Relation between the Fourier transform of Y and X

So, I watched a video from MIT Signals and Systems Open course on youtube, and there is a moment when he substitutes $$\ x(t)e^{jw_ct} \$$ for y(t), and then: Shouldn't it be just $$\ 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

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.

• 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. – joe electro Nov 6 '18 at 4:43

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.