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Topic: Differential Eqn & Transform Probken (Read 1328 times) |
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svs14
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Posts: 8
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Hi there, im trying to study for my exam and ive encountered a few problems that i cant seem to solve. Itll be helpful if any of you can help me out as its examinable material. Thanks. Ive also attached a .doc copy of it for convinience
1. The velocity, v, of an object of mass, m, falling in the Earth’s atmosphere, with air
resistance proportional to the velocity, satsifies the differential equation
m(dv/dt)+ pv = mg
where g is the the acceleration due to gravity (a constant) and p is a positive constant.
Without finding an expression for v(t), show that the body reaches a terminal
velocity and find this velocity irrespective of its initial velocity. [Note: You will lose
marks if your solution depends on the knowledge of the explicit solution for v(t).]
It is frequently convenient to be able to compute a specific coefficient in the partial fraction
decomposition of the Laplace transform. Suppose r is a simple pole of Y(s). Thus
Y(s) =A/(s-r)+ R(s)
where the “remainder” R(s) is finite at s = r. Rearranging this equation, we obtain
A = (s - r) Y(s) - (s - r) R(s)
and so
A = lim(s - r) Y(s)
s-r
since
lim(s - r) R(s) = 0
s-r
.
2. Consider the square wave of period 2a
f2a(t) = H(t) - H(t - a) + H(t - 2a) - H(t - 3a) + • • •
(a) Find the coefficient of the 1/s term in the partial fraction decomposition of L{f2a(t)}.
(b) Find the coefficients of the remaining terms in the partial fraction decomposition
of L{f2a(t)}.
Note these coefficients are the Fourier coefficients of f2a. Upon taking the inverse
Laplace transform, we obtain the Fourier series representation of f2a.
6. Consider the damped harmonic oscillator
y"+ 2by’ + y = delta(t)
with y(0) = y0(0) = 0. This could, for example, be describing the suspension of a car
(y measuring vertical displacement) when the car hits a pothole in the road (described
by the delta fuction). The damping b is provided by the car’s shock absorbers. The
design goal is to choose b so that the half life of the transient solution is as short as
possible (that this, the effect of pothole on the performance of the car is minimized).
(a) Find the half life of the transient solution (for b > 0) and plot th as a function
of b.
(b) Find the value of b which minimizes th. For this value of b, we say the oscillator
is critically damped.
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asterix0
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Re: Differential Eqn & Transform Probken
« Reply #1 on: Aug 27th, 2005, 8:59am » |
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In problem 6, what does "th" refer to? - as in "(b) Find the value of b which minimizes th."
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svs14
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Re: Differential Eqn & Transform Probken
« Reply #2 on: Aug 27th, 2005, 11:57pm » |
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the half life time i presume
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K Strom
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Re: Differential Eqn & Transform Probken
« Reply #3 on: Sep 12th, 2005, 10:50pm » |
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These look interesting.
It appears that the diff equ persons do not visit this board much anymore or else someone would have contributed.
In the second problem, the infiinite sum is difficult to vision.
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Icarus
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Boldly going where even angels fear to tread.
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Re: Differential Eqn & Transform Probken
« Reply #4 on: Sep 14th, 2005, 6:35pm » |
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I can't speak for anyone else, but I haven't done anything serious with the Laplace transform in 20 years, and was not particularly eager to go hunt up my references and relearn it, so I left this one to others. Apparently, those who knew Laplace transforms felt much the same way, so no one ended up answering. At this point I am sure svs14 has long since taken the test he (she?) was studying for, and has also learned another lesson: don't depend on this forum for answers you need quickly. It may be weeks before you post receives an answer.
In the first problem, one approach that does exactly what was asked, but is definitely not what the teacher wanted (since it has nothing to do with Laplace transforms) is:
Let vT = mg/p, and rewrite the equation as dv/dt = -p(v - vT).
Note that this means that when v > vT, dv/dt is negative, so the value of v is decreasing (tending towards vT). When v < vT, dv/dt > 0, so v is increasing (once again tending towards vT). v can never cross vT, because it cannot pull away on the other side, so either v <= vT for all t, and is increasing, or else v >= vT for all t, and decreasing.
This alone is enough to guarantee that v has some terminal velocity. To see that the velocity must be vT, note that if v has a different limit, then v is bounded away from vT, and so dv/dt is also bounded away from 0. This means that v must increase or decrease without bound, in contradiction to the fact that v is bounded by vT. Hence the limit of v is vT.
