MHB Discovering the Inverse Fourier Transform

  • Thread starter Thread starter Prove It
  • Start date Start date
  • Tags Tags
    Email
AI Thread Summary
The discussion centers on finding the Inverse Fourier Transform of the function F(ω) = -5(iω + 4) / (16 + (iω + 4)²). It is noted that Miko initially attempted to apply a shift incorrectly. The correct approach involves recognizing that the function can be expressed in terms of known Fourier Transform pairs from a reference table. After algebraic manipulation, the Inverse Fourier Transform is determined to be -5e^(-4t)e^(4it)H(t) + 5ie^(-4t)sin(4t)H(t). This solution is shared for Miko and others with similar inquiries.
Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
Miko asks:

What is the Inverse Fourier Transform of [math]\displaystyle \begin{align*} F(\omega) = \frac{-5(i\omega + 4)}{16+(i\omega + 4)^2} \end{align*}[/math]?
 
Last edited by a moderator:
Mathematics news on Phys.org
Miko asks:

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

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

[math]\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*}[/math]

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

[math]\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*}[/math]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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top