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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

enter image description here

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?

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  • $\begingroup$ Reason for the downvote? $\endgroup$
    – KMC
    Oct 29 '21 at 2:47
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Your $X(\omega)$ is incorrect. It should use

$$e^{j2\times\pi\times 60 t} \leftrightarrow 2\pi \delta(\omega-2\times\pi\times 60).$$ See this table.

See this table for the same thing in $f$.

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  • $\begingroup$ $ X(\omega) $ has two spectral lines each with amplitude 0.5. How do I write it in terms of $ 2\pi \delta(\omega-2\pi 60 $? $$ X(\omega) = 0.5 \cdot 2\pi \delta(\omega -2\pi 60) \text{???} $$ But that would give me $$ 0.5 \cdot 2\pi \cdot 2 = 2\pi V $$ $ \approx $ six times the voltage magnitude of the original signal ($ 1V$) $\endgroup$
    – KMC
    Oct 28 '21 at 18:31
  • $\begingroup$ @KMC Sorry! My bad. I updated my answer to use the correct Fourier transform pair. Yours will have a positive and negative frequency exponential with a coefficient of 0.5. So each will generate a term of $\pi\delta(\omega-2\times\pi\times 60)$. $\endgroup$
    – Peter K.
    Oct 28 '21 at 19:06
  • $\begingroup$ I'm still not getting this ... so $$ x(t) = 0.5[\pi \delta(\omega-2 \pi 60)t] + 0.5[\pi \delta(-\omega-2\pi60)t] $$ does not look like $$ x(t) = \cos(2\pi 60t) $$ $\endgroup$
    – KMC
    Oct 29 '21 at 1:53
  • $\begingroup$ @KMC But it is precisely that. That's what the Fourier transform tables I've posted say. $$0.5[2\pi \delta(\omega-2 \pi 60)] + 0.5[2\pi \delta(-\omega-2\pi60)] \leftrightarrow \cos(2\pi 60t)$$. $\endgroup$
    – Peter K.
    Oct 29 '21 at 2:32
  • $\begingroup$ I always thought (and I was taught that) the height of the spectral line in $ X(\omega) $ IS the value of the amplitude of the time-domain sinusoid of a corresponding $ \omega $, but it is wrong. So if I want to correlated the scalar value of the transformed function, I really need to normalize height by dividing the value by $ 2\pi $. If I plot my spectral lines in $ X(\omega) $ I have two impulse each with height $ 0.5 \cdot 2\pi $. Whereas in $ X(f) $ their impulse height are exactly $ 0.5 $ which sum up to $ 1V $ for the sinusoid in the time domain. $\endgroup$
    – KMC
    Oct 29 '21 at 2:45

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