Archive for March, 2010

黒澤 明

March 24, 2010

100 years ago today, legendary filmmaker Kurosawa Akira was born.

Monday Math 113

March 22, 2010

The Laplace Transform
Part 14: Gaussians and Error Function

Now, let’s look at the Laplace transform of a Gaussian function, specifically the function , for t≥0. From our transform definition,

We proceed by completing the square:

To simplify from here, we avoid complications of trajectory in the complex plane by assuming s real (the result can be extended to the region of convergence ). Then, by substituting , we have
.
In terms of the error function , and using , we see
,
and so
.

We can also use this to find the transform of the error function, via our integration rule:

(also with ).

Physics Friday 113

March 19, 2010

We return again to our driven pendulum. This time, we consider what happens when the pendulum is near the upright vertical position, θ=π, the unstable equilibrium of the undriven pendulum. To do this, let us denote a new angle measure , so that measures the angle from the upright vertical. Then , , and in terms of , the equation of motion

becomes
;
and for small , this is approximated by
.

Now, one might expect that the pendulum will fall over, as it does in the undriven case. After all, the tangential acceleration, , appears like it should average over a cycle to , a positive quantity. However, this reasoning is incorrect in one important way: recall the “wiggle” in θ, and thus in , due to the driving, and also proportional to a . By examination, we see that when the pivot is accelerating upward, will be driven to increase, and a downward acceleration will produce the part of the wiggle decreasing (or slowing the increase in) .


Thus, the in the driving term of the acceleration will contain a cosine term, so the above will have a cosine squared term, which, like in our previous rapid ω, small α approximation, will have a nonzero average. In fact, we see that due to the wiggle, will be slightly larger when the driving acceleration is towards lower , and thus this will have a greater effect than when the driving is in the opposite direction, and is slightly smaller. Thus, if ω is large enough, then the net effect of the driving is a restoring force stronger than gravity, and one gets the pendulum oscillating about the upright vertical, with a small wiggle superimposed!

Working as we did in our approximation here, we try for an approximate solution of the form , where b(t) and c(t) are functions whose variation over time is much slower that the “wiggle”, so that on a time-scale of , with , we can ignore their time dependence and treat them like constants; we also expect b to be much smaller than c (as it represents the “wiggle”).
As before, we substitute into our equation of motion:
.
Now, with , and rapid enough driving that , we can, as before, ignore the term to get
.
Which, to leading order, tels us that , so , and so our approximate solution is
.
From our equation of motion, we have acceleration of φ given by
.
Now, taking the average over a period of the driving (), we see that in the product , we will have four terms; the constant term will obviously average to itself, while the two cosine terms will each average to zero, and since the average of over one of its periods is 1/2, the fourth term, will have average . Lastly, c(t) is approximately constant on this timescale.
Therefore,
.
Suppose that , so that the quantity in brackets is positive. Assuming this is true, then define . Then
.
But from , we see that on the timescale we are averaging, , and so we have
,
which is (in the approximation) simple harmonic motion of frequency
.
Now, for this to happen, we noted that must be real. With , we see then that to have , we need , as was assumed earlier.
Note that if (instead of just ), then .

Thus, in this approximation, our pendulum swings with a frequency ; with a small wiggle of frequency superimposed. We also note that the amplitude of the “wiggle” is proportional to the current angle about which it is wiggling, being smaller by a factor of , creating behavior similar to that here, except with a distinctly different , and with a wiggle of opposite phase.

Monday Math 112

March 15, 2010

The Laplace Transform
Part 13: Transform of the Logarithm

Now, let us find the Laplace transform of the natural logarithm. From the definition of the Laplace transform,
.
Making the u-substitution u=st, we get
.

Now, one may recall that we found the remaining integral here:
, where γ is the Euler-Mascheroni constant. Thus,
.

Happy Pi Day

March 14, 2010

It’s 3-14, so, have a happy Pi Day!

Physics Friday 112

March 12, 2010

Continuing from last time, we noted that the equation of motion for our driven pendulum,

cannot be solved analytically even in the small angle approximation.

