. So, how do you represent this in graphically? So, this z 1, z 2 etcetera are called the 0s of h of s to be precise minus z 1, minus z 2 and so on and this minus p 1, minus p 2 etcetera are called the poles of h of s. Now, in addition we will often have terms like s and s square and so on in the numerator and in it can things can get actually more complicated because some of these 0s or poles can come in complex conjugate pairs, but in this particular presentation we will take a stick to the simple case where these are real and distinct. So, what does the construction of body plot involve? It involves computing approximate contribution from each of these terms like this term and this term and this term and so on and then adding them up together, combining the various contributions to obtain h and mod of h and angle of h, okay. So, always good to take this simple example and show that show how it comes about. So, here is a contribution of a single pole, okay or so h of 0 omega is that and mod of h is 1 over square root of 1 plus omega by p square. So, body plot basically come through asymptotic approximation and so we need to consider the two different asymptotes in this case, asymptote 1 is when omega is much smaller than 1, okay. What happens then is this becomes almost 0 compared to this 1 and therefore h of 0 omega becomes constant equal to 1 and when if you take log of that then you get 0 to p. So, that is 1 asymptote and that is shown in over here in the red line which you may not be able to see right now, but you can see it later. Asymptote 2 is when omega becomes much larger than this pole value, okay. So, in that case your transfer function looks like 1 over omega over p because this one is then negligible or that is the same as p over omega. If you take mod, if you take h and db then you get 20 log of p minus 20 log of omega. Now, very often people students do not quite understand why this 20 db per decade where it comes about, okay. So, now this is an explanation of that. So, if I change omega from omega 1 to 10 times omega 1, okay, the mod of h will change by 20 db and that is why this 20 db per decade are slow. Also important to point out that this asymptotes actually differ from the actual exact calculation at omega 0 by 1 over square root 2 and that corresponds to minus 3 db and that is why that point is also called 3 db point, alright. So, well there is a contribution to phase as well and we will skip this, we will just take an example of a function which has multiple 0s and poles and then see how that is that is constructed and then after that I just wanted to give you, show you a tool which we can use in class, okay. So, the rest of the presentation just shows what is the contribution of s and s squared and so on and finally there is an example in this example. So, in this example what we have is term s in the numerator and we have two poles, one at 10 raise to 2, one at 10 raise to pi and how basically how to go about constructing the body plot systematically. So, here is the contribution of each of these terms, this is just the constant 10, this is the term s in the numerator, this is the term, this is the first pole, this is the second pole and then how all of these combine in the log plot to give you this final picture and then finally we also show how the exact one differs from the asymptotic one, okay. Then there is a similar plot plot for the page, that is right. So, I suggest that you can go through all of these later but we find that body plot is not something that is well understood. So, when you are teaching in a class it will be good to actually spend some time and give some examples, all right. So, for today I think we will stop here but I want to show you one's tool that you can use for body plot in particular. So, can we share the desktop, okay. If you go to the SQL URL that we announced earlier, that we wrote on the paper here earlier, you will find the link for E101 and there are some exercises there and then there are some animations there. So, you can look at body plots one, okay and you will get this, so they can see this now, all right. So, this actually can help the students to check out if their understanding is correct or not. For example, if you select this particular form of H of S and let us say I have A0 equal to A1 equal to 0.0 or let us say A0 is equal to, yeah. Let us say the pole is at 100 radians per second. So, that means A1 would be 0.01, okay. If I do that, I can plot the magnitude and I can also plot the phase. Now, what this does is can they, they should be able to see this, okay. So, what this shows is actually the asymptotic plot as well as the exact plot, okay and that is something that most of these packages like MATLAB will not show you. For example, the phase here is the asymptotic plot like that and here is the exact plot like that. So, let us take something more complicated. Let us say we select this function here, 1 0 and 1 pole. Let us say the pole is at 100 and sorry the 0 is at 100 and then the pole is at some other higher frequency. And then let us see what happens then, okay. There is some information, let me take some other example, sorry I am typing at the wrong, sorry I was making a mistake there. So, let me take this one, okay. So, if I have a 1 0 and 1 pole, this is what the asymptotic function looks like and that is the exact one. So, plot the phase, that is what the phase looks like. So, this will also help the students to know how accurate the approximation is for the given transfer function. So, the phase for example, you can see that this there is a considerable error in this particular case where you use the phase approximation. You can also use this