# "Simple" Fourier transform problem

• schniefen
In summary, the question asks for an estimate for when a decaying exponential will be "small". They want an answer that is within 1% of "normal" or "maximum".

#### schniefen

Homework Statement
Consider the Fourier transform ##\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}##. Estimate a time ##t_m## above which you expect ##f(t)## to be small.
Relevant Equations
Unsure.
I am unsure about what is being asked for in the question. At first I thought the question asks one to calculate the inverse Fourier transform and then to analyze its depends on ##t##, however, the "estimate" makes me think otherwise.

If you do not recognize a way to easily estimate that number then you probably need ( for your edification) to work it out exactly. Are you happy in the complex plane?

• topsquark
Most Fourier Transform problems in homework (as opposed to real data) are based on recognizing standard forms (i.e. transform pairs of common functions) plus some properties of the transform, like the shifting and scaling properties. Does the general form of this transform look like something you've seen before?

OK, the printed book is old school (like me, LOL), but your life will be easier if you study this stuff ASAP. This is really what the question is about. • topsquark
DaveE said:
Most Fourier Transform problems in homework (as opposed to real data) are based on recognizing standard forms (i.e. transform pairs of common functions) plus some properties of the transform, like the shifting and scaling properties. Does the general form of this transform look like something you've seen before?

It kind of looks like something I've seen before. Consider the function ##f(t)=e^{-\gamma t}e^{-\omega_0 t}\Theta (t)##, where ##\gamma>0## and ##\Theta (t)## is the Heaviside step function. Then, Fourier transforming it using the convention
##\hat{f}(\omega)=\int_{-\infty}^{\infty} f(t)e^{i\omega t}\mathrm{d}t,##​
one obtains
##\hat{f}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}.##​
The magnitude of ##\hat{f}## is
##|\hat{f}(\omega)|=\frac{1}{(\omega-\omega_0)^2+\gamma^2}.##​

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• DaveE
Good! So that seems sufficient to estimate the answer. When is a decaying exponential ##\exp(-\gamma t )## "small" (hint: ##\gamma ^{-1} ##is called the "decay time"...) . They just want an estimate.

• schniefen and topsquark
hutchphd said:
Good! So that seems sufficient to estimate the answer. When is a decaying exponential ##\exp(-\gamma t )## "small" (hint: ##\gamma ^{-1} ##is called the "decay time"...) . They just want an estimate.
What I find confusing is though that in the problem they talk about the Fourier transform
##\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}##, whereas in the example I gave it is the magnitude of the Fourier transform that shows similarities.

Write your result with a real denominator and you will see that real part will look familiar. They asked for an "estimate" for when it will be "small". Some wiggle room there. So not a problem (there may be a factor of two lurking somewhere). But what does "small" mean for a complex valued function? Your point is well taken.

• schniefen
The progress on this problem so far is that I have identified the function in the problem, that is ##
\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}##, as almost the real part of the function

##\hat{g}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}.##​

I know the inverse Fourier transform of ##\hat{g}(\omega)## (as stated above), but not of ##\hat{f}(\omega)##. Hence I am unsure about estimating anything.
hutchphd said:
But what does "small" mean for a complex valued function?
I guess it means for the magnitude to be small.

Actually ##\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}## is the real part of

##\hat{g}(\omega)=\frac{2}{\gamma/2-i(\omega-\omega_0)}.##​

schniefen said:
The progress on this problem so far is that I have identified the function in the problem, that is ##
\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}##, as almost the real part of the function

##\hat{g}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}.##​

I know the inverse Fourier transform of ##\hat{g}(\omega)## (as stated above), but not of ##\hat{f}(\omega)##. Hence I am unsure about estimating anything.
Do you know how to calculate the inverse Fourier transform using contour integration?

schniefen said:
I guess it means for the magnitude to be small.
In my engineering world, I would have assumed "small" in this context would be <1% of "normal" or "maximum". But I have no idea what your instructor thinks small means. If you don't know, I would just tell them what your version of "small" is and then do the math to answer that.

• schniefen
vela said:
Do you know how to calculate the inverse Fourier transform using contour integration?
Unfortunately not.

DaveE said:
In my engineering world, I would have assumed "small" in this context would be <1% of "normal" or "maximum". But I have no idea what your instructor thinks small means. If you don't know, I would just tell them what your version of "small" is and then do the math to answer that.
The answer given is: ##\pi/\gamma## (where one can argue about an additional factor of 2 or so).

