 So, one will come to the 33rd session of the second module of the core signal and systems. We shall now take up the example that we had just introduced in the previous session namely the example where we took two rectangular pulses we wanted to convolve them and then find out the Fourier transform. What I said is that this particular example illustrates some ideas. So, let us first carry out the task of calculating the Fourier transform of each of the rectangular pulses. I know we have done it before, no harm in just doing a couple of steps again. So, let us calculate this is x1 t and x2 t and its Fourier transform essentially minus infinity plus infinity either x1 or x2 as you like, e raised to power minus j omega t dt which is essentially integral from minus t to t a e raised to power minus j omega t dt and that is a times e raised to power minus j omega t divided by minus j omega integrated from minus t to plus t and that gives us a e raised to power minus j omega t minus a e raised to power j omega t by minus j omega which can now be simplified and we will do that. So, it simplifies to 2 a j sin omega t divided by j omega and as you know we have done this before you can always multiply and divide by t. So, I can put a t there and cut off the j you have 2 t times a sin omega t by omega t. Now, this is the Fourier transform of any one of those rectangular pulses. So, if we convolve them, let us convolve them, this convolution is easy to calculate. In fact, I would not carry out the whole calculation, I leave it to you to complete. So, I will just write down the convolution, you know you could let me just indicate the main step. So, you have the train platform analogy, you have tau, you have written x1 of tau and x2 t minus tau and you could multiply these 2 together and integrate. So, point by point you need to multiply these 2 together and find the area under the product. I would not do this whole thing, we have done it before and I will straight away write down the result. The result you will follow, of course the first thing is that the result would extend from minus 2 t to plus 2, it will also result in a straight line and at the maximum you are going to have a total area of a squared over a length of 2 t, so a squared into 2 t, that is the area contained at the maximum and it is going to change linearly starting from minus 2 t up to 0 and then fall off as you go to 2 t. This is how it is going to look, the convolution is going to be essentially a triangle like this, let me darken it, essentially a symmetric triangle. Now, if we were to try and calculate the Fourier transform of this directly, so if I wanted to evaluate the Fourier transform of this, it is going to be a bit tedious. You would require to evaluate something like t times or you know 1 minus or some constant minus t times e raised to power minus t omega t over t and so on. But now with the convolution theorem, calculating the Fourier transform of this is very easy. The Fourier transform of this is simply the product, so it is x1 omega into x2 omega and that is essentially the square, the square of x1 omega because both of them are identical. So it is 2 t into a sin omega t by omega t, the whole square as simple as that. So you see you notice the convolution theorem helps you to calculate Fourier transform very conveniently here. Now, could the Fourier transform also help you to calculate the convolution more conveniently? That is what we need to now establish. Now what we will do is to establish a theorem or a very important principle whatever you might want to call it before we do that and we will take the same example and illustrate how by using that principle or that theorem, it is very useful to use the Fourier transform to calculate a convolution. Now let us establish the principle first and that principle which we are going to establish is called the principle of duality. It is the next property of the Fourier transform. The idea is very simple. Begin with the expression for the Fourier transform. So I have capital X omega is integral xt e raised to the power minus j omega t dt. Now write down the expression for the inverse Fourier transform. xt is 1 by 2 pi integral, of course all these integrals are from minus to plus infinity. Capital X omega times e raised to the power j omega t d omega. Now all that I need to do is to interchange the variables. So you know after all what is a variable, what is, it is essentially just a name. I could write omega here and I could write t there, this will become t omega, this will become t. So essentially what I would have is an expression x omega, small x omega is 1 by 2 pi integral minus to plus infinity. Capital X t e raised to the power j omega t again. And now I could simply write x of minus omega instead of x omega, make that little change in the integral. A very interesting integral. Now what are we saying here? We are saying essentially this is the Fourier transform of this except for the factor of 2 pi. So if I make that correction then we will simply write down a Fourier transform you can take the 2 pi to the other side. We can carry it here and we have a very interesting conclusion. 2 pi x small x of minus omega is minus to plus infinity capital X t e raised to the power minus j omega t dt. Now formally we are saying if small x t has the Fourier transform capital X omega then capital X of t has the Fourier transform 2 pi small x evaluated at minus omega. A very powerful result. It says that this Fourier transform pair X t and capital X omega can be inverted. So if you looked at the Fourier transform and thought of it as a function of time instead of frequency. Then each Fourier transform comes out to be very similar to the original time function except that it is reversed. There is a reversal or a mirroring around 0 and a multiplication by 2 pi that multiplication by 2 pi is just a scaling. Now what is the consequence of this? The consequence of this is that a rectangular pulse has a sine x by x kind of pattern for its Fourier transform. And therefore a sine x by x kind of pattern would have a rectangular pulse for its Fourier transform. So convolving 2 sine x by x kind of pulses is better done by taking the Fourier transforms of each of them, multiplying the Fourier transforms and then taking the inverse Fourier transform. It is not a good idea to convolve 2 sine x by x pulses directly. It is better to take their Fourier transform, multiply the Fourier transform and then take the inverse Fourier transform. We will see more about the property of duality in the next session. Thank you.