 Today in our 28th lecture we will be continuing with universal active filter that we have synthesized in our earlier lecture and we will be seeing the effect of the active device parameter gain mandate product on the performance of these filters. Let us briefly review what we did in the last class we discussed first, second, third order state space filter realizations systematically we have synthesized these circuits then the op-amp realizations were also discussed the universal active filter block why it is universal because of its ability to give important second order filters like low pass, high pass, band pass and band stop at the different outputs of the op-amps and a combination of low pass high pass band pass can be generated by using an summing amplifier generalized summing amplifier to locate the zero of the second order system anywhere on the S plane when the poles are located always on the left of plane as complex conjugate pass or a pair of poles on the negative real axis. Now the magnitude and phase plots were indicated for all these types of filters and the significance of phase plot in tuning filter how it is important why magnitude plot cannot be strictly used when it is high Q filter because peak indicates no variation of magnitude around that frequency whereas phase variation is maximum and we have also shown theoretically that the phase variation of a second order system is directly proportional to Q which is 2Q by omega naught that is the change of phase for frequency variation. State response coupled with the steady state response gives us a quick method of measuring the effect of Q variation with respect to gain band width product etc. So we had seen the qualitative measure of the effect of gain band width product on the Q and the notch output today we will be discussing the effect of gain band width product quantitatively. Let us now consider the by quad resonator block that we had used in forming the pole forming loop remember this this contains what we had discussed earlier an integrator followed by another integrator than an inverter forming a closed loop what does this give obviously this I had earlier pointed out in the last class that this gives us del squared V naught by DT squared plus K naught V naught equal to 0 there is no VI applied and there is no addition of delta V naught by delta T to this summing output. So that means this is nothing but a harmonic oscillator which I had indicated is something that you can formulate using a capacitor and an inductor this is nothing but the harmonic oscillator if you write this as V naught and right on the ahh ahh current loop current in this or in terms of node equation here C DV naught by DT is the current is equal to minus 1 over L integral V DT is the current. So you get this as a second order harmonic equation like this where K naught is equal to 1 by LC root K naught is the frequency of oscillation or 1 by root LC is the omega naught of this. So that is a resonator block for you how will this look like a resonator that is what we are going to explain to you today. So we have the capacitor now let us therefore prove how this is nothing but similar to this so called tank circuit. So let us start with this let us say this is the capacitor and let us see what the capacitor sees because of this loop. So this is virtual ground okay so V naught 3 in this case is the capacity voltage okay so we will call it VC let us see voltage across the capacitor. So if you take the voltage across the capacitor as VC then that is equal to V naught 3. So the current in this is VC divided by R this current goes to this and develops a potential which is VC this way right because R is the same so VC by R into R. So this V naught 1 is equal to minus VC that you know that it is an inverting amplifier here so that minus VC comes gets integrated so this voltage is plus VC divided by SCR inversion at the integration and the current in this is now VC divided by SCR square. So the current in this goes through this like this that means this voltage is seeing a current being given out by the capacitor which is actually equal to VC divided by SCR that means it is feeding its current to an inductor of magnitude L equal to CR square that means the capacitor across this you see an inductor getting simulated because of this loop. Same way if you work out the voltage across this capacitor as VC you will see that there is an inductor L equal to CR square across this. So at every port in this network you will see a resonant circuit getting formulated because of this resonant loop that is why this is called a resonator block. Now you might question me regarding this what does resistance see across it so I would like you to work it out put the voltage across the resistance as we are and then proceed further and you will see that you can prove the same thing that it is also seeing an element such that it is also going to resonate