Now there is an exception to our rule. There's no rules without exception, right, Melodie? So here some risk exceptions. So, there are quite simple circuits that cannot be analyzed by the method of repeated application of the formulas for series and parallel impedances. To analyze such circuits, we must write down the current and voltage equations from Kirchhoff's rule. So, here is an example. There is just one current equation here, as you can see. All of the currents will meet at the node. Therefore, the summation of the current should be zero. Therefore, I sub three should be minus of the summation of I_1 and I_2. There are two independent voltage equations as are written there. If I solve this one, as you can see, you will have I_1 and I_2 like this, complicated one. These one you cannot use the repeated applications of series and parallel impedances. Now let's move on to the bridge circuit. This is a very popular circuit that is used in many measurement system, and we also have some experience to use it. Now, bridge circuit cannot be analyzed by a repeated application of the formulas for series and parallel impedances as well. So it appears in many instrument used for measuring impedances, as I mentioned. But this unfortunately, you cannot use what we learned before. However, let's think about this question. How must the various impedances be related if the current through the impedance Z_3 is to be zero? So that will be your homework. But, in fact, there is a solution to your homework in this slide. So this is extracted from Wikipedia, and you can see this will be your solution. If you are more interested, you can read what is being done with the bridge circuit, but I'm going to show you one of our own example, which is the resistive probe in an AFM, where you can have source and drain and channel, where this channel is a sensor of the charge underneath the tip. The resistance of this channel will be modulated by the electric field exerted from the materials, and we need to measure the resistance in real-time. So, you can see we made a bridge circuit to measure the resistance is real time and map that out as a function of position, where we have artificially decorated the surface of the materials with the later resistive probe. That was published in Nano Lett in 2011. Those who are interested more in this research, you can google this one and read it further. Now, let's move on to equivalent circuits. Suppose we connect a generator Epsilon to a circuit containing complicated interconnection of impedances. As all of the equations we get from Kirchhoff's rule are linear, the current I through the generator will satisfy the following equation. I is equal to Epsilon over Z sub effective, where Z sub effective is a complex number, an algebraic function of all the elements in the circuit. So this you can see it's very complicated. It can be simplified to this very simple box. Any two-terminal network of passive elements can be replaced by a single impedance Z sub effective without changing the currents and voltages in the rest of the circuit. So imagine you have a human mind, very complicated and there you can simplify it to a very simple elements, how interesting will that be? Melodie, right? Yeah. Okay. So here's an example. One generator plus one impedance. So the idea can be generalized to a circuit that contains generators as well as impedances. An arbitrary circuit can always be replaced by an equivalent combination of a generator here in series with an impedance. We learned that for real cell. Real cell is a generator plus internal resistance. Suppose we look at such a circuit from the point of view of Z sub n, n is arbitrary number. If we were to solve the equation for the whole circuit, we would find that the voltage V sub n between the two terminals A and B is a linear function of I. Therefore, V sub n is equal to A minus B times I sub n, where A and B depends on the generators and impedances in the circuit to the left of the terminals. So, from the point of view of the Z sub n, everything is very simple. So, we had this different approach, Melodie, when we are approaching the Maxwell equations using the field concept. We were thinking the world of interactions can be boiled down to field. Like one particle exerted to the system will suddenly interact with multiple particles, multiple charge, but that interaction can be simplified into field concept. That what we learnt, and we are using the similar approach here to simplify our solution. So, for the circuit on the right, we find V sub one is equal to Z sub one I sub one, here, you see. Then, V sub one is equal to this one minus this one. If you take a look more closely, this one is like an equivalent generator composed of Epsilon two and Epsilon one. This one is composed of Z_2 and Z_3 impedances. So indeed, we can simplify this circuit into these two. This circuit was an exception to the repeated application of impedances in series and impedances in parallel. But, it can be analyzed using this simple diagram of effective impedance plus effective generators. So that's what you are now dealing with. So from this equation, you go to this equation, you go to this equation, you go to this equation, and then you will be able to derive this. So check yourself if this is correct. Now