Fourier Transform Reference – Chapter 2.2 – 2.5, Carlson, Communication Systems Using the Fourier series, a signal over a finite interval can be represented in terms of a complex exponential series. If the function is periodic, this representation can be extended over the entire interval (− ∞, ∞ ) . f (t ) =
∞
∑ Fne jnωot
n = −∞
Fn =
1 T
∫
T
f (t )e − jnω ot dt
0
Fourier transforms.1
Fourier Transform On the other hand, Fourier transform provides the link between the time-domain and frequency domain descriptions of a signal. Fourier transform can be used for both periodic and non-periodic signals.
∞
f (t ) = ∫ F (ω )e jωt dω −∞
F (ω ) = ∫
∞
−∞
f (t )e − jωt dt
Fourier transforms.2
1
Example The the last example of the previous lecture shows that if the period (T) of a periodic signal increases, the fundamental frequency (ωo=2π/T) becomes smaller and the frequency spectrum becomes more dense while the amplitude of each frequency component decreases. The shape of the spectrum, however, remains unchanged with varying T. Now, we will consider a signal with period approaching infinity. Suppose we are given a non-periodic signal f(t). In order to applying Fourier series to the signal f(t), we construct a new periodic signal fT(t) with period T. Fourier transforms.3
f (t ) fT (t )
t t
The original signal can be obtained back again by letting the period T → ∞ , that is, f (t ) = lim fT (t ) T →∞
Fourier transforms.4
2
The periodic function fT(t) can be represented by an exponential Fourier series. fT (t ) =
∞
∑F e
n = −∞
Fn =
1 T
n
∫
jnω o t
T /2
−T / 2
where
fT (t )e − jnω ot dt and ω o = 2π / T
As the magnitude of the Fourier coefficients go to zero when the period is increased, we define . n ω n≡ nω o and F (ω n ) ≡ TF The Fourier series pair become fT (t ) =
∞
1 ∑ T F (ω )e ω j
n
n = −∞
nt
− jω t where F (ω n ) = ∫−T / 2 fT (t )e n dt T /2
Fourier transforms.5
The spacing between adjacent lines (∆ω) in the line spectrum of fT(t) is ∆ω = 2π / T
Therefore, we have fT (t ) =
∞
∑ F (ω )e ω
n = −∞
j
n
nt
∆ω 2π
(1)
Now as T becomes very large, ∆ω becomes smaller and the spectrum becomes denser. In the limit, the discrete lines in the spectrum of fT(t) merge and the frequency spectrum becomes continuous. Fourier transforms.6
3
Mathematically, the infinite sum (1) becomes an integral 1 ∞ lim fT (t ) = lim F (ω n )e jω nt ∆ω ∑ T →∞ T → ∞ 2π n = −∞ ∞ 1 ⇒ f (t ) = F (ω )e jωt dω ∫ 2π −∞ Inverse Fourier transform of F(ω)
Similarly, Fn =
1 T
∫
T /2
fT (t )e − jnω ot dt
−T / 2
⇒ F (ω n ) = ∫
T /2
−T / 2
fT (t )e − jω nt dt
⇒ lim F (ω n ) = lim ∫
T /2
T → ∞ −T / 2
T →∞
⇒ F (ω ) = ∫
T /2
−T / 2
Qω n ≡ nω o and F (ω n ) ≡ TFn
fT (t )e − jnω ot dt
fT (t )e − jnω ot dt
Fourier transform of f(t) Fourier transforms.7
Operators are often used to denote the transform pair. ℑ{ f (t )}
Fourier transform of f(t)
ℑ−1{F (ω )}
Inverse Fourier transform of F (ω )
f (t ) = ℑ−1 [ℑ{ f (t )}]
Fourier transforms.8
4
Singularity functions – This is a particular class of functions which are useful in signal analysis. – They are mathematical idealization and, strictly speaking, do not occur in physical systems. – Good approximation to certain limiting condition in physical systems. For example, a very narrow pulse.
