 Hello everyone. In the last lecture, we had started discussion about filters. So, let us start with quickly what we had discussed in the last lecture. So, we had started with the ideal filter response. Here is a response of ideal low pass filter which passes all the frequencies up to omega c and blocks all the other frequency. A high pass filter passes all the upper frequencies beyond omega c, a band pass filter passes all the frequencies between omega c 1 and omega c 2 and a band reject filter or band stop filter blocks all the frequencies between omega c 1 and omega c 2. Then we had seen a simple example of RC filter and I did mention that we do not use RC filter at microwave frequency because this R here gives additional losses. So, instead of using RC filter we in general use inductors and capacitors. So, one can start with a capacitor and then go to inductor or one can start with the inductor and then go to the capacitor. So, the objective is now to find out these G parameters G 1, G 2, G 3 and so on. Then we looked at 4 different types of realization. One was maximally flat or Butterworth filter, the response of that was given by this curve here. Then we looked into equi ripple or Chebyshev filter, the response of that has equi ripple in the pass band and then relatively sharper transition compared to the Butterworth filter. Then we had also looked at the Bessel filter, but we saw that the response is very very poor as far as the amplitude response is concerned. However, phase response is nearly perfect, but I just want to tell you very rarely people use Bessel filter ok because of this particular reason that the response is very very slow. Electric filters in general have equi ripple in the pass band as well as they have a equi ripple in the stop band also, but the advantage of electric filter is that the response transition from pass band to stop band is very very fast. So, suppose for the same fifth order, this is the response for maximally flat, this is the response for equi ripple, but for electric filter the response will be even sharper and that will be something like this and then there will be some ripples in this particular region here. And from here we actually defined another thing which is attenuation at frequency omega 1. So, if the cutoff frequency is omega c, so it is generally desired that at a given frequency omega 1, how much is the attenuation ok because there may be a some interfering signal over here which needs to be attenuated by maybe 20 dB or 30 dB or more. And the order of the filter can be obtained by using this particular equation which is obtained by taking log on both the sides and solving for n. Then we looked at the g parameters, so how to find out the g parameters for Butterworth. So, we saw that g 0 is equal to g n plus 1 equal to 1, these are nothing but corresponding to input and output source as well as load impedances. And g parameters can be found by using this simple expression and we had seen that these are the typical values for a fifth order filter. And instead of using this particular expression, one can use the table to find out what are the different g parameters. So, after that what we need to do? We need to do impedance scaling as well as frequency scaling. Why we have to do that? Because we had taken input and output impedances to be equal to 1, but in reality that will never be the case. And we had taken a normalized frequency which is omega equal to 1, but we need to design filter at the desired frequency. So, ultimately we combine impedance as well as frequency scaling. And what we had seen that the g parameters which are given by L k or C k, we will see the example in a short while. So, those g parameters have to be modified for inductor and capacitors slightly differently. For case of g parameters which realize inductor, we have to multiply by r 0 divide by omega c. Whereas for the g parameters which will represent capacitance, those things have to be divided by r 0 and omega c to realize the real capacitance. So, let us take now an example. So, we will start with an example that we need to design a low pass filter with a cutoff frequency of 400 megahertz. And we want attenuation of 20 dB at 1 gigahertz. And it is given that the source and load impedances are 50 ohm. So, the first thing what we need to do is we need to find out what is the order of the filter. So, we had seen the expression to find out the order of the filter n given by this particular expression. So, let us just substitute the various values. So, log 10 10 to the power a what is a 20 dB. So, you do not convert into numeric value that conversion is done by this particular expression here ok. So, 20 comes right over here then for omega 1. So, we want attenuation of 20 dB at 1 gigahertz. So, that is 1000 in terms of megahertz divided by omega c which is 400 megahertz. So, that comes over here. So, this comes out to be 2.51 of course, we cannot take real number