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Erich Poppitz

Lecture 13: Fourier Series ▯ Today’s topic: Fourier Series, is not covered in Morin. Your main resource for this topic should be these lecture notes. Fourier Series ▯ Last time we discussed the possibility of the driving force not being a single sinusoid (but still periodic on T). We found that if our driving force is a sum of sinusoids, then because of the linearity of the equation and the principle of superposition, we can solve F = ma for each single sinusoid, then add all the solutions to get the total solution. ▯ If the driving force doesn’t look like a sum of sinusoids (for example we looked at a square wave), then we can use Fourier’s theorem to represent the function as a sum of sinusoids of speci▯c periods and amplitudes. Once we’ve found this representation, then we can solve our equation the same way we did for the case above. ▯ We also found last time that the periods of the sinusoids we use must be integer divisions of the period of the function itself (i.e. T=n; n = 0;1;2;::: integer, where T is the period of the function we are repre- senting). This is to ensure that the sinusoids themselves are periodic on T and hence when we add them, the sum will be periodic on T. ▯ So, using the above info, we can write a \Fourier Series" representation of the function we are considering (call it f(t)): X ▯ ▯2▯n ▯ ▯2▯n ▯▯ f(t) = Ancos t + B nin t n=0 T T where Anand B ane the amplitudes of the sinusoids for each n value. ▯ ▯ ▯ Slight Technical Detail: Notice that for n = 0, sTnt = 0 which means that B can be anything. To avoid this arbitrariness, we separate 0 the n = 0 term from the others in the sum and write the sum starting from n = 1: X1 ▯ ▯ ▯ ▯ ▯▯ f(t) = A + A cos 2▯n t + B sin 2▯nt (1) 0 n T n T n=1 1 This is the general form for a Fourier Series. Now we have to ▯nd expressions for the amplitudes of the sinusoids0 An, A andnB and we will be all set. Orthogonality Properties of Sine and Cosine ▯ We are going to use \orthogonality properties" of sinusoids. In order to use them, we should ▯gure out what they are, so here is a \math break" where we develop them. The idea is to integrate a product of 2 sinusoids of di▯erent periods, that are periodic in T over a single period. There are three options for the products: (1) 2 cosines, (2) 2 sines, (3) 1 sine, 1 cosine. We will look at each of these: 1. We will be evaluating the integral of the product of 2 cosines: Z ▯ ▯ ▯ ▯ T 2▯n 2▯m cos t cos t dt (2) 0 T T where n and m are any integers. In order to integrate, we can make use of the sum of angle formulas to rewrite the integrand: cos(A + B) = cosAcosB ▯ sinAsinB cos(A ▯ B) = cosAcosB + sinAsinB Adding the equations results in: cosAcosB = 1[cos(A + B) + cos(A ▯ B)] (3) 2 We can use equation (3) to rewrite the integral in equation (2) as: 2 Z T ▯ ▯ ▯ ▯ cos 2▯n t cos 2▯m t dt 0 T T Z ! ! 1 T 2▯(n + m) 2▯(n ▯ m) = cos t + cos t dt (4) 2 0 T T This can be integrated immediately to give: 2 3 ! ▯ !▯T = 1 4 T sin 2▯(n + m) t ▯ + T sin 2▯(n ▯ m) t ▯ 5 = 0 2 2▯(n + m) T ▯ 2▯(n ▯ m) T ▯ 0 0 Notice that this is only correct if n 6= m, otherwise the 2nd term has a division by 0. We will do the n = m case next. Notice that plugging in the bounds resulted in sin’s whose arguments were all multiples of 2▯ which means they are all 0. So this integral became 0. Now lets do the n = m case. Starting from equation (4) and plugging in n = m gives the integral: Z T ▯2▯n ▯ ▯2▯n ▯ cos t cos t dt 0 T T ▯ ▯ ! 1 ZT 4▯n 2▯(0) = cos t + cos t dt (5) 2 0 T T Notice that the second integrand is equal to 1. Integrating gives: 2 3 ▯ ▯▯T = 14 T sin 4▯n t ▯ + tjT5 = T 2 4▯(n) T ▯ 0 2 0 The non-zero term comes from the integration of \1" to get t. So we have now determined the integral in equation (2) for all possible n and m’s. We notice that for most cases, the integral is 0, and only for the case where n = m do we get a non-zero result. We can write this succinctly using the \delta" function de▯ned as follows: ▯nm = 0 for n 6= m = 1 for n = m 3 Using this notation we can write equation (2) as: ZT ▯ ▯ ▯ ▯ 2▯n 2▯m T 0 cos T t cos T t dt = 2 ▯nm (6) This is the ▯rst \orthogonality relation" we need. Its called an "orthogonality relation" because, in analogy to unit vectors, multiplying 2 cosines gives 0 unless the cosines have the same frequency. The analogy to unit vectors is that dotting 2 unit vectors is 0 unless they have the same direction. Okay, we need to do 2 more. The next is: 2. We will be evaluating the integral of the product of 2 sines: Z ▯ ▯ ▯ ▯ T 2▯n 2▯m sin t sin t dt (7) 0 T T Just like in the cosine case, we will start by using the sum of angle formulas (given above, right before equation (3)) to rewrite the integral as: Z T ▯ 2▯n ▯ ▯ 2▯m ▯ sin t sin t dt 0 T T 1 ZT 2▯(n ▯ m) ! 2▯(n + m) ! = cos t ▯ cos t dt (8) 2 0 T T You can now do the integration, but notice that this integral is identical to the case above except for the minus sign between the functions. The result ends up being identical to that in case 1 (do the integral to convince yourself): Z ▯ ▯ ▯ ▯ T 2▯n 2▯m T sin T t sin T t dt = 2 ▯nm (9) 0 So we now have our second "orthogonality relation". Next... 3. We will be evaluating the integral of the product of a sine and a cosine: ZT ▯2▯n ▯ ▯ 2▯m ▯ sin t cos t dt (10) 0 T T In this case, we get no help from the cosine sum of angle formulas given above. Instead, we need the sine sum of angle formulas: 4 sin(A + B) = sinAcosB + cosAsinB sin(A ▯ B) = sinAcosB ▯ cosAsinB Adding the equations results in: 1 sinAcosB = 2[sin(A + B) + sin(A ▯ B)] (11) Plugging (11) into (10) gives: ZT ▯2▯n ▯ ▯ 2▯m ▯ sin t cos t dt 0 T T Z T ! ! 1 2▯(n + m) 2▯(n ▯ m) = 2 0 sin T t + sin T t dt (12) If you integrate the above, for both cases n = m and n 6= m, you will ▯nd that the integral becomes 0. In your next set of tutorial problems you will show this. So this orthogonality relation becomes: ▯ ▯ ▯ ▯
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