at omega naught okay and the admittance is going to become equal to 0 at omega naught just as in this case when it resonates the admittance of this network goes to 0 at omega naught equal to 1 over root L C. So let us see what the resistance is so we can just do that so this V naught 3 is the voltage across the resistance V R. So V R by R is okay this voltage now becomes minus V so that minus V R becomes plus V R divided by SCR and another integrator makes this voltage V naught 3 now become equal to that is the current in this right is going to be developing a voltage across this which is going to be again divided by SCR right. So V R divided by SCR square okay that into R is the voltage across the resistance that we are getting so we have that as current through the resistance as V R okay by SCR whole square into R is the voltage across the resistance. So what it simulates is nothing but 1 by S square C square R which is a negative resistance that you are seeing across this resistance R right so a frequency dependent negative resistance okay which is cancelling with the positive resistance is what happens when it resonates. So if you see the resistance the resistance is a negative resistance okay of magnitude 1 by S square C square R okay as the negative resistance across it frequency dependent negative resistance across it which also resonates at omega naught equal to 1 by RC. So this is the thing we start with voltage across the capacitor across the resistance and get the voltage in terms of the current and get what the resistance is as the impedance okay shunting the resistance so it is either resistance okay with the negative resistance which is frequency dependent okay that is what we see in this structure. So this is the resonator block. So now what I am doing is I am just connecting this app with certain impulse given at the beginning to one of the capacitors so as to make it have some charge initially then what happens some 100 millivolts charge is given so that 100 millivolts keeps on building up you can see because of the negative resistance right there is due to the finite gain bandwidth product this we saw in the inductor simulation circuit also because of and also the Q enhancement due to give in negative feedback structures. So that is simulating a negative resistance okay across the inductor and therefore it is growing the oscillation that or the energy that has been put initially across the capacitor is causing the whole thing to oscillate at F naught okay and that F naught is equal to for this time constant that we have chosen okay is nothing but 1.59 kilohertz or this time interval if you measure it comes out as 630 microseconds is the time period. So it is oscillating at that frequency and the oscillation amplitude will keep on building up until it gets limited by the power supply. So this is what happens if you make the Q go to infinity the Q infinity means if you put an energy that amplitude is sustained now because the negative resistance appears across this capacitance it is growing okay exponentially. So this is the effect of finite gain banded product it is enhancing the Q again we will measure this effect quantitatively later using the circuits that is a net of blocks. Now commercially we have a universal active filter block available TI makes it and TAF 42 is the number it is an analog AC that serves as a second order active filter building block that can be used to realize any filter. So it has 2 integrators and one summing amplifier in forming a loop and then another summing amplifier which is the Q forming loop that we had shown earlier. So that Q forming loop has been removed here. So this entire thing 2 integrators and a summing amplifier and the Q forming loop. So 4 op amps are there in that. So realization of any filter can be done therefore cascading the second order blocks and one first order block AF 42 therefore can synthesize any higher order filter. Now resistors can be connected or even capacitors can be connected external to the integrator it already has capacitors inside 1 nanofarad capacitors. We will see that circuit these are the 2 capacitors forming the 2 integrators. So this is one integrator this is normally connected to the ground so this is normally connected to the ground. So both integrators are grounded and then resistance can be connected here. So this is R this is C so R2 C2 and say R1 C1 form the time constants of the integrators so this R2 and R1 are external time constants so R2 is equal to R1 equal to R C2 equal to C1 equal to C equal to 1 nanofarad in this case. If you want to increase the capacitor you can connect capacitor across this as well as across this increasing the capacitor you cannot decrease it further. So 1 nano is going to give you the highest frequency that is possible with this with the lowest value of R2 you can connect at this point. So let us see how and then this is the inverter okay 