Lecture 03 : Module 01

Welcome back, this is lecture 3. In the last couple of lectures we saw the need for a network that can amplify power and we also briefly saw one example of where a feedback network can help us in saving power, right. So, as it turns out there are several more applications of feedback and we will see that going forward in the course and you will also see that feedback helps in improving accuracy of the output as well. So, this is too early to talk about it but in the previous couple of lectures I wanted to lay the foundation and the motivation behind what you will be expecting throughout the course. So, let us start off with the first thing that is trying to build a network that can give us something which can amplify power, right ok. So, first things first we have a we have a voltage source and some internal resistance associated with it and we have a we have a resistance R L. What do you want to do? We want to put a network in between, we want to put a network in between such that whatever power that goes in, right whatever power that goes in P in and whatever power that comes out if we call it P out, right we would expect we want P out to be greater than, ok. So, this is what we want. Now, in order to go in order to go forward and do an analysis of this let me be a bit more rigorous and say that this whatever device whatever stuff that I have put between the load R L and the source V i or V in and R s does not seem to have any ground. One might argue that the ground is inside. So, what can I what I what I will do is in this case is I will just say that let us say everybody has a common ground and let me let me take the ground out, ok. A ground is nothing but a common terminal that everybody shares or in other words a ground is nothing but a reference terminal between which all the node voltages with reference to which all the node voltages are measured, ok. So, let me redraw this circuit a bit more carefully. So, let us say I have a network n, right. So, I have a network n and we need to figure out we we we are in the quest of figuring out what are the properties of this network such that the power that goes out that comes out of the network is more than the power that goes in, ok. So, what can we what can we comment? Now, clearly now clearly it is should be quite evident that this is not possible simply because simply because the law of power conservation must hold, right. We saw the example of of a transformer couple of lectures back. In fact, in the first lecture where we saw that a transformer can give us voltage amplification, but it cannot give us power amplification. Similarly, similarly if we have any passive network, right if this network is passive, if this network is passive we will not get any power amplification. So, passive network cannot power amplification. So, this is really not possible. So, then what is the what is the other possible solution which means that we need we need some additional power supply, right. So, we we need an additional power supply to boost the output power, right. So, so let us say now I say that I will put an additional power supply which means maybe it has a battery inside, correct. So, next thing what I will do is because now this battery is battery is supposed to give me some additional power. I will take the battery out. I will I will say that let us say there is another additional port and I take the battery out and this battery is also it is assumed is referenced to ground. Now, it is not necessary that a battery must be referenced to ground. It can be referenced to some some arbitrary nodes inside the network as well, but to make our life easy to make this point without loss of generality I am taking taking the battery out. I mean you can you can argue after we finish the argument you will see that this does not really cause any loss of generality, ok. Let me call this battery V dc. So, now, what are we after? We are trying to see if this can give us power amplification and let me poseanother impose another constraint. Let me say that this network inside is L di, right, ok. Let us assume that this network inside is L di linear time invariant and let us see whether this can help us in our code, ok. So, what is what is what is my input and output power? How should I define them? So, let us say the voltage across this port I call is V 2, ok. Let us say the current that goes in this port I call it I 2. Similarly, let us say the voltage at the input port I call it V 1. The current that is going in I call it I 1. The third port to which across which the dc voltage supply is attached the voltage is V dc and let me call this current to be I dc, ok, fine, ok. So, let me go back go to the next page and let me catch the circuit once again, ok. So, now, how do we go about analyzing this, analyzing this network? Now, note that this is a general generalized network, but we took an important, we can we can make an important analysis here because because we made an important assumption that this is an L di network, right. So, what is what is the fundamental property of an L di network? Firstly, it is linear which means superposition holds and it is time invariant, right. It is time invariant means whatever happens if I apply an input now and observe it after a certain time today and if I apply an input after one day and observe it after the same difference that difference in time there will be absolutely no difference in, no difference in response, right. So, the network does not change with time. That is all, that is all time invariance means. One of the properties of time invariance is that it can cause a frequency shift of the input, ok. So, the even though that is not a part of this course, it is it is instructive to be aware of one of the properties of time variant circuits. A time variant circuit can cause frequency shifts, right. For example, a mixer is a time variant circuit. However, a time invariant circuit cannot cause a shift in frequency of the signal that it is processing, right. You will see the importance of this statement in a few minutes, in a few minutes, ok. So, if it is an L di system and we say that superposition holds, what can we say? How can we go about figuring out what is V2? So, since superposition holds, we can analyze the network with one