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I learned Fourier transforms completely in $ f $ and never needed to convert to $ \omega $ until now, and I'm very confused on the translation between variables specifically in the context of the Fourier transform table of rules. I understand that $ \delta(ay) = \frac{1}{|a|} \delta(y)$ and that $ \delta(\omega) = \delta(2\pi f) = \frac{1}{2\pi} \delta(f)$. However, I'm having trouble actually applying this principle, especially in the cases of rectangular and triangular functions. For example, given this triangular function in the frequency ($ \omega $) domain:

$ H(\omega) = \begin{cases} 1 - \frac{T}{\pi}|\omega|, & |\omega| \leq \frac{\pi}{T}, \\ 0, & |\omega| > \frac{\pi}{T}. \end{cases} $

If I convert the bounds to $ f $ and apply the transformation rule $\Delta \left( \frac{f}{\tau} \right) = \frac{\tau}{2} \cdot \text{sinc}^2 \left( \frac{\pi t \tau}{2} \right)$ I get the time domain solution of $ h(t) = \frac{1}{2 \tau} \cdot \text{sinc}^2 \left( \frac{\pi t}{2 \tau} \right)$. However, if I leave the bounds in terms of $ \omega $ and apply the transformation rule $\Delta \left( \frac{\omega}{\tau} \right) = \frac{\tau}{2} \cdot \text{sinc}^2 \left( \frac{ t \tau}{4} \right)$, I get the time domain solution as $ h(t) = \frac{\pi}{\tau} \cdot \text{sinc}^2 \left( \frac{\pi t}{2 \tau} \right)$. I can see clearly that the former solution is just the latter solution divided by $ 2\pi $, but I don't understand why. The Fourier transform definitions I am using are: Fourier Transforms

And the “shortcut” formulas I’m referring to are: FT of triangle FT of triangle

Applying both the definitions of the inverse FT gives me a time domain solution of $ h(t) = \frac{1}{2 \tau} \cdot \text{sinc}^2 \left( \frac{\pi t}{2 \tau} \right)$. However, applying the $ f $ “shortcut rule” gives me this same answer, whereas applying the $ \omega $ “shortcut rule” gives me $ h(t) = \frac{\pi}{\tau} \cdot \text{sinc}^2 \left( \frac{\pi t}{2 \tau} \right)$. I don’t understand why this is.

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  • $\begingroup$ Perhaps formulas (15) and (16) at Wolfram MathWorld: FourierTransform will provide insight with respect to your confusion. You need to clarify the specific values you're using for the Fourier parameters $a$ and $b$. $\endgroup$ Commented Oct 20, 2024 at 4:53

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Assuming the alternate definition of the triangle function (which has a base width of $1$ instead of $2$)

$$\Delta(t)=\left\{\begin{array}{cc} 1-2 |t| & |t| \leq \frac{1}{2} \\ 0 & |t| > \frac{1}{2} \\ \end{array}\right.\tag{1}$$

your linked shortcut formula

$$\hat{\Delta}_{1,\tau}(f)=\frac{\tau}{2}\, \text{sinc}\left(\frac{\pi \tau f}{2}\right)^2\tag{2}$$

seems consistent with your linked definition of the Fourier transform

$$\hat{g}_1(f)=\mathcal{F}_{1, t}[g(t)](f)=\int\limits_{-\infty}^{\infty} g(t)\, e^{-i 2 \pi f t} \, dt\tag{3}$$

with corresponding inverse Fourier transform

$$g(t)=\mathcal{F}_{1, f}^{-1}[\hat{g}_1(f)](t)=\int\limits_{-\infty}^{\infty} \hat{g}_1(f)\, e^{i 2 \pi t f} \, df\tag{4}.$$

Likewise your linked shortcut formula

$$\hat{\Delta}_{2,\tau}(\omega)=\frac{\tau}{2}\, \text{sinc}\left(\frac{\tau \omega}{4}\right)^2\tag{5}$$

seems consistent with your linked definition of the Fourier transform

$$\hat{g}_2(\omega)=\mathcal{F}_{2, t}[g(t)](f)=\int\limits_{-\infty}^{\infty} g(t)\, e^{-i \omega t} \, dt\tag{6}$$

with corresponding inverse Fourier transform

$$g(t)=\mathcal{F}_{2, \omega}^{-1}[\hat{g}_2(\omega)](t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} \hat{g}_2(\omega)\, e^{i t \omega} \, d\omega\tag{7}.$$

Note that $\hat{g}_1(f)$ in formula (3) above is equivalent to $\hat{g}_2(2 \pi f)$ in formula (6) above, and consequently evaluating the right-side of formula (7) above with the substitution $\omega\to 2 \pi f$ is equivalent to the right-side of formula (4) above (since the $2 \pi$ factor in $d\omega=2 \pi\, df$ cancels the $\frac{1}{2 \pi}$ factor at the beginning of the right-side of formula (7) above).

