Introduction
In this part we’ll define and understand Fourier transforms. In the last part we covered the concept of Fourier series.
Which was a way of describing periodic functions as a linear combination (sum) of simple trigonometric functions. The Fourier transform takes this one step further, it’s a way to extend the Fourier series of non-periodic functions.
Definition 1 (Fourier transform) Given a function in the time-domain, the equivalent function in the frequency domain is:
F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t}\ dt F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t The other way around:
f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) e j ω t d ω f(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t}\ d\omega f ( t ) = 2 π 1 ∫ − ∞ ∞ F ( ω ) e j ω t d ω We call these functions the Fourier pair. Meaning they are the same functions, in different domains. They represent the same signal.
f ( t ) ⟺ F ( ω ) f(t) \iff F(\omega) f ( t ) ⟺ F ( ω ) So the notation for the Fourier transform is:
F ( ω ) = F [ f ( t ) ] F(\omega) = \mathcal{F}[f(t)] F ( ω ) = F [ f ( t )] f ( t ) = F − 1 [ F ( ω ) ] f(t) = \mathcal{F}^{-1}[F(\omega)] f ( t ) = F − 1 [ F ( ω )] Let’s do our first Fourier transform!
Example 1 Let’s do this on a simple unit gate function, let’s do it on a general unit gate function.
f ( t ) = r e c t ( t τ ) f(t) = rect\left(\dfrac{t}{\tau}\right) f ( t ) = r ec t ( τ t ) F ( ω ) = ∫ − ∞ ∞ r e c t ( t τ ) e − j ω t d t F(\omega) = \int_{-\infty}^{\infty} rect\left(\dfrac{t}{\tau}\right) e^{-j\omega t}\ dt F ( ω ) = ∫ − ∞ ∞ r ec t ( τ t ) e − j ω t d t Since the gate function is only defined at t = − τ 2 t = -\dfrac{\tau}{2} t = − 2 τ t = τ 2 t = \dfrac{\tau}{2} t = 2 τ
F ( ω ) = ∫ − τ 2 τ 2 e − j ω t d t F(\omega) = \int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} e^{-j\omega t}\ dt F ( ω ) = ∫ − 2 τ 2 τ e − j ω t d t F ( ω ) = − 1 j ω e − j ω t ∣ τ 2 − τ 2 F(\omega) = -\dfrac{1}{j\omega} e^{-j\omega t} \bigg\rvert_{\frac{\tau}{2}}^{-\frac{\tau}{2}} F ( ω ) = − j ω 1 e − j ω t 2 τ − 2 τ F ( ω ) = − 1 j ω ( e − j ω τ 2 − e j ω t τ 2 ) F(\omega) = -\dfrac{1}{j\omega} \left(e^{-j\omega \dfrac{\tau}{2}} - e^{j\omega t \dfrac{\tau}{2}}\right) F ( ω ) = − j ω 1 ( e − j ω 2 τ − e j ω t 2 τ ) Using Euler’s formula:
F ( ω ) = 2 s i n ( ω τ 2 ) ω F(\omega) = \dfrac{2 sin\left(\dfrac{\omega \tau}{2}\right)}{\omega} F ( ω ) = ω 2 s in ( 2 ω τ ) Let’s rewrite it as:
F ( ω ) = τ s i n ( ω τ 2 ) ω τ 2 F(\omega) = \tau \dfrac{sin\left(\dfrac{\omega \tau}{2}\right)}{\dfrac{\omega \tau}{2}} F ( ω ) = τ 2 ω τ s in ( 2 ω τ ) We can rewrite using s i n c ( x ) = s i n ( π x ) π x sinc(x) = \dfrac{sin(\pi x)}{\pi x} s in c ( x ) = π x s in ( π x )
F ( ω ) = τ s i n c ( ω τ 2 ) F(\omega) = \tau sinc\left(\dfrac{\omega \tau}{2}\right) F ( ω ) = τ s in c ( 2 ω τ ) So, this means that:
r e c t ( t τ ) ⟺ τ s i n c ( ω τ 2 ) rect\left(\dfrac{t}{\tau}\right) \iff \tau sinc\left(\dfrac{\omega \tau}{2}\right) r ec t ( τ t ) ⟺ τ s in c ( 2 ω τ ) Now, we can do this manually each time, but that’s tedious, that’s why we’ll usually use a Fourier table.
Let’s cover some important Fourier transforms.
c o s ( ω 0 t ) ⟺ π [ δ ( ω + ω 0 ) + δ ( ω − ω 0 ) ] cos(\omega_0 t) \iff \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)] cos ( ω 0 t ) ⟺ π [ δ ( ω + ω 0 ) + δ ( ω − ω 0 )] s i n ( ω 0 t ) ⟺ j π [ δ ( ω + ω 0 ) + δ ( ω − ω 0 ) ] sin(\omega_0 t) \iff j\pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)] s in ( ω 0 t ) ⟺ j π [ δ ( ω + ω 0 ) + δ ( ω − ω 0 )] We’ll use these often.