50K, 50K forms the inverter okay actually let us call this R1A and R2A make it distinctly different from these R2s and R1s. Now this is the inverter integrator integrator loop okay. Next the possibility of using this plus for adding also is there to make what is called the KHN network but if you want to get the same universal active filter that we have used there which can give you low pass, high pass, band pass and notch at the output of the op-amps then we can use this as the Q forming loop okay. So this particular output can be taken from the band pass output here and then this is going to be Q times R and R here that is the Q forming network and input can be added so this is R by H0 VI. So and this has to be connected to this through another resistance which is adding resistance. So this is going to be 50K so that this adds to this this 50K adds to this forming loop this is the Q forming loop 50K that we are having. So now this exactly represents the universal active filter block that we had used earlier and let us see how a Butterworth low pass filter can be designed using this universal active filter block 42. So design a Butterworth low pass filter with cutoff frequency of 4 kHz and again of 2 using by 1. The quality factor of Butterworth filter is 1 over root 2 so we know that this is the Q forming resistance this is put as R this is Q 1 over root 2 times R. So if this is 1K this is 700 ohms so that decides that. If you use universal active filter 42 the integrating capacitor is 1 NFL so R equal to 1K R1 can be evaluated 1 by 2 pi C 1 F naught so 2 pi 10 power minus 9 Farads 4 KHz so gives you 39.8 kilo ohm as R to be connected externally so that is what it is 39.8 kilo ohm and 1 nano Farad from the time constant of the two integrators and R is 1K this is 1K by root 2 K which is 700 ohms right R by H naught if H naught is 2 for 1K this is 500 ohms so that is the low pass Butterworth filter design over that is simulated the response of the 4 kilo Hertz low pass filter so you can see that gain is 2 so that is going to be the low pass filter gain and then it is coming down it is 6 dB and then coming down to 0 or almost minus 10 dB at about 10 kilo Hertz 3 dB point is 4 kilo Hertz. Design a band pass filter using by quad this has been simulated using TINATI which is the software where all these models are given for universal active filter block 42 so we do not have to we just connect the external resistances and obtain the output of the low pass so the low pass is got at this point so this is the second integrator so this is the low pass okay and this is the band stop okay and this is the adder so this is the high pass this is the band pass design a band pass filter using by quad with bandwidth of 120 Hertz bandwidth is 120 Hertz and we have here is the thing at center frequency of 1.2 kilo Hertz and 3 dB bandwidth is given as 120 Hertz that means Q is 10 center frequency by bandwidth the quality factor is 10 H naught the gain at the center frequency you want it to be 10 for a feedback resistance of 10 kilo ohm so that is the resistance that is put okay Q determining resistance is equal to that is R equal to 10 kilo ohm okay Q determining resistance is Q times R so that is 100 kilo ohm gain determining resistance is 1 kilo ohm okay because Q is 10 H naught is 10 so that gain is 100 H naught into Q at this point it is H naught into Q so we have gain determining resistance equal to 1 key okay we use universal active filter 42 the integrating capacitor is 1 nanofarad so R1 comes out as 132 for this frequency of 1.2 kilo Hertz 1.2 kilo Hertz 1 nanofarad so the circuit is given as follows 132 kilo ohm and 1 nanofarad 132 and 1 nanofarad and gain determining resistance 1 K 10 K 1 K 100 K Q of 100 okay so this points are to be grounded okay and this is connected as 50 K summing amplifier so output is taken because it is band pass output output is taken at this point here so that output is simulated here so you can see the band pass output with center frequency exactly at 1.2 kilo Hertz 1.2 kilo Hertz and bandwidth of 120 Hertz 1.2 kilo Hertz Q of 10 okay band pass filter circuit using universal active filter we have 132 K 1 nanofarad and all that we have already seen notch filter now so these are the important designs that we can carry out design a notch filter with Q of 100 that means you now want very narrow notch at 50 Hertz this power line removal okay H naught is 1 so this is going to be very narrow 200 that means 50 by 100 about 0.5 Hertz bandwidth so Q determining resistance R is 100 K gain determining resistance is 100 K gain is 1 okay feedback resistance of 1 K so Q is fixed at H naught into Q okay this is H naught H naught is 1 okay so we can calculate this this is 50 Hertz output 1 by 2 pi C 1 has been enhanced to higher value because 1 nanofarad will give 2 higher value of resistance for 50 Hertz so we have me connected across 1 nanofarad 100 nanofarads extra