 Because in a certain sense, when we go from low pass to high pass what we are doing is to reverse the roles of frequencies in a way. I will explain what I mean, high pass transformation. Now you see in a high pass transform what we wish is that as the frequency increases you know in a high pass filter we have a specification something like this. So, you will expect that you will have a stop band in a pass band like this. So, what we are saying is as we go from 0 towards plus infinity and of course the same thing is mirrored on the negative side. We do not need to keep drawing the negative side. As we go from 0 to infinity on the omega axis we expect that we first encounter the stop band and then the pass band. So, it is very clear that if I want to go from here to a response like this as I had for the low pass filter then I should be traversing the frequencies in the opposite direction. So, when I move this way on the high pass frequency axis I should be moving this way on the low pass frequency axis. That is what it is it sort of suggests to us that we want an inversion of behavior and then it is very easy to think of an admittance which gives me inversion. Either the admittance of an inductor or the impedance of a capacitor gives me inversion. So, all that we need to do is to consider a transform like this f high pass s is of the form omega p by s simple inspired by inductance admittance of an inductance or impedance of capacitance. Now, let us verify what this does to us. Let us start with the last property first. Let us see what it does to the imaginary axis. Let us put s equal to j omega. So, what we are doing essentially is we are going to replace omega or rather we are going to replace s l of the low pass filter by omega p by s. So, what would we get? When we put s equal to j omega we would get omega p by j omega l by j omega which is minus j omega p by omega and this is of the form j omega l with omega l equal to minus omega p by omega. So, indeed the sinusoidal frequency behaves as we expected to and that is not surprising you have taken an LC impedance anyway. It is bound to have an imaginary impedance, but the good thing is if you now look at the way the frequency changes. So, let us draw what we call the frequency plot the frequency transform the sinusoidal frequency transform or the frequency mapping these are different names that are used for this. So, the sinusoidal frequency mapping. So, what we do in a sinusoidal frequency map is we take omega and map it to omega l. Of course, we know that omega l is minus omega p by omega. So, what we are saying essentially is as omega goes you see you want this is the kind of response that you want in the high pass filter. Now, omega p would map to minus 1 omega s is less than omega p and therefore, minus omega p by omega s is a quantity less than minus 1. So, it is behind minus 1. So, there we are omega s would map there and of course, 0 now you know we must be careful about 0. We do not really go to 0 we say 0 plus 0 plus means a very small positive value and you can make that small as small as you desire to move to 0. So, very small positive value here would go to minus infinity. So, in fact, 0 plus takes you to minus infinity. So, we can draw that this goes to minus infinity. So, there we are this passband now moves from let us let us call this omega s l this this point you know where the edge of the stop panned maps. So, stop panned now maps between minus infinity and minus omega s l and minus omega s l is of course, behind minus 1 further what happens to plus infinity now plus infinity maps clearly to 0 minus. So, plus infinity maps to 0 minus 0 minus means you know. So, when you say plus infinity you mean a positive number that grows without bound. So, when you do minus omega p by omega if omega is a positive number that grows without bound this would tend to a negative number that becomes as small as you desire in magnitude that is what you mean by 0 minus. And therefore, the passband is brought between minus 1 and 0 the stop panned is taken between minus infinity and omega s l and that indeed is what you expect for a low pass filter. That means, we have transformed the low pass filter is thus transformed is transformed to a low pass filter as follows plus 1 of course, there is symmetry. So, the passband is brought between minus 1 and 0 and therefore, also between 0 and 1 the stop panned is brought between omega s l and plus infinity and minus omega s l and minus infinity. Remember the passband and stop panned are carried as they are the dependent variable is carried as it is that is not affected. So, when you bring the frequencies from omega to omega l no effect is seen on the dependent variable that is carried. That means the tolerances are carried as they are that those are unaffected. So, if you have if you desire a certain set of tolerances in the high pass filter the same set of tolerances are carried into the low pass filter. Tolerances carried as they are carried as is and of course, omega s l is omega p by omega s for the high pass filter where omega p is more than omega s yes is a question. So, the question is would the maximum value of omega be pi. Now, please remember that we have already come to the analog domain. So, what was from 0 to pi has now been brought to 0 to infinity by the bilinear transform. So, now we are no longer dealing with the discrete angular normalized frequency. We are now dealing with an analog frequency which has emerged from the bilinear transform. So, so much so for the high pass. Now, you see therefore, the high pass filter design problem is easy to formulate. Now, design a low pass a corresponding low pass filter with passpan edge equal to 1 stoppan edge equal to omega p by omega s passpan stoppan tolerances and nature carried as they are as are. What I mean by that is if the passpan is monotonic let it continue to be monotonic in the low pass filter the pass pan were equi-ripple let it be equi-ripple here. So, 2 for the stoppan and that means you now have the tools to do it you can design it using the Butterworth approximation or the Chebyshev approximation or if you are trying out any of the other 2 inverse Chebyshev or elliptic you could do that and then all that you need to do is to. So, having designed the low pass filter the next step would be to make a frequency transformation. So, transform s l that is the complex frequency variable of low pass filter or Laplace variable of the low pass filter using omega p by s replace s l by omega p by s and you get the high pass filter that you desire yes to the question. The question is how do we prove the tolerance are carried as we do not need to prove anything there. You see what we are moving is the independent variable. Now, what value the function takes at that independent variable is unchanged all that you are saying is that I am evaluating the function at this value of the independent variable and that is anyway specified in the original filter. So, I am saying the whole dependent variable is carried as it is nothing to be proved there you just have to note that you are making a transformation on the independent variable and there is no change of the dependent variable at every given point that is carried as it is all right. Now, having made this transformation omega s l replaced by omega p by s we now need to check whether we have obeyed the other 2 requirements of rationality and stability. Now, rationality is obvious this is the rational transformation. So, if the original low pass filter is rational as it is expected to be then this transformation would retain the rational character. So, rationality is not a problem, but stability we need to check. So, we need to check what this does to the real part. So, let us consider s sigma to be s to be sigma plus j omega where upon s l becomes omega p divided by sigma plus j omega which can be rewritten as omega p into sigma minus j omega divided by sigma squared plus omega squared. And therefore, the real part of s l the real part of s l is omega p sigma divided by sigma squared plus omega squared. Now, it is very obvious that if sigma is positive so is the real part of s l because omega p is a positive quantity and sigma squared plus omega squared is bound to be positive. If sigma is positive then sigma squared plus omega squared has no choice, but to be positive it cannot be 0 and omega p is positive. So, the sign of sigma is carried to the real part of s l. If sigma is positive the real part of s l is positive. If sigma is negative the real part of s l is negative. And therefore, the left half plane goes to left half plane and the right half plane goes to right half plane. What it means is you cannot possibly have poles coming to the right half plane if you did not have them in the first place in the low pass filter. That is because a point of the right half plane in the filter of desired nature goes to a point in the right half plane of the low pass filter and that could not possibly have been a pole. And therefore, the poles can only be in the left half plane. In fact, we have also excluded the imaginary axis because the imaginary axis goes to the imaginary axis. So, the danger of poles going on to the imaginary axis also has been precluded in this process. Anyway, so that completes the design of high pass filters.