# Frequency shift property of Fourier transform

Which one of the following is actually a definition of frequency shift property $$e^{j\omega_0 t}x(t)\leftrightarrow X(j\omega - \omega_0) \tag{1}$$ $$e^{j\omega_0 t}x(t)\leftrightarrow X(j(\omega - \omega_0)) \tag{2}$$

Is frequency shift applicaple in all the situations?

Case I
$$1 \leftrightarrow 2\pi \delta(\omega)$$ if we apply frequency shift property we may obtain $$e^{j\omega_0 t} \leftrightarrow 2\pi \delta(\omega -\omega_0)$$ which works according to result 2

Case II $$u(t) \leftrightarrow \frac{1}{j\omega} +\pi \delta(\omega)$$ $$e^{-at} u(t)\leftrightarrow \frac{1}{a+j\omega}$$ which exactly isn't 1 or 2

• (1) is not correct. – robert bristow-johnson Nov 30 '15 at 7:10
• @robertbristow-johnson why can't we apply this on $u(t)$ – mahes Nov 30 '15 at 8:02
• The Fourier transform pair $e^{at}u(t)\leftrightarrow \frac{1}{a+j\omega}$ is wrong; the given frequency domain function is the Fourier transform of $e^{-at}u(t)$. – Matt L. Nov 30 '15 at 8:32
• @MattL.ok, corrected it – mahes Nov 30 '15 at 9:15
• @user2332665, can't you see that right-hand side of (1) and (2) are different (while the left-hand side is the same)? they could both be wrong, but they can't both be correct. – robert bristow-johnson Nov 30 '15 at 20:32

Your Eq. $(2)$ is the frequency shift property; Eq. $(1)$ is wrong. This is easy to show:
\begin{align}\mathcal{F}\left\{e^{j\omega_0t}x(t)\right\}=&\int_{-\infty}^{\infty}x(t)e^{j\omega_0t}e^{-j\omega t}dt\\&=\int_{-\infty}^{\infty}x(t)e^{-j(\omega-\omega_0)t}dt\\&=X(j(\omega-\omega_0)) \end{align}
Your 'Case II' has nothing to do with the frequency shift property, because it is no frequency shift. For a frequency shift you need to multiply by a complex exponential $e^{j\omega_0t}$, and not by a real-valued exponential $e^{at}$.
You can come up with a rule for the Fourier transform of $x(t)e^{-at}$, $a\in\mathbb{R}$, but this is more tricky than the frequency shift property. For the frequency shift property, if you know that $X(j\omega)$ exists, then you know for sure that also the Fourier transform of $x(t)e^{j\omega_0t}$ exists. On the other hand, if $X(j\omega)$ exists, it is not certain that also the Fourier transform of $x(t)e^{-at}$ exists. This has everything to do with the Laplace transform and its region of convergence:
\begin{align}\mathcal{F}\left\{x(t)e^{-at}\right\}&=\int_{-\infty}^{\infty}x(t)e^{-at}e^{-j\omega t}dt\\&=\int_{-\infty}^{\infty}x(t)e^{-(a+j\omega)t}dt\stackrel{?}{=}X(a+j\omega)\end{align}
The last equality only holds if the integral converges, i.e. if $s=a$ is inside the region of convergence of the Laplace transform of $x(t)$, and if the Laplace transform of $x(t)$ has no singularities on the imaginary axis, i.e. if its Fourier transform has no Dirac delta impulses. The latter explains why the above 'rule' doesn't work for $x(t)=u(t)$ (as in your example).