# Fourier transforms

1. Mar 11, 2010

### dnp33

1. The problem statement, all variables and given/known data
3. Consider the narrow triangular voltage pulse V(t) shown in the figure.
i can't paste the function but its a piecewise function of the form:
V(t)={1+(1/a)t from -a <=t<=0
{1-(1/a)t from 0<t<=a
{ 0 otherwise
(a) Find and sketch the spectral content g(ω), i.e. determine the Fourier transform of V.
i got this part, its the next i'm having trouble with but the equation i have is
2(1-cos(a w)/(sqrt(2pi)*a*w^2)

(b) Show that a relationship of the type holds. delta w * delta t ~ 1

3. The attempt at a solution

i don't really know where to start to be honest. any help would be appreciated. and sorry i don't know how to use latex properly.

2. Mar 11, 2010

### collinsmark

$$\Delta w$$ and $$\Delta t$$ are typically the standard deviation (essentially/similarly the root mean square around the mean) of each waveform. Find the standard deviation (root mean square deviation) of each waveform and multiply those together.

3. Mar 11, 2010

### dnp33

okay i did know that, but i don't know how to find the root mean square of each waveform.
the problem is in my lack of knowledge of statistics it seems...

4. Mar 11, 2010

### dnp33

would i take the integral of each function squared and the square root of that?

5. Mar 11, 2010

### collinsmark

First Normalize the waveform. We'll call the normalization constant "A". Before calculating the standard deviation, we must make sure that the area under the curve is 1, when performing our calculations.

$$1 = A \int _{-\infty} ^{\infty} f(x) dx$$

So evaluate the integral, and solve for A.

This problem is a little easier, since in both cases in your particular problem the mean (i.e. expectation value) is going to end up being zero, given the symmetry of each waveform around 0. If you didn't happen to know this, calculating the expectation value (mean), $$\mu$$ would be your next step.

$$\mu = A\int _{-\infty}^{\infty}x f(x) dx$$

But you can probably skip that part here, and assume $$\mu$$ is zero.

You can calculate the variation $$\sigma^2$$ for some function f(x) by using

$$\sigma^2 = A\int _{-\infty}^{\infty}(x - \mu)^2f(x)dx$$

Then get to the standard deviation $$\sigma$$ by

$$\sigma = \sqrt{\sigma^2}$$

So in your cases, repace $$x$$ with either $$\omega$$ or $$t$$ to find the respective standard deviations $$\sigma _{\omega}$$ and $$\sigma _t$$; and replace f(x) with your frequency based waveform and time based waveform as appropriate. (And since you know that $$\mu$$ is going to end up being zero anyway, you might start out with that assumption.)

[Edit: I almost forgot. You need to normalize your waveform first, before finding the standard deviation! Sorry about that. I've made a few edits to the above material, with this in mind]

Last edited: Mar 11, 2010
6. Mar 11, 2010

### collinsmark

Okay, after giving it some thought, you might want to scrap some of that last post, depending on how your text and instructor treats the uncertainty principle. It seems that different forms/interpretations of the uncertainty principle are out there, and you should use the interpretation of your text and/or instructor.

It could be that your instructor is just asking you to eyeball the waveforms to pick the $$\Delta t$$ and $$\Delta \omega$$. I'm not sure it depends on how your text and/or instructor defines the uncertainty principle.

On the other hand, it could be very detailed. Here is a different interpretation of the uncertainty principle, more akin to its use in quantum mechanics (except without planks constant). This is different than the previous post, because the probability distribution function we will use is the magnitude squared of the particular function, rather than the function itself.

First Normalize the waveform. We'll call the normalization constant "A". Before calculating the standard deviation, we must make sure that the area under the curve of f*(x)f(x) is 1, when performing our calculations, where f*(x) is the complex conjugate of f(x).

$$1 = A \int _{-\infty} ^{\infty} f^*(x)f(x) dx$$

So evaluate the integral, and solve for A.

Then calculate the expectation value (mean).

$$\mu = A\int _{-\infty}^{\infty}x f^*(x)f(x) dx$$

Then the variation,

$$\sigma^2 = A\int _{-\infty}^{\infty}(x - \mu)^2 f^*(x)f(x)dx$$

Then finally get to the standard deviation by

$$\sigma = \sqrt{\sigma^2}$$

Using this approach it is guaranteed that $$\sigma _{\omega} \sigma _t$$ will always be greater than or equal to 1/2, no matter what the waveforms look like. This is more formal way of handling the uncertainty principle.

But again, it seems there are many ways to approach what is meant by $$\Delta t$$ and $$\Delta \omega$$. So you might be better off checking with your instructor on how he or she wants the uncertainty principle handled.