 a warm welcome to the 40th session of the second module of the course Signals and Systems. We have been talking about the interpretation of the passables theorem in the previous session and in fact, underlying all this was the idea of duality. Duality where one domain and the other domain could be interchanged both in terms of properties and in terms of equations. In this session, we are going to take up a new pair of properties to establish duality and this set of properties relates to an operation to be performed on the signal or correspondingly on the Fourier transform. So, let us get down to a very important operation that we do when we deal with continuous time systems namely the operation of differentiation taking the derivative. Now, before we continue to derive the mathematical principles involved, I would like to say a few words as to why the derivative is so important, why should we worry about what happens when we take the derivative of a signal. Remember, if you looked at the typical system, typical linear shift invariant continuous time system particularly one which is rational or realizable, then we said it has a typical system description given as follows y of t is summation k going from 1 to n the kth derivative of yt multiplied by ak plus a similar summation on the derivatives of the input. So, when we write d0 dt0, we essentially mean the identity operator, there is no derivative involved. Now, here all over we have derivatives as you can see, there are derivatives here and there are derivatives here and therefore, it is very important that we understand what happens to the Fourier transform when we take a derivative. So, let us now proceed to bring out the connection between what happens to the derivative of a time signal and what happens to the corresponding Fourier transform. Let us now derive the property. So, let xt as usual have the Fourier transform and first work with the angular frequency domain. So, let me call it capital omega x capital omega in terms of angular frequency. We want to find out what is the Fourier transform of dx t dt and we answer the question by writing down x t as an inverse Fourier transform. So, here is an inverse Fourier transform and now we take the derivative on both sides. When we do that, we get dx t dt is d dt of the integral and now under reasonable conditions, we can interchange the order of the derivative and the integral. Now, you may wonder what reasonable conditions are. Reasonable conditions for the moment I will just say in most situations of practical importance. I am not being too rigorous mathematically here, it does not really matter that much. So, when we so interchange, what do we get? We get dx t dt is well you can bring the 1 by 2 pi outside anyway and the integral keeps as it is and the d dt operator is going to operate only on the term which has time and there we are. So, it reads dx t dt is equal to 1 by 2 pi integral x omega times j omega e raise the power j omega t d omega and now let me aggregate that term together. Clearly this is a Fourier transform in its own right and if you look at this whole expression here which I am marking now in green, this whole thing is the inverse Fourier transform of that quantity and clearly since there is an equation here must be equal to this and therefore what we are saying is that what we have underlined in red here is really the Fourier transform of dx t dt. So, let us write that down. We made a very simple conclusion by a very simple line of reasoning. So, thus the Fourier transform of dx t dt must be j omega times x omega as we know. So, we have answered the question dx t dt has the Fourier transform j omega x omega. In fact, it shows us that there is a pair of operations here, there is differentiation in time and there is multiplication by frequency in the frequency domain or rather not just multiplication by frequency, multiplication by frequency and by j, we should not ignore that j. In fact, what does it really denote if we think about it, let us interpret this. So, multiplication by j is essentially a phase shift of 90 degrees and multiplication by omega scaling the amplitude by omega. So, it is not very surprising actually if you look at a sinusoid, take a sine wave say for example, take x t equal to a times cos omega t plus phi and let us find out dx t dt. It is of course, equal to minus a omega sine omega t plus phi. So, what happens when we take a derivative, two things have happened. One is the phase is changed by 90 degrees, that is why you go from cos to sine and the second is the amplitude is multiplied by omega. So, it is very clear that when we differentiate a sine wave, a real sine wave, these two things do indeed happen, there is a phase change of 90 degrees and there is a multiplication of the amplitude by omega. Of course, it is very obvious that that happens when you differentiate e raised to power j omega t. That is what this property is really saying, differentiation in time means thinking of the signal as a set of either sinusoidal components if it is a real signal or even if it is a real signal, you could think of it as complex exponential components or rotating phasor components and of course, if it is a complex signal, you have no choice but to think of it as rotating phasor components and the effect of differentiation is to replace each of those components by a 90 degree shifted component and an amplitude multiplied by omega component. That is what we are really saying. It is very important in a course like this to interpret every property that we derive, to interpret every equation that we write. We can write down the algebra but the interpretation is extremely important. In the absence of an interpretation, our understanding would be incomplete anyway, so much so for this interpretation and now we would like to ask, what is the dual of this property? In other words, can we reverse the roles of time and frequency here? Now to answer that question, all that we need to do is to write down the Fourier transform expression and to differentiate it. Let us do that next. So we write down capital X omega is equal to minus 2 plus infinity x t e raised to the power minus j omega t dt. This is the Fourier transform expression and we differentiate both sides with respect to infinity with respect to omega rather to the variable. So once again you have dx omega d omega and here I can write down d d omega of this integral but then as before I could argue that it is reasonable to interchange this in most practical situations which means I can now write down first integrate and then differentiate and the differential will act only on the quantities having omega. So there we go. Once again we see a very simple relationship like we did in the previous example or the previous property that the Fourier transform of this is this. That is what this relation essentially says and that is what the property is. So let us write down the two properties together. In fact, before we do that we will just summarize, we have two properties here. One involving derivative in time and the other involving derivative in frequency. We need to see in greater depth what these properties give us and that will require a whole session. So we will do it in the next session. Thank you.