source at a time while desensitizing all other sources. Right. So, this is this is crucial, right. So, since this is an L di system, since this is a linear system, we can desensitize with all but one source and analyze the circuit and we can do it for all available sources and find out the individual responses and in the end, we can club together all the responses and the final output that we will get by clubbing together will be equivalent to the case where all the all the sources were in play together, right. Note that this is only possible in an L di system, right, ok, ok, great. So, let us start off by saying that we will desensitize V dc, ok. So, if we desensitize V dc, what we get? If we desensitize V dc, so desensitizing V dc means V dc goes to 0, which means this becomes a short circuit since V in, right and this is let us say V 2 under the case of V dc is equal to 0, ok and let us further assume that let us further assume that V in is a sinusoidal signal, right. So, it makes sense that an input is a time varying signal, right. So, if there is no change in input which means there is no essential, no essential information since we are in the business of processing information, right which means that which means that there is some time variability, right of the signal and one of the fundamental signals that we are aware of which changes with time, right is a sinusoidal. So, let us let us start off by using a sinusoidal input, ok. So, what will be what will be my output V 2? V 2 under the condition V dc equal to 0, right will be let us say some alpha times V p sin omega naught t, right where alpha is some factor which is dependent on which is dependent on this network, ok, fine. So, let us say my V 2 when V dc was 0 is some alpha times V p sin omega t, ok. So, where this peak is alpha V p, ok. So, what is the power delivered, what is the power that is getting delivered to the output? So, p out, p out is alpha squared V p squared by 2 R L, correct. Now, can you comment on, can you comment on whether this power is greater than the power that is being fed from my source? Clearly, clearly this is lesser than the power that is coming in that is coming in to the source, correct because the same old story that output power cannot be more than the input power, ok. So, let me call it less than equal to V, fine, no issues, ok. So, now let us say, now let us say I added, let us say I added the source V dc, right, I added back the source V dc and I now shorted the, I shorted the input source, ok. We are taking superposition, so one source at a time. So, what will be my new output V 2 when V in equal to 0, what will be my new output? So, V 2 when V in is equal to 0 will be some, will be some constant dc, right because there is no other source, no other time varying source in my network. Since there is no other time varying source in my network, the output will be, output will be, ok, of some value, correct. Let me call this value beta V dc, ok, fine. So, what is superposition telling us? Superposition is telling us that now if I put back both the sources, if I put back both the sources and observe this voltage V 2, right, what will V 2 be? V 2 will be the summation of the voltages of the previous two cases where we observed the output individually while desensitizing the other source, right. So, now if I add these two waveforms up, what do I get? This is V 2 total, what should I get? I should get this sinusoid which is, which is essentially riding on top of a dc voltage. So, it looks like the sinusoid has been shifted up by this value of beta V dc, right and what is the amplitude of the sinusoid? This will still be alpha times V. Now can you comment, can you comment on the power, can you comment on the power of the sinusoid that is getting delivered at the output? Note that I am not interested in the power of the dc signal, I am only interested in the power of the sinusoid because that is the signal that we are trying to amplify, right. So, what is the power of the sinusoid that is, that is getting delivered to the output? So, my, this is my new signal, right. So, power of the sinusoid at the output is equal to, it remains the same, right. So, this is alpha times V p. So, the power of the sinusoid that is getting delivered to the output is still alpha squared V squared by, by twice R L that is still less than the power that you are extracting from the input, ok. So, so what is the conclusion, what is the conclusion of this entire exercise? The conclusion of this entire exercise is that, is that the power, in order to get a power amplification, in order to get a power amplification, you, it is not sufficient to have an additional power source in the network, you also need to do something about the network. As it looks, as it turns out a linear time invariant network with as many power sources as possible is not able to amplify the power of the sinusoidal signal that you want to amplify, right. So, what is the conclusion? The conclusion is an LTI system cannot amplify the power of the sinusoidal source even if there is an additional DC voltage source available, ok. Note that, note that we are here talking about the average power, right. So, these are all average power, right. So, these are average power, ok. Now, if you, if you, if you think how do we go about, how did we go about finding this, how did we go about finding this? This can be, you can, you can easily figure it out by doing the following. What will be the average power if this is the total voltage, right, if this is the total voltage, if V2 is equal to alpha Vp sin omega t plus beta Vdc. So, the P out will be average of alpha Vp sin omega t plus beta Vdc whole squared by, by, by Rf, right. So, once you do this, once you do this, you will see that there will be one DC term, right, there will be one term which will be squared of sin squared omega, which will be, which will be proportional to sin squared omega t, there will be another term which will be the cross product of Vdc and sin omega t, right. Now, if you average that out, if you, if you do the average of the entire term, the average of Vdc sin omega t will become 0, you will be ending up with two terms out of which only one will give you the power of the sinusoid, ok. So, I encourage you to do this exercise yourself, ok.