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So, I watched a video from MIT Signals and Systems Open course on youtube, and there is a moment when he substitutes \$ x(t)e^{jw_ct} \$ for y(t), and then:

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Shouldn't it be just \$ X(w - w_c) \$ ? The theorem about the shift in the frequency domain says so. So why there is an "j"?

Youtube video with the exact moment:

https://youtu.be/OT04cEdpK-M?t=759

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I too have struggled with this notation. It seems to defy logic that a j would appear in the brackets, it almost appears that j is an argument as opposed to just a number. Regardless, this notation is seen frequently in the literature so try to get used to it. It almost seems like they are taking the Laplace transform and evaluating it at s=jw which would make sense. I don't know exactly the reason for the different notations. Traditionally, as a function, the only argument is w and not jw so in that sense the j shouldn't be there. But just try to get used to it as it is written like this a lot.

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  • \$\begingroup\$ The argument of \$Y(j\omega)\$ indeed contains \$j\$ so it's no surprise that some \$j\$ also appears in the argument of the transform. Must be some kind of convention. \$\endgroup\$ – joe electro Nov 6 '18 at 4:43
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I think it's consistency with the Laplace transform that uses \$s=\sigma+j\:\omega\$ and \$F_s=\int f_t\: e^{-s\:t}\:\text{d}t\$. Note the absence of \$j\$ (or \$i\$) in the Laplace transform? That's because it's part of \$s\$, by convention.

I don't know if you recall, but complex domain multiplication involves two actions, scaling and rotation, in a single operation. In electronics, you are often not interested in scaling but only rotation (the frequency part.) So \$\sigma=0\$, by fiat. This leaves only the \$j\:\omega\$ part remaining in \$s\$.

So the Laplace transform is reduced to \$F_{j\:\omega}=\int f_t\: e^{-j\:\omega\:t}\:\text{d}t\$, by simple substitution of \$s=0+j\:\omega\$, where \$\sigma=0\$.

Therefore, to remain consistent and true to the Laplace notation, it must be the case that \$F_{j\left(\omega-\omega_c\right)}=\int f_t\:e^{-j\left(\omega-\omega_c\right)t}\:\text{d}t\$. Or, put another way, if \$y_t=x_t\:e^{\:j\:\omega_c\:t}\$ and when using Laplace notation, you get:

$$\begin{align*} Y_{j\:\omega}&=\int y_t\:e^{-j\:\omega\:t}\:\text{d}t\\\\ &=\int x_t\:e^{\:j\:\omega_c\:t}\:e^{-j\:\omega\:t}\:\text{d}t\\\\ &=\int x_t\:e^{\:j\:\left(\omega_c-\omega\right)\:t}\:\text{d}t\\\\ &=\int x_t\:e^{-j\:\left(\omega-\omega_c\right)\:t}\:\text{d}t \end{align*}$$

The only possible consistent notation for this, in keeping with Laplace notation of \$F_s=\int f_t\: e^{-s\:t}\:\text{d}t\$ and where \$\sigma=0\$, would be:

$$\begin{align*} Y_{j\:\omega}&=X_{j\:\left(\omega-\omega_c\right)} \end{align*}$$

Fourier notation (I think of it as a subset of Laplace) is often expressed a little differently, implying \$j\$ (or \$i\$), without making it explicit in the parameter. But that's just a different convention for Fourier notation.

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