Symmetry
After we have obtained one representation of the signal, it doesn’t stop us from just replacing t t t ω \omega ω
For example:
r e c t ( t τ ) ⟺ τ s i n c ( ω τ 2 ) rect\left(\dfrac{t}{\tau}\right) \iff \tau sinc\left(\dfrac{\omega \tau}{2}\right) r ec t ( τ t ) ⟺ τ s in c ( 2 ω τ ) We can do:
F ( t ) = τ s i n c ( t τ 2 ) F(t) = \tau sinc\left(\dfrac{t \tau}{2}\right) F ( t ) = τ s in c ( 2 t τ ) We can utilize:
F ( t ) ⟺ 2 π f ( − ω ) F(t) \iff 2\pi f(-\omega) F ( t ) ⟺ 2 π f ( − ω ) Now this means:
F ( t ) ⟺ 2 π f ( − ω ) = 2 π r e c t ( − ω τ ) F(t) \iff 2\pi f(-\omega) = 2\pi \ rect\left(\dfrac{-\omega}{\tau}\right) F ( t ) ⟺ 2 π f ( − ω ) = 2 π r ec t ( τ − ω ) Since the unit gate function is an even function:
F ( t ) ⟺ 2 π f ( − ω ) = 2 π r e c t ( ω τ ) F(t) \iff 2\pi f(-\omega) = 2\pi \ rect\left(\dfrac{\omega}{\tau}\right) F ( t ) ⟺ 2 π f ( − ω ) = 2 π r ec t ( τ ω ) Scaling
f ( a t ) ⟺ 1 ∣ a ∣ F ( ω a ) f(at) \iff \dfrac{1}{|a|} F\left(\dfrac{\omega}{a}\right) f ( a t ) ⟺ ∣ a ∣ 1 F ( a ω ) As we can see, an expansion in the time domain, means a compression in the frequency domain and vice-versa.
From this we can also gain the time and frequency inversion property:
f ( − t ) ⟺ F ( − ω ) f(-t) \iff F(-\omega) f ( − t ) ⟺ F ( − ω ) Time and frequency shifting
f ( t − t 0 ) ⟺ F ( ω ) e − j ω t 0 f(t - t_0) \iff F(\omega) e^{-j\omega t_0} f ( t − t 0 ) ⟺ F ( ω ) e − j ω t 0 This also means:
f ( t ) e j ω 0 t ⟺ F ( ω − ω 0 ) f(t) e^{j\omega_0 t} \iff F(\omega - \omega_0) f ( t ) e j ω 0 t ⟺ F ( ω − ω 0 ) Time differentiation and integration
d f d t ⟺ j ω F ( ω ) \dfrac{df}{dt} \iff j\omega F(\omega) d t df ⟺ j ω F ( ω ) ∫ − ∞ t f ( τ ) d τ ⟺ F ( ω ) j ω + π F ( 0 ) δ ( ω ) \int_{-\infty}^{t} f(\tau)\ d\tau \iff \dfrac{F(\omega)}{j\omega} + \pi F(0)\delta(\omega) ∫ − ∞ t f ( τ ) d τ ⟺ j ω F ( ω ) + π F ( 0 ) δ ( ω ) Now, for the most powerful and beautiful property, which we’ll use.
Convolution
Given f 1 ( t ) ⟺ F 1 ( ω ) f_1(t) \iff F_1(\omega) f 1 ( t ) ⟺ F 1 ( ω ) f 2 ( t ) ⟺ F 2 ( ω ) f_2(t) \iff F_2(\omega) f 2 ( t ) ⟺ F 2 ( ω )
Time convolution:
f 1 ( t ) ∗ f 2 ( t ) ⟺ F 1 ( ω ) F 2 ( ω ) f_1(t) * f_2(t) \iff F_1(\omega) F_2(\omega) f 1 ( t ) ∗ f 2 ( t ) ⟺ F 1 ( ω ) F 2 ( ω ) Frequency convolution
f 1 ( t ) f 2 ( t ) ⟺ 1 2 π F 1 ( ω ) ∗ F 2 ( ω ) f_1(t) f_2(t) \iff \dfrac{1}{2\pi} F_1(\omega) * F_2(\omega) f 1 ( t ) f 2 ( t ) ⟺ 2 π 1 F 1 ( ω ) ∗ F 2 ( ω ) The Fourier transform exist, if:
∫ − ∞ ∞ ∣ f ( t ) ∣ d t < ∞ \int_{-\infty}^{\infty} |f(t)|\ dt < \infty ∫ − ∞ ∞ ∣ f ( t ) ∣ d t < ∞