Properties of the Fourier Transform

Sources:

1. B. P. Lathi & Roger Green. (2018). Chapter 7: Continuous-Time Signal Analysis: The Fourier Transform. Signal Processing and Linear Systems (3nd ed., pp. 701-720). Oxford University Press.

For a quick reference table, see Wikipedia page on the Fourier transform.

Linearity

The Fourier transform is linear; that is, if $$x_1(t)⟺X_1(\omega )\text{ and }{x}_2(t)⟺X_2(\omega )$$ then $$a_1{x}_1(t)+a_2{x}_2(t)⟺a_1{X}_1(\omega )+a_2{X}_2(\omega )$$

The proof is trivial.

conjugation property

The conjugation property states that if $x(t)⟺X(\omega )$, then $$x^{\ast }(t)⟺X^{\ast }(-\omega )$$

From this property follows the conjugate symmetry property, also introduced earlier, which states that if $x(t)$ is real, then $$X(-\omega )=X^{\ast }(\omega )$$

Duality

The duality property states that if $$x(t)⟺X(\omega )$$ then $$\begin{array}{}\text{(1)}& X(t)⟺2\pi x(-\omega )\end{array}$$

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Proof:

From the deginition of Fourier transform, we can write $$x(t)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X(u)e^{jut}du$$

Hence, $$2\pi x(-t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X(u)e^{-jut}du$$

Changing $t$ to $\omega$ yields $\text{(1)}$

The scaling property

If $$x(t)⟺X(\omega )$$ then, for any real constant $a$, $$\begin{array}{}\text{(2)}& x(at)⟺\frac{1}{|a|}X\left(\frac{\omega }{a}\right)\end{array}$$

Proof. For a positive real constant $a$, $$\mathcal{F}[x(at)]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(at)e^{-j\omega t}dt=\frac{1}{a}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(u)e^{(-j\omega /a)u}du=\frac{1}{a}X\left(\frac{\omega }{a}\right)$$

Similarly, we can demonstrate that if $a<0$, $$x(at)⟺\frac{-1}{a}X\left(\frac{\omega }{a}\right)$$

Hence follows $\text{(2)}$.

The time-shifting property

If $$x(t)⟺X(\omega )$$ then $$x(t-t_0)⟺X(\omega )e^{-j\omega t_0}$$

Proof. By definition, $$\mathcal{F}\left[x(t-t_0)\right]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t-t_0)e^{-j\omega t}dt$$

Letting $t-t_0=u$, we have $$\mathcal{F}\left[x(t-t_0)\right]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(u)e^{-j\omega (u+t_0)}du=e^{-j\omega t_0}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(u)e^{-j\omega u}du=X(\omega )e^{-j\omega t_0}$$

This result shows that delaying a signal by $t_0$ seconds does not change its amplitude spectrum. The phase spectrum, however, is changed by $-\omega t_0$.

The frequency-shifting property

Figure 7.25

If $$x(t)⟺X(\omega )$$ then $$x(t)e^{j{\omega }_{0}t}⟺X(\omega -{\omega }_0)$$

Proof

By definition, $$\mathcal{F}[x(t)e^{j{\omega }_{0}t}]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{j{\omega }_{0}t}{e}^{-j\omega t}dt={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j(\omega -{\omega }_0)t}dt=X(\omega -{\omega }_0)$$

According to this property, the multiplication of a signal by a factor $e^{j{\omega }_{0}t}$ shifts the spectrum of that signal by $\omega ={\omega }_0$. Note the duality between the time-shifting and the frequency-shifting properties. Changing ${\omega }_0$ to $-{\omega }_0$ yields $$x(t)e^{-j{\omega }_{0}t}⟺X(\omega +{\omega }_0)$$

Because $e^{j{\omega }_{0}t}$ is not a real function that can be generated, frequency shifting in practice is achieved by multiplying $x(t)$ by a sinusoid. Observe that $$x(t)\cos {\omega }_{0}t=\frac{1}{2}[x(t)e^{j{\omega }_{0}t}+x(t)e^{-j{\omega }_{0}t}]$$

we obtain $$x(t)\cos {\omega }_{0}t⟺\frac{1}{2}[X(\omega -{\omega }_0)+X(\omega +{\omega }_0)]$$

This result shows that the multiplication of a signal $x(t)$ by a sinusoid of frequency ${\omega }_0$ shifts the spectrum $X(\omega )$ by $\pm {\omega }_0$, as depicted in Fig. 7.25.