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2250-4 12:55pm Lecture Record Week 11 S2010

Last Modified: April 04, 2010, 12:41 MDT.    Today: October 23, 2017, 14:40 MDT.

Week 11, Mar 30 and Apr 1: Sections 10.5,EPbvp7.6,5.5,5.6,,EPbvp3.7,7.1

30 Mar: Section 10.5 and EPbvp supplement 7.6. Delta function and hammer hits.

 Hammer hits and the Delta function
   Definition 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 models
   Short 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
           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))  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 examples
    x'' + 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).
    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.

30 Mar: Problem session. Sections 10.4, 10.5.

 Forward and Backward Table Applications
   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 Applications
    Problem 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 Applications
     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 thm
   S-differentiation theorem
     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.
   Convolution theorem
     THEOREM. 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)
   Second Shifting Theorem Applications
     Problem 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 Theorems
     e^{-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 Applications
     Staircase or floor function
     Sawtooth wave
     Square wave
   Dirac Applications
     x''+x=5 Delta(t-1), x(0)=0,x'(0)=1
     THEOREM. The Laplace integral has limit zero at t=infinity.
 Piecewise Functions
   Unit 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 theorem
    Laplace of the square wave. Problem 10.5-25.
      Answer: (1/s)tanh(as/2)

01 Apr: Laplace Resolvent Method. Undetermined Coefficients. Sections 10.5,7.1,5.5

Wine Glass Experiment
   The lab table setup
      Speaker.
      Frequency generator with adjustment knob.
      Amplifier with volume knob.
      Wine glass.
   x(t)=deflection from equilibrium of the radial component of the
      glass rim, represented in polar coordinates, orthogonal to
      the speaker front.
   mx'' + cx' + kx = F_0 cos(omega t)  The model of the wine glass
      m,c,k are properties of the glass sample itself
      F_0 = volume knob adjustment
      omega = frequency generator knob adjustment
 Projection: glass-breaking video.

Video: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 31 Mar 2008)
Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
Slides: Resonance and undetermined coefficients (143.3 K, pdf, 07 Nov 2009)
Theory of Practical Resonance
   The 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. The practical resonance frequency is the number omega
               defined by the above square root expression.
Engineering models
   The job-site cable hoist example [delayed]
   Sliding plates example  [delayed]
   Home heating example  [delayed]

31 Mar, 01 Apr:Murphy

Review starts for Exam 3, using the 7:30 exam key from F2009. Solved laplace theory problems from chapter 10 dailies.

01 Apr: Variation of parameters. Undetermined Coefficients. Sections 5.5, 10.4

 More Laplace Examples
   Continuing 10.3, 10.5 examples from last lecture.
 Intro to the Laplace resolvent method for 2x2 systems
   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'=Au
   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 Terminology
   Input
   Output
   Transfer Function

 Variation of parameters
   The second order formula.
   Application to y''=1+x
   Application to y''+y=sec(x) [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.

 
 
Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
Wine Glass Experiment
   The lab table setup
      Speaker.
      Frequency generator with adjustment knob.
      Amplifier with volume knob.
      Wine glass.
   x(t)=deflection from equilibrium of the radial component of the
      glass rim, represented in polar coordinates, orthogonal to
      the speaker front.
   mx'' + cx' + kx = F_0 cos(omega t)  The model of the wine glass
      m,c,k are properties of the glass sample itself
      F_0 = volume knob adjustment
      omega = frequency generator knob adjustment
Theory of Practical Resonance
   The 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. The practical resonance frequency is the number omega
               defined by the above square root expression.
 Projection: glass-breaking video. Wine glass experiment. Tacoma narrows.

Video: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 31 Mar 2008)
Video: Wine glass experiment (12mb mpg, 2min) (12493.8 K, mpg, 01 Apr 2008)
Video: Tacoma Narrows Bridge Nov 7, 1940 (18mb mpg, 4min) (18185.8 K, mpg, 01 Apr 2008)

Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
Slides: Resonance and undetermined coefficients (143.3 K, pdf, 07 Nov 2009)
Extra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.
    Laplace theory references
    Slides: Laplace and Newton calculus. Photos. (145.3 K, pdf, 01 Nov 2009)
    Slides: Intro to Laplace theory. Calculus assumed. (109.5 K, pdf, 01 Nov 2009)
    Slides: Laplace rules (112.2 K, pdf, 01 Nov 2009)
    Slides: Laplace table proofs (130.3 K, pdf, 01 Nov 2009)
    Slides: Laplace examples (101.2 K, pdf, 07 Nov 2009)
    Slides: Piecewise functions and Laplace theory (64.7 K, pdf, 01 Nov 2009)
    MAPLE: Maple Lab 7. Laplace applications (0.0 K, pdf, 31 Dec 1969)
    Manuscript: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (56.4 K, pdf, 01 Nov 2009)
    Slides: Laplace second order systems (248.9 K, pdf, 01 Nov 2009)
    Slides: Home heating, attic, main floor, basement (73.8 K, pdf, 30 Nov 2009)
    Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
    Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)
    Manuscript: Heaviside's method 2008 (186.8 K, pdf, 20 Oct 2009)
    Manuscript: Laplace theory 2008 (350.5 K, pdf, 06 Mar 2009)
    Transparencies: Ch10 Laplace solutions 10.1 to 10.4 (1968.3 K, pdf, 13 Nov 2003)
    Text: Laplace theory problem notes F2008 (8.9 K, txt, 31 Dec 2009)
    Text: Final exam study guide (7.6 K, txt, 12 Dec 2009)
    Variation of Parameters and Undetermined Coefficients references
    Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
    Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
    Systems of Differential Equations references
    Manuscript: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (56.4 K, pdf, 01 Nov 2009)
    Slides: Laplace second order systems (248.9 K, pdf, 01 Nov 2009)
    Slides: Home heating, attic, main floor, basement (73.8 K, pdf, 30 Nov 2009)
    Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
    Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)
    Oscillations. Mechanical and Electrical.
    Slides: Electrical circuits (87.1 K, pdf, 11 Oct 2009)
    Slides: Forced damped vibrations (235.0 K, pdf, 11 Oct 2009)
    Slides: Forced vibrations and resonance (185.3 K, pdf, 11 Oct 2009)
    Slides: Forced undamped vibrations (174.7 K, pdf, 11 Oct 2009)
    Slides: Resonance and undetermined coefficients (143.3 K, pdf, 07 Nov 2009)
    Slides: Unforced vibrations 2008 (620.4 K, pdf, 11 Oct 2009)
    Eigenanalysis and Systems of Differential Equations.
    Manuscript: Eigenanalysis S2010, 46 pages (345.3 K, pdf, 31 Mar 2010)
    Manuscript: Algebraic eigenanalysis 2008 (127.8 K, pdf, 23 Nov 2009)
    Text: Lawrence Page's pagerank algorithm (0.7 K, txt, 06 Oct 2008)
    Text: History of telecom companies (1.4 K, txt, 30 Dec 2009)
    Manuscript: What's eigenanalysis, draft 1 (152.2 K, pdf, 01 Apr 2008)
    Manuscript: What's eigenanalysis, draft 2 (124.0 K, pdf, 14 Nov 2007)
    Manuscript: What's eigenanalysis 2008 (126.8 K, pdf, 11 Apr 2010)
    Slides: Cayley-Hamilton method for solving vector-matrix system u'=Au. (111.4 K, pdf, 30 Nov 2009)