with that, we are interested only in what happens to the right of the terminals A and B. Any arbitrary circuits can be replaced by an equivalent combination of a generator in series with an impedance. So Melodie, do you think this will make our life easier? Yeah, I hope so. Yeah, and I hope you think the same way as we do. Now let think about energy of an ideal inductance. So, we learned energy is conserved. So to build up the current I in an inductance, the energy U is equal to one over two LI squared, which must be provided by the external circuit. When the current falls back to zero, this energy is delivered back to the external circuit. Why? Because for ideal inductance there is no loss. So if I give something, I have to get it back, to make both of us happy, right? Yeah. Give and take. So when there is an alternating currents for an inductance, energy flows back and forth between it and the rest of the circuit, but the average rate at which energy is delivered to the circuit is zero. Therefore, an inductance is non-dissipative element. Again, it's a non-dissipative element, and that has to do with the fact that L, the inductance, is an imaginary number. So, in e to the I Omega t, you always have rotation. There is no falling back. So an inductance is a non-dissipative element and no electric energy is dissipated. How about capacitor? The capacitor has an energy U, which is equal to one over two CV squared. It's not curriculum vitae, it is CV squared, capacitance times voltage squared. It is returned to the external circuit when it's capacitor is discharged. When a capacitor is in an AC circuit, energy flows in and out of it but the net energy flow in each cycle is zero, the same. So an ideal capacitor is a non-dissipative element and it has to do with the fact capacitance is also pure imaginary number. How about energy of an ideal generator? An EMF is a source of energy, as we mentioned. When a current I flows in the direction of the EMF, energy is delivered to the external circuit at the rate dU over dt is equal to Epsilon I. It's like voltage times current. It's like power. If current, I, is driven against the EMF by other generators in the circuit, the EMF will absorb energy at the rate Epsilon I. Since I is negative, dU over dt will also be negative. So giving minus energy to the circuit is equivalent to getting positive energy from the circuit. Again, giving minus energy to the circuit is equivalent to getting positive energy from the circuit. So if a generator is connected to a resistor R, the current through the resistor is I is equal to Epsilon over R, and the energy being supplied by the generator at the rate Epsilon I is being absorbed by the resistor. Resistor has a real number, meaning you have a decade. You have a loss of energy. This energy goes into heat. So it's like a heater. If you have a heater in your home, then this is how you spend energy, and that's your electricity bill. So the resistor, and is lost from the electric energy of the circuit. Electrical energy is dissipated in a resistor. The rate at which energy is dissipated in the resistor is dU over dt, which is RI squared or, in other words, I square R. But IR is voltage, so this is VI as well, voltage times current, and that's power. So, Melodie. Yes. What was the average rate of energy lost to a resistor in this circuit? It's right here, RI squared. Exactly. It's RI squared. So, since I is equal to I current times e to the I Omega t, by which we mean that I varies as cosine Omega t, the average of I square over one cycle is the absolute value of I current squared over two. Since peak current is the absolute value of I current, and the average of cosine squared Omega t is one half. Therefore, in an AC circuit the average rate of energy lost to a resistor is R times absolute value of I current square over two. That will be the average rate of energy lost to a resistor. Now, any impedance Z can be written as the sum of its real and imaginary parts as you can see on the right side here. It's just next to Melodie. So, impedance can be resistance plus reactants, and reactants includes capacitance and inductance, where you have imagined number here. From the point of view of equivalent circuits, any impedance is equivalent to a resistance in series with a pure imaginary impedance called reactance. Any circuit containing only L's and C's has an impedance with a pure imaginary number and will have no energy loss. So, if a generator with the EMF epsilon, is connected to the impedance Z, the EMF must be related to the current I from the generator by epsilon is equal to I times this is Z, R plus iX. To find the average rate at which energy is delivered, we want the average of the product epsilon I. When dealing with such product, we must deal with the real quantities, epsilon t and I of t. Suppose we choose our origin of t so that the amplitude I cared is a real number I naught. Then, the actual time variation is given by I is equal to I naught times cosine omega t. The EMF of epsilon, which is equal to I times R plus iX, is really the real part of this equation where we have now replace I by I naught e to the i omega t. If you remember, e to i omega t is cosine omega t plus I sine omega t, and expand this and only take the real part, you will have this epsilon is