Fourier Transform Properties

Last section we calculated some fourier transform. Now we can have some useful properties.

Let's put the formula here,

$$F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-j \omega t} \mathrm{d}t$$

And inversely,

$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(\omega) e^{j \omega t} \mathrm{d}\omega$$

Linear

Because, integrate is linear, thus fourier transform is linear.

That is to say,

$$af(t) + bg(t) \xrightarrow{\mathcal{F}} aF(\omega) + bG(\omega)$$

Duality

If we assume,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$

Let's consider the fourier transform of $F(t)$.

$$F^*(\omega) = \int_{-\infty}^{+\infty} F(t) e^{-j \omega t} \mathrm{d}t$$

We also know that,

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} F(\omega) e^{j \omega t} \mathrm{d} \omega$$

We can switch $\omega$ and $t$,

$$f(\omega) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} F(t) e^{-j \omega t} \mathrm{d}t$$

Then we replace $\omega$ with $-\omega$.

$$f(-\omega) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} F(t) e^{j \omega t} \mathrm{d}t$$

Now look at,

$$F^*(\omega) = \int_{-\infty}^{+\infty} F(t) e^{-j \omega t} \mathrm{d}t$$

We can conclude that,

$$F^*(\omega) = 2\pi f(-\omega)$$

In all, if,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$

Then,

$$F(t) \xrightarrow{\mathcal{F}} 2\pi f(-\omega)$$

We've seen a real case before, that is,

$$\delta(t) \xrightarrow{\mathcal{F}} 1$$

And

$$1 \xrightarrow{\mathcal{F}} 2\pi \delta(\omega)$$

Shifting Time is Shifting Phase

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$

Then,

$$\int_{-\infty}^{+\infty} f(t - t_0) e^{-j\omega t} \mathrm{d} t \\ = e^{-j\omega t_0} \int_{-\infty}^{+\infty} f(t - t_0) e^{-j\omega (t-t_0)} d(t - t_0) \\ = e^{-j\omega t_0} F(\omega)$$

Or,

$$f(t-t_0) \xrightarrow{\mathcal{F}} e^{-j\omega t_0} F(\omega)$$

Shifting Phase is Shifting Frequency

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$ $$\int_{-\infty}^{+\infty} f(t) e^{-j\omega_0t} e^{-j\omega t} \mathrm{d} t \\ = F(\omega + \omega_0) \\$$

Or,

$$f(t) e^{-j\omega_0t} \xrightarrow{\mathcal{F}} F(\omega + \omega_0)$$

Scale

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$ $$\int_{-\infty}^{+\infty} f(\alpha t) e^{-j\omega t} \mathrm{d}t \\ = \frac{1}{||\alpha||} \int_{-\infty}^{+\infty} f(\alpha t) e^{-j \frac{\omega}{\alpha} \alpha|t} \mathrm{d} \alpha t \\ = \frac{1}{||\alpha||} F(\frac{\omega}{\alpha})$$

Or,

$$f(\alpha t) \xrightarrow{\mathcal{F}} \frac{1}{||\alpha||} F(\frac{\omega}{\alpha})$$

Derivative

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$

Then,

$$\int_{-\infty}^{+\infty} f'(t) e^{-j\omega t} \mathrm{d}t \\ = -j\omega \int_{-\infty}^{+\infty} f(t) e^{-j\omega t} \mathrm{d}t \\ = -j\omega F(\omega)$$

Or,

$$f'(t) \xrightarrow{\mathcal{F}} -j\omega F(\omega)$$

Integral

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega)$$

Because,

$$\int_{-\infty}^{+\infty} (\int_{-\infty}^{t}f(\tau)\mathrm{d}\tau) e^{-j\omega t} \mathrm{d}t \\ = \int_{-\infty}^{+\infty} (\int_{-\infty}^{t}e^{-j\omega \tau}\mathrm{d}\tau) f(t) \mathrm{d}t = \frac{F(\omega)}{-j\omega}$$

Or,

$$\int_{-\infty}^{t}f(\tau) \mathrm{d}\tau \xrightarrow{\mathcal{F}} \frac{F(\omega)}{-j\omega}$$

Convolution

Suppose,

$$f(t) \xrightarrow{\mathcal{F}} F(\omega) \\ g(t) \xrightarrow{\mathcal{F}} G(\omega)$$

Then,

$$f(t) \ast g(t) = \int_{-\infty}^{+\infty} f(\tau) g(t - \tau) \mathrm{d}\tau$$ $$\int_{-\infty}^{+\infty} (f(t) \ast g(t)) e^{-j\omega t}\mathrm{d}t \\ = \int_{-\infty}^{+\infty} (\int_{-\infty}^{+\infty} f(\tau) g(t - \tau) \mathrm{d}\tau) e^{-j\omega t}\mathrm{d}t \\ = \int_{-\infty}^{+\infty} (\int_{-\infty}^{+\infty} f(\tau) e^{-j\omega \tau} \mathrm{d}\tau) g(t-\tau) e^{-j \omega (t-\tau)} \mathrm{d}t \\ = \int_{-\infty}^{+\infty} F(\omega) g(t-\tau) e^{-j \omega (t-\tau)} \mathrm{d}t \\ = F(\omega)G(\omega)$$

So,

$$f(t) \ast g(t) \xrightarrow{\mathcal{F}} F(\omega)G(\omega)$$

Multiplication

Similarly, we can prove,

$$f(t)g(t) \xrightarrow{\mathcal{F}} \frac{1}{2\pi}F(\omega) \ast G(\omega)$$

With fourier inverse transform.