Fourier transform (FT)
Valentina Hubeika, Jan Černocký
DCGM FIT BUT Brno, {ihubeika,cernocky}@fit.vutbr.cz
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Fourier transform
Properties of spectral function
Spectral function of important signals
Hints on spectra
Energy and Parseval theorem.
1
Reasons for developement of FT:
• We want to do frequency analysis of signals other than periodic
• Nonperiodic signals will be also expressed as a sum of harmonic signals (system
responce is nicely calculated for input ejωt , etc). It will be a little more difficult to
imagine as we will obtain an infinite number of components that are infinitelly small.
s(t)
a)
t
b)
t
c)
t
d)
t
2
|ck|
a)
ω
b)
ω
c)
ω
d)
ω
3
From FS to FT
Coefficients of FS:
1
ck =
T1
Z
+
−
T1
2
T1
2
x(t)e−jkω1 t dt.
Now, we will be ”stretching“ the period to infinity
T1 → ∞,
ω1 =
2π
→ dω,
T1
kω1 → ω
dω
1
→
ck → dc,
,
T1
2π
Derive a new equation for the coefficients’ calculation:
Z
dω +∞
x(t)e−jωt dt.
dc =
2π −∞
dc
is an infinitelly small coefficient increment on an infinitelly small increment of angular
2π dω
frequency multiplied by 2π. Rather we will introduce term: Spectral function X(jω).
4
X(jω) =
Z
+∞
x(t)e−jωt dt,
−∞
Function X(jω) will be called Fourier projection/image or simply image of signal x(t).
Spectral function X(jω) can also be called spectrum. Fourier transform is sometimes
F
denoted as F: x(t) → X(jω).
5
Fundamental properties of spectral function
If projection exists, then:
X(jω) = X ⋆ (−jω)
which follows from,
X(jω) =
Z
+∞
x(t) cos(ωt)dt − j
−∞
Z
+∞
x(t) sin(ωt)dt.
−∞
Other properties are special cases. Even signal has only real spectrum:
x(t) = x(−t) ⇒ X(jω) = ℜ{X(jω)}
Odd signal has only imaginary spectrum:
x(t) = −x(−t) ⇒ X(jω) = jℑ{X(jω)}
6
Inverse Fourier transform
Signal synthesis from FS coefficients:
x(t) =
+∞
X
ck ejkω1 t
k=−∞
+∞
1 X
ck jkω1 t
2π e
=
ω1 .
2π
ω1
k=−∞
By transition T1 → ∞, we obtain:
1
x(t) =
2π
Z
+∞
X(jω)e+jωt dω.
−∞
7
Convergence of FT
If we assume that spectral function will consist of impulse signals, for FT to converge it
only needs to satisfy the following restrictions:
• the signal has to be bounded with a constant M < ∞,
|x(t)| < M
Its energy thus does not have to be finite. It allows us to calculate FT of a direct
current of a periodic signal!
• Signal can be composed from unit impulses - shifted and scaled. This property will
help us during sampling, where we will be looking for a spectrum of a periodic
sequence of unit impulses.
8
s(t)
sa(t)
+M
t
-M
9
SPECTRAL FUNCTION OF IMPORTANT SIGNALS
Unit impuls
Unit impuls is an infinitely sharp peak bounding unit area, that is a derivation of the unit
∆ (t)
, where ∆ goes to 0:
step function: δ∆ (t) = dσdt
Unit impulse can be used in sampling, it satisfies:
Z +∞
x(t)δ(t − τ )dt = x(τ )
−∞
10
Why? If we multiply signal x by δ∆ (t − τ ) and integrate, we obtain an area of the size
1
∆∆
x(τ ) = x(τ ). If ∆ is sufficiently short, we can assume the signal to be a constant
within the area it bounds. Thus the equation holds for ∆ → 0.
Z
+∞
∞
x(t)δ∆ (t−τ )dt =
Z
τ +∆
x(t)δ∆ (t−τ )dt ≈
τ
Z
τ
11
τ +∆
x(τ )δ∆ (t−τ )dt = ∆
1
x(τ ) = x(τ )
∆
Spectral function of unit impuls
X(jω) =
Z
+∞
δ(t − τ )e−jωt dt = e−jωτ
−∞
|S(ω)|
1
δ(t −τ)
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
1
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
t
0
0
arg S(ω)
ω
0
ω
−τω
12
Direct signal
X(jω) = 2πAδ(ω)
Prove by inverse FT:
1
2πA
x(t) =
2π
Z
+∞
δ(ω)ejωt dω = A.
−∞
|S(ω)|
s(t)
A
1111111111
0000000000
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
2πΑ
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
0000000000
1111111111
ω
t 0
13
Periodic signal described using FS
x(t) =
+∞
X
ck ejkω1 t
k=−∞
First, lets see what a signal with the following spectral function looks like:
X(jω) = 2πδ(ω − ω0 )
Z +∞
1
x(t) =
2π
δ(ω − ω0 )ejωt dω = . . .
