18.03 Topic 24: More calculation tricks; Linear ODE’s with periodic input.Author: Jeremy OrloffReading: EP §8.4.Calculation shortcuts1. Use even-odd (discussed before).2. Make new series from old ones (discussed in previous topic).3. Use differentiation and integration.More examples of differentiation and integrationExample: f(t) = 1 + cos t +cos 2t2+cos 3t3+ . . ..⇒ h(t) =Zt0f(u) du = t + sin t +sin 2t22+sin 3t32+ . . ..Note: h(t) is not periodic (because of the t term).Example: (Very cool, but will only do this in class if there is time at the end)Discontinuous sawtooth: f(t) =tπ2ππ2−π2f(t) is odd, with period 2π (so L = π). On [0, 2π], f (t) =π − t2⇒ An= 0 and Bn=1πZ2π0π − t2sin nt dt =1n.(It’s slightly easier to do this calculation over a full period, rather than over [0, π].)⇒ f(t) = sin t +sin 2t2+sin 3t3+ . . ..Let h(t) =Zt0f(u) du =Zt0sin u +sin 2u2+sin 3u3+ . . . du= (1 − cos t) +1 − cos 2t22+1 − cos 3t32+ . . .=∞X11n2−∞X1cos ntn2.The DC term isA02=X1n2, which we can compute directly.On [0, 2π], h(t) =Zt0π2−u2du =πt2−t24.⇒ A0=1πZ2π0πt2−t24dt =π23. ⇒A02=π26=X1n2.We’ve summed an infinite series!Example: Differentiation of discontinuous functions.Let f(t) = square wave =tπ2π1=4πXn oddsin ntn.118.03 topic 24 2What is f0(t)? Answer:f0(t) = . . . −2δ(t+π)+2δ(t) − 2δ(t − π)+2δ(t−2π)+. . . =tJTJT−ππ2πTerm-by-term differentiation ⇒ f0(t) =4πXn o ddcos nt.We can check at t = 0, f0(t) = ∞ and at t = π, f0(t) = −∞.Applications to DE’sExample: Let f(t) = square wave of period 2π and height 1.Solve x00+ 8x = f(t).Using Fourier series: x00+ 8x =4πsin t +sin 3t3+sin 5t5+ . . .=4πXn oddsin ntnSide work: (using exponential input theorem)Solve x00n+ 8xn=sin ntn⇒ xn,p= Imsin(nt)P (in)=sin ntn(8 − n2).Putting it together using superposition:xsp=4πsin t8 − 1+sin 3t3(8 − 9)+sin 5t5(8 − 25)+ . . .=4πXn oddsin ntn(8 − n2)This is called the steady periodic solution.Near resonance: In the previous example the n = 3 term has the biggest amplitude.ResonanceExample: Same f as in previous example. Solve x00+ 9x = f.Side work: solve x00n+ 9xn=sin ntn.Same as before except when n = 3:n 6= 3 ⇒ xn,p=sin ntn(9 − n2)n = 3 ⇒ x3,p= −t cos 3t18(resonance)⇒ xsp=4πsin t9 − 1−t cos 3t18+sin 5t5(9 − 25)+ . . .=4πsin t8−t cos 3t18+4πXn>3 oddsin ntn(9 − n2)Note: The driver has angular frequency 1 but it contains a frequency 3 componentwhich causes the resonance.Example: Same f as in previous examples. Solve x00+ 16x = f.There is no resonance because f does not have a frequency 4 component.⇒ xsp=4πXn o ddsin ntn(16 − n2)No term in the Fourier series is near the resonant frequency.18.03 topic 24 3Notes: (as before with the exponential input theorem)1. No damping ⇒ exponential input theorm is easy, but can have resonance.2. Damping ⇒ exponential input theorem is messier but there is no resonance.More examples (probably won’t do in class)Example 1: Solve x00+ 2x0+ 9x = f, wheref(t) = triangle wave =t−ππ2π 3π 4π1Fourier series: f(t) =12−4π2cos t +cos 3t32+cos 5t52+ . . ..As usual do calculations in polar coordinates.Side work in pieces:x00n+ 2x0n+ 9xn= cos ntn = 0 ⇒ (x0)p=19.n 6= 0Complexify: ˜x0n+ 2˜xn+ 9˜xn= eint, xn= Re(˜xn)E.I.T.: ˜xn,p=eintP (in)=eint9 − n2+ 2in.Polar coords: 9 − n2+ 2in = Rneiφn.Where Rn= |P (ni)| =p(9 − n2)2+ 4n2and φn= Arg(P (ni)) = tan−12n9−n2with φnin the 1st or 2nd quadrants.⇒ ˜xn,p=1Rneint−φn.⇒ xn,p=1Rncos(nt − φn)⇒ xsp=118−4π2(1R1cos(t − φ1) +132R3cos(3t − φ3) + . . .).Example 2: x0+ kx = f is similarEnd topic 24
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