2250-1 7:30am Lecture Record Week 11 S2010

Last Modified: April 04, 2010, 09:27 MDT.    Today: October 19, 2017, 17:49 MDT.

29 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.
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));
```

31 Mar: Laplace Resolvent Method. Undetermined Coefficients. Sections 10.5,7.1,5.5

```Wine Glass Experiment
The lab table setup
Speaker.
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
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.
```
```    Undetermined Coefficients
Which 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_p
```

Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
```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.

02 Apr: Variation of parameters. 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
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.
```
```Wine Glass Experiment
The lab table setup
Speaker.
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
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.
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)