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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?

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  • \$\begingroup\$ 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) \$\endgroup\$
    – The Photon
    Commented May 20 at 14:52
  • \$\begingroup\$ 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. \$\endgroup\$
    – The Photon
    Commented May 20 at 14:54
  • \$\begingroup\$ @ThePhoton What would fourier transform look like for \$x(t^2 + t_0)\$? \$\endgroup\$
    – kile
    Commented May 20 at 16:30
  • \$\begingroup\$ 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. \$\endgroup\$
    – The Photon
    Commented May 20 at 18:51

2 Answers 2

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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.

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

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  • \$\begingroup\$ What would fourier transform look like for \$x(t^2 - t_0)\$? ? \$\endgroup\$
    – kile
    Commented May 20 at 17:05
  • \$\begingroup\$ 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}) \$. \$\endgroup\$
    – Amit M
    Commented May 21 at 3:23
  • \$\begingroup\$ Let's assume function is what you said. Can you proceed with this? \$\endgroup\$
    – kile
    Commented May 21 at 7:05
  • \$\begingroup\$ math.stackexchange.com/questions/2656200/… . This is a derivation for fourier transform of \$ t^{2} \$. \$\endgroup\$
    – Amit M
    Commented May 21 at 7:56

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