equal to I naught R cosine omega t minus I naught X sine omega t. You now see the real part of this equation has two parts which is off by 90 degrees in phase. Now, the two terms represent the voltage drop across R and X. The voltage drop across R is in phase with I while the voltage drop across X is out of phase with I, and the phase difference is 90 degrees. The average rate of energy loss, this one P of average, from the generator is the integral of the product EI over one cycle divided by the period capital T. So you do this, one over capital T zero to T EIdt. If you do this math, following this equations with me, you will end up having this I naught square R over two, which is similar to what we have done with the resistance only. So, this will be our average power loss, and you can now see all of the power loss is induced by the resistance. So, the average energy loss in an impedance Z is equal to R plus iX depends only on the real part of Z and that is this one. This is in an agreement with the result for the energy loss in a resistor. So there's no energy loss in the reactive part. So now, lets take a look at the ladder network. Let's consider an interesting circuit that can be analyzed in terms of a series and parallel combinations. Now look at this, for a, the impedance from terminal a to b is simply addition of those two. For b, replace two impedances on the right by a single impedance Z3 like here. Then, the two impedances Z2 and Z3 can be replaced by Z4 as in part d, as you can see here. Finally, Z1 and Z4 are equivalent to a single impedance Z5 as you can see here. So you can have Z5 as a function of Z1, Z2. What would happen if we kept on adding more sections forever as shown in this a? What would happen if I add this ladder more and more to make the ladder infinitely long? First, such an infinite network is unchanged if we add one more section at the front end. That's the beauty of infinity. Infinity plus one is infinity here. One plus infinity is infinity. So, if we add one more section to an infinite network, it is still the same infinite network. Now, let's call the impedance between the two terminals a and b of the infinite network Z naught. Then, the impedance of all the stuff to the right of the two terminal c and d is also Z naught. So you can see if this whole part is Z naught, then this one is also Z naught because adding one doesn't affect Z naught. Therefore, so far as the front end is concerned, we can represent the network as shown in b. You can see here. Combining the parallel combinations Z2Z0 and adding the results in series with Z1, we can write down the impedance of this combination like this. So you can see Z is equal to Z1 plus Z2Z0 over Z2Z0. Can you see that? So, if you think of this circuit, you have this one in parallel, and that's this one, and this one in series with this parallel circuit. That's why total Z is zero. However, this total Z is also equal to Z0. So, we replace this by Z0. So we have Z0 on the left, Z0 on the right. Then you can rearrange that to get Z0 in terms of Z1 and Z2. This will be the characteristic impedance of an infinite network. We would like to now specifically understand this ladder network consisting of L and C only. So we would like to consider a specific example where the series element is an inductance L and the shunt element is a capacitance C. So now you may think, if I only have L and C, do you think we can have energy loss here? No. No. But you will see, at some point, even with this, you may have energy lost. Which is intriguing. We will try to understand what's happening. So, for a, we find the impedance of the infinite network by setting Z sub one is equal to i omega L. This is i omega L. Z sub two, which is the capacitance is one over i omega C. If we replace all Z1 and Z2 by this one, so you can see also the first term Z1 over two is just one-half of the impedance of the first element. This one is just one-half of this one, one-half. So as such it would be simpler if we were to draw our infinite network as shown in this one. But cutting it half, half, and half, and then starting from after the half of this inductance. Looking at the infinite network from the terminal a prime, we would see the characteristic impedance Z naught as below. So, if I start from a prime instead of a, then we can get rid of this term. So we only have this one. If I replace Z sub one by i omega L, then the square will be minus omega squared L squared. That's why we have this one, Z1Z2. You will see is i omega L over one over i omega C. Therefore, it is L over C. If I rearrange that, that is Z naught is equal to square root of L over C minus omega squared L squared over four. So now you can see this might be larger than zero, but if this is smaller than zero, then this becomes imaginary. If this is larger than zero, impedance is real. If it is smaller than zero, it's imaginary. So Melodie, if impedance is imaginary, do we have energy loss?. No. If it is real, do we have energy loss? Yes. So you can see, even in this ladder network, we can have energy loss. Isn't that fascinating? So that's fascinating. So there are two interesting cases depending on the frequency omega. As you can see, omega square is less than four over LC or omega square is larger than that. So depending on this