# Fourier Transform with both Time Delay and Frequency Shift

I know that the Fourier transform of a function with time delay can be written as: $$\mathscr{F}\big\{x(t-t_0)\big\}=X(f)e^{-j2\pi f t_0}$$

The Fourier transform of a function with frequency shift can also be written as: $$\mathscr{F}\Big\{x(t)e^{j2\pi f_0 t}\Big\}=X(f-f_0)$$

So what if we have both shift and delay at the time domain, what will be the result in the frequency domain? E.g.:

$$\mathscr{F}\Big\{x(t-t_0)e^{j2\pi f_0 (t-t_0)}\Big\}$$

Will the result be: $$X(f-f_0)e^{-j 2 \pi f (t-t_0)}$$

Also what will be the result of:

$$\mathscr{F}\Big\{x(t-t_0)e^{j2\pi f_0 t)}\Big\}$$

Is there an order to apply these properties?

• it's really nice to see new contributors use $\LaTeX$. thanks. – robert bristow-johnson Feb 5 at 20:37

As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:

$$\mathcal{F}\left\{x(t-t_0)e^{j2\pi f_0(t-t_0)}\right\}=\mathcal{F}\left\{x(t)e^{j2\pi f_0t}\right\}e^{-j2\pi ft_0}=X(f-f_0)e^{-j2\pi ft_0}$$

where $$X(f)$$ is the Fourier transform of $$x(t)$$.

And, for your second example, with $$\tilde{X}(f)=\mathcal{F}\{x(t-t_0)\}=X(f)e^{-j2\pi ft_0}$$ you get

$$\mathcal{F}\left\{x(t-t_0)e^{j2\pi f_0t}\right\}=\tilde{X}(f-f_0)=X(f-f_0)e^{-j2\pi (f-f_0)t_0}$$

Of course, recognizing that the function in the second example just equals the first function scaled by $$e^{j2\pi f_0t_0}$$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.

If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:

\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)}\right\} &= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)} e^{-j2\pi f t}dt\\ \\ &= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\ \\ &= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\ \\ &= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\ \\ &= e^{-j2\pi ft_0} X(f-f_0)\\ \\ \end{align*}

\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0 t}\right\} &= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0t} e^{-j2\pi f t}dt\\ \\ &= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0(\tau+t_0)} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\ \\ &= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\ \\ &= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\ \\ &= e^{-j2\pi (f-f_0)t_0} X(f-f_0)\\ \\ \end{align*}