and made it 101 nanofarads and 50 Hertz okay is going to result in a reasonable value of resistance which is 31.5 kilo ohms okay that is what it is C2 is shunted by 100 nanofarads so 101 100 101 31.5 kilo that we have calculated and Q of 100 okay and the gain okay is going to be actually H naught which is 1 by 100 K into Q H naught into Q is 1 that has been chosen here right H naught is 1 by 100 in this case if you make this equal to also 1 K then the gain is H naught is equal to 1 okay let us see what has been done that is what is done for 100 K you have 40 dB attenuation there okay because it is H naught is 1 by 100 that has been designed for that so this goes very low right that itself is 40 dB at low frequency and high frequency this is going still lower so it is acting as an attenuator right so if you select this as 1 K on the other hand it will go to 0 dB at low and high frequencies and therefore let us see what the design is H naught is supposed to be 1 whereas here H naught into Q has been made equal to 1 so H naught is actually in this case we have designed it for 1 by 100 okay that is what is simulated say this has gone to less than 80 dB at this point so almost nearly 40 dB attenuation is got around this even better than that commercial available by quads consist of 4 op amps and several precision resistors and capacitors are more expensive you have 42 cos 8.25 dollars per unit procured in thousands several discrete passive components have to be used to achieve the required filter parameters an alternative you have 42 obviously is like a quad op amp TL084 or LM324 okay it is available for only 0.2 dollars so this can be quickly built okay using op amps so Butterworth second order high pass filter is designed design a high pass filter with lower cutoff frequency of 4 kHz using quad op amp the quality factor of Butterworth filter is 1 over root 2 we choose C1 as 1 nano farad R1 comes out as 39.8 kilo ohms which is same one that we use for low pass filter design so 39.8 1 nano 39.8 so this imitates the earlier design using a that is universal filter 42 and 1K 1K 1K and this 1K this Q times R that is 700 ohms and gain at H naught is going to be gain at high frequency is 2 so I can see that going to 60B at high frequencies and the 3 dB point is 4 kilohertz exactly this limitation comes about due to the finite gain bandwidth product of 741 which is about megahertz or so so this is the cumulative effect of gain bandwidth product in the loop used by 7 using 741 okay this will study later right so it seems to be almost acting like a sort of high pass all the way up to about say this frequency about 100 and all 50 kilohertz so if lm 741 is replaced by better op amp like TL0 A2 or 81 or 84 that you can get this shifted all the way up to about 1.2 megahertz that is what is done LF356 or TL0 A2 if it is used the same thing gives you the 4 kilohertz with upper cutoff frequency going up to about 1.2 megahertz so what is the effect of GB finite GB on the back part filter that is what we are studying here right so that is the high pass filter built using 741 or 356 TL0 84 so and we had seen the effect of gain bandwidth product at the high end of the high pass. Now in general how to evaluate the effect of this quantitatively the by cut filter has two feedback loops now the feedback loop formed by two integrators and an inverting amplifier ABC in this whole thing okay just look at this ABC forms one loop this is what is called the pole forming loop resonator block then the Q forming loop these two loops are involved so what is the effect the effect is primarily due to this pole forming loop this for high Q as very little influence on the performance of this we will see that presently what happens to this pole forming loop the pole forming loop the loop gain is this is from here to here is an integrator omega naught by S okay minus omega naught by S into another omega naught by S plus omega naught squared by S squared and this is inverted so it becomes minus omega naught squared S squared if the loop is broken so that is the what is called loop gain of this so it has two integrators and an inverter this only gets added but for high Q this resistance will be R times Q so this is going to be very large let us say it is very large because it is high Q that we are considering so in such situation what happens this contribution is only for adding to this minus omega naught by S into Q that is all as the loop gain and this quantity for high Q is very small compared to this so the contribution is only due to the finite gainment product of all these three things that comes into picture and alters this we had seen that integrator in the earlier case when we discussed feedback amplifiers the integrator what the effect of gainment product is each integrator will give you so S by omega naught divided by 1 plus 1 over loop gain that part of it is coming out as omega naught by GB a