Discovering the Inverse Fourier Transform

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In summary, Miko asked about the inverse Fourier Transform of a given function. After some algebraic manipulation, it was determined that the function is a combination of two functions found in a table of Fourier Transforms. Thus, the inverse Fourier Transform can be written as -5e^(-4t)e^(4it)H(t) + 5ie^(-4t)sin(4t)H(t).
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Miko asks:

What is the Inverse Fourier Transform of \(\displaystyle \displaystyle \begin{align*} F(\omega) = \frac{-5(i\omega + 4)}{16+(i\omega + 4)^2} \end{align*}\)?
 
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Miko asks:

What is the Inverse Fourier Transform of \(\displaystyle \displaystyle \begin{align*} F(\omega) = \frac{-5(i\omega + 4)}{16+(i\omega + 4)^2} \end{align*}\)?

Miko has sent me a scan with a great attempt of this question, but has tried to use a shift when a shift is not necessary.

From a table of Fourier Transforms, we can see that \(\displaystyle \displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{a + i(\omega + k)}{k^2 + (a + i\omega)} \right\} = e^{-at}e^{ikt}H(t) \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{k}{k^2 + (a + i\omega)^2} \right\} = e^{-at}\sin{(kt)} H(t) \end{align*}\). The function given has an identical denominator (with \(\displaystyle \displaystyle \begin{align*} a = 4 \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} k = 4 \end{align*}\)) and so it would suggest that the function given is a combination of the functions given in the tables. So doing some algebraic manipulation...

\(\displaystyle \displaystyle \begin{align*} F(\omega) &= \frac{-5(i\omega + 4)}{16 + (i\omega + 4)^2} \\ &= \frac{-5i\omega - 20}{16 + (i\omega + 4)^2} \\ &= \frac{-5i(\omega + 4) + 20i - 20}{16 + (i\omega + 4)^2} \\ &= -5 \left[ \frac{4 + i(\omega + 4)}{16 + (i\omega + 4)^2} \right] + 5i \left[ \frac{4}{16 + (i\omega + 4)^2} \right] \end{align*}\)

which is now in the forms given in the tables. Therefore

\(\displaystyle \displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{-5(i\omega + 4)}{16 + (i\omega + 4)^2} \right\} &= \mathcal{F}^{-1} \left\{ -5 \left[ \frac{4 + i(\omega + 4)}{16 + (i\omega + 4)^2} \right] + 5i \left[ \frac{4}{16 + (i\omega + 4)^2} \right] \right\} \\ &= -5e^{-4t}e^{4it}H(t) + 5i\,e^{-4t}\sin{(4t)}H(t) \end{align*}\)For Miko and anyone else viewing this and having similar questions, I invite you to read and post in the http://www.mathhelpboards.com/f16/ subforum.
 

Related to Discovering the Inverse Fourier Transform

1. What was Miko's question via email?

Miko's question via email was: "What is the chemical formula for water?"

2. Why did Miko ask this question?

Miko asked this question because they are conducting an experiment and need to know the chemical formula for water in order to accurately record and analyze their results.

3. Is Miko a student or a professional?

This information was not provided in the email. However, based on the nature of the question, it is possible that Miko is a student studying science or a professional working in a scientific field.

4. How did you respond to Miko's question?

I responded to Miko's question by providing the chemical formula for water, which is H2O. I also offered to provide additional information or resources if needed.

5. Did Miko have any follow-up questions?

No, Miko did not have any follow-up questions. They simply replied with a thank you for the quick response.

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