frequency, you can see either you will have a real number for your impedance or imaginary number for impedance. So now you can guess if my frequency is lower than this number, all of a sudden, this one will be real. You will have energy loss. If my frequency is larger than just one, then this one will be like ordinary inductance and capacitance. No loss of energy. So we have said earlier, that a circuit that contains only imaginary impedances, such as L and C, will have an impedance that is purely imaginary. Then, how is it possible for the impedance of an infinite ladder network of L and C to have a pure resistance for omega squared is less than four over LC or, in other words, frequency below square root of four over LC. That is the big question. To be or not to be. So, think about this. Let's think about this question again and we're going to state it again. How is it possible for the impedance of an infinite ladder network of L and C to have a pure resistance for low frequency? What is your guess, Melodie? I am not quite sure. It's very difficult. Right? Yeah. So let's figure this out. For higher frequencies, the impedance is purely imaginary, in agreement with our earlier statement. For lower frequency, the impedance is a pure resistance and will therefore absorb energy. But how can the circuit continuously absorb energy, as a resistance does, if it is made only of inductances and capacitances? We are repeating the same question again and again like it's torturing you. Now, let's think about this. Because there is an infinite number of L's and C's, so that when a source is connected to the circuit, it supplies energy to the first L and C, then to the second, then to the third, and so on. So, think about a bucket is infinitely connected and you're pouring water to the bucket. So, even there is no loss of water as you pour, but there's infinite demand. So, from your side, it's feeling like there's only give, no take. So, that's how it is absorbing. Now, this idea suggests an interesting point about what is happening in the circuit. We would expect that if we connect a source to the front end, the effects of the source will be propagated through the network towards the infinite end. The propagation of the waves down the line is much like the radiation from an antenna which absorbs energy from its driving source. So, we expect such a propagation to occur when the impedance is real, which occurs if omega is less than this one, the square root of four over LC. We wouldn't expect to see any such propagation when the impedance is purely imaginary, which happens for omega greater than square root of four over LC. So what it means, if it is lower than that frequency, all the energy we give will propagate. It will never come back. But if it is above certain frequency, it will always be reflected. So, we will always have give-and-take. But that means that there would be no propagation away. So, you can think despite your filter, you can set a filter which one will pass and which one will not. Only with this simple sets of L and C's. So, that kind of filter we learn in the previous lecture, if you think about ionosphere, which is the plasma, if you shake the plasma with electromagnetic wave you learn that, below a certain frequency, it will be reflected, but above certain frequency, it will go through. That's satellite communication or AM-FM radio type of broadcasting that we already learned. In this case, it was high pass filter, but in our case, it was low pass filter. In our case, we will also deal with high-pass filter like in this case as well. So, the infinite ladder network absorbs energy continuously if it is driven at a frequency below a certain critical frequency that's square root of four over LC, which we will call cut-off frequency, omega note. This effect could be understood in terms of a continuous transport of energy down the line. On the other hand, at high frequencies, for omega is larger than omega note, there is no continuous absorption of energy. We should then expect that the currents do not ''penetrate'' very far down the line. It should be reflected. Sometimes we call this impedance matching for RF's potray, for example, to make sure the energy is propagated to the system so we can supply the energy to have the right sputtering rate. If it is all reflected, then there will be no deposition of fillers in the section. Now, suppose we have the front end of the ladder connected to some AC generator and we ask what the voltage looks like at, say, the 826th section of the ladder. Since the network is infinite, whatever happens to the voltage from one section to the next is always the same. So, let's look at what happens when we go from some section, say, the nth to the next. We will define the currents I_n and voltages V_n as shown in a. So, you will see you have 1,2,3, and then we will take an arbitrary nth element. We can get the voltage V_n plus one from V_n by remembering that we can always replace the rest of the ladder after the nth section by its characteristic impedance Z note. So, you can see here after this one, and after that one is all Z note. Notice that any V_n, since it is across Z sub zero, must equal to I_n Z zero. From here to everything else on the right side is equal to zero while from here to everything else to