 A warm welcome to the 17th session in the second module of the course Signals and Systems. In the last session, we had worked out one full example. Let me write down that final solution once again, it is nice to see it in one place written down in company, let us see. One full decommission. We had a symmetric square wave of period t. Of course, you say that by writing x t plus capital T is equal to x t for all t and we will show x t in one period. It is 1 for the first half period and minus 1 for the second half period and then periodically repeat it. This is how x t looks. Now, we calculated the components. We said the zero frequency component is equal to 0. The kth frequency component would have two orthogonal components again, a cosine component and a sine component, k greater than 0 of course. The cosine component is always 0. The sine component is 0 for even k. For odd k, it is non-zero, present. And therefore, we wrote down the overall expression x t is only a sum of odd components which are essentially 2t by pi into 1 by k sine 2 pi by t times kt summed over k odd. And that of course, we could write down as summation k going from 1 to infinity 2t by pi into 1 divided by 2k minus 1 sine 2 pi by t into 2k minus 1 small t. Of course, remember, you need to divide by t fine. That normalization is required. So, I need to insert the 1 by t term here as also here. But the point anyway is that you see that the components are present only for odd integers. There are only odd sine components and these sine components decay as per the reciprocal of the frequency index. So, let us make a note of that. Components decay according to the reciprocal of the frequency as per 1 by 3k or 1 by k like. Now, I am going to leave you with a few more exercises in this session. And I am going to give you hints to work those exercises out. I am not going to work the exercises out for you completely. I am only going to give you hints. And what I would definitely like all of you to do is to work those exercises out and come with your feelings. Let me put down those exercises. Exercise 1, work out the decomposition for an asymmetric square. That means the period remains at t. However, the transition is somewhere other than the middle. So, let us say the transition is at t1. This is how xt looks. t1 is between 0 and t, but not equal to t by 2. And of course, as usual, x of t plus t, capital T is equal to x of t for all t. This, please remember again, is the symbol for all. Now, few remarks. When you work this out, you will realize two things. One is that you cannot now have that degenerate situation of the cosine components being absent. The cosine components will be present as also possibly the sine components at even k. You will have to work out both the cosine and sine components completely. Moreover, the mean over a cycle is not 0. So, now you have a full house. You have to work out the mean, you have to work out the cosine components and you have to work out the sine components and that too for all values of k. Now, what you must also do is to look at the expression that you get here in general. And then of course, there is nothing wrong in substituting t1 equal to t by 2. You must check that when you substitute that as exercise 2 for you now. Exercise 2. In the answer of exercise 1, put t1 equal to t by 2 and verify that you get what you got for the symmetric square wave. That is exercise 2. Exercise 3 will now be a slightly more complex wave problem. Let me draw the wave problem. Exercise 3. We will have a symmetric triangular wave. I am showing you the wave over 1p. So, it rises to 1, falls to 0, then goes down to minus 1 and comes back to 0. This is shown over 1p. Of course, xt is equal to xt plus capital T for all t. Now, let me work this example for you in a few steps. I am not going to work out the full expression. That is left to you to do. I am going to work out a few of the steps. So, let us do that. I will show it to you graphically what is going to happen. Let me draw the wave form. Let me draw one complete period and let me draw upon it a sine wave. So, let us draw just the first sinusoid for k equal to 1. So, let us draw sine 2 pi by t times kt for k equal to 1. That is easy. That will look like this. Of course, my drawing is not terribly accurate. I am just trying to show it schematically. Do not take my drawing literally. The point is, however, that again you have a symmetry in the integral. Whatever integral you get between 0 and t by 2 is the same as what you get between 0 and t. So, in fact, here when you take xt sine 2 pi by t kt integrated from 0 to t, you can see it is two times what you get from 0 to t by 2. I leave this for you to verify. That is one thing that will make your job easier. Now, the other thing that will happen is with regard to the odd and even components. So, here you need to see what happens to the odd sine components and the even sine components. In fact, you will again see that probably only one of them are going to be present, either the odd components or the even components. Find out which one. And also see what happens to the cosine components. I have a feeling that some of them will vanish or maybe all of them will vanish. Work it out. And my recommendation to you is do not start evaluating the integral suddenly. Try and look at what is happening graphically and then try and put down expressions that will help you show that some of those integrals vanish and therefore, you need to evaluate only some of those integrals. So, I will also just show you how to evaluate an integral of this kind. So, let us do that evaluation. I will just show you the indefinite integral that we need to evaluate. Typically, the indefinite integral that we need is something like this, integral t times sine 2 pi by t times kt dt. Now, the approach to evaluating this integral is by parts. So, essentially, you keep the first term as it is and integrate the second and then differentiate the first and integrate the second and then integrate the respectivity. That is the approach you will need to take to evaluate the integral here and of course, substitute the limits and so on. I am just giving you hints. I am not solving the full problem. You should do that. Also, my recommendation to you is compare the answer that you got for the symmetric triangular wave and what answer you got for the symmetric square wave. What is the relation between the coefficients? You see some way in which the coefficients decay here as a function of k as opposed to what you saw in the symmetric square wave. Is there a difference? Why do you think there is that difference? Try and reason it out. And finally, this is the most difficult of the problems that I am giving you today and that is exercise 4 for you. Evaluate the decomposition of an asymmetric triangular wave which would look something like this. This is a difficult problem. This rises to 1 and falls back and then goes down to minus 1 and comes back. But t1 is not equal to t by 2. And as usual, you would then substitute t1 equal to t by 2 once you evaluated this and verify you get what you got for the symmetric triangular wave. You know, there is one more symmetric triangular wave that we can consider. Let me draw that thing. So, exercise 5, a variant of the symmetric triangular wave. That looks like this. In one period, it only rises and falls. It does not go negative. Find its decomposition. Compare. That will be the interesting part. Compare with exercise 3. What differences do you see? And the last exercise, a variant of the asymmetric triangular wave. So, it rises to 1 but then falls and the rise and fall are on different subintervals. By the way, there is a little correction that you should see in the previous exercise. I just noticed it. This should be for all small. I hope you noticed it. That is one way of taking you away. Anyway, so here we are. A variant of an asymmetric triangular wave. Here, t1 is between 0 and t but t1 is not equal to t by 2. And here again, you could do the same thing. Substitute t1 equal to t by 2 and see if you got what you got for exercise 5. And also compare this with what you got for the asymmetric triangular wave shown earlier. Try and explain the differences. That is the challenge for you. Try and explain the differences. Well, these are the exercises that you need to pursue. Please pursue them and we hope to have a lively discussion sometime. Thank you.