2π
−∞
after variable substitution:
Z +∞
Z
x(t) =
δ(r)ej(r+ω0 )t dr = ejω0 t
+∞
δ(r)ejrt dr = ejω0 t
−∞
−∞
It is a complex exponential rotating on frequency ω0 . FT of a periodic signal defined by
means of FS is thus:
+∞
X
2πck δ(ω − kω1 ).
X(jω) =
k=−∞
14
Example of a harmonic signal x(t) = C1 cos(ω1 t + φ1 ) = c−1 e−jω1 t + c1 ejω1 t
15
Rb
Square impuls We know: −b e±jxy dy = 2b sinc(bx). Let’s define b = ϑ2 , y = t, x = ω,
we obtain:
µ ¶
µ ¶
Z + ϑ2
ϑ
ϑ
ϑ
−jωt
ω = Dϑsinc
ω
X(jω) = D
e
dt = D2 sinc
ϑ
2
2
2
−2
|S(ω)|
−2π/υ 0
arg S(ω)
π
s(t)
D
−υ/2
Dυ
υ/2
0
t
2π/υ
4π/υ 6π/υ
ω
ω
−π
16
Inverse projection of a square spectral function
S(ω)
H
−ωc
1
x(t) =
2π
0
Z
ωc
∞
ω
1
+jωt
X(jω)e
dω =
2π
−∞
=
Z
ωc
H
+jωt
He
dω =
2π
−ωc
Z
ωc
e+jωt dω =
−ωc
Hωc
H
2ωc sinc(ωc t) =
sinc(ωc t)
2π
π
What is the maximum of the function? Points where the function crosses time axis?
17
Hints to the spetcra of non periodic signals
x(t)
X(jω)
linearity
axa (t) + bxb (t)
aXa (jω) + bXb (jω)
shift in time
x(t − τ )
X(jω)e−jωτ
s(mt) m > 0
1 ³ω´
X
m
m
change of scale
convolution
x1 (t) ⋆ x2 (t) =
Z
∞
x1 (τ )x2 (t − τ )dτ
−∞
18
X1 (jω)X2 (jω)
Time-shift
Same as for FS, only argument/phase of spectral function will be changed, by: −ωτ .
Example:
arg S(ω)
π
π/2
s(t)
−π/2
0
−π/2
−π
-2
2
t
19
π/2
π
ω
1 s delay: X(jω) multiplied by function e−jω , thus ω will be subtracted from the original
argument:
arg S(ω)
π
π/2
s(t)
−π/2
0
−π/2
−π
1
3
t
20
π/2
π
ω
1 s advance: X(jω) multiplied by function ejω , thus ω will be added to the original
argument:
arg S(ω)
π
π/2
s(t)
−π/2
0
−π/2
−π
-3
1
t
21
π/2
π
ω
Examples of time axis scale
S(ω)
s(t)
D
2D
-1
s(t)
t
1
0
π
ω
π
ω
S(ω)
6D
D
-3
3
t
0
22
Spectrum of convolution
Z
∞
−∞
=
·Z
∞
x1 (τ )x2 (t − τ )dτ e−jωt dt =
−∞
Z
∞
−∞
¸
Z
∞
x1 (τ )
−∞
£
¤
−jωτ
x1 (τ ) X2 (jω)e
dτ = X2 (jω)
23
·Z
∞
−∞
Z
¸
x2 (t − τ )e−jωt dt dτ =
∞
−∞
x1 (τ )e−jωτ dτ = X1 (ω)X2 (ω)
Example about spectrum of convolution
S1(ω)
s1(t)
0,8
-1
s2(t)
1
1,6
t
0
π
ω
1,2
1
t
2
0
Y(ω)
0,96
-2
ω
S2(ω)
0,6
-1
y(t)
π
t
0
24
1,92
π
ω
Parseval theorem - absolute energy of signal using spectral function
Z
∞
s2 (t)dt =
−∞
1
=
2π
Z
∞
·Z
Z
∞
x(t)
−∞
∞
·
1
2π
¸
1
jωt
X(jω)
x(t)e dt dω =
2π
−∞
−∞
{z
}
|
X(−jω)
Z
∞
¸
X(jω)ejωt dω dt =
−∞
Z ∞
X(jω)X(−jω)dω =
−∞
Z
∞
−∞
Ld (ω) is called (double sided) spectral density function of energy
|X(jω)|2
Ld (ω) =
2π
Ld(ω)
2 2
Dυ
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
2π
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
000
111
90% energie
00000000000000000
11111111111111111
000
111
00000000000000000
11111111111111111
000
111
00000000000000000
11111111111111111
000
111
00000000000000000
11111111111111111
000
111
00000000000000000
11111111111111111
000
111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
00000000000000000
11111111111111111
ω
−2π/υ 0
2π/υ 4π/υ 6π/υ
25
Ld (ω)dω