 A warm welcome to the 41st session of the second module in the course signals and systems and we continue to discuss the differentiation with multiplication by independent variable duality here. In fact, we had not quite shown the duality in the previous session, but now we will do so explicitly. So, we had promised to write down the two properties that we had discussed in the previous session together to see them at once together and to identify the duality there and we will do that first. So, we said that if xt has the Fourier transform capital X of omega, then dx t dt we derived all this in the previous session. dx t dt has the Fourier transform j omega x omega and minus j t xt has the Fourier transform dx omega d omega. Of course, we can always write down corresponding expressions or properties with the cycles per second frequency and I leave it to you as an exercise to do that. I am going to now give you that exercise, but before that let us identify the duality here. Let us look at this and this. Here you have a differentiation in time and here you have a differentiation in frequency and correspondingly, here you have a multiplication by the frequency and you have a multiplication here point twice by time. So, there is a duality here, duality meaning reversal of roles. When you differentiate in time, you multiply point wise in frequency. When you differentiate in frequency, you multiply point wise in time. Of course, there are small adjustments, there is a minus sign there and so on, but those are minor. In fact, now I am going to give you as I said two exercises to do. Let me write down these two exercises. Exercise one is derive corresponding properties for these cycles per second frequency and the hint is begin from the equation x t is of course, the inverse Fourier transform, but written in terms of the cycles per second frequency. Now, please remember I must emphasize this again. When we write down x f, do not interpret this as x omega with omega replaced by f. It is in fact, x omega with omega replaced by 2 pi f. So, as I said it is a notational abuse which we shall do somewhat freely in this subject because we sometimes need to talk about cycles per second frequency and sometimes about angular frequency. So, we do not want to confuse, but it is little bit of liberty that we are taking. Now, of course, you can continue and complete this exercise. So, I leave it to you. So, you can do the same thing. So, you can differentiate both sides with respect to t and carry out all the subsequent working and the other exercise I am going to give you is exercise two, derive one property from the other by duality in the following. So, this is one and this is the other. So, what I mean in this exercise is you of course, have these two properties derived independently that is all right, but you know the basic principle of duality. If let me write that principle down once again to remind you. So, duality said that if x t has the Fourier transform capital X of omega, then capital X of t has the Fourier transform 2 pi small x minus omega. Now, what I am saying is use this to derive one from the other. Now, anyway we have seen it in formally you know we have seen it intuitively there. There are two operations on one side you have a differentiation operation and you have a multiplication by the independent variable operation and they are equivalent. So, if I do one operation in one domain, the other operation gets done in the other domain that we have seen that is what duality is all about. You can of course, formally derive it by duality as I have asked you to do in this particular exercise, but once you do that you will be convinced even mathematically that they are dual, but informally it is clear. Now, let us take an example to see how the differentiation property can simplify our work at times. So, for example, let us go back to the famous rectangular pulse and let us see what happens when we take the derivative property there. So, let us find out again the Fourier transform of our famous rectangular pulse with an amplitude of a going from minus capital T to plus capital T this is x t. So, its Fourier transform is of course, as usual minus t I am skipping a couple of steps a times e raised to power minus t omega t dt. So, you can simplify this it becomes 2 j a sin omega t divided by j omega and I can multiply and divide by t and cancel the j's and that leaves me with 2 a t sin omega t by omega t. Now, let us take the derivative of this. So, in fact, I will show the derivative right there first in a different color. Let me show you the derivative in green. How would dx t dt look? You see of course, in the conventional sense dx t dt would not exist because the right in the left derivative would not be equal, but then here you could use the idea of impulses. So, dx t dt would essentially have 2 impulses 1 impulse located here of strength a and 1 impulse located here of strength minus a. Let me draw this derivative now and impulse of strength a at minus t and an impulse of strength minus a at plus t this is dx t dt. It is very easy to calculate its Fourier transform. Well, let us call this equal to g t, g t is equal to dx t dt. So, what is its Fourier transform minus infinity to plus infinity as usual g t e raise to the power minus j omega t, but then g t just has 2 impulses and the impulse is simply pick up the values of those functions remaining functions remaining integrand at the points where the impulse lie. So, you have a times e raise to the power minus j omega into minus t minus a times e raise to the power minus j omega into t. This is the Fourier transform which we can now simplify. So, it is a times 2 j sin omega t so simple and now you can see what is happening here. So, if I multiply this by j omega indeed gives this you know any way the t's would cancel as they are we just introduce them for convenience and the 2 a is there as it is here and the j comes in and the omega goes away because of this. So, you can see that multiplying this by j omega gives this. So, we verified the differentiation property for the rectangular parts. Now, in this case of course differentiation gave you impulses. So, you have to do a little bit of work to think what would happen, but now let me give you an example where use of the differentiation property will actually make your calculation of the Fourier transform easy and I am going to leave some of the details to you as an exercise. Let me take a triangular pulse. So, there we go. Now, there are 2 ways in which we can find the Fourier transform. We can find the Fourier transform by the differentiation property. So, let us call this x t as usual and let us sketch d x t d t. In fact, let me do it on the same graph, but in a different color. So, this rises from 0 to a in time t. So, its slope is a by t that is easy to see. There is a positive slope of a by t here and a negative slope of a by t here I have sketched the derivative on the same graph and you can see it is very easy to calculate this for the Fourier transform of the green function. Let me do it for you. It is essentially integral from minus t to 0 a by t that is a constant e raised to the power minus j omega t d t plus integral from 0 to capital T minus a by t times e raised to the power minus j omega t d t easy to evaluate. You can do it. I leave it to you as an exercise. And therefore, the Fourier transform of the triangular pulse is essentially 1 by j omega times this. So, Fourier transform of triangular pulse multiplied by j omega is equal to this Fourier transform. So, we can easily calculate the Fourier transform of the triangular pulse. We will see more in the next session. Thank you.