right side is Z zero. In addition, the difference between V_n and V_n plus one is I_n times Z_1. So, difference between this one and this one is only this part, voltage drop across this flow which is I_n times Z_1. So, this is the equation V_n minus V_n plus one is I_n plus Z_1, and I_n is V_n over Z note. So, that's why we get this one. So, we get the ratio here. This one is Z_1 over Z note. If you rearrange this, V_n plus one over V_1 is equal to one minus Z_1 over Z note, which is Z note minus Z_1 over Z note. That's simple. So, the ratio of this voltage is called the propagation factor alpha for one section of this ladder. So, Alpha is Z note minus Z_1 over Z note. So now because we have this general equation, you can understand V_1 also has this Alpha, V_2 also have this Alpha, V_3, V_4, and so on and so forth. So, you can just multiply this n times. So, since V_1 is Epsilon, and V_n plus one over V_n is equal to Alpha or V_n minus one is equal to Alpha or V_2 over V_1 is Alpha. If I multiply V_2 over V_1, V_3 over V_2, V_4 over V_3, V_n minus two over V_n minus three up to V_n minus V_n over V_n minus one, you can see all of them disappears and only that remains is V_n over V_1 which is Alpha to the power n. So, therefore, V_n is equal to Alpha to the power n times Epsilon. So, if I want to find the voltage after 826 sections, what would be the voltage, Melodie? You have it here. It's the Alpha raised to that power. 826. Exactly. If Alpha is less than one, you can see this will be very small. Suppose we want to get Alpha for the L-C ladder of a, this one, I want to see Alpha her. We already solved the characteristic impedance of this ladder. So we can already know what the Alpha is. So if I recall this, this is i Omega l over 2 plus square root L over C minus Omega squared L squared over 4 plus the fact that Z sub 1 is i Omega L, is i Omega L, then Z0 minus Z1 is 0 minus Z1 over Z Naught. So you can see if I subtract this one from this, then you will see this is minus i Omega L over 2. Z Naught is just this one, so you have plus. So this will be Alpha. Now, Alpha here is this one. If the driving frequency is below cutoff frequency, Omega Naught, which is equal to square root of 4 over LC. The radical here is real number and the magnitude of the complex number in the numerator and denominator are equal. Can you see that? They are equal. Therefore, the absolute value of Alpha will be one. Let's take a look. So here, you have A minus BI, which is here, A minus BI. Here, you have A plus BI, which is here. So A minus BI is here. A plus BI is here. But in this real complex space, you see the magnitude of each vector is the same. Therefore, if I take the absolute value of this one or if I think about this, it's just e to the i Delta, and Delta is minus 2 Theta. Because you have here double the Theta and you can see how we do this math. So it means there is no loss. So you can see Delta is the phase change, a negative number which represents the delay of the voltage as it passes along the network, and the magnitude will be one. So meaning all the voltage will be the same in the ladder. That is below cut off frequency. If it is above, then this part radical becomes imaginary. If these are all now imaginary, then you see it becomes real number. Because it's real number and less than one, it will decay. The voltage will decay as you go node to node. So the voltage you apply here will never propagate to the end. So you can see this is the basic principle of filter. So the propagational factor Alpha is now a real number and a number less than one. So voltage at any section is always less than the voltage at the preceding section by the factor Alpha. For any frequency above Omega Naught, which is equal to square root of 4 over LC, the voltage dies away rapidly as we go along the network. So we have explained to you low-pass filters. Low-pass filter is used in every equipment you can imagine. For example, in atomic force microscopy, the feedback circuit has a low-pass filters from the photo diode that detects the movement of cantilever. Because the movement of cantilever induced by topography has a maximum frequency. Above that, they're all artifacts or noise. Therefore, they put low-pass filter to operate in a reliable frequency range, to feed it back to the actuators, so you can maintain the constant bending. Likewise, they will use this kind of low-pass filters to make that happen. So if Omega is less than Omega Naught, which is square root of 4 over LC, the magnitude of voltage is the same at every section as the magnitude of Alpha is one. Now, Melodie, if I want to increase this cutoff frequency, what should I do to L and C? So you want to increase the cutoff frequency? Yeah. I want to increase Omega Naught. What should I do to L and C? Then, you should decrease L and C. Exactly. We need to decrease L and C. We need to decrease the inductance. We need to decrease the capacitance. So if I wanted to decrease my capacitance, what would be the easiest way? I think you can decrease the area. Exactly. If I decrease the area of the plate, your cutoff frequency goes up. So you can see if you're using AFM tip, which is really really small, then your cutoff frequency will go up. So in other case, if Omega is larger than Omega Naught, which is equal to