The CTFT of a signal as a function of $ \omega $ and $ f $ is identical:
$$ X(j\omega) = \int_{-\infty}^{\infty} x(t) \, e^{-j \omega t} \operatorname{dt} \;\;\;\;\bigg|\;\;\;\; X(j f) = \int_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \operatorname{dt} $$
But the Inverse CTFT has a scaling factor of $ 1 / (2 \pi) $ if $ f(t) $ is transformed from $ X(\omega) $:
$$ f(t) = {1 \over {2\pi}} \int_{-\infty}^{\infty} X(j\omega) \, e^{j \omega t} \operatorname{d\omega} \;\;\;\;\bigg|\;\;\;\; f(t) = \int_{-\infty}^{\infty} X(j f) \, e^{j 2 \pi f t} \operatorname{df} $$
1. What is the conceptual implication of having the scaling factor?
Say if if I have a signal that is a pure sinusoid ( $ 1V $ in voltage amplitude and $ 60 \operatorname{Hz} $ in frequency):
$$ x(t) = \cos(2\pi \cdot 60t) $$
The spectrums between $ X(j\omega) $ and $ X(jf) $ is practically identical
If I convert the $ X(\omega)$ and $ X(f) $ spectrum back into its time domain function $ f(t) $, I yield different results. I can reconstruct $ x(t) $ from $ X(jf) $ but there is this (attenuating) scaling factor of $ 1 / (2\pi) $ from $ X(j\omega) $:
$$ \begin{align} \text{from } X(\omega) \implies f(t) &= \frac{1}{2\pi} \big[ 0.5e^{j2\pi60t} + 0.5e^{-j2\pi60t} \big] = \frac{1}{2\pi} \big[ 0.5\cos(2\pi60t) + 0.5\cos(2\pi60t) \big] \\ &= \frac{1}{2\pi} \cdot \cos(2 \pi 60t) \neq x(t) \\ &\implies 0.16V \text{ in voltage amplitude and frequency of } 60 \operatorname{Hz} \end{align} $$
whereas,
$$ \begin{align} \text{from } X(f) \implies f(t) &= 0.5e^{j2\pi60t} + 0.5e^{-j2\pi60t} = 0.5\cos(2\pi60t) + 0.5\cos(2\pi60t) \\ &= \cos(2 \pi 60t) = x(t) \end{align} $$
2. Does that imply that I have to normalized whichever value I read on a $ X(\omega) $ vs $ \omega $ plot by multiplying it by $ 2\pi $ or else the reconstructed voltage amplitude would be lower than what it really is?