So what is the issue. This is a reasonable answer to an approximate question. I don't know why there is a pi in the answer but why not? Do you understand what you are doing? If not please ask a specific question.

hutchphd said:
So what is the issue. This is a reasonable answer to an approximate question. I don't know why there is a pi in the answer but why not? Do you understand what you are doing? If not please ask a specific question.
I do not understand how to estimate the function in the problem, that is

##\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}.##​

##\hat{g}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}##​

then I know the inverse Fourier transform of it, namely ##f(t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)##. This is small for large ##t## I would say. I know that ##\hat{f}(\omega)## is approximately the real part of ##\hat{g}(\omega)##, but they could have different Fourier transforms altogether, unless I am missing some property about Fourier transforms. As you may notice, my strategy is still to find out the inverse Fourier transform of ##\hat{f}(\omega)## somehow.

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schniefen said:
I do not understand how to estimate the function in the problem, that is

##\hat{f}(\omega)=\frac{\gamma}{(\omega-\omega_0)^2+\gamma^2/4}.##​

##\hat{g}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}##​

then I know the inverse Fourier transform of it, namely ##f(t)=e^{-\gamma t}e^{-\omega_0 t}\Theta (t)##. This is small for large ##t## I would say. I know that ##\hat{f}(\omega)## is approximately the real part of ##\hat{g}(\omega)##, but they could have different Fourier transforms altogether, unless I am missing some property about Fourier transforms. As you may notice, my strategy is still to find out the inverse Fourier transform of ##\hat{f}(\omega)## somehow.
First I believe you mean $$g(t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)$$ Notice the i in the exponent. Then $$\hat{g}(\omega)=\frac{1}{\gamma-i(\omega-\omega_0)}$$
And $$\hat f(\omega)=\frac 1 2 \left[ \hat{g}(\omega)+{\hat g}^*(\omega)\right]$$

But notice (following @DaveE above) that ##{\hat g}^*(\omega)## is the transform of g(-t) and so $$f(t)=\frac 1 2 \left[ g(t)+g(-t) \right]$$ This is the transform you desire. There is a factor of 2 floating around in gamma but I don't care. Otherwise correct I believe

The question is poorly worded IMHO.

• schniefen
Thank you @hutchphd , clarified it a lot. So the next task is to determine when ##g(t)+g(-t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)+e^{\gamma t}e^{i\omega_0 t}\Theta (-t)## is small, right? I would just say for large negative and positive ##t##s.

For large negative ##t##s, we have ##g(t)+g(-t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)+e^{\gamma t}e^{i\omega_0 t}\Theta (-t)## reducing to ##e^{\gamma t}e^{i\omega_0 t}##, which will be small due to ##e^{\gamma t}## (##\gamma## is positive). For large positive ##t##s, we have ##e^{-\gamma t}e^{-i\omega_0 t}##, which again will be small due to ##e^{-\gamma t}##.

Given that ##t## typically represents time and time generally moves in one direction, the word above in the problem statement would rule out negative ##t##'s, right?

• schniefen
vela said:
Given that ##t## typically represents time and time generally moves in one direction, the word above in the problem statement would rule out negative ##t##'s, right?
Maybe the words imply that. But the transform function is clear and it defines t<0 in the domain. I'll always choose equations over words.

Negative time is maybe the strangest thing about these transforms. It takes getting used to translating from math to the Real world.

vela said:
Given that ##t## typically represents time and time generally moves in one direction, the word above in the problem statement would rule out negative ##t##'s, right?
If ##t## is only positive, we are left with ##g(t)+g(-t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)=g(t)##, which is small for large ##t##. Anyway, this is a bit confusing. I do not understand why the answer is given in terms of ##\gamma^{-1}##.

schniefen said:
I do not understand why the answer is given in terms of γ−1.
This is a common convention in the engineering world, where we tend to normalize things to the circuit "time constant", which is 1/γ, in this case. In this way, we don't have to worry about whether you're system is designed for the 455KHz, 250MHz, or 2.2GHz bands. This allows me to recall from memory that a 10%-90% step response is the same as 2.2 time constants in the exponential, or that for <1% settling, I need 5 time constants.

This is because there really isn't a fundamental difference between a filter designed for 1GHz or 2.2 GHz. A pendulum swinging with a 1Kg weight isn't really different from one with a 2Kg weight. First we learn to solve the general problem, then the last step in the analysis/design is to figure out the values needed for a particular application.

• schniefen and hutchphd
schniefen said:
If ##t## is only positive, we are left with ##g(t)+g(-t)=e^{-\gamma t}e^{-i\omega_0 t}\Theta (t)=g(t)##, which is small for large ##t##. Anyway, this is a bit confusing. I do not understand why the answer is given in terms of ##\gamma^{-1}##.
The question is what do you mean by large? Is ##t=86400~\rm s## large? It's not if ##1/\gamma = 10^{10}~\rm s##, but it is if ##1/\gamma=1~\rm s##. ##\gamma## sets the time scale of the exponential decay.

• schniefen