square root of 4 over LC, the voltage dies away rapidly as we go along the network as the magnitude of Alpha is less than one. So you can see schematically, this low-pass filter is working like this. For this region, this is the Alpha. For this region, this is the Alpha. So see, there is no loss but here you have a drastic loss. So in this way, you can make a low-pass filter. Now, you may think of this from what we have learned from energy propagation standpoint. So when we learned about the ladder network of L and C, we remember that the characteristic impedance is like resistance when the frequency is low. When the frequency is high it is like imaginary number and that was what we learned here. The propagation, so why is it like energy absorption? Because the gift that I give is endless. But in other words, you are propagating the wave. When they give and you have a take right away, then is reflected. So in that point of view, we were understanding this ladder network. Now, we are using that fact to design a filter. So now you can connect the knowledge that we have discussed before with the knowledge that we are discussing here. So that was about the low-pass filter, where you can see that the network passes low frequencies and rejects or filters the high frequencies, and we call that low-pass filter. Now, you may argue, look, this is infinite. In real-world, we cannot have infinite, right? We only have finite network. So, what is the point here? Okay the point is, that the same characteristics are found in a finite network, if we finish it off at the end with an impedance equal to the characteristic impedance Z naught. So, in that way, you can have artificial infinite that way. But in practice, it is not possible to exactly reproduce the characteristic impedance with a few simple elements, like R, L and Cs. However, it's often possible to do so with a fair approximation for a certain range of frequencies. In this way, one can make a finite filter network whose properties are similar to those for the infinite case. For example, the L-C ladder behaves as much as we have described if it is terminated in a pure resistance R is equal to square root of L over C, and you can see it, this one will be the part that we will use, and this one which we will ignore, okay? Now, what about high-pass filters? What about high pass filters? So, if you think a little bit further, if you replace inductance by capacitance and capacitance by inductance, everything is changed and you can see only thing that change is omega becomes one over omega. So, now you're reversing your frequency term. So, now the same graph has different tendency. So, as you increase the frequency, it will go from right to left instead of left to right. Therefore, you will only permit high-frequency wave to pass, and low-frequency to die out. So, this is how you can make a high-pass filter like ionized sphere that we just discussed before. So, you can see whenever we change an L to a C and vice versa, we also change every i_omega to one over i_omega. So, whatever happen to omega will happen at one over omega. In that way, you make low high, high low. Now, let's think about applications. The low-pass and high-pass filters have various technical applications. If we want to manufacturer DC power from an AC source, we begin with a rectifier that permits current to flow in one direction. So, in rectifier, we usually have diodes as you can see here. With diodes you can make this sinusoidal wave to something like the absolute sinusoidal waves. If I put a low-pass filter here, then you will have something close to a DC, right? It's a smoothing filter. So, we know that the time function above can be represented as superposition of a constant voltage plus a sine wave, plus the higher frequency sine wave, plus a still higher stick frequency sine wave, that's a Fourier series, right? If our filter is linear, if the L's and C's don't vary with the currents and voltage, then what comes out of the filter is the superposition of the outputs for each component at the input. So, if we arranged that the cutoff frequency, omega naught of our filters well below the lowest frequency in the function V of t, which is omega zero, goes fine, but the amplitude of the first harmonic will be cut down a lot, and the amplitudes of the higher harmonics will be cut down even more. So, if we put that frequency low enough, so everything else is being caught, only the DC goes out, then we will have DC voltage. Also, there are band-pass filters. So, Melodie, where can you find band-pass filters in our daily lives? Maybe the radio. Exactly, you see this one, radio like a bumblebee. Okay. It is possible to make band-pass filters that reject frequency below some frequency omega one, and above another frequency, omega two, omega one and omega two. This can be done simply by putting together a high-pass and low-pass filter, but it's more usually done by making a ladder in which the impedances Z1 and Z2 are more complicated, being each a combination of L's and C. It can be used to tune the broadcasting radio frequency of the program of interest, like the one here. So, which radio station is one of your favorite? I don't listen to the radio. You don't listened to the radio. Okay, in the states? I don't have a. Remember. Yes. For