for order of the filter we have to take the higher value compared to this one here. So, we choose n equal to 3. So, the next step would be is to find out the g parameters. So, we know that g 0 is equal to g n plus 1 here n is 3. So, input and the output source impedances are equal to 1. We will do the impedance scaling as well as frequency scaling in the next slide. The next step is now to find out g parameter. So, g 1, g 2, g 3 n here is equal to 3 we put the value we find out the corresponding g parameters which are 1, 2, 1. Now, let us do the impedance and frequency scaling. Here we have taken the first component as inductor, second capacitor, third inductor. We will also show you the design where we take first component as capacitor and then inductor and capacitor, but let us just look into this design now. So, l 1 is equal to l 3. So, there I had written r 0, but here r 0 is equal to z 0 which is given as 50 and we have seen that g 1 is equal to g 3. So, that is equal to 1 what is omega c 2 pi into f c. So, 2 pi into 400 megahertz. So, simplify this that comes out to be 19.9 nano Henry. For capacitance again for the g parameter g 2 if you recall g 2 was equal to 2 z 0 is 50 omega c is 2 pi f c which is 400 into 10 to the power 6. So, that comes out to be 15.9 pico ferrate. So, this is the actual circuit which will meet the requirement of the 400 megahertz cutoff frequency and at least 20 dB attenuation at 1 gigahertz. So, let us see the response of this particular filter. So, here is the response of S 2 1 which is a low pass filter and this is the reflection coefficient for this given low pass filter. So, let us first look at S 2 1. So, we had designed this particular filter with a cutoff frequency of 400 megahertz which is 0.4 gigahertz. So, if you look at from here 0.4 gigahertz we go up here. So, this is the crossover point and that corresponds to about minus 3 dB. I just want to mention here if S 2 1 is equal to minus 3 dB correspondingly S 1 1 also will be equal to minus 3 dB and remember the circuit consist of only inductors and capacitors. Hence, there is a no insertion loss in this particular filter. Of course, this is an ideal response practically inductors and capacitors will have some losses. So, the response will not be exactly 0 dB, but it may be a 0.1 dB or 0.2 dB depending upon the losses in the inductors or capacitors. Let us see what is the response at 1 gigahertz. So, at 1 gigahertz if you see this value that is about 24 dB. Now, you might wonder we had actually designed it for minus 20 dB, but this is giving us minus 24 dB. The reason for that is minus 20 dB was obtained for a filter order n equal to 2.5 ok, but you cannot have a 2.5. So, we took a higher value which means we took third order filter. So, for third order filter response will be little more steeper. Had we taken suppose fifth order then this response will be steeper ok. Here had we taken n equal to 2 then the response would be like this here we would not have got a 20 dB attenuation which was desired. So, if the attenuation desired is 20 and if you are getting 25 or 30 or 35 it is acceptable all the time ok. So, now let us just look at alternate way of realizing that means here first element has been taken as capacitor then inductor and then capacitor. Just recall previously we had taken first element as inductor then capacitor then inductor. So, in this case what happens now G 1 is equal to G 3 and we are now realizing capacitance. So, for capacitance what we do we start with the G parameter we divide that by Z 0 and omega c. So, G 1 is equal to G 3 equal to 1. So, Z 0 is 52 pi into f c. So, that comes out to be 7.95 picofarad. Similarly, we find the value of the inductor we start with the G 2 parameter multiply this with Z 0 or r 0 divided by omega c. So, 2 into 50 divided by omega that comes out to be 39.8 nano Henry ok. In fact, I would like you people to compare this particular circuit with the previous circuit and you will actually see that if you look at the equivalent inductance and equivalent capacitance they are coming out to be similar ok. So, I want you to check that let me just tell you for one case inductor we had seen that this was 19.9 and this was 19.9. So, 19.9 plus 19.9 is equal to 39.8 nano Henry you can verify the same thing for the capacitance value ok. So, let us see the response. So, you can see that the response is exactly same there is a no difference at all. Now, I just want to also mention. So, how do we get these responses ok? Of course, this particular thing has been done using ADS software, but in the beginning we had used a freely available software the name of the software is RFCM99.exe. You can download from the internet it is freely available and in fact, what you do in that particular software you can just give that analysis and then you could look at the design part you can do the simulation of filter. In fact, it would be very very easy you simply have to say