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) L(step(t-a)) = (1/s) exp(-as) [for a >= 0 only]Integral TheoremL(int(g(x),x=0..t)) = s L(g(t)) Applications to computing ramp(t-a) L(ramp(t-a)) = (1/s^2) exp(-as) [for a >= 0 only]Second 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.Periodic function theoremProof Application to the square wave.Convolution theoremApplication: L(cos t)L(sin t) = L(0.5 t sin(t))

Hammer hits and the Delta functionDefinition of delta(t) delta(t) = idealized injection of energy into a system at time t=0 of impulse=1. Hammer hit models How delta functions enter into circuit calculations 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. Why int( f(t) delta(t-a), t=-infinity .. infinity) = f(a) The symbol delta(t-a) makes sense only under an integral sign.Engineering modelsShort duration impulses: Injection of energy into a mechanical or electrical model.Definition:The 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 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)); 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)) Previously 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.Resonance examplesx'' + x = cos(t) Pure resonance, unbounded solution x(t) = 0.5 t sin(t) mx'' + cx' + kx = F_0 cos(omega t) Practical resonance, all solutions bounded, but x(t) can have extremely large amplitude when omega is tuned to the frequency omega = sqrt(k/m - c^2/(2m^2)) LQ'' + RQ' + (1/C)Q = E_0 sin(omega t) Practical resonance, all solutions bounded, but the current I(t)=dQ/dt can have large amplitude when omega is tuned to the resonant frequency omega = 1/sqrt(LC). later lecture, with slides and video: Soldiers marching in cadence, Tacoma narrows bridge, Wine Glass Experiment. Theodore Von Karman and vortex shedding. Cable model of the Tacoma bridge, year 2000. Resonance explanations.

Forward and Backward Table ApplicationsReview of previously solved problems. Problem 10.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta) used for L(sin(3t)cos(3t)). Problem 10.1-28. Splitting a fraction into backward table entries.Partial Fractions and Backward Table ApplicationsProblem 10.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for partial fractions: sampling, atom method, Heaviside cover-up. Problem 10.2-9. Solve x''+3x'+2x=t, x(0)=0, x'(0)=2. Get resolvent equation (s^2+3s+2)L(x)=2+L(t) L(x)=(1+2s^2)/(s^2(s+2)(s+1)) L(x)=A/s + B/s^2 + C/(s+2) + D(s+1) L(x)=L(A+Bt+C e^{-2t} +D e^{-t}) Solve for A,B,C,D by the sampling method.Shifting Theorem and u-substitution ApplicationsProblem 10.3-8. L(f)=(s-1)/(s+1)^3 See #18 details for a similar problem. Problem 10.3-18. L(f)=s^3/(s-4)^4. L(f) = (u+4)^3/u^4 where u=s-4 L(f) = (u^3+12u^2+48u+64)/u^4 L(f) = (1/s + 12/s^2 + 48/s^3 + 64/s^4) where s --> (s-4) L(f)=L(e^{4t}(1+12t+48t^2/2+64t^3/6)) by shifting thm Problem 10.3-8. L(f)=(s+2)/(s^2+4s+5) L(f) = (s+2)/((s+2)^2 + 1) L(f) = u/(u^2 + 1) where u=s+2 L(f) = s/(s^2 + 1) where s --> s+2 L(f) = L(e^{-2t} cos(t)) by shifting thmS-differentiation theoremProblem 10.4-21. Similar to Problem 10.4-22. 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.Convolution theoremTHEOREM. L(f(t)) L(g(t)) = L(convolution of f and g) Example. L(cos t)L(sin t) = L(0.5 t sin t) Example: 10.4-36. x''+4x=f(t), x(0)=x'(0)=0 has solution x(t)=0.5 int(sin(2u)f(t-u),u=0..t)Periodic function theoremLaplace of the square wave. Problem 10.5-25. Answer: (1/s)tanh(as/2) Problem 10.5-28. Find L(f(t)) where f(t) =t on 0 <= t < a and f(t)=0 on a <= t < 2a, with f(t) 2a-periodic [f(t+2a)=f(t)]. Details According to the periodic function theorem, the answer is found from maple integration: L(f) = 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))

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)Periodic function theoremLaplace of the square wave. Problem 10.5-25. Done earlier. 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 f(t)=t on 0 <= t <= a, f(t)=0 on a <= t <= 2a 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)) A better way to solve the problem is to write a formula for f'(t) and use the s-differentiation rule. We get for a=1 f'(t) = (1/2)(1+sqw(t)) and then sL(f(t)) = (1/(2s))(1+tanh(s/2)) L(f(t)) = (1/(2s^2))(1+tanh(s/2))

Second Shifting Theorem ApplicationsProblem 10.5-3. L(f)=e^{-s}/(s+2) Problem 10.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1) Problem 10.5-22. f(t)=t^3 pulse(t,1,2)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)) [2nd shifting theorem] 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) 2nd shifting theorem L(t^3 step(t-2)) = exp(-2s)L((t+2)^3) 2nd shifting theorem Details were finished in class. Pascal's triangle and (a+b)^3. Function notation and dummy variables.Piecewise ApplicationsStaircase or floor function Sawtooth wave Square waveDirac Applicationsx''+x=5 Delta(t-1), x(0)=0,x'(0)=1 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)Theory of Practical ResonanceThe equation is mx''+cx'+kx=F_0 cos(omega t) THEOREM. The limit of x_h(t) is zero at t=infinity THEOREM. x_p(t) = C(omega) cos(omega t - phi) C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the undetermined coefficient answers for trial solution x(t) = A cos(omega t) + B sin(omega t). THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically just x_p(t) = C(omega) cos(omega t - phi) for large t. Therefore, x_p(t) is the OBSERVABLE output. THEOREM. The amplitude C(omega) is maximized over all possible input frequencies omega>0 by the single choice omega = sqrt(k/m - c^2/(2m^2)). DEFINITION. Thepractical resonance frequencyis the number omega defined by the above square root expression.

Undetermined CoefficientsWhich equations can be solved Intro to the basic trial solution method Laplace solution of x'' + 9x = 30 sin(2t) 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_p

More Laplace ExamplesContinuing 10.3, 10.5 examples from last lecture.Intro to the Laplace resolvent method for 2x2 systemsProblem: 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'=Au Laplace of u(t) = Resolvent x u(0) Resolvent = inverse(sI - A) Solve 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). Remarks on problem 10.2-16, a 3x3 system that can be solved with the resolvent equation shortcut. RULE: Use Cramer's rule or matrix inversion to solve the resolvent equation for the vector of components L(x), L(y), L(z). Any linear algebra problem Bu=c where B contains symbols should be solved this way, unless B is triangular.Transform TerminologyInput Output Transfer FunctionVariation of parametersThe second order formula. Application to y''=1+x Application to y''+y=sec(x) [see also slides] How to calculate y_p(x) from the five parameters y1(x) y2(x) W(x) = y1(x)y2'(x)-y1'(x)y2(x) A(x) = coefficient in the DE of y'' f(x) = input or forcing term, the RHS of the DE See (33) in section 5.5.