Week 10: 02 Nov, |
03 Nov, | 04 Nov, | 05 Nov, | 06 Nov, |

Piecewise FunctionsUnit Step Pulse RampSecond shifting Theoremse^{-as}L(f(t))=L(f(t-a)step(t-a)) L(g(t)step(t-a))=e^{-as}L(g(t+a))Piecewise defined periodic wavesSquare wave Triangular wave Sawtooth Rectified sine Half-wave rectified sine Parabolic wavePeriodic function theoremLaplace of the square wave, tanh function.Hammer hits and the Delta functionDefinition of delta(t) Hammer hit models Paul Dirac (1905-1985) and impulses Laurent Schwartz (1915-2002) and distribution theory Riemann Stieltjes integration theory: making sense of the Dirac delta. Def: RS-integral equals the limit of RS-sums as N-->infinity and mesh-->zero. RS-sum = sum of terms f(x_i)(alpha(x_i)-alpha(x_{i-1})) where alpha(x) is the monotonic RS integrator.

Engineering modelsShort duration impulses Definition: impulse of force f(t) on [a.b] equals the integral of f(t) over [a,b] An example when f(t) has impulse 5, defined by f(t) = (5/(2h))pulse(t,-h,h) Laplace integral of f(t) and its limit as h --> 0. Answer is the Dirac delta. The delta function model x''(t) + 4x(t) = 5 delta(t-t0), x(0)=0, x'(0)=0 The delta function model from EPbvp 7.6 x''(t) + 4x(t) = 8 delta(t-2 pi), x(0)=3, x'(0)=0 Phase amplitude conversion [see EP 5.4] x(t) = 3 cos(2t) until hammer hit at t=2Pi. It has amplitude 3. x(t) = 3 cos(2t)+4 sin(2t) after the hit. It has amplitude 5. = 5 cos(2t - arctan(4/3)) Discussed how to convert in class. See 5.4. An RLC circuit model Q'' + 110 Q' + 1000 Q = E(t) Differentiate to get [see EPbvp 3.7] I'' + 100 I' + 1000 I = E'(t) When E(t) is a step, then E'(t) is a Dirac delta.

Engineering modelsThe delta function model x''(t) + 4x(t) = 8 delta(t-2 pi), x(0)=3, x'(0)=0 How to solve it with dsolve in maple. de:=diff(x(t),t,t)+4*x(t)=f(t);f:=t->8*Dirac(t-2*Pi); ic:=x(0)=3,D(x)(0)=0; dsolve({de,ic},x(t)); convert(%,piecewise,t); Details of the Laplace calculus in maple: inttrans package. with(inttrans): f:=x->cos(omega*t)+8*Dirac(t-2*Pi); G:=laplace(f(t),t,s); invlaplace(G,s,t); de:=diff(x(t),t,t)+4*x(t)=f(t); laplace(de,t,s); subs(ic,%); solve(%,laplace(x(t),t,s)); Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au. Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au. The general vector-matrix DE Model u'=AuUndetermined CoefficientsWhich equations can be solved Intro to the basic trial solution method Laplace solution of y'' + y = 1+x [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0] How to find the atoms in y_p(x). How to find the atoms in y_h(x) THEOREM. Solution y_h(x) is a linear combination of atoms. THEOREM. Solution y_p(x) is a linear combination of atoms. THEOREM. (superposition) y = y_h + y_pIntro to the Laplace resolvent method for 2x2 systemsSolve the systems by ch1 methods for x(t), y(t): x' = 2x, x(0)=100, y' = 3y, y(0)=50. Answer: x = 100 exp(2t), y = 50 exp(3t) x' = 2x+y, x(0)=1, y' = 3y, y(0)=2. Answer: y(t) = 2 exp(3t) and x(t) is the solution of the linear integrating factor problem x'(t)=2x(t)+2 exp(3t).

S-differentiation theoremProblem 10.4-22. Due next week. Problem 10.4-21. Clear fractions, multiply by (-1), then: (-t)f(t) = -exp(3t)+1 L((-t)f(t)) = -1/(s-3) + 1/s (d/ds)F(s) = -1/(s-3) + 1/s F(s) = ln(|s|/|s-3|)+c To show c=0, use this theorem: THEOREM. The Laplace integral has limit zero at t=infinity.Piecewise FunctionsUnit Step: step(t)=1 for t>=0, step(t)=0 for t<0. Pulse: pulse(t,a,b)=step(t-a)-step(t-b) Ramp: ramp(t-a)=(t-a)step(t-a)Second shifting Theoremse^{-as}L(f(t))=L(f(t-a)step(t-a)) Requires a>=0. L(g(t)step(t-a))=e^{-as}L(g(t+a)) Requires a>=0. Problem 10.5-4. F_1(s) = exp(-s)/(s-1) = exp(-as)L(exp(t)) with a=1 = L(exp(t-1)step(t-1)) by the second shifting theorem F_2(s) = exp(2-2s)/(s-1) = exp(-2s)/s with shift s --> s-1 = L(1 step(t-2)) [by the second shifting theorem] with shift s --> s-1 = L( exp(t) 1 step(t-2)) by the first shifting theorem F=F_1 - F_2 = L(exp(t-1)step(t-1)-exp(t)step(t-2)) f(t) = exp(t-1)step(t-1)-exp(t)step(t-2) Problem 10.5-22. f(t)=t^3 pulse(t,1,2) = t^3 step(t-1) - t^3 step(t-2) L(t^3 step(t-1)) = exp(-s)L((t+1)^3) by the second shifting theorem L(t^3 step(t-2)) = exp(-2s)L((t+2)^3) by the second shifting theorem Details were finished in class. Pascal's triangle and (a+b)^3. Function notation and dummy variables.Periodic function theoremLaplace of the square wave. Problem 10.5-25. Answer: (1/s)tanh(as/2) Laplace of the sawtooth wave. Problem 10.5-26. Answer: (1/s^2)tanh(as/2) Method: (d/dt) sawtooth = square wave The use the parts theorem. Laplace of the staircase function. Problem 10.5-27. This is floor(t/a). The Laplace answer is L(floor(t/a))=1/(s(exp(as)-1)) This answer can be verified in maple by the code with(inttrans): laplace(floor(t/a),t,s); Problem 10.5-28. Details in class. According to the periodic function theorem, the answer is found from maple integration: int(t*exp(-s*t),t=0..a)/(1-exp(-2*a*s)); # answer == (-1+exp(-s*a)+exp(-s*a)*s*a)/s^2/(-1+exp(-2*s*a))