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Multiplication in the time domain corresponds to convolution in the frequency domain:

$$ f(t) \cdot x(t) \iff F(j \omega) * X( j \omega) \tag*{No scaling factor} $$ I know the fourier transform of the dirac comb is: $$ \mathcal{F} \big \{ \text{III}_{T_{s}} (t) \big \} = \omega_{s} \cdot \text{III}_{T_{s} } (j \omega) $$ But according to oppenheim and others:

$$ \mathcal{F} \big \{ f(t) \cdot \text{III}_{T_{s}} (t) \} = \underbrace{\dfrac{1}{T_{s}}}_{\text{scaling factor}} \cdot \displaystyle \sum_{k \to - \infty}^{ k \to \infty} X(j( \omega - k \omega_{s} )) \text{ (Scaling factor)} $$ So my question is, for the above, why does the scaling factor not remain $$ \omega_{s} $$ why does it become $$ \dfrac{1}{T_{s}} $$ Which means that $$ x(t) \cdot y(t) \iff \dfrac{1}{2 \pi} X(j \omega) * Y(j \omega) \tag{ ??? } $$ Why does the scaling factor not remain the sampling angular velocity?

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    $\begingroup$ That's really only up to your definition of the Fourier transform. some authors opt for minimum numbers of magic factors in the Fourier transform and its inverse, some opt for unitarity with the square root of that prefactor applied to both the forward and inverse Fourier transform, some only apply it to either. $\endgroup$ – Marcus Müller May 1 at 20:57
  • $\begingroup$ Hi, thanks, Ok so if I use this definition with the scaling factor as $$ \omega_{s} $$ I cant find this form of the frequency domain of the sampled signal. $\endgroup$ – Nash Brewer May 1 at 21:00
  • $\begingroup$ The version that we were taught is: $$ \mathcal{F} \{ f(t) \} = \displaystyle \int_{- \infty}^{\infty} f(t) \cdot e^{-j(\omega t)} \,\,\, \text{d}t $$ and the inverse as: $$ f(t) = \dfrac{1}{2 \pi} \cdot \displaystyle \int_{-\infty}^{\infty} X(j \omega) e^{j(\omega t)} \,\,\, \text{d}\omega \tag{Scaling factor here} $$ SO i assume that oppenheim is using the definition of the fourier transform as: $\endgroup$ – Nash Brewer May 1 at 21:02
  • $\begingroup$ SO i assume that oppenheim is using the definition of the fourier transform as: $$ X(j \omega) = \dfrac{1}{2 \pi} \displaystyle \int_{ - \infty}^{ \infty} f(t) e^{- j( \omega t) } \,\,\,\, \text{d}t $$ $\endgroup$ – Nash Brewer May 1 at 21:08
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Your first formula is wrong. If you use angular frequency ($\omega$), then multiplication in the time domain corresponds to $1/(2\pi)$ times convolution in the frequency domain, just as in the last formula in your question:

$$\mathcal{F}\{x(t)\cdot y(t)\}=\frac{1}{2\pi}X(j\omega)\star Y(j\omega)\tag{1}$$

If you use frequency $f=\omega/(2\pi)$, then you get

$$\mathcal{F}\{u(t)\cdot v(t)\}=U(f)\star V(f)\tag{2}$$

where I've used the definitions

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{3}$$

$$U(f)=\int_{-\infty}^{\infty}u(t)e^{-j2\pi f t}dt\tag{4}$$

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  • $\begingroup$ Only thing I might add is this is the very reason i prefer the "ordinary frequency, $f$" definition of the continuous-time Fourier Transform. but, for reasons that don't seem consistent, i prefer the "angular frequency, $\omega$" definition of the DTFT and keep it consistent with the Z-Transform. if i do the angular frequency $\Omega$ definition for the continuous-time FT, then i would keep it consistent with the Laplace Transform as Matt has above with $X(j\Omega)$ (but i would use the capital "$\Omega$" and leave "$\omega$" for normalized radian frequency in the DTFT). $\endgroup$ – robert bristow-johnson May 1 at 21:30

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