 Hi everyone, I am Sushrut Thakar, a student from IIT Bombay and also TA for this course. In this video, we will be looking at Fourier series and Fourier transform, which we have learnt in the lectures. And it is actually very interesting to know that this concept was not at all developed for signal and systems. And actually it was developed while Fourier was exploring the phenomena of heat conduction, as in how a heat moves from one place to another. And we will also look at a little bit of history, which is also very interesting. So Fourier series is a trigonometric series. And trigonometric series are actually very old, like as old as Babylonian civilization. And in modern times, Euler and Bernoulli used it for the problem of plucked strings and analyzing that situation. But it was also criticized very much, especially by a great mathematician Lagrange, who said that such series could not represent a wide array of signals and wider of functions and were limited to only a few functions. So in this background, Fourier was analyzing the problem of heat conduction. And he developed the series while doing that. And it was not mathematical, not at all. The mathematical work was actually done by his student Dirichlet, who completed it 20 years after. That was in 1829. So Fourier was going to publish his paper containing this Fourier series in 1807. And it was to be reviewed by Lagrange and three other scientists before publishing. Now Lagrange was very much against it. So he did not approve it and it was not published at all. And later Fourier wrote his theory in his book. So we will see what Fourier did. So Fourier was studying the problem of heat conduction. And we will consider a very simple case, a 1D case. So he was studying time dependent heat conduction. Let us say we have a wall here. You can say a rod, 1D heat conduction. And we are cooling it on both sides by putting ice on it. So t equal to 0. Actually we do not need to cool it because we can anyway set the temperature scale in which this wall temperature will be 0. So here we will put x is equal to 0 and x is equal to l for the wall thickness. And we know these two boundary conditions as well as the initial temperature t at x comma t equal to 0. Actually let us call time instead of t tau so that we do not confuse between the variables. So Fourier empirically formulated the equation governing this by using heat balance. So basically it is balancing the heat content going out plus the change in the heat content of the wall. So he got the following equation. So he got the equation given by this tau t of x comma tau with respect to tau tau is equal to tau 2 t of x comma tau with respect to tau x square. So this equation has two variables the space and the time. And in differential equations we usually try to separate these variables. So there is a technique for that and we will assume that the time and space part are independent and we will put the following substitution. This is t of x comma tau is equal to g of tau the time part and r of x the space part. So after putting this equation and simplifying we will get 1 by g of tau dg of tau with respect to d tau. Note that this is an exact differential here because g is only a function of tau the time is equal to 1 by r of x into d 2 r of x with respect to dx square. Now the left hand side of this equation is a function of time and the right hand side is a function of space. So both must be equal to constants in order to have this equation satisfied for all the time and for all the x values. So we will put it equal to minus lambda square a constant. And now there are two separate equations a time equation and the space equation by equating the corresponding terms. The time equation is very easy to solve actually I will just write it here and by integrating once its solution can be found of the form e raise to minus lambda square tau is equal to g tau. Of course there will be a constant multiplying this and we will denote it by a and we will determine this constant later on. I will highlight this because we will be using this equation later on. So we have solved the time part now I will write the space part it is 1 by r x d 2 r x with respect to dx square is equal to minus lambda square. Now this is a very standard equation called the simple harmonic oscillator equation and it is known that this equation has a solution of the type this which is C sin of lambda x plus d cos of lambda x. So we will need to determine these constants and eventually. So this is called Eigen solution of this equation and we will substitute now the boundary conditions which we had the temperature at both ends x is equal to 0 and x is equal to l r 0. So the space part must be 0 because the temperature is 0 at any point of time. So we will substitute r is equal to 0 at x is equal to 0 as well as at x is equal to l and by substituting we will get d equal to 0 you can easily see that if you substitute x is equal to 0 the sin term automatically goes away. Also when we substitute r is equal to 0 when x is equal to l we will get a condition on what values lambda can take. So lambda will have some discrete allowed values for which this equation will be satisfied. So we have this before substituting x is equal to l and when we substitute x is equal to l then r becomes 0. So we have sin lambda l equals 0. So lambda turns out to be pi by l into some integer let us call it k where k takes the values 1 2 so on. In general we can write any solution to this differential equation as linear combination of these solutions. So this is property of the differential equation that you can write any general solution in terms of its basis solutions or Eigen solutions. Let me write the complete equation involving both time and space parts and we will write it as a linear combination of the individual solutions given by individual case. So here we can see we have written both the time and space parts and we can combine this constants a and c into one constant say c k to denote the constant coming out from the k th term as c k. So the constants determined by different case can be different that is why we are writing c k to denote the k th constant and we know the initial temperature because we want to eventually determine these c k's to find the temperature and from initial temperature we can calculate these. We have the initial temperature as a function of x only because the time is fixed time t equal to 0. So we can write it as so we can write it as a function of x say f x and if we substitute it above we will get summation k going from 1 to infinity c k sin of pi k x by L and the time part will vanish because e raise to 0 will be 1. So we have this Fourier series here a special case indeed where all the cosine terms are vanishing and the constant is also not there. So we solved a very simple problem here and we can always go forward we can add multiple dimensions say solve a 2D problem or 3D problem or add a source in between where we will have to take the Fourier transform of the source as well that is one of the methods and you are encouraged to try it at home please post your solutions on the discussion forum. You can always look at the books regarding heat and mass transfer or heat conduction if you want to explore further. In this video we saw how Fourier developed this Fourier series and also a little bit of history regarding that and I will see you in another video perhaps next time. Thank you for watching.