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

# 2250 Lecture Record Week 10 F2009

Last Modified: November 07, 2009, 08:06 MST.    Today: September 20, 2018, 08:46 MDT.

## 02 Nov: Piecewise Functions. Shifting. Delta function. Section 10.5 and EPbvp supplement 7.6.

Piecewise Functions
Unit Step
Pulse
Ramp
Second shifting Theorems
e^{-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 waves
Square wave
Triangular wave
Sawtooth
Rectified sine
Half-wave rectified sine
Parabolic wave
Periodic function theorem
Laplace of the square wave, tanh function.
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.

## 03 Nov: Delta function and hammer hits. Periodic piecewise functions. Second shifting theorem.

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
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.

## 04 Nov: Intro Laplace Resolvent. Undetermined Coefficients. Section 5.5, EPbvp7.6, 3.7.

Engineering models
The 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'=Au

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

Intro to the Laplace resolvent method for 2x2 systems
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).

## 05nov: Fusi and Richins

Exam 2 at 6:50am to 8:30am in WEB 103 on 05 Nov

## 06 Nov: Problem session. Sections 10.4, 10.5.

S-differentiation theorem
Problem 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 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)
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)) [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 theorem
Laplace of the square wave. Problem 10.5-25.
Laplace of the sawtooth wave. Problem 10.5-26.