# CTFT of $X(j\omega)$ vs $X(jf)$

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?

• Reason for the downvote?
– KMC
Oct 29 '21 at 2:47

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 $$f$$.

• $X(\omega)$ 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 ($1V$)
– KMC
Oct 28 '21 at 18:31
• @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 $\pi\delta(\omega-2\times\pi\times 60)$.
– Peter K.
Oct 28 '21 at 19:06
• 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 $$x(t) = \cos(2\pi 60t)$$
– KMC
Oct 29 '21 at 1:53
• @KMC But it is precisely that. That's what the Fourier transform tables I've posted say. $$0.5[2\pi \delta(\omega-2 \pi 60)] + 0.5[2\pi \delta(-\omega-2\pi60)] \leftrightarrow \cos(2\pi 60t)$$.
– Peter K.
Oct 29 '21 at 2:32
• 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.
– KMC
Oct 29 '21 at 2:45