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I'm trying to use the duality property:

$$g(t) \leftrightarrow G(j\omega) \implies G(t) \leftrightarrow 2\pi g(-j\omega)$$

and the time-shifting property:

$$x(t-t_0) \leftrightarrow e^{-j\omega t_0}X(j\omega)$$

to prove the frequency shifting property: $$e^{j\omega_0t}X(t) \leftrightarrow \hat{X}(j(\omega-\omega_0))$$ (where $\hat X(t)$ is the fourier transform of $X(t)$ to avoid confusion).

Applying the duality property to the time-shifting property

$$e^{-j\omega_0 \tau}X(j\tau) \leftrightarrow 2\pi x(-(\omega - \omega_0))$$

Then replacing the RHS by $\hat X(j\omega) = 2\pi x(-j\omega)$:

$$e^{-j\omega_0 \tau}X(j\tau) \leftrightarrow \hat X(j(\omega-\omega_0))$$

Now replacing $j\tau$ with $t$ (s.t. $\tau = -jt$) we have:

$$e^{\omega_0 t}X(t) \leftrightarrow \hat X(j(\omega-\omega_0))$$

But now I've lost my $j$. What have I done wrong?

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  • $\begingroup$ a) you need to be careful with you Transform variables. The FT is defined between two variables $t$ and $\omega$ which can both be complex. Writing the frequency side as $X*(j \omega)$ is a somewhat questionable convention.. Try $x(t) \leftrightarrow X(\omega)$ instead. b) This one is easy to proof directly from the definition. so no need to derive it from other properties. $\endgroup$ Commented Aug 17, 2024 at 8:02

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You got the duality property wrong. Using $X(j\omega)$ to denote the Fourier transform of $x(t)$, the duality property is

$$x(t) \leftrightarrow X(j\omega) \implies X(jt) \leftrightarrow 2\pi x(-\omega)\tag{1}$$

Note that with this notation there's always a $j$ in the argument of $X(\cdot)$, and never in the argument of $x(\cdot)$, no matter if the independent variable is time or frequency. So you always just exchange the variables $t$ and $\omega$, possibly with a negative sign, but never with a multiplication of $j$.

Now you can show the modulation property as the dual of the time shifting property:

From

$$x(t-t_0)\leftrightarrow e^{-j\omega t_0}X(j\omega)\tag{2}$$

we obtain via $(1)$

$$e^{-j\omega_0 t}X(jt)\leftrightarrow 2\pi x(-\omega-\omega_0)=2\pi x(-(\omega+\omega_0))\tag{3}$$

If we use $\hat{X}(j\omega)$ to denote the Fourier transform of $X(jt)$ we obtain from $(1)$

$$\hat{X}(j\omega)=2\pi x(-\omega)\tag{4}$$

Plugging $(4)$ into $(3)$ we get

$$e^{-j\omega_0 t}X(jt)\leftrightarrow \hat{X}(j(\omega+\omega_0))\tag{5}$$

Of course we can change the sign of $\omega_0$ to obtain the following more common form of the modulation property:

$$e^{j\omega_0 t}X(jt)\leftrightarrow \hat{X}(j(\omega-\omega_0))\tag{6}$$

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