# Practice Question on the Properties of the Continuous-time Fourier Transform

Let x(t) be a continuous time signal with Fourier transform ${\mathcal X} (\omega)$.

We have seen the time-shifting property of the Fourier transform:

${\mathcal F} \left( x(t-t_0) \right) = e^{-j \omega t_0} {\mathcal X} (\omega)$

which gives us an expression for the Fourier transform of $y(t)=x(t-t_0)$ in terms of ${\mathcal X} (\omega)$. Note that, to prove this property, one can proceed as follows:

\begin{align} {\mathcal F} \left( x(t-t_0) \right) &= \int_{-\infty}^\infty x(t-t_0) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega (u+t_0)} du, \text{ (letting }u=t-t_0), \\ &= e^{-j\omega t_0} \int_{-\infty}^\infty x(u) e^{-j\omega u} du \\ &= e^{-j\omega t_0} {\mathcal X} (\omega). \end{align}

Using a similar approach as above, derive an expression for the Fourier transform of y(t)=x(3t+7) in terms of ${\mathcal X} (\omega)$.

Instructor's hint: You should start like this

\begin{align} {\mathcal F} \left( x(3t+7) \right) &= \int_{-\infty}^\infty x(3t+7) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega \frac{(u-7)}{3}} \frac{du}{3}, \text{ (letting }u=3t+7), \\ &= ... \end{align}

\begin{align} {\mathcal F} \left( x(3t+7) \right) &= \int_{-\infty}^\infty x(3t+7) e^{-j\omega t} dt \\ &= \int_{-\infty}^\infty x(u) e^{-j\omega \frac{(u-7)}{3}} \frac{du}{3}, \text{ (letting }u=3t+7), \\ &= \frac{1}{3} \int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}}e^{\frac{7j\omega}{3}} du\\ &= \frac{e^{\frac{7j\omega}{3}}}{3}\int_{-\infty}^\infty x(u)e^{-\frac{j\omega u}{3}} du \\ &= \frac{e^{\frac{7j\omega}{3}}}{3} \mathcal{X}(\omega) \end{align} --Ekhall 11:45, 2 March 2011 (UTC)

Instructor's note: You should replace $\mathcal{X}(\omega)$ by $\mathcal{X}(\frac{\omega}{3})$.