which type of filter you want to design low pass high pass or band pass and then what you do you give the frequency what is the cutoff frequency of course, there you have to define what is the order of the filter that software will not calculate for you order of the filter and then it will give you the response which is very similar to this. It will also give you the inductors and capacitors value it will also give the circuit diagram and then you can actually go and edit those components also. I just to mention to you for example, if you go to the market you would not get any capacitor which has a value of 7.95 picofarad ok. You may get 8 picofarad or 7 picofarad or may be 9 picofarad or 10 picofarad. Similarly, you may not get 39.8 nano Henry you may get 40 nano Henry or may be 39 nano Henry and so on. So, what you need to do then these are the design values, but these are not the practical values. So, you go in that particular software you can click on that particular component double click on that and then you can change the value of the inductor or change the value of the capacitor and then simulate. You will know what is the response and then check those things and depending upon whether it meets your requirement or not you can keep on doing the modification and sometimes maybe you can add inductors in series or in parallel to realize the desired value of inductor. Same thing you can do for the capacitor, you can use two capacitors in series or parallel to realize the desired capacitance value. So, now, let us see the response for Chebyshev filter which is also known as equi ripple filter. Now, for Lopez filter the response of the Chebyshev filter looks something like this over here. Now, try to compare this with the previous configuration which was for Butterworth filter. So, for the Butterworth filter if you see this is a minus 1 by 2 this can be written as 1 divided by square root 2 ok. So, in the case of Butterworth this expression was 1 plus you can say omega by omega c to the power 2 n, but here we have a F 0 component as well as C n square component which is C n square of omega by omega c ok. So, this is not really coming as omega by omega c, but it is the function over here. Let us just look at what are these different things here. So, C n x is Chebyshev polynomial of order n and let me just tell you this Chebyshev polynomial actually is defined in two different limits. One limit is for omega less than omega c ok or if it is normalized then omega less than 1. So, in that particular region the variation is cosine ok and in the region of stopband which is for omega greater than omega c or you can say normalized case omega greater than 1 this Chebyshev polynomial is nothing, but cos hyperbolic function ok. So, n of course, is order of the filter as before omega c is cutoff frequency and what is F 0? F 0 is constant related to passband ripple. In fact, you can actually choose how much ripple you want in the passband. So, generally speaking if you look at the low frequency component most of the time they actually define minus 3 dB as passband ripple, but I want to tell you please at microwave frequency never design a Chebyshev filter for minus 3 dB passband ok. The reason for that is minus 3 dB for S 2 1 will also correspond to minus 3 dB for S 1 1. So, that means, in the passband half the power will reflect back only half the power will transmit ok. So, majority of the time when we design microwave filter we never ever design a Chebyshev response for higher than 0.5 dB passband ripple. Why cut off a 0.5 dB? In fact, 0.5 dB passband ripple corresponds to S 1 1 approximately equal to minus 10 dB or approximately equal to V SWR equal to 2 ok. So, here is the expression for F 0 this is 10 L R by 10 L R is the ripple attenuation in passband. And as I mentioned please do not take L R as 3 dB which is generally done at lower frequency at microwave frequency please never ever do majority of the time we take this L R value to be may be from 0.1 dB to about 0.5 dB. So, how these Chebyshev polynomials are simplified in fact, this is the one of the simplified version of that which actually speaking does not even mention anything about cosine variation or cos hyperbolic variation. So, let us see what is this very very simple thing C 0 x is equal to 1. So, that is the C 0 value then what is C n 1 which is equal to 1 which is corresponding to a general term again the last term. And then let us see how are the other parameters here . So, C 1 x what is x here x is defined by this particular term over here. And this simplification is please remember it is for omega equal to omega c. So, when omega is equal to omega c this term will be equal to 1. So, you can say x is equal to 1. So, we know now what is C 0 which is 1 we can use C 1 x which is nothing, but equal to x and then all the other C n parameters can be obtained by using this particular expression. I will just make it simple for you people