So the relationship between your two shortcut formulas (2) and (5) above is

$$\hat{\Delta}_{1,\tau}(f)=\hat{\Delta}_{2,\tau}(2 \pi f)=\frac{\tau}{2}\, \text{sinc}\left(\frac{\pi f \tau}{2}\right)^2\tag{8}$$

Your function

$$H(\omega)=\left\{\begin{array}{cc} 1-\frac{T | \omega | }{\pi } & -\frac{\pi }{T}\leq \omega \leq \frac{\pi }{T} \\ 0 & \text{True} \\ \end{array}\right.\tag{9}$$

is equivalent to $\Delta\left(\frac{T \omega}{2 \pi}\right)$ and setting $\tau=\frac{2 \pi}{T}$ in formulas (2) and (5) above results in

$$\hat{H}_1(f)=\hat{\Delta}_{1,\frac{2 \pi}{T}}(f)=\frac{\pi\, \text{sinc}\left(\frac{\pi^2 f}{T}\right)^2}{T}\tag{10}$$

$$\hat{H}_2(\omega)=\hat{\Delta}_{2,\frac{2 \pi}{T}}(\omega)=\frac{\pi\, \text{sinc}\left(\frac{\pi \omega}{2 T}\right)^2}{T}\tag{11}$$

which are consistent with the Fourier transforms of $H(t)$ evaluated via formulas (3) and (6) above. Note formula (10) above is equivalent to formula (11) above evaluated at $\omega=2 \pi f$.

But you're asking about inverse Fourier transforms whereas the shortcuts in formulas (2) and (5) above are for Fourier transforms. Its possible to evaluate the inverse Fourier transforms using formulas (2) and (5) above because formulas (3) and (4) above, and likewise the integrals in formulas (6) and (7) above, exhibit a symmetry for even functions such as $H(\omega)$ which have an even Fourier transform.

Formula (2) above can be used to evaluate the inverse Fourier transform of $H(2 \pi f)$ as

$$\mathcal{F}_{1, 2 \pi f}^{-1}[H(2 \pi f)](t)=\hat{\Delta}_{1,\frac{1}{T}}(t)=\frac{\text{sinc}\left(\frac{\pi t}{2 T}\right)^2}{2 T}\tag{12}$$

Formula (5) above can be used to evaluate the inverse Fourier transform of $H(\omega)$ as

$$\mathcal{F}_{2, \omega}^{-1}[H(\omega)](t)=\frac{1}{2 \pi} \hat{\Delta}_{2,\frac{2 \pi}{T}}(t)=\frac{\text{sinc}\left(\frac{\pi t}{2 T}\right)^2}{2 T}\tag{13}$$

which accounts for the $\frac{1}{2 \pi}$ factor preceding the integral in formula (7) above.

Note the results in formulas (12) and (13) above are consistent.

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  • $\begingroup$ I edited my post to include the definitions of the Fourier transforms I’m using; as far as I can tell they align with what you’re saying in that the omega version of the IFT has a 1/2pi whereas the f version does not. I guess my confusion mainly lies in the “shortcut formulas” that I added in my edit; I don’t understand why they don’t both lead to the same time domain result as they’re given. $\endgroup$ Commented Oct 20, 2024 at 19:49
  • $\begingroup$ @arim I think the problem is you're using two different and non-equivalent definitions of the Fourier/inverse Fourier transform pairs. Note that $$g(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} G(\omega) e^{i t \omega} \, d\omega=\int\limits_{-\infty}^{\infty} G(2 \pi f)\, e^{i t 2 \pi f} \, df\neq \int\limits_{-\infty}^{\infty} G(f)\, e^{i t 2 \pi f} \, df.$$ $\endgroup$ Commented Oct 20, 2024 at 20:38
  • $\begingroup$ @arim The equivalencies $$g(t)=\mathcal{F}_{\omega}^{-1}[\mathcal{F}_{t}[g(t)](\omega)](t)$$ and $$G(\omega)=\mathcal{F}_{t}[\mathcal{F}^{-1}_{\omega}[G(\omega)](t)](\omega)$$ are only valid when you use the same Fourier parameters $a$ and $b$ to evaluate both the Fourier transform $$G(\omega)=\mathcal{F}_{t}[g(t)](\omega)=\sqrt{\frac{|b|}{(2 \pi)^{1-a}}} \int\limits_{-\infty}^{\infty} g(t)\, e^{i b \omega t}\, dt$$ and the inverse Fourier transform $$g(t)=\mathcal{F}_{\omega}^{-1}[G(\omega)](t)=\sqrt{\frac{|b|}{(2 \pi )^{a+1}}} \int_{-\infty}^{\infty} G(\omega ) e^{-i b t \omega}\, d\omega.$$ $\endgroup$ Commented Oct 20, 2024 at 20:41
  • $\begingroup$ This is good to know, I didn’t realize about the false equivalency before. However, I’m still confused because assuming the tables I took the “shortcut formulas” from correctly derived their formulas from their respective definitions of the Fourier transform, shouldn’t the final answer of the time domain solutions still match? Also, my professor has said that the 1/2T solution was the correct answer, but it seems you’re saying that the $\pi$/T solution is the correct answer.. which solution should I be using as reference to sort out my confusion? $\endgroup$ Commented Oct 20, 2024 at 22:03

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