# Why is $e^{-j2\pi f ~t_0}$ instead of $e^{-j2\pi f ~(t - t_0)}$ for time shifting in the Fourier transform?

In signal processing and the study of Fourier transforms, the time-shifting property is a fundamental concept. This property describes how a shift in the time domain of a signal affects its Fourier transform in the frequency domain.

### Time-Shifting Property

If $$\ x(t) \$$ is a continuous-time signal with the Fourier transform $$\ X(f) \$$, then the time-shifted signal $$\ x(t - t_0) \$$ has the Fourier transform $$\ X(f) e^{-j2\pi f t_0} \$$.

Mathematically, if: $$x(t) \xrightarrow{\mathcal{F}} X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$$

then: $$x(t - t_0) \xrightarrow{\mathcal{F}} X(f) e^{-j2\pi f t_0}.$$

### Proof

To prove this property, we start with the definition of the Fourier transform of $$\ x(t - t_0) \$$:

$$x(t - t_0) \xrightarrow{\mathcal{F}} X(f) = \mathcal{F}\{x(t - t_0)\} = \int_{-\infty}^{\infty} x(t - t_0) e^{-j2\pi ft} dt$$

I think it should follow this:

$$x(t - t_0) \xrightarrow{\mathcal{F}} X(f) = \mathcal{F}\{x(t - t_0)\} = \int_{-\infty}^{\infty} x(t - t_0) e^{-j2\pi f (t - t_0)} dt$$

Where am I wrong?

• Snarky answer: Because t isn't a variable in the Fourier domain. (i.e. because X(f) simply doesn't depend on t and if you ever calculate X(f) and it looks like it does, that means you did something wrong) Commented May 20 at 14:52
• I adjusted the title to (I think) better reflect the content of the question. If I got it wrong, feel free to revert my edit. Commented May 20 at 14:54
• @ThePhoton What would fourier transform look like for $x(t^2 + t_0)$?
– kile
Commented May 20 at 16:30
• I don't know if there's any tabulated rule for that...depending what x(t) is you might be able to simplify it some other way. Commented May 20 at 18:51

Only $$\x(t)\$$ is being shifted, not the transform. By shifting the transform as well, the shift is essentially removed.

Let $$\t^{'}=t-t_0\$$. Edit: [Then the OP's last equation becomes (noting that $$\dt^{'}=dt\$$)] $$x(t^{'}) \xrightarrow{\mathcal{F}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft^{'}} dt^{'}=X(f)$$ not $$X(f) e^{-j2\pi f t_0}$$

as it should be

So understand that only $$\x(t)\$$ is being shifted, not the transform.

• if you replace $t'$ with $t - t_0$. $$x(t^{'}) \xrightarrow{\mathcal{F}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft^{'}} dt^{'}= \int_{-\infty}^{\infty} x(t - t_0) e^{-j2\pi f (t-t_0)} d (t- t_0)$$
– kile
Commented May 20 at 16:16
• So what is your concern? @kile Commented May 20 at 17:41
• Is my formula in the comment here correct?
– kile
Commented May 20 at 20:45

Let $$\t^{'}=t-t_0\$$. Then dt' = dt $$x(t^{'}) \xrightarrow{\mathcal{F}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft} dt^{'}$$

$$X^{'}(f) = \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft} dt^{'}$$ $$= \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi f (t^{'}+ t_{o})} dt^{'}$$ $$= \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi f (t_{0}} e^{-j2\pi f (t^{'}} dt^{'}$$ $$= e^{-j2\pi f t_{0}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft^{'}} dt^{'}$$

$$X^{'}(f)= e^{-j2\pi ft_{0}} X(f)$$ This proves the time shifting property of Fourier transform. Always remember that when you are doing fourier transform, whatever the function of t, you have to integrate from -∞ to ∞ with the multiplier $$\ e^{-j2\pi ft} \$$ inside the integral which is not $$\ e^{-j2\pi f(t-t_{0})} \$$.

• What would fourier transform look like for $x(t^2 - t_0)$? ?
– kile
Commented May 20 at 17:05
• It would be easier if you can tell what the function $x(t^2 - t_{0})$ is as an algebraic expression such as $3t^{2} +t - t_{0})$. Commented May 21 at 3:23
• Let's assume function is what you said. Can you proceed with this?
– kile
Commented May 21 at 7:05
• math.stackexchange.com/questions/2656200/… . This is a derivation for fourier transform of $t^{2}$. Commented May 21 at 7:56