other one, which has got similar format, but it has got the poles and 0s explicitly written like this, s minus z 1, s minus z 2 and so on. So, you can also use this and take it out. Alright. So, I think we will stop at this point. I will take a few questions. NIT Warangal, you have a question. We prefer NAND gate implementation rather than NOR gate. Can you explain the reason? NAND gate implementation rather than NOR gate. The question is whether you prefer NAND gate implementation over NOR gate implementation. Actually, there is no such thing. I mean it, for some functions the number of NOR gates required will be less than number of NAND gates. So, for some other functions, it will be just the reverse. So, in a realistic situation, actually you do not have a choice. You have a chip which has got thousands of NOR gates or you have a chip which has got thousands of NAND gates and you have no choice but to use NAND gates or NOR gates. That is how it is in real life. So, you normally do not choose, no make this choice between NOR and NAND depending on the function, but you just use, do all your functions with either NOR gates or NAND gates because you want to use one chip only. Is there any power consumption difference while implementing with NAND or NOR? Is there any power consumption difference in implementing NAND or NOR? This would actually depend on the, first of all it will depend on the number of gates that you require and therefore the number of function. Secondly, it will also require, it will also depend on which technology you are using, whether you are using CMOS, whether you are using TTL, etc. So, I think this question, let us raise this question again when Professor John covers these gates and if it is not covered at that time, let us just put it up on Moodle and then we will see. We can try to give a generic answer to that question, but it will depend on the technology. Mufakam Jha College, go ahead with your question. Sir, my question is, what is the application of Bode plot and what is the importance? Bode plot, okay. What is the question is, what is the application of Bode plots? Applications are many. For example, if you have an amplifier, let us say you have two amplifiers and you are designing a two-stage amplifier to get a larger gain, for example, and now you want to find out whether this combined amplifier, the amplifier with larger gain is going to be stable or not stable, right. So, that is where Bode plots can be immediately useful. For example, if you know the pole zero information about both of these amplifiers, then you can plot Bode plots and get the pole zero and you can get the gain margin, phase margin of the combined amplifier. That will immediately tell you whether the combined amplifier is going to be stable or not stable. So, that is a very important application. Control systems, any control system Bode plots are always useful, okay. We will go to the next center. Srinivas Institute, Takenada, go ahead with your question. My question is, in most logic gates, why the low state is approximately zero volts, why the high is approximately five volts? Okay. The question is why is the low level approximately zero volts and why is the high level approximately five volts in most logic gates? First of all, this is not true anymore. It used to be five volts like five, ten years ago. Now, these days people are talking about 1.5 volts, even about one volt of VDD. So, these voltages actually are coming down and there is a good reason for doing that because if you bring down the voltage, the power dissipation will reduce dramatically and that is how we can increase the packing density and all kinds of related advantages. But these levels will actually depend on, again, technology, whether you are using TTL or ECL or MOS and these days, most of these digital gates will use MOS because they are obvious packing density advantages. But the voltages have come down. Over. Okay. Jayachand Rajendra, Karnataka College, go ahead with your question. Over. Hello, sir. How to calculate the delay between input and output signals of digital logic circuit? Can you explain with an example, sir? Okay. The question is how to calculate the delay between an input and output of a digital circuit? And can you explain with an example? So, let us take a very simple example. So, there is an inverter and you always have a, in most cases, you always have a capacity load at the output and you have an input here. So, there is an input, there is a gate and there is a capacitor. So, now let us say that your input is perfect, that it is making a very sharp transition at t equal to 0. Now, the output if you plot, it will be high in the beginning and then it might become low and it might take some time to do that. And so, therefore, there is some delay involved between these two and that is the question. How do you calculate this delay? Okay. Now, this delay will of course, definitely depend on the technology and if you are familiar with CMOS technology, let us say what, let us see what happens in a CMOS technology. So, this inverter in fact is nothing but this circuit and let us see now what is happening if your, if this input voltage goes from 0 to 1. So, when this input voltage was 0, this transistor was on and this transistor was off. Now, when this goes high, this is going to turn on and this is going to turn off. What happens to the capacitor? The capacitor was earlier at logic level 1 that is let us say VDD and then this is going down to 0 volts. So, where does this charge go? This charge will actually go through this transistor. And also there is an in-between period when both of these transistors are conducted. So, now in order to, so