 A warm welcome to the 36th session of the second module of the core signals and systems. So we will be talking about duality. Duality as I said is a very interesting and a very important property of the Fourier transform and we saw a very important consequence of duality in the previous 35th session. Namely, you could see what happens to the product of two functions. When I take a product of two functions the Fourier transforms are convolved. Now, in this session we are going to see a very important consequence of that property again. Namely, when I take a product of two functions what happens to the Fourier transform and what consequences does it have with a particular evaluation of that and that leads us to a very important property again in signals and systems called passable theorem. We will see the details. Let us first put down what we learnt in the last session. So we have seen if y1 t has the Fourier transform capital Y1 omega and if y2 t also has the Fourier transform capital Y2 omega then provided all the Fourier transforms exist y1 t into y2 t has the Fourier transform 1 by 2 pi capital Y1 convolved with capital Y2 evaluated at omega. Again, let us write this down explicitly. y1 t into y2 t has the Fourier transform 1 by 2 pi integral from minus to plus infinity capital Y1 lambda capital Y2 omega minus lambda d lambda. Essentially, this is the convolution. Now, let us consider a special case. Suppose I took y2 conjugate instead of y1. So let us see what happens when I complex conjugate a function. What happens to its Fourier transform? Suppose we know x t has the Fourier transform capital X of omega. We will ask what is the Fourier transform of x bar t. In fact, that is not very easy to get if we try and do it directly in the Fourier domain but let us take the inverse Fourier transform. So let us write down we will obtain x t in terms of capital X omega by taking the inverse Fourier transform. Capital X omega e raised to the power j omega t d omega integrated from minus to plus infinity divided by 2 pi. We can now take a complex conjugate here. In fact, let me do it right here first and then operate it separately too. So I have x bar t. I will write a bar everywhere as required. Now, complex conjugation of an integral is the integral of the complex conjugation and then I could disintegrate that complex conjugation into the integral of a product. So I have x bar t is 1 by 2 pi. Complex conjugate of x omega multiplied by complex conjugate of e raised to the power j omega t and of course d omega does not have to be complex conjugated because omega is real. So let me write down the overall expression. X bar t is 1 by 2 pi. Capital X omega bar e raised to the power minus j omega t d omega and now I could put omega equal to minus alpha here and play the same trick. So d omega is minus d alpha and of course when omega goes from minus to plus infinity, alpha goes from plus to minus infinity. So this minus sign can be absorbed with the reversal of the direction of integration here and together therefore we have x bar t is 1 by 2 pi minus to plus infinity. Capital X minus alpha the whole bar e raised to the power j alpha t d alpha. So essentially what it says is that x bar t is the inverse Fourier transform of capital X minus omega bar that is what we are saying effectively. So essentially when I complex conjugate a function two things happen in the Fourier domain. There is a complex conjugation and there is also a time reversal. Now you know in fact it is not too difficult to interpret this. When you complex conjugate a function what are you really doing? You know remember that complex conjugation you see what really is the inverse Fourier transform. It is a way of thinking of the function in terms of rotating complex exponentials. So you are saying when I complex conjugate the function you are doing two things. Inverting the role of the phasor rotating at omega and the phasor rotating at minus omega you are inverting their roles and you are also reversing. So that is what you mean by saying omega is replaced by minus omega and you are also complex conjugating the Fourier transform. So you are also complex conjugating the coefficient associated with that rotating phasor. So in place of the rotating complex number at angular frequency omega you have taken the rotating complex number at angular frequency minus omega that means rotating in the opposite direction with the same magnitude of the angular velocity and you have also taken the complex conjugate of the coefficient. This is what you have done to complex conjugate the whole function. If you think about it is not too difficult to appreciate. We have of course done it algebraically. One can also reason this out intuitively in terms of its interpretation. Whatever it be we have now come to the conclusion that we know how to calculate the Fourier transform once we have complex conjugates. So let us now write down something interesting. So now let us find out if y t rather y 1 t has the Fourier transform capital y 1 omega and let us now find out. So let us take y 2 t let y 2 t have the Fourier transform y 2 omega and then we ask what is the Fourier transform of y 2 bar t y 2 bar t would have the Fourier transform as we know y 2 minus omega bar and now we can find out what is the Fourier transform of y 1 t into y 2 bar t. We will take the product of these two and not y 1 and y 2 and we will apply this theorem that we just derived in the previous session. We call it the multiplication theorem. So apply the multiplication theorem. So what would we get? Well simple it is going to be essentially 1 by 2 pi the convolution of y 1 and y 2 bar at minus omega and this we can write down simple. Now let us evaluate this. So let me write down the whole story. I have y 1 t multiplied by y 2 bar t has the Fourier transform 1 by 2 pi integral minus 2 plus infinity capital y 2 minus lambda bar y 1 omega minus lambda now let us write down the statement explicitly in terms of evaluating the Fourier transform. What I mean is we have used duality and we have got this relationship but let us evaluate the Fourier transform of y 1 y t or rather y 1 t times y 2 bar t ab initio. What does that mean? It essentially means the Fourier transform of y 1 t y 2 bar t is essentially minus 2 plus infinity y 1 t y 2 t bar e raise to the power minus j omega t dt and this we have found to be this expression. Now let us make a few transformations. Let us put capital omega equal to 0 here and also here and also let us make the transformation of variable minus lambda equal to beta and play the same tricks. Let us do that. First let us do the minus lambda equal to beta part. So minus lambda equal to beta so as usual minus t lambda is d beta and therefore d lambda is minus t beta and then you have lambda going from minus to plus infinity implies beta goes from minus plus infinity to minus infinity but we can absorb again. So we can absorb this negative sign in this reversal of integral and therefore together we have minus to plus infinity capital y 2 minus lambda bar y 1 0 minus lambda we have put omega equal to 0 remember is essentially again minus to plus infinity y 2 beta bar y 1 beta and this gives us a very interesting conclusion. Let me write that conclusion now minus to plus infinity y 1 t y 2 bar t e raise to the power minus j 0 t dt is essentially minus to plus infinity y 1 t y 2 bar t dt after all that e raise to the power 0 is 1 and this is equal to minus to plus infinity capital y 2 beta bar capital y 1 beta d beta. Let us put them down together and this is a very powerful conclusion we have drawn here. Let us look at the conclusion in fact maybe I should just rewrite it a little bit by the way when I wrote down this I had forgotten the factor of 1 by 2 pi before that is alright I have taken care of it here let me reorder them. So I have 1 by integral minus to plus infinity y 1 t y 2 t bar dt is minus to plus infinity divided by 2 pi y 1 beta capital y 2 bar beta d beta. There is a beautiful symmetry on both sides what are we saying here if we look at carefully essentially this is the inner product of y 1 and y 2 and this is if you ignore the factor of 2 pi this is essentially an inner product of capital y 1 and capital y 2. So we have a beautiful conclusion here there is an equivalence between two inner products and inner product in time and an inner product in frequency. This is a very very important result and we shall have a great deal to say about this result as we go along. For the time being let us appreciate that we have come to an important conclusion based on the multiplication theorem and we will take up more consequences of this conclusion in the coming session. Thank you.