let us say we want to find for n equal to 2 let us say we want to find what is C 2. So, for n equal to 2 let us see what will be that this will be 2 into 2 minus 1 C 1. So, C 1 x what is C 1 x is equal to x. So, this term will become 2 x minus C n minus 2 will be 2 minus 2 is 0. So, that will be 1. So, we can say that C 2 x is nothing, but 2 x minus 1. Then when you want to find C 3 what you do now for C 3 this will be 2 and this will be 1. And I have just mentioned how to calculate C 2. So, this is the way you can find out all the C n parameters. Now, in this particular case again when you want to find out the G parameters it is now not as simple as finding G parameters for maximally flat. Here many more steps are required to find out the G parameters ok. So, let us see what are these steps here. So, G 1 expression is given by a 1 divided by f 2. So, what is a 1? You can see from here you can find the value of a 1 from here. What is f 2? f 2 you have to use this particular expression, but in this expression there is a f 1. So, you have to calculate f 1. And in f 1 you can see that there are losses in the ripple ok or you can say attenuation in the ripple ok. And you can see that there is a cot hyperbolic function cot hyperbolic function is nothing, but cos hyperbolic divided by sin hyperbolic. So, you have to do these calculations in general to find out the G 1 parameter as well as other G parameters. So, you can see from here if you want to find G k it requires a k minus 1 into a k divided by b k minus 1 into G k minus 1. So, you can find the values of a k from here and b k from here ok. So, let us just look at a quick design of a third order Jb-Shap low pass filter. So, here the order is defined which may not be always given to you ok, but let us say we want to design a third order Jb-Shap low pass filter that has a ripple of 0.05 dB. You can see that it is a very very small ripple value the cutoff frequency is 1 gigahertz. So, the formulas which have been given on the previous slide you use that. So, first we calculate F 1, then we calculate F 2, then we calculate A 1, A 2, B 1, then G 1, G 3 which are equal and G 2. Once you find these G parameters then after that we can use frequency and impedance scaling. So, we know that for inductor L 1 equal to L 3 which will be corresponding to G 1 equal to G 3. So, here this value is 0.8794 and to find the value of inductor we have to multiply with the impedance which is 50 ohm here divide by the frequency which is 1 gigahertz. So, that will be 2 pi into 1 gigahertz this comes out to be 7 nano Henry. To find the value of the capacitance you use G 2 parameter. So, this is G 2 divided by R 0 or you can say Z 0 which is equal to 50 ohm multiplied by omega c which is 2 pi into 10 to the power 9. So, that comes out to be this value over here. So, this is the actual circuit simulation for third order Jb-Shap filter with the response of 0.05 dB ripple in the passband and cut off frequency of 1 gigahertz. So, let us see the response of this particular filter. You can see that here this is the frequency versus S parameter plot here. So, this is the plot for S 21. Now you can see here there is a very little blip over here actually speaking that blip here corresponds to 0.05 dB it may not be even visible over here and then this is the response. So, that is S 21 ok. Now this is the response for S 11 you can see that this response is slightly different than the maximally flat. In case of maximally flat S 11 actually looked like this here, but over here in fact, this value corresponds to which is minus 20 dB this minus 20 dB actually speaking corresponds to the ripple at this particular point over here ok. So, corresponding to this little dip over here S 11 is about minus 20 dB. Now just to tell you had we taken this as a minus 3 dB then S 11 will also would have been minus 3 dB, had this been 0.5 dB then S 11 would have gone up to about minus 10 dB ok. Now again I want to mention here that we had designed this filter at 1 gigahertz. Now do not look at the crossover point because the crossover point corresponds to about minus 3 dB ok. Remember this we had designed for 0.05 dB ripple ok. So, corresponding to this 1 gigahertz. So, this is where that 0.05 dB ripple would come into picture and that is how the response is there ok. Now of course, you can do the calculations by using the formulas given to you or alternatively you can use these tables. So, here is a table given for 0.5 dB ripple n parameters are given from n equal to 1 to 10 corresponding g parameters are given over here. So, you can actually do the checking. So, depending upon the order of the filter let us say you are designing for a fifth order filter you can calculate g parameters from here ok. G 6 is equal to 1 ok which is the terminating impedance. Now I just also want to mention that there is a book Methai, Young and Jones book which has given many of these tables for different values