Fourier transforms.9
Singularity functions Impulse function – This function has the property exhibited by the following integral:
∫
b
a
f (t ) a < to < b f (t )δ (t − to )dt = o elsewhere 0
(2)
for any f(t) continuous at t = to, to is finite. All the properties can be derived from this definition. ∞
tFourier o transforms.10
5
Singularity functions Properties of the impulse function Amplitude All values of δ(t) for t ≠ to are zero. The amplitude at the point t = to is undefined. Area (strength) If f(t) =1, (2) becomes b
∫ δ (t − t )dt = 1 a
o
a < to < b
Therefore δ(t) has unit area. Similarly, Aδ(t) has an area of A units. Fourier transforms.11
Graphic representation To display the impulse function at t = to, an arrow is used to avoid to display the amplitude. The area of the impulse is designated by a quantity in parentheses beside the arrow or by the height of the arrow. An arrow pointing down 2 Aδ (t − t1 ) indicates negative area. Aδ (t − t ) o
Relation to the unit step function The unit step function is defined by 1
to
to
t1
1 t > to u (t − t0 ) = 0 t < to Fourier transforms.12
6
Using (2) and letting
∫
b
a
a = - ∞, b = t , and f(t) = 1
f (t )δ (t − t o )dt
t t 1 t > t o ⇒ ∫ δ (τ − t o )dτ = u (t − t o ) ⇒ ∫ δ (τ − t o )dτ = −∞ −∞ 0 t < t o
Therefore the integral of the unit impulse function is the unit step function. The converse can also be shown by differentiating both sides of the above equation. t ∫ δ (τ − to )dτ = u (t − to ) −∞
d t d δ (τ − to )dτ = u (t − to ) ∫ dt −∞ dt d ⇒ δ (t − to ) = u (t − to ) dt ⇒
Fourier transforms.13
Spectral density functionF (ω ) f (t ) =
1 2π
∫
∞
F (ω )e jωt dω
represents f(t) as a continuous sum of exponential functions with frequencies lying in the interval (−∞, ∞) . The relative amplitude of the components at any frequency ω is proportional to F(ω). F(ω) is called the spectral density function of f(t). −∞
– Each point on the F(ω) curve contributes nothing to the representation of f(t); it is the area that contributes.
t Fourier transforms.14
7
Spectral density functionF (ω ) Example – Consider a rectangular pulse train – The line spectrum of the Fourier series of the signal is
The spectral density of the signal is Impulse function
Fourier transforms.15
Existence of the Fourier transform We may ignore the question of the existence of the Fourier transform of a time function when it is an accurately specified description of a physically realizable signal. In other words, physical realizability is a sufficient condition for the existence of a Fourier transform.