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Icarus
wu::riddles Moderator
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Re: Differential Eqn & Transform Probken
« Reply #5 on: Sep 19th, 2005, 4:54pm » |
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Concerning problem 2. The start of the information for problem 2 begins after the note in problem 1. I.e., all that stuff about partial fractions has nothing to do with problem 1 at all.
Some deduction from the problem statement leads me to believe that H must be the unit step function at 0:
H(t) = 1 for t > 0; H(t) = 0 for t < 0.
How the teacher chooses to define H(0) is unclear, but this isn't needed to solve the problem.
The infinite series is not all that hard to understand. For any particular value of t, only a finite number of the expressions H(t), H(t-a), H(t-2a), H(t-3a), ... is non-zero. So the infinite sum is actually finite for each fixed t. Adding them up shows that the result is
For na < t < (n+1)a, f(t) = 1 if n is odd, f(t) = 0 if n is even. (I will call the resultant function f, rather than f2a, for simplicity.) I.e. f is a square wave - a squared-off version of cos(2[pi]t/a). (Note that f is only defined for t >= 0. What happens below zero is ignored by the Laplace transform.)
The Laplace transform of any function g is defined by L[g](s) = [int]0oo e-stg(t) dt. In particular L[H(t-na)](s) = e-nas/s.
Because the sum defining f is very well-behaved, we can exchange the transform and the summation to get L[f](s) = [sum]n=0oo L[H(t-na)](s) = [sum]n=0oo e-nas/s = 1/s(1-e-as), provided that real(s) > 0.
So the problem is to find the partial fraction decomposition of F(s) = 1/s(1-e-as), where for convenience, we drop the restriction that real(s) > 0. e-as = 1 iff s = 2n[pi]i/a. Let r = 2n[pi]i/a, n <> 0, and use L'Hospital's rules to get
lims->r (s-r)/s(1-e-as)
= (1/r) lim (s-r)/(1-e-as)
= (1/r) lim 1/(1+ae-as)
= (1/r)(1/(1+a))
= a/(1+a)2n[pi]i.
For n = 0 (so r = 0), we have a slightly more complicated situation. The denominator has a double root here. Fortunately, it is easy enough to find that sF(s) has partial fraction coefficient 1/(1+a) at s = 0, for F(s) = 1/(1+a)s2 + R(s), where R(s) is some function which is finite about 0.
Since two holomorphic functions with the same residues about the same poles must be equal, we have
F(s) = 1/(1+a)s2 + (a/(1+a)2[pi]i) [sum]n<>0 1/n(s - 2n[pi]i/a).
----------------------------------------------------
I think I must have made a mistake in this, since I do not come up with a partial fraction coefficient for 1/s at all (I get a second order pole at 0). It may be that I am mistaken about what the function H is supposed to be. Or maybe I am just missing something, since my knowledge of Laplace transforms is so rusty. (Of course, it couldn't be that I made an algebra error somewhere -- anything that looks wrong to you is because I am using a higher order of mathematics!  ).
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And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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SWF
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Posts: 879
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Re: Differential Eqn & Transform Probken
« Reply #6 on: Sep 22nd, 2005, 5:41pm » |
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Icarus, you are forgetting it is an alternating series. The transform should be 1/s/(1+exp(-a*s)). Another way to get that is to note that the function f(t) (I'll call it that instead of f2a(t)) satisifies: f(t) + f(t-a) = H(t). Take Laplace Transform to get F(s) + exp(-a*s)*F(s) = 1/s, and solve for F(s) giving the same.
The poles of this are at s=0, s=n*[pi]*i/a, and s=-n*[pi]*i/a for n being all odd integers. To find the coefficient for any of these roots r, take limit (s-r)*F(s) as s approaches r. That gives (sum is over odd n=1,3,5,7,...):
F(s)= 1/(2*s) + sum [ i/(s + n[pi]*i/a) - i/(s - n[pi]*i/a) ]/(n[pi])
= 1/(2*s) + (2/[pi])*sum (n[pi]/a)/(s2 + (n[pi]/a)2) / n
Invert the transform ( use L(sin(a*t)) = a/(s*s+a*a) ):
f(t) = 1/2 + (2/[pi]) * sum (1/n)*sin(n[pi]*t/a)
This is the Fourier series mentioned in the problem statement.
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