Fourier Series – Sawtooth step function

What if we were charged with the task of modeling this function mathematically? We see that it’s a Sawtooth periodic function and then it has a wavelength of 2Pi We see the starts here and then goes up to one of y-value of one or a function value of one rather and then Indiscreetly jumps back down to zero a function value of zero at 2PI, so we see that the wavelength is in fact 2pi The Wavelength of the period if you will okay So the first thing we need to do is to find a description of this function over one period Okay? So we can look at it, and we need to basically find the equation for the line that goes from here to here We’re going to start at the origin and integrate from zero to 2pI that’s going to be that’s usually the easiest thing to integrate over so the function value over that interval f of x is equal to The rise over the run the rise is it Rises 1 and it runs 2 Pi So 1? over 2 Pi times x and this is good from 0 Listen it X and then less than 2 Pi It’s good over that right? But we want to model this function globally and we want to model it using the fourier series Because that’s what you’re going to be asked to do on your exam, and so this is this of course is modeling the function but it’s only being modeled over 2 PI and we want to model it over all of the function values and so let’s see if We can come up with a mathematical description using the fourier series So here is the fourier series I’ve seen this before for a 0 we’ve discussed previously that really all we need to find is the Midline of the graph that is the center line of the graph Instead of actually performing this integration because this integrations is to integrate the function over one wavelength is from here to here So you take this area and then divide by the way? towel or here towel So you find this area divided by two PI and you? Would find it right you would find what that is But it’s a much easier way to do that is just to find the midline of the graph the center line on the graph which is just f max plus f min divided by 2 So we see that the maximum is 1 the minimum is 0? the x axis so the Center line is just 1/2 right 1 plus 0 is 1 divided by 2 is 1/2, so Right away without even doing an integration. We can just say that this is equal to 1/2 Okay for a sub n We write 2 over the wavelength which is 2 Pi So that’s why it’s so convenient to write this notation with the towel Right because tau could be a wavelength or it could be period for time for integrating over A position space or for a time space that doesn’t matter which is a convenient notation We’re going to integrate from 0 to 2 PI and integrate over 1 wavelength f of x over that period is 1 over 2 PI or over that wavelength rather 1 over 2 PI times x times of cosine of 2 Pi over 2 PI is just 1 so this is just a cosine of n. X and that’s Dx and Then we’re going to Grade N. Equals 1 2 3 So on if we perform those integrations, we will see that in fact those will come out to be equal to 0 ok so you remember that we said that those will come out to 0 0 whenever it’s an odd function that they even From the even coefficients which of these will all come out to be equal to 0 I? Did not check this graph to see if it was even or odd But it must be odd because all the even code come out to zero because I’ve done this this Calculation already, I make him out to be zero The odd coefficients do not come out to be a zero and so therefore the function must be odd But I’m not hanging my hat on that fact. I didn’t actually even test the function to see if it’s odd I just saw that these integrations came out to be zero and these integrations did not Okay, so this is 2 Pi 2 over 2 Pi You can do the odd-even test for yourself if you like It’s here to 2 pi the beauty of the math is the math always works out you just perform the integrations You know it’s just going to tell you what 0 and what’s not zero ok 1 over 2 PI times x Sine once again 2 Pi over 2 PI cancels, and we just get sine of Nx Dx and we evaluated at N. Equals 1 2 3 those will not come out to be equal to 0 those will come out to be actual values okay, so be 1 and I’ve done these integrations. It’s basically just a calculus 2 problem to do the integrations and so we won’t bore you with integrations, but you let n equal 1 and then you perform the integration and Then you let n equal 2 and you perform the integration you let vehicle-3 and you perform the integration You can use this involve symbolic manipulating calculator to perform those integrations Or you can do that by hand either way the answers will come out to be the following b. 1 will be negative 1 over Pi B to be negative 1 over 2 PI B 3 may be negative 1 over 3 PI and you see that Do you see the pattern 1 over 1 Pi 1 over 2 Pi 1 over 3 Pi? So B sub N In General is going to be negative 1 Over N. Pi All right Ok so now let’s write down the fourier series that corresponds to these coefficients, so here’s the the Final result is going to come from this right So f of x Is approximately equal to? A0 is 1/2 so we have 1/2? All of the a sub ends are 0 so these go away, and then we have these? You see you have 2 pi over 2 pI tau is equal to 2 pI so these are going to cancel so it’s just going To be sine of in x each time so what we’re gonna. Have is B1 times sine of 1 x Plus B2 times the sine of 2 x plus B3 times the sine of 3x And so on and you can just keep adding them. We’re just going to do three terms for the sake of speed But you could add as many as you want to make the 4-h series as accurate as possible ok so this is equal to substituting the values in one half Plus b. 1 is negative 1 over Pi Okay, so we won’t put the plus. We’ll just put minus 1 over Pi times The sine of 1 x which is just sine of x and then it’s minus again 1 over 2 Pi times the sign Of 2x, and you see what this is go on minus one over three Pi times the sine of 3x That’s it, and it’s the for you series for the Sawtooth function. I’m gonna step aside and give you your clear screen shot See you next time