me, it was. 93.9 MHz. Yes, but that was a while ago. So, you might have your own favorite frequency, and you can now see the band-pass filter is enabling this to happen. Now, similar filtering can be done using the selectivity of an ordinary resonance curve as shown in B here. You can compare this characteristics as with the band-pass filter. When a carrier of a frequency omega C is modulated with single frequency omega S, the total signal contains not only the carrier frequency, but also the two side band frequency, which is omega C plus omega S, and omega C minus omega S. With a resonant filter, the side-bands are always attenuated. As you can see, they're attenuated, which leads to a poor frequency response. With a band-pass filter designed so that the width omega two minus omega one is at least twice omega S, the frequency response will be flat as you can see here, for the signals wanted. So, it might be better to use band-pass filter than resonant filter. Now, think about transmission line. You may have seen this biaxial or coaxial cable when you connected your TV to the antenna, all right? This can be represented by L-C ladder, and we call this transmission line. If we have a long conductor that runs parallel to another conductor, such as a wire and coaxial cable as I just mentioned, or wire suspended above the earth, there will be some capacitance between the two conductors, and some inductance due to the magnet and that magnetic field between them. We have covered this cylinder type of conductor and capacitance in the previous lecture. You can revisit the slide if you don't remember. If we imagine the line as broken up into small lengths delta L, each length will look like one section of the L-C ladder. With a series inductance delta L and a shunt capacitance delta C. If we take the limit as delta L goes to zero, we have a good description of a transmission line. Now, notice that as delta L is made smaller, both delta L and delta C decrease in the same proportion, so that the ratio delta L over delta C remains constant. So, you can see here that the ratio here will remain constant. Now, if we take the limit of Z naught is equal to square root of L over C minus omega square L square over 4 as delta L and delta C equal to 0, we find that the characteristic impedance Z naught is a pure resistance with a magnitude of square root of delta L over delta C. Look at this, as we change delta L and delta C here and change this one as well and make this go to zero. But here both of them goes to zero, so this will remain. But this one will go to zero. Then you see Z naught is just pure resistance of magnitude with magnitude of square root of delta L over delta C, and if we replaced that with L per unit length which is L naught and C per unit length which is C naught, then Z naught is equal to square root of L naught or C naught. So we can see that as delta L and delta C equal to 0, the cutoff frequency omega naught which is this one, goes to infinity. Because omega naught is square root of 4 over Delta L delta C. What does it mean going to infinity? It means that your plot will extend and then there is no decay region. Exactly. So there is no decade. So there's no cut off frequency for an ideal transmission line. So transmission lines should transmit everybody. It's not a filter and that served the purpose. So there is no cutoff frequency for an ideal transmission line. Now, let's think about other circuit elements. We would like to show that outer circuit elements such as, mutual inductances or transistors or vacuum tubes can be described by using only the same basic elements. Suppose we have two coils by two coils and that some flux from one of the coils links the other. Then the two coils will have a mutual inductance M such that when the current varies in one of the coils, there will be a voltage generated in the other. We learned that before, right? Yes. So can we take into account such an effect in our equivalent circuit, Melodie. Yeah, probably. Yeah. So let's see how we can do that. So take a look at some of the previous slides we had before; mutual inductance M2 sub 21 use barely status but you see the electromotive force in coil two is a function of the current through coil one and the coefficient is mutual inductance. Also we have self-inductance and if we combine those two effects we can describe all of the EMF in each coil as a function of both currents in coil one and coil two. So if we think of this equation again and take the self-inductance as normal inductance and mutual inductance as this generator, then we can describe this mutual inductance in terms of the equivalent components in our circuits. So as for an ideal inductor as the potential difference across the terminals of each coil is equal to EMF in the coil. So now, you can see how we can describe our mutual inductance in terms of equivalent circuits. Now the two equations above are the same as the ones we would get from the equivalent circuit on the lower right side of the EMF and each of the two circuits depends on the current in the opposite circuit according to the relations here. As you can see. If epsilon 1 prime is equal to plus minus i omega Mi2 epsilon 2 prime is equal to plus minus i omega Mi1. If you think about the relationship between v and EMF and L, this