magnitude error and a phase error remember this that we have discussed earlier this also contributes to the same thing so it will be minus this will be squared and this will be squared so and that into this inverter so inverter is going to cause ideally minus 1 and effect due to gainment product is no magnitude error only the phase error which is 1K 1K here so 1 by 1 plus 2S now if you have one more 1K here this will be becoming ahh sort of again 3S okay this will also contribute to one more addition so it is this by this plus 1 if this is not there otherwise it is this by this plus this by this plus 1 which is 3 ahh 1 plus 3S so effectively if you put for this the contribution for delay is S by GB here by one integrator S by GB here why the another integrator and 3S by GB by the summing amplifier so effectively there is a delay error of 5S by GB if you know this this will only change the omega naught slightly right so this 2 integrators will cause 1 plus 2 omega naught by GB that means total loop gain is going to get altered from this okay minus omega naught squared by S squared okay by this amount what is it 1 by 1 plus ahh 2 omega naught by GB plus 5S by GB is this clear okay so this loop gain is going to get altered here what this also gets altered slightly due to this but that is of no consequence because that magnitude contribution itself is small for high Q so let us therefore see what happens to the overall transfer function this what I have mentioned here the overall loop gain is primarily minus omega naught squared by S squared apart from that there is an addition of minus omega naught by QS if effectively it is minus omega naught squared by S squared minus omega naught by SQ so the loop gain which was ideally GL equal to this becomes make the denominator 1 minus GL so 1 plus omega naught by SQ plus omega naught squared by S squared it is the alteration of 1 minus GL that is of interest ahh for us in evaluating the performance of the feedback loop so now that 1 minus GL now becomes because of the modification that we have mentioned earlier right because of the effect of omega minus omega naught squared by S squared changing to minus omega naught squared by S squared into 1 by this which is approximately equal to approximately equal to minus omega naught squared by S squared into 1 minus 2 omega naught by GB minus 5S by GB that is all the alteration is let us therefore see what happens when we replace omega naught squared by S squared in our expression by that okay so 1 plus omega naught squared by S squared now gets replaced by omega naught squared by S squared into 1 minus this factor all these things can be ignored because their contribution is very small this we have ignored otherwise that is the exact value so it is only remaining as omega naught by SQ so main alteration is in the coefficient of omega naught squared by S squared by the phase error here 1S by GB by 1 integrator 2 integrator 2S by GB the summing amplifier contributing 3S by GB totally 5S by GB phase lag error this is the magnitude error this only changes omega naught to omega naught dash okay which corresponds to 1 omega naught into 1 minus half of this omega naught by GB that means omega naught shifts slightly downwards by that but that is of no consequence compared to what happens because this is already a small quantity the coefficient of S gets added to this this S gets cancelled with this and a negative coefficient of S gets added to this so this becomes equal to 1 plus omega naught squared by S squared into 1 minus 2 omega naught by GB plus omega naught by SQ okay into 1 minus this S cancels this so 1 omega naught by S okay into okay we get here 5 omega naught okay by GB into Q because Q also has been taken out so that results in a new value of Q and of course a slightly shifted value of omega naught which is omega naught dash is equal to omega naught into 1 minus omega naught by GB. So if GB is very large compared to omega naught this is very close to omega naught itself however here we have Q actual due to the gain made product becoming equal to Q ideal divided by 1 minus 5 omega naught by GB that is the phase error phase lag error into Q that gets magnified by a factor of Q. So F naught into Q becomes an important parameter which has to be concerned okay which has to be of our great concern to us in deciding what type of op-amp should be used in order to prevent this sensitivity to the gain made product. So let us therefore continue with our discussion omega actual becomes this okay omega naught just a minute I think there is a mistake I think it is written wrongly it is omega naught okay not divided by into square root of 1 minus 2 omega naught by GB or omega naught okay into 1 minus omega naught by GB. So omega actual gets reduced that is of not much significance Q gets enhanced. So this is the final effect Q enhancement occurs due to finite gain made product by what amount Q