this, this current is finite, the current that this transistor can conduct is finite and therefore there is some delay involved in discharging this capacitor and that is what basically we are going to see over here. Now, the details of these calculations Professor John will cover in this lecture on CMOS and TTL and so on. So, let us wait until then and if it is not clear at that time, we can just take it upon model again. So, essentially it has to do with the discharging or charging of this capacitor through the transistors. Or if you have any other questions, go ahead. Can you please explain fan in and fan out of a logic circuit with an example? So, the question is, can we explain fan in and fan out of a logic gate? Alright, now this question actually is a little difficult to answer in general. Again we should take an example and let us take the same CMOS example. Let us say there is an inverter which is driving some other inverters or some other gates. Let us just take for simplicity again inverters. Let us say this is driving two different, two other inverters. So, we in E O and E O 1 and E O 2 etcetera. Alright, so what is the impact of this connecting one more load here or let us say 2, 3, 4, 5 more different loads? Essentially what happens is that this the node, this output node here is seeing an effective capacitance and this capacitance will depend on how many gates you connected the output. Because these gate capacitances are actually coming or just adding up and then they are giving you all this entire, this total capacitance. And if your capacitance becomes large, then obviously you are going to take larger time to charge or discharge this capacitor and your delays are going to increase. So, that is one consideration. In CMOS I think that would be a dominant consideration. Other in other technologies like TTL for example, they will also be DC current. That means even if this, even in steady state all of these gates will actually consume some, will take some current and then your output stage of this gate will have to be able to supply all that current. So, that is another consideration that can come in. And CMOS of course, the DC current is not there. So, that we do not need to worry about. Again this will also be something that will be addressed in the next week's lectures. Over. Sir, while manufacturing the gates, which type of logic family is more used, sir? And is there that TTL is more advantageous compared to other logic families? Okay, the question is while manufacturing gates which type of technology is used and whether TTL is advantageous. Okay, the answer is that these days all complex digital circuits are essentially fabricated with CMOS technology because they have this big advantage of low power and very high packing density and that beats the bipolar technology way down. TTL gates are still used, especially if you have an application where you just need, you know, a few gates, then the TTL can be considered. And sometimes these decisions are based on convenience. For example, what is the voltage level you are using and that kind of thing. And how much current you want to drive, want the gate to drive. If you are going to drive some LEDs, a large number of LEDs, for example, and whether the CMOS gate will be able to deliver that much current or not. In that sense, it is TTL advantages and so on. So all of these considerations will be used. But for all large circuits these days CMOS is the only technology that is used. Okay, over. Do you have a question? So my question is regarding the combination logic that we actually say that we are having the mean terms and we call it, it is a product. Whereas it's not in actual, it is an end, end of the terms. But we call it as product. And in the similar way, the max terms, we call they are in sum, but they are actually in OR. Doesn't it create learning ambiguity? Awesome. All right. The question is that why are we using the terms product and sum to describe logical operations like AND and OR. I am not sure about the historical development of this topic, but I think it is just that people felt comfortable with this terminology. And but really one has to go into history and see what, for example, in the Boolean algebra what they used and if there is an explanation given to that. So let's see, I'll try to find an answer and then I'll post it on model. As of now, it's just that some of these things like we have seen some of the laws, they look very similar to the laws that you would write it for real numbers if you use dot for the AND and plus for OR. So in that sense, things look familiar if you use those symbols. I think that may be right now the only explanation, but if I find more information, I'll take it over. How to explain the difference between demultiplexer and decoder with IC74138, which is used as a 3S28 decoder. How to explain the difference between, I think the best way is to, I mean just go back to the basic definitions of this demultiplexer and decoder and a demultiplexer will have this control pins, whereas decoder will not. So that is, I mean in that sense there is a major difference between these two and therefore if you just explain that to the students, they will know that there is something external which is controlling the information flow in a demultiplexer, whereas there is nothing external that is controlling information flow in a decoder, because the information flow itself depends on the input there. That would be my answer, but just post this on model and if we have some other views from Professor Sherma or Professor John, they will also add to it. They are not here right now. Over. Thank you, sir. Over. Thank you. So I will see you tomorrow.