was already discussed in our own lecture here and you can see V is equal to minus epsilon which is equal to L times di over dt. So in that sense, so what we can do is represent the effect of the self-inductance in a normal way but replace the effect of the mutual inductance by an auxiliary ideal voltage generator. So we must have the equation that relates this EMF to the current in some other part of the circuit. With that in mind we can do that. So in addition to mutual inductances, terminals will be mutual capacitance. In many situations there may be many electrodes close to each other. For example, field effect transistors. This is an excerpt from one of our collaborators paper where we were one of the co-authors where you have junctions ferroelectric FinFET source, drain, and gate wrap-around gate and you can see you have mutual capacitance between S and G, G and D, and S and D. So this is always the problem of interests in our semiconductor industry. Consider arrangement are placed shown below. Suppose these four plates are connected to external circuits by means of the wires A, B, C, and D. As far as we are only worried about electrostatic effects the equivalent circuit of such an arrangement of electrons is shown in the right side of the figure. So this two where this big capacitor is eating the small capacitor can be described by this equivalent circuit. The electrostatic interaction of any electrode with each of the others is equivalent to capacity between two electrodes. With that in mind you can construct this circuit and solve it. So let's consider how we should represent complicated devices like transistors and vacuum tubes in an AC circuit. So, Melodie where can we find transistors in our daily life? Probably in the computer. Yeah in computer like in DRAM or SRAM logic devices transistors are switches that enable memory and logic functions. Now those devices are off often operate it where the relationship between the currents and voltages is non-linear. The statements we have made that depend on the linearity of the equations are no longer correct. However, in many applications the operating characteristics are sufficiently linear that we may consider the transistors and vacuum tubes to be linear devices. So this is approximation and this is the limit. When we have such linear relationships, we can cooperate the device into our equivalent circuit representation. Our representation will include auxiliary voltage generators that describe the influence of the voltages or currents in one part of the device under currents or voltages in another part. The plate circuit of a trial you can see here can be represented by a resistance in series with an ideal voltage generator whose source strength is proportional to the grid voltage. So you see here is the grit and the source strength is proportional to the grid voltage. The equivalent circuit is correct only for low frequencies for high frequencies the current circuit gets more complicated and includes various parasitic C's and L's uses. This will be the gatekeeper. So depending on the voltage you will have more current or less current. In that way you will have a switch and switch can always be an amplifier. The collector circuit of a transistor is conveniently represented as a resistor in series with an ideal voltage generator whose strength is proportional to the current from the emitter to the base of the transistor as you can see here. So one of the example that we can find from our own research is this one where we have published resistive probe in applied physics letters. You can see you have source, drain, and channel made by this implantation process and reusing this implantation mass to make it that tip mass and S set away to align channel on top of the tip. Then put that tip on top of a silicon oxide cut silicon wafer using semiconducting processing parameter analyzers to vary the voltage across the source and drain and measure the current and vary the gate voltage from minus 10 to plus 10 volts. You can see the I-V curve is changing and that is characteristic of a transistor where you can see source drain and gate which can also be represented by an equivalent circuit on the right side where you can see the epsilon is equal to minus kVG meaning the gate voltage can modulate the strengths of the generator which means the current flow across the channel. So there is one remarkable thing about transistor and vacuum tube circus that is different from circuits containing only impedances. The real part of the effective impedance Z sub effective can become negative. Again, it can become negative is negative resistance. This means that transistors and tubes can supply energy to the circuit as the real part of Z represents the loss of energy. This energy comes from the DC circuits of the power supplies which is converted to AC energy. So it is possible to have a circuit with a negative resistance. If we connect it to a positive resistance to make the sum of the real part is zero. There is no loss of energy and any AC voltage is one started will remain forever. Give and take, give and take, give and take, give and take, and that is oscillator. This can be part of your digital watch as well. With that would like to conclude our final lecture and congratulations. Thank you. Bye bye.