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:

enter image description here

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:



2 Answers 2


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.

  • \$\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\$ Nov 6, 2018 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.