of ripple ok. And then either you can use that book to find out what are the g parameters or you can use the formulas given in the previous slides to find out what are the g parameters ok. Now we are going to look at a one comparison thing. So, for example, what is the order required for a low pass filter realized using Butterworth or Chebyshev configuration. So, here we are taking a little different problem in a sense that a larger attenuation is required at relatively closer frequency to the cutoff frequency. So, let us say the desired value is 30 dB attenuation at omega by omega c equal to 1.2. I just to mention in our earlier slides we have mentioned this as omega 1 ok. So, please do not get confused it is the same thing omega by omega c or you can say omega 1 by omega c. So, this is what is the frequency ratio equal to 1.2. You can think this way omega c may be 1 gigahertz then this will be 1.2 gigahertz. If it is 2 gigahertz that means, omega will be equal to 2.4 gigahertz ok. So, for now 30 dB attenuation we can now look at step by step. Earlier I had given you the formula for n, but now we will show you all the steps how that formula comes into picture and you need not remember the formula all the time. You can use this particular simple expression to do the derivation. So, let us say we want H j omega square to be equal to 30 dB which is in a reality minus. So, you calculate the value that comes out to be 0.001 and please note this is a square term. Now this term here is given by this particular expression over here. So, now substitute different values. So, this one here is 0.001 1 divided by omega by omega c is 1.2 to the power 2 n. Now what you do? You take log on both the sides, but before you take log you take this on this side this divided over here. So, 1 divided by 0.001 will be 1000 and then this one comes on this side which will become triple 9 then take a log ok. So, that will be 2 n log 1.2 and the right hand side that time would be 999 take that and now you simplify this n comes out to be 18.94. So, that means, to realize this particular thing you need a 19th order Butterworth filter. Let us see what happens if you want to realize using Chebyshev filter. Here in this particular example we have assumed 1 dB ripple in the pass band. So, L r is equal to 1 dB. So, now first calculate f 0 f 0 is given by this expression substitute the value of L r 1 you will get f 0 equal to this. Now in the stop band I had told you c n function is nothing but cos hyperbolic function. So, this is the expression for that. So, what we do now we write the same thing H j omega square is given by this particular expression we substitute the value 0.001 this is 1 divided by 1 plus f 0 just the expression of c n has been written over here which is right here. Now we try to simplify this particular thing over here right. So, substitute the value of f 0.2589 and this term is transferred over here with simplification you get over here and then what you do for this particular term next simplification is you take cos hyperbolic inverse of this particular term after taking a square root. So, this is what the term is that is on the left hand side simplify for n you will get n equal to 8. So, you can actually see that to realize this particular response if we use Butterworth concept we would need an order of filter which is 19 whereas, if we use a Chebyshev filter we would require an order of only 8. Of course, I just want to mention here we had assumed ripple to be equal to 1 dB. Now if it is 0.5 dB you will see that the number of elements required will be more ok. So, this is an exercise for you people please find out take this as 0.5 dB ripple and see what would be the value of n, but you can see the big difference. If we use n equal to 8 that means, total number of inductors and capacitors required will be only 8 here total number of components required will be 19. So, you can say that the cost of this particular thing will be much larger even the circuit realization will be larger. So, in the next lecture I will show you what is the response of these two filters ok. So, just to summarize today's lecture we started discussion about the Butterworth filter we use that concept which we had discussed in the previous lecture of frequency scaling as well as impedance scaling and then we designed a low pass filter using Butterworth and we saw that whether you take first element as inductor or capacitance response remains exactly same. Then we looked at the Chebyshev filter and for Chebyshev filter we saw that the finding g parameters required more steps, but however Chebyshev filter has an advantage that for this particular specification given here you would require only 8 number of elements compared to 19 element. So, in the next lecture we will see what is the response how the two responses compare with each other. So, till then thank you very much enjoy yourself work hard we will see you next time. Bye.