divided by 1 minus the error in the phase lag that has happened okay due to gain made product into Q. So sensitivity of Q to gain made product is delta QA by QA delta GB by GB which is 5 omega naught Q by GB 1 minus 5 omega naught Q by GB. So this quantity much less than 1 in order to desensitize the whole design okay with respect to Q variation with respect to GB. The manufacture of bypass usually specify what is called F naught Q product. So please look at this F naught Q product and see whether for your design this is satisfactory that F naught Q is much less than the gain made product is what should be of concern for universal active filter for 42 therefore it is about 100 kilohertz this product. So you should maintain this okay if it is 10 kilohertz you can only go for a Q of about 10 okay if it is 1 kilohertz on the other hand you can go up to 100 Q of 100 so on and so forth in order to desensitize it with respect to gain made product. Now this is quantitatively judged here by us okay as follows this is the earlier simulation of this band pass filter with F naught equal to 1.59 and there is however a major difference in simulation that I have done I used near ideal op-amps as integrators only for the summing amplifier I used this practical op-amp 741 okay in simulation which means actually the phase error due to the two integrators are not to be taken into account because they are near ideal integrators have been designed using a DC gain of 2000 okay for these op-amps so that is all it is near ideal op-amp it is gain made product is infinity only the summing amplifier has thing so that means actually the phase error is just that of the summing amplifier which is 3 omega naught by GB so that is why the Q gets enhanced as Q divided by 1 minus 3 omega naught by GB that is the phase error into Q. So for a Q of 5 okay 3 times omega naught which is 1.59 kilo hertz divided by gain bandwidth product is 10 to power 6 comes out as 5 by 1 minus this factor is about 0.024 it results in okay 0.024 also the gain getting enhanced to or Q getting enhanced to 5.12 since H naught into Q is the gain at the center frequency of this H naught being equal to 1 gain at center frequency H naught into Q is equal to in this case 5 should have been and it has changed to 5.12 in practice it became 5.081 you can see the accuracy with which and see frequency is nearly 1.59 so it is 1.595 okay here okay. Now as the frequency is 10 fold increase the frequency F naught is changed now from 1.59 to 15.9 by making the capacitor 10 times less than the previous circuit from 0.1 micro far to 0.01 micro far. So what happens then Q is 5 so this factor now becomes 1 minus 5 times Q of 5 into 3 times omega naught is 15.9 10 fold more divided by 10 to power 6 that means this factor is changing point 0.024 to 0.24. So 5 divided by 1 minus 0.24 which is 5 by 0.76 is 6.57 whereas actually we got it in simulation as 6.7 6.68 so 6.7 so that is 6.6 pretty close agreement for the approximation that we are making. Please remember that all these approximations assume that this factor is very small compared to 1 okay that is why 1 by 1 plus x is changed to 1 by 1 minus x. So within that limitation this is pretty accurate as far as we are concerned. So what happens to the notch filter effect of finite product. Nose filter output is H naught into 1 plus S square by omega naught square ideally divided by the denominator S square by omega naught square plus S by omega naught Q plus 1. Now with S square by omega naught square being replaced by S square by omega naught square into 1 minus 5S by GB the denominator okay has changed the Q to this which is used earlier in the band pass filter to estimate the change in gain. Here now what happens the numerator has earlier no S coefficient now it has an S coefficient which is dependent upon GB and GB is infinity this goes to 0. So because of the finite GB the numerator under resonance S equal to j omega omega equal to omega naught this gets cancelled with this factor is 5 omega naught by GB. So this factor becomes 5 omega naught by GB that remains at omega equal to omega naught this gets cancelled with this and we get this as Q divided by 1 minus 5 omega naught Q. So 5 omega naught Q by GB divided by 1 minus 5 omega naught Q by GB into H naught original H naught in this case is 1. So we can evaluate the effect of the gain which should have gone to 0 no transmission at omega equal to omega naught because of the notch because of the 0 here but because of the non-ideality it is not going to 0 it is going to be finite and you can see that as Q increases it increases. So making high Q for exact removal of the 50 hertz okay and not losing the nearby signal okay it is disadvantageous as far as 50 hertz transmission itself is concerned 50 hertz comes out without getting removed because of this high Q in the notch. So you can look at this Q of 5 H naught of 1 and F naught of 1.59 this