Recall that we required that the driving motion of the pivot be much smaller in amplitude than the length of the pendulum, so that . Now, let us assume that the driving frequency ω of pivot is much, much faster than the natural (small-angle) frequency of the pendulum ω0; so that ,
and thus the term in the equation of motion will dominate the term for all but a tiny fraction of the driving cycle.
Now, we note that, superimposed upon any larger-scale motion of the pendulum will be a “wiggle” of frequency ω. Now, to compare that wiggle to the driving motion, we note that when the pivot has upward acceleration, that means an upward force acts on the pivot, a force whose torque on the pendulum would tend to reduce θ, while a downward acceleration of the pivot will tend to increase θ.

Since the vertical acceleration of our pivot is , we expect the wiggle in θ to be proportional to . So, let us look for an approximate solution of the form , where b(t) and c(t) are functions whose variation over time is much slower that the “wiggle”, so that on a time-scale of , with , we can ignore their time dependence and treat them like constants.
Plugging into the equation of motion;
,
Now, neglecting the ,

,
which means that we see that to leading order, we need , so that , and our approximate solution is thus
.
From our equation of motion, we have acceleration of θ given by
.
Now, taking the average over a period of the driving (), we see that in the product , we will have four terms; the constant term will obviously average to itself, while the two cosine terms will each average to zero, and since the average of over one of its periods is 1/2, the fourth term, will have average . Lastly, c(t) is approximately constant on this timescale.
Thus, we see:
.
Now define . Then
.
But from , we see that on the timescale we are averaging, , and so we have
,
which is (in the approximation) simple harmonic motion of frequency
.

Thus, in this approximation, our pendulum swings with a frequency , faster than the natural frequency ; with a small wiggle of frequency superimposed. We also note that the amplitude of the “wiggle” is proportional to the current angle about which it is wiggling, being smaller by a factor of .


Plot of the approximate solution (blue) with L=1 m, a0=0.01 m, and ω=100 s-1. The undriven oscillation of similar overall amplitude (gray) is shown for comparison. (Note: Click on graph for link to full size image, as it antialiases weirdly when scaled down).

[Note—Major edit on 3-15: Corrected a sign error that had greatly complicated the analysis and caused an error in comparing driven vs. undriven frequencies].

Monday Math 111

March 8, 2010

The Laplace Transform
Part 12: Combining Rules

We can combine many of the previous rules we’ve found to determine the Laplace transform of some more complicated functions.

For example, what is the Laplace transform of f(t)=t^2e^{-t}\cos{t}? We know from here that the Laplace transform for cosine is \mathcal{L}\left[\cos(\omega{t})\right](s)=\frac{s}{s^2+\omega^2}, so \mathcal{L}\left[\cos{t}\right](s)=\frac{s}{s^2+1} Thus, via the frequency shift formula here, \mathcal{L}\left[e^{-t}\cos{t}\right](s)=\frac{(s+1)}{(s+1)^2+1}.
Considering, then, our frequency differentiation formula,
\begin{array}{rcl}\mathcal{L}\left[t^2e^{-t}\cos{t}\right](s)&=&\frac{d^2}{ds^2}\left(\frac{(s+1)}{(s+1)^2+1}\right)\\\mathcal{L}\left[t^2e^{-t}\cos{t}\right](s)&=&\frac{2(s+1)\left((s+1)^2-3\right)}{\left((s+1)^2+1\right)^3}\\&=&\frac{2(s+1)\left(s^2+2s-2\right)}{\left(s^2+2s+2\right)^3}\\&=&\frac{2\left(s^3+3s^2-2\right)}{\left(s^2+2s+2\right)^3}\end{array}.

Now consider the “sine cardinal” or “sinc” function \mathrm{sinc}(t)=\frac{\sin{t}}t. What is the Laplace transform of sinc(kt)?
We recall from here that the Laplace transform for the sine is
\mathcal{L}\left[\sin(\omega{t})\right](s)=\frac{\omega}{s^2+\omega^2}. We apply to this our frequency integration formula to get the transform of sinc(kt)
\begin{array}{rcl}\mathcal{L}\left[\mathrm{sinc}(kt)\right](s)&=&\mathcal{L}\left[\frac{\sin(kt)}{kt}\right](s)\\&=&\frac{1}{k}\mathcal{L}\left[\frac{\sin(kt)}{t}\right](s)\\&=&\frac{1}{k}\int_s^{\infty}\mathcal{L}\left[\sin(kt)\right](\sigma)\,d\sigma\\&=&\frac{1}{k}\int_s^{\infty}\frac{k}{\sigma^2+k^2}\,d\sigma\\&=&\int_s^{\infty}\frac{d\sigma}{\sigma^2+k^2}\\&=&\left.\frac{1}{k}\tan^{-1}\left(\frac{\sigma}k\right)\right|_s^{\infty}\\&=&\frac{1}{k}\left(\frac{\pi}2-\tan^{-1}\left(\frac{s}k\right)\right)\\\mathcal{L}\left[\mathrm{sinc}(kt)\right](s)&=&\frac{1}{k}\tan^{-1}\left(\frac{k}s\right)\end{array},
\Re(s)>0.

