Discrete Fourier Transform – Trig Parameters in Practice

Preliminaries

This post is a follow-up to the previous one, Discrete Fourier Transform – Trig Parameters in Principle. The titles are deliberately similar – but you will want to distinguish them.

Having shown you how to find trigonometric parameters of a sine wave, I want to show you a real example. I found this in Peter Bloomfield’s “Fourier analysis of time series: an introduction”, 978-0-471-88948-9.

The data, however, I found by searching the Internet. Although the book provides a couple of sources – the data isn’t there any longer.

According to Bloomfield, these numbers are the magnitude of a variable star at midnight on 600 consecutive nights. They have been rescaled.

Before we get started, let me point out that I am a rookie at using the Discrete Fourier Transform. I knew, for example, to expect bin leakage in the previous post… but it hasn’t been all that long that I’ve known about it. Still, what we’re about to do is pretty simple and everything works out fine.

Here are the data…

Discrete Fourier Transform – Trig Parameters in Principle

pure sine

I want to look at the Discrete Fourier Transform (henceforth the DFT) of a perfect sine wave. How the heck can I interpret the DFT of some data, if I don’t understand what it tells me in the simplest possible case?

As usual, I am using Mathematica®… in fact I’m using version 7.

Suppose we have the following function: a pure sine wave, with mean 3, amplitude 5, and period 7. (I habitually use prime numbers in examples so that I can see where they end up in any answers.)

Actually, let me do this in stages. First let the mean be zero:

.

Over the range from 0 to 21, we have the following:

We see that the period is, indeed, 7. Now let me construct the function I really want, with mean 3:

Regression 1 – Multicollinearity in Review

As I draft this, I plan to do four things in this post.

1. Summarize the methods I’ve used to analyze multicollinearity.
2. Suggest that multicollinearity is a continuum with no clear-cut boundaries.
3. Summarize the conventional wisdom on its diagnosis and treatment.
4. Flag significant points made in my posts.

Let me say up front that there is one more thing I know of that I want to learn about multicollinearity – but it won’t happen this time around. I would like to know what economists did to get around the multicollinearity involved in estimating production functions, such as the Cobb-Douglas.

An Overview of Truss Designs

Introduction

I want to close the recent examples of trusses by providing a sampler of truss designs. This is far from encyclopedic. In fact, this post is limited to planar trusses.

First, however, let me give you a link to an online calculator. I checked it out on the Howe truss with a snow load.

As you can see, I scaled the loads by a factor of 10. I had to, for the program – not a big deal.

There are plenty of websites with information. You can search for yourself… you could start with the usual wiki article – the merit of which is that it has a lot of external links.

Personally, I have kept a link to bridge trusses in western PA and a link to roof trusses by an Australian contractor.

Okay, what do we have?

Trusses – Snow Load on Howe, Fink, and Pratt trusses

Intoduction

I want to work another snow load problem… using three different trusses. I found a pair of these, for a Fink truss and a Howe truss, on a professor’s university website – his name is Zig Herzog and his main statics page is this. Individual links for the two problems will follow.

Each of these trusses is 12 meters across the bottom, and 4 meters high. Each has a total snow load of 2400 Newtons. As ever, I have used Mathematica® for the computations and graphics.

One of the things I liked was that Herzog asked us to find the maximum values of stress (both compression and tension), and the total length of the beams used in each truss. In addition to using Fink and Howe trusses, I will do the Pratt truss again, with the parameters of this problem.

The purpose of this post was to see how different the trusses are, under the same load. In one respect they are the same: the maximum values of both compression and tension are the same for all three trusses. And that’s the summary!

Here are screenshots of the author’s assigned problem… for which he does provide solutions, so I’m not giving away any secrets.

For comparison, here’s the Pratt truss I used in the previous post.

Trusses – Example 4, Pratt roof truss with snow

This problem, like the Howe truss, comes from J.L. Meriam, “Statics”, John Wiley & Sons, 1966.

This diagram actually shows concentrated forces at the joints – which is where they must be for what we’re doing. Nevertheless, the problem as posed says that we have a total weight of 4000 pounds distributed uniformly across the top of the truss. We take the 1000 pounds on each of these four beams, and assign 500 pounds to each of its endpoints. Points A and E have 500 pounds, but the other three joints connect two loaded beams and get 1000 pounds.

Trusses – Example 3, a Howe truss

Introduction

This is a sample problem from “Statics”, by J.L. Meriam, John Wiley & Sons, 1966; #105, p. 88.

It is called a Howe truss – or, sometimes, a double Howe truss. Here’s a link that will give you the names of some trusses so you can look for more information. There’s a lot to be had, but it doesn’t seem all that standardized.

And here’s a link to some class notes, in case you want more explanation than I have been providing. And, since this is the third post about trusses, you could go look at my first truss post and my second truss post.

Anyway, we see that it is fixed at points A and G… and it has three external loads, each of 2 kips (a kip is a kilogram-force), at points B, C, I.

Trusses – Example 2

Let’s try a slightly more complicated truss. You may want to read the previous post, if you have not already.

Trusses – Example 1

Given the following picture of three beams… with a given force applied at the apex… let’s see if we can work out the internal forces in the beams, and the reaction forces at the two bottom points.

This is an example of a “statically determinate” problem. We will be able to solve this assuming that the three beams do not bend or compress or stretch.

Using the QR Decomposition to orthogonalize data

This is going to be a very short post, illustrating one idea with one example (yes, one, not five).

It turns out that there is another way to have Mathematica® orthogonalize a matrix: it’s called the QR decomposition. The matrix Q will contain the orthogonalized data… and the matrix R will specify the relationship between the original data and the orthogonalized.

That means we do not have to do the laborious computations described in this post. Understand, if we do not care about the relationship between the original data and the orthogonalized data, then I see no advantage in Mathematica to using the QR over using the Orthogonalize command.