factor was 0.024 as earlier determined. So it will now become 0.024 divided by 1 minus 0.0 or it is 0.025 25 milli right you get it as actually very nearly the same as practical simulation result shows 25 milli this is 1 and this is 1 this is gone to 25 milli this is what is practically shown in simulation. And again as the frequency is 10 fold increased 15.9 cos frequency is 10 fold increase this is going to increase okay in this manner it will become 0.24 divided by 1 minus 0.4 which is 0.32 so you can see it is about 0.336 actually milli 0.336 pretty close to 0.32 that has been evaluated. So this is a very useful information for the designer to decide what kind of gain bandwidth product he has to select okay that has become important in most of the designs today okay of filters. There is another way instead of selecting a higher gain bandwidth product perhaps it is possible to get this delay error compensated by lead error instead of lag error right. So the double integrator loop is called resonator block as it simulates the second order system with coefficient of first order term 0. Normalizing frequency is solely determined by the integrator time constants and invert again in the resonator block. Q determining loop is independent of this Q and omega naught can be adjusted separately conveniently so that is the great advantage of state space filter. H naught gain at low frequency for low pass gain at high frequency for high pass that is H naught gain at center frequency for band pass is H naught into Q gain at low and high frequency for notch is H naught again. Can all be independently set these independent adjustments of filter parameters are not always possible with other active filter realizations may be other than inductor simulation. This constitute basic building blocks of all present day IC filters that is why this is a very significant piece of structure that is a must for every analog engineer to understand. Voltage control state space filters integrators in by called have fixed resistance and capacitors filter is tuned by changing the resistances and capacitance are both by replacing the integrator with the fixed integrator and a multiplier it is possible to create a voltage control filter in the following manner. So this integrator which is used in the double integrator block each integrator is replaced by an integrator okay preceded by a multiplier this way. So what happens this VI which was directly getting applied here and constituting current of VI by R now has converted the input to the resistance to VC VI by 10 okay 10 is the reference voltage of the multiplier. So the current in this now changes to VC VI by 10R it is equivalent to saying that the resistance now can be changed okay to 10R by VC in all our expressions that means VC by that is VI by SCR which changes to VI VC by 10 SCR. So the frequency omega naught the normalizing frequency of the filter now is nothing but VC by 10 RC which is fantastic because it is linearly related to the control voltage. So we can change the tune the whole thing because the RC product is modified by this VC because of this same thing can be done using feedback loop having a multiplier there is an alternative but this is a better scheme. Now what is done is the filter is designed okay using such an arrangement of integrator with a multiplier in between inserted here multiplier here and a multiplier here or these multipliers are having connection to VC the control voltage the following. So the RC time constant of both are ganked because of this two multipliers a single VC and this becomes a voltage control filter that filter is simulated now okay and VC is varied from 1 volt to 10 volts so that the omega naught varies by 10 volt from here to here say 1.59 to 15.9 it is changing okay in steps of maybe 2 volts or so. So we get for different VC okay the center frequency is waning linearly please remember that this is log plot so this is changing linearly and the notch and the band pass okay simultaneously changing for different values of VC finally this. So going all the way up to nearly I think some 18 kilohertz or so. So this is the voltage control filter more about this why this is needed in what is called tuning filter when the RC time constants are not accurately fixable tuning of filters okay why this is needed is going to be the topic of the next lecture. So we had discussed the effect of finite gain bandwidth product on the Q of the filters and the notch output how it is very sensitive to these Q gain bandwidth product if the Q is high okay we have concluded that the total phase lag error is the one that causes a shift in Q by Q divided by 1 minus Q into delta phi that is the factor by which the Q gets altered Q due to the gain bandwidth product and this delta phi depends okay there are inversely proportional to gain bandwidth product higher the gain bandwidth product lesser is the error thank you very much.