Fourier Transform with both Time Delay and Frequency Shift

Question

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?

Answer 1

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

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

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

Answer 2

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.