Fourier transforms.16
8
Parseval’s theorem for energy signals Using the Parseval’s theorem, we can find the energy of a signal in either the time domain or the frequency domain. E=∫
∞
−∞
2
f (t ) dt =
1 2π
∫
∞
−∞
2
F (ω ) dω
Example Energy contained in the frequency band ω1 < ω < ω 2 real-valued signal is
of a
ω2 1 −ω1 2 2 F d + F dω ( ω ) ω ( ω ) ∫ ∫ − ω ω 1 2π 2 1 ω2 2 = ∫ F (ω ) dω
π
ω1
Fourier transforms.17
Fourier transforms of some signals Impulse function δ(t) The Fourier transform of a unit impulse δ(t) is ∞ ∞ ℑ{δ (t )} = ∫ δ (t )e − jωt dt = e j 0 = 1 Q∫ δ (t ) f (t )dt = f (0) −∞
−∞
It shows that an impulse function has a uniform spectral density over the entire frequency spectrum. In practice, a narrow pulse in time domain has a very wide bandwidth in frequency domain. Example If we increase the transmission rate of digital signal, a wider frequency bandwidth is needed. Fourier transforms.18
9
Fourier transforms of some signals Complex exponential function The Fourier transform of a complex exponential function is
{
}
∞
ℑ e ± jω ot = ∫ e ± jω ot e − jωt dt = ? −∞
On the other hand, the inverse Fourier transform of δ (ω ± ω o ) is 1 ∞ 1 ± jω t ℑ−1{δ (ω ± ω o )} = δ (ω ± ω o )e jωt dt = e ∫ 2π −∞ 2π ∴ ℑ{e ± jω t } = ℑ{ℑ−1{δ (ω ± ω o )}} = 2πδ (ω ± ω o ) o
o
Fourier transforms.19
Sinusoidal signals The sinusoidal signals can be written as e jx + e − jx cos x = 2
e jx − e − jx sin x = 2j
The Fourier transform of these signals are e jω o t + e − jω o t ℑ{cos ω ot} = ℑ = πδ (ω − ω o ) + πδ (ω + ω o ) 2 e jω o t − e − jω o t π π ℑ{cos ω ot} = ℑ = δ (ω − ω o ) − δ (ω + ω o ) 2j j j Fourier transforms.20
10
sin ω ot
t
jF (ω ) (π )
− ωo
ωo
( −π )
Fourier transforms.21
Periodic signal A periodic signal f(t) can be represented by its exponential ∞ Fourier series. f (t ) = F e jnω t whereω = 2π / T
∑
n = −∞
n
o
o
∞
The Fourier transform is ℑ{ f (t )} = ℑ ∑ Fn e jnω t o
n = −∞
=
∑ F ℑ{e ∞
n = −∞
= 2π
n
jnω o t
}
∞
∑ F δ (ω − nω
n = −∞
n
o
)
Thus the spectral density of a periodic signal consists of a set of impulses located at the harmonic frequencies of the signal. The area of each impulse is 2π times the values of its corresponding coefficient in the exponential Fourier series. Fourier transforms.22
11
Some properties of the Fourier transform Linearity (superposition) The Fourier transform is a linear operation based on the properties of integration and therefore superposition applies. ℑ{af (t ) + bg (t )} = aF (ω ) + bG (ω )
Fourier transforms.23
Some properties of the Fourier transform Duality If ℑ{ f (t )} = F (ω )
, then ℑ{F (t )} = 2πF (−ω )
Example 1 1 0.5
2π
t
1
ω
2π
2π t
0.5
ω
Fourier transforms.24
12
Coordinate scaling The expansion or compression of a time waveform affects the spectral density of the waveform. For a real-valued scaling constant α and any signal f(t) 1 ω ℑ{ f (αt )} = F ( ) α α Example 2π e −ω / 2 −t 2
e
2
e − ( 2t )
2π e − (ω / 2 )
2
2
/2
Fourier transforms.25
Time shifting (delay) ℑ{ f (t − to )} = F (ω )e − jωto
If a signal is delayed in time by to, its magnitude spectral density remains unchanged and a negative phase -ωto is added to each frequency component. e jω o t
Fourier transforms.26
13
Frequency shifting (Modulation)
{
}
ℑ f (t )e jω ot = F (ω − ω o )
Therefore multiplying a time function by e jω t causes its spectral density to be translated in frequency by ωo. o
Example
ℑ{ f (t ) cos ω ot} =
1 [F (ω + ω o ) + F (ω − ω o )] 2
F (ω )
F (ω + ω o )
F (ω − ω o )
Fourier transforms.27
Differentiation d ℑ f (t ) = jωF (ω ) dt
Time differentiation enhances the high frequency components of a signal. Integration
{
t
}
ℑ ∫ f (τ )dτ = −∞
1 F (ω ) + πF (0)δ (ω ) jω
∞
where F (0) = ∫ f (t )dt −∞
Integration in time suppresses the high-frequency components of a signal. Fourier transforms.28
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