Lastly, let’s consider the square wave with period 2π given by f(t)=\mathrm{sgn}(\sin{t})=\left\{\begin{array}{rcl}1&\,\text{if}&\sin{t}>{0}\\-1&\,\text{if}&\;\sin{t}<{0}\end{array}\right.. As this is a periodic function, we use our periodic function Laplace transform
\mathcal{L}\left[f(t)\right](s)=\frac{1}{1-e^{-sT}}\int_0^{T}f(u)e^{-su}\,du,
where T is the period.
For our current function, this gives us:
\begin{array}{rcl}\mathcal{L}\left[f(t)\right](s)&=&\frac{1}{1-e^{-sT}}\int_0^{T}f(u)e^{-su}\,du\\&=&\frac{1}{1-e^{-s(2\pi)}}\int_0^{2\pi}f(u)e^{-su}\,du\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\int_0^{\pi}e^{-su}\,du-\int_{\pi}^{2\pi}e^{-su}\,du\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\left[-\frac{e^{-su}}s\right]_{u=0}^{\pi}-\left[-\frac{e^{-su}}s\right]_{u=\pi}^{2\pi}\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\frac1s-\frac{e^{-\pi{s}}}s+\frac{e^{-2\pi{s}}}s-\frac{e^{-\pi{s}}}s\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\frac1s\left(1-2e^{-\pi{s}}+e^{-2\pi{s}}\right)\\&=&\frac1s\frac{\left(1-e^{-\pi{s}}\right)^2}{1-e^{-2\pi{s}}}\\&=&\frac1s\frac{1-e^{-\pi{s}}}{1+e^{-\pi{s}}}\\&=&\frac1s\frac{e^{\pi{s}/2}-e^{-\pi{s}/2}}{e^{\pi{s}/2}+e^{-\pi{s}/2}}\\\mathcal{L}\left[f(t)\right](s)&=&\frac1s\tanh\left(\frac{\pi}2s\right)\end{array}.

Physics Friday 111

March 5, 2010

Consider a simple pendulum, with a mass m on the end of a rod of length l. In terms of the angle θ from the downward direction, the equation of motion for this pendulum is .

Now, suppose the pivot to which this pendulum is attached moves vertically in a sinusoidal motion of small amplitude, so that its vertical position is given by , with . What then is the new equation of motion? What happens for small-angle oscillations ()?
(more…)

Monday Math 110

March 1, 2010

The Laplace Transforn
Part 11: Frequency Derivatives and Integrals

We found previously what happens when one takes the Laplace transform of the derivative of a function,
,
and what happens when we take the Laplace transform of the integral
.
But what happens if we take the derivative with respect to s of the Laplace transform? Namely, what is F’(s)?
Since F(s) is the Laplace transform of f(f),
.
Taking the derivative of both sides with respect to s,
.
This “frequency differentiation” formula is usually written
,
and gives us the transform of a function multiplied by t in terms of the function’s transform. For example, consider the ramp function, . From here, we know , and so the above tells us that
,
as expected from here.
In fact, we can apply the formula repeatedly
,
and in general,
;
compare to our earlier results for the transforms of positive integer powers of t.

From our previous discussion of limits, we found that for any Laplace transform F(s); thus the integral makes a more reasonable antiderivative of F(s) over the region of convergence. Note that if we define , then G’(s)=-F(s), and G(∞)=0 (as needed for it to be the Laplace transform of some g(t)).
Now, applying our frequency differentiation rule to G(s),
